Find each measure. 1.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

2.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, 11-4 Inscribed Angles Therefore,

Find each measure. 4. SCIENCE The diagram shows how light bends in a 1. raindrop to make the colors of the rainbow. If , what is ?

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

2. intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 3. arc or congruent arcs, then the angles are congruent. So,

Therefore,

6. SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, SOLUTION:

Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 4. SCIENCE The diagram shows how light bends in a So, raindrop to make the colors of the rainbow. If , what is ? eSolutions Manual - Powered by Cognero Page 1 Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, SOLUTION: ALGEBRA Find each measure. Given: bisects . 5. Prove:

Proof: SOLUTION: Statements (Reasons) If two inscribed angles of a circle intercept the same 1. bisects . (Given) arc or congruent arcs, then the angles are congruent. 2. (Def. of segment bisector) So, 3. intercepts . intercepts .

(Def. of intercepted arc) 4. (Inscribed of same arc are Therefore, .) 5. (Vertical are .) 6. 6. (AAS)

CCSS STRUCTURE Find each value. 8.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: So, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle Therefore, is 180. So, 7. PROOF Write a two-column proof. Given: bisects .

Prove: Therefore,

9. x

SOLUTION: Given: bisects . Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle Proof: Statements (Reasons) is 180. So, 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . 10. and (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y.

Use the values of the variables to find and SOLUTION: . An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle ALGEBRA Find each measure.

is 180. So, 11.

Therefore,

9. x

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

SOLUTION: 12. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So, SOLUTION: If an angle is inscribed in a circle, then the measure

10. and of the angle equals one half the measure of its

intercepted arc. Therefore,

13.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, Use the values of the variables to find and .

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

ALGEBRA Find each measure. intercepted arc. Therefore, 11. 14.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc. Therefore, of the angle equals one half the measure of its intercepted arc. Therefore, 12. 15.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure Here, of the angle equals one half the measure of its The arc is a semicircle. So, intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 13. intercepted arc. So, Therefore,

16.

SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: Here, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its The arc is a semicircle. So, Then, intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

14. intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, SOLUTION: 15. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 16. arc or congruent arcs, then the angles are congruent. So,

Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 17. ALGEBRA Find each measure. arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

SOLUTION: If two inscribed angles of a circle intercept the same Therefore, m∠C = 5(10) – 3 or 47. arc or congruent arcs, then the angles are congruent. PROOF Write the specified type of proof. So, 21. paragraph proof

Given: Therefore, Prove: 19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

SOLUTION: . Since and If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. , the equation becomes Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. . Multiplying each side of the

equation by 2 results in . Therefore, m∠A = 5(4) or 20. 22. two-column proof 20. ALGEBRA Find each measure. Given:

Prove:

SOLUTION: Statements (Reasons): SOLUTION: 1. (Given) If two inscribed angles of a circle intercept the same 2. (Inscribed intercepting same arc arc or congruent arcs, then the angles are congruent. are .) Since ∠C and ∠D both intercept arc AB, m∠C = 3. (Vertical are .) m∠D. 4. (AA Similarity)

ALGEBRA Find each value. Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given: 23. x Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, Proof: Given means that

. Since and 24. SOLUTION: , the equation becomes An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. . Multiplying each side of the So, The sum of the measures of the angles of a triangle equation by 2 results in . is 180. So, 22. two-column proof Given: Therefore, Prove: 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So,

The sum of the measures of the angles of a triangle SOLUTION: Statements (Reasons): is 180. So, 1. (Given) 2. (Inscribed intercepting same arc are .) 26. 3. (Vertical are .) 4. (AA Similarity) SOLUTION: An inscribed angle of a triangle intercepts a diameter ALGEBRA Find each value. or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

23. x Therefore,

CCSS STRUCTURE Find each measure. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 27. SOLUTION: If a quadrilateral is inscribed in a circle, then its 24. opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, 28. The sum of the measures of the angles of a triangle SOLUTION: is 180. So, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, 25. x Therefore, SOLUTION: An inscribed angle of a triangle intercepts a diameter CCSS STRUCTURE Find each measure. or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

26. 29. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter If a quadrilateral is inscribed in a circle, then its or semicircle if and only if the angle is a right angle. opposite angles are supplementary. So, The sum of the measures of the angles of a triangle

is 180. So, Therefore,

Therefore, 30.

CCSS STRUCTURE Find each measure. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, 27. SOLUTION: 31. PROOF Write a paragraph proof for Theorem 11.9. If a quadrilateral is inscribed in a circle, then its SOLUTION: opposite angles are supplementary. Given: Quadrilateral ABCD is inscribed in .

Prove: and are supplementary.

Therefore, and are supplementary.

28. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Proof: By arc addition and the definitions of arc

Therefore, measure and the sum of central angles, . Since by Theorem 10.6 CCSS STRUCTURE Find each measure. and , but

, so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, 29. . But SOLUTION: , so , making them supplementary also. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore,

30.

SOLUTION: measure. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 32. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are Therefore, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 31. PROOF Write a paragraph proof for Theorem 11.9. of each arc joining consecutive vertices is of 360

SOLUTION: or 45. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. Therefore, measure of arc NPQ is 135. and are supplementary. 33. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 Proof: By arc addition and the definitions of arc measure and the sum of central angles, or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its . Since by Theorem 10.6 intercepted arc QR. and , but

, so or 180. This makes ∠C and ∠A supplementary. Therefore, the measure of ∠RLQ is 22.5. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But 34. , so , making them supplementary also. SOLUTION: If an angle is inscribed in a circle, then the measure SIGNS: A stop sign in the shape of a regular of the angle equals one half the measure of its octagon is inscribed in a circle. Find each intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

measure. Then,

32. Therefore, SOLUTION: Since all the sides of the stop sign are congruent 35. chords of the circle, the corresponding arcs are SOLUTION: congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of each arc joining consecutive vertices is of 360 or 45. intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, Therefore, measure of arc NPQ is 135. 360 is divided into 8 equal arcs and each arc

33. measures SOLUTION: Then, Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are Therefore, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 36. ART Four different string art star patterns are of each arc joining consecutive vertices is of 360 shown. If all of the inscribed angles of each star or 45. Since ∠RLQ is inscribed in the circle, its shown are congruent, find the measure of each measure equals one half the measure of its inscribed angle. intercepted arc QR. a.

Therefore, the measure of ∠RLQ is 22.5. b.

34. SOLUTION: If an angle is inscribed in a circle, then the measure c. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc d. measures Then,

Therefore,

35. SOLUTION: SOLUTION: a. The sum of the measures of the central angles of If an angle is inscribed in a circle, then the measure a circle with no interior points in common is 360. of the angle equals one half the measure of its Here, 360 is divided into 5 equal arcs and each arc

intercepted arc. So, measures The sum of the measures of the central angles of a If an angle is inscribed in a circle, then the measure circle with no interior points in common is 360. Here, of the angle equals one half the measure of its 360 is divided into 8 equal arcs and each arc intercepted arc. So, the measure of each inscribed angle is 36. measures b. Here, 360 is divided into 6 equal arcs and each arc

Then, measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each Therefore, intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one 36. ART Four different string art star patterns are half the measure of its intercepted arc. So, the shown. If all of the inscribed angles of each star measure of each inscribed angle is 60. shown are congruent, find the measure of each c. Here, 360 is divided into 7 equal arcs and each arc inscribed angle. measures If an angle is inscribed in a a. circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360.

b. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the c. measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2

Given: P lies inside . is a diameter. d. Prove:

SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc SOLUTION: Proof: measures Statements (Reasons) If an angle is inscribed in a circle, then the measure 1. m∠ABC = m∠ABD + m∠DBC ( Addition of the angle equals one half the measure of its Postulate) intercepted arc. So, the measure of each inscribed 2. m∠ABD = m(arc AD) angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the measures Each inscribed angle is formed measure of the intercepted arc (Case 1)). by intercepting alternate vertices of the star. So, each 3. m∠ABC = m(arc AD) + m(arc DC) intercepted arc measures 120. If an angle is inscribed (Substitution) in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) measures If an angle is inscribed in a 6. m∠ABC = m(arc AC) (Substitution) circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure 38. Case 3 of each inscribed angle is about 25.7.

d. The sum of the measures of the central angles of Given: P lies outside . is a diameter.

a circle with no interior points in common is 360. Prove: Here, 360 is divided into 8 equal arcs and each arc

measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. SOLUTION: 37. Case 2 Proof: Given: P lies inside . is a diameter. Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Prove: Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) SOLUTION:

Proof: Statements (Reasons) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 1. m ABC = m ABD + m DBC ( Addition ∠ ∠ ∠ 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Postulate) Property of Equality) 2. m∠ABD = m(arc AD) 7. m∠ABC = m(arc AC) (Substitution) m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). PROOF Write the specified proof for each theorem. 3. m∠ABC = m(arc AD) + m(arc DC) 39. Theorem 11.7, two-column proof (Substitution) SOLUTION: 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) Given: and are inscribed; 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Prove: Addition Postulate) Proof: 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter.

Prove:

Statements (Reasons) 1. and are inscribed; (Given) 2. ; (Measure of an inscribed = half measure of intercepted arc.) SOLUTION: 3. (Def. of arcs) Proof: Statements (Reasons) 4. (Mult. Prop. of Equality) 1. m∠ABC = m∠DBC – m∠DBA ( Addition 5. (Substitution) Postulate, Subtraction Property of Equality) 6. (Def. of ) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an 40. Theorem 11.8, paragraph proof inscribed whose side is a diameter is half the SOLUTION: measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction

Property of Equality) Part I: Given: is a semicircle.

7. m∠ABC = m(arc AC) (Substitution) Prove: is a right angle. Proof: Since is a semicircle, then PROOF Write the specified proof for each . Since is an inscribed angle, theorem. then or 90. So, by definition, 39. Theorem 11.7, two-column proof SOLUTION: is a right angle.

Given: and are inscribed; Part II: Given: is a right angle. Prove: Prove Proof: : is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

Statements (Reasons) 41. MULTIPLE REPRESENTATIONS In this 1. and are inscribed; problem, you will investigate the relationship between (Given) the arcs of a circle that are cut by two parallel chords. 2. ; (Measure a. GEOMETRIC Use a compass to draw a circle of an inscribed = half measure of intercepted arc.) with parallel chords and . Connect points A

3. (Def. of arcs) and D by drawing segment . b. NUMERICAL Use a protractor to find 4. (Mult. Prop. of Equality) and . Then determine and . 5. (Substitution) What is true about these arcs? Explain. 6. (Def. of ) c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. 40. Theorem 11.8, paragraph proof d. ANALYTICAL Use your conjecture to find SOLUTION: and in the figure. Verify by using inscribed angles to find the measures of the arcs.

SOLUTION: Part I: Given: is a semicircle. a. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition,

is a right angle. b. Sample answer: , ; , ; The arcs are congruent Part II: Given: is a right angle. because they have equal measures. Prove: is a semicircle. c. Sample answer:

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then

m = 2(90) or 180. So by definition, is a m∠A = 15, m∠D = 15, = 30, and = 30. semicircle. The arcs are congruent. In a circle, two parallel chords cut congruent arcs. 41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between d. In a circle, two parallel chords cut congruent arcs. the arcs of a circle that are cut by two parallel So, = . chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A

and D by drawing segment . = 4(16) + 6 or 70 b. NUMERICAL Use a protractor to find = 6(16) – 26 or 70 and . Then determine and . Therefore, the measure of arcs QS and PR are each

What is true about these arcs? Explain. 70. c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that CCSS ARGUMENTS Determine whether the are cut by two parallel chords. quadrilateral can always, sometimes, or never d. ANALYTICAL Use your conjecture to find be inscribed in a circle. Explain your reasoning. 42. square and in the figure. Verify by using inscribed angles to find the measures of the arcs. SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: a. SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram

b. Sample answer: , ; SOLUTION: The opposite angles of a parallelogram are always , ; The arcs are congruent congruent. They will only be supplementary when because they have equal measures. they are right angles. So, a parallelogram can be c. Sample answer: inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed m∠A = 15, m∠D = 15, = 30, and = 30. in a circle as long as it is a square. Since the opposite The arcs are congruent. angles of rhombi that are not squares are not In a circle, two parallel chords cut congruent arcs. supplementary, they can not be inscribed in a circle. d. In a circle, two parallel chords cut congruent arcs. Therefore, the statement is sometimes true. So, = . 46. kite

SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be = 4(16) + 6 or 70 a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must = 6(16) – 26 or 70 also be supplementary. So, as long as the angles that Therefore, the measure of arcs QS and PR are each compose the pair of congruent opposite angles are 70. right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never 47. CHALLENGE A square is inscribed in a circle. be inscribed in a circle. Explain your reasoning. What is the ratio of the area of the circle to the area 42. square of the square? SOLUTION: SOLUTION: Squares have right angles at each vertex, therefore A square with side s is inscribed in a circle with each pair of opposite angles will be supplementary radius r. and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex,

therefore each pair of opposite angles will be Using the Pythagorean Theorem, supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram SOLUTION: 2 The opposite angles of a parallelogram are always The area of a square of side s is A = s and the area congruent. They will only be supplementary when of a circle of radius r is A = . they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when Therefore, the ratio of the area of the circle to the they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite area of the inscribed square is angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. 48. WRITING IN MATH A right Therefore, the statement is sometimes true. triangle is inscribed in a circle. If the radius of the circle is given, explain how to find the lengths of the 46. kite right triangle’s legs. SOLUTION: SOLUTION: Exactly one pair of opposite angles of a kite are Sample answer: A triangle will have congruent. To be supplementary, they must each be two inscribed angles of 45 and one of 90. The a right angle. If one pair of opposite angles for a hypotenuse is across from the 90, or right angle. quadrilateral is supplementary, the other pair must According to theorem 10.8, an inscribed angle of a also be supplementary. So, as long as the angles that triangle intercepts a diameter if the angle is a right compose the pair of congruent opposite angles are angle. Therefore, the hypotenuse is a diameter and right angles, a kite can be inscribed in a circle. has a length of 2r. Use trigonometry to find the Therefore, the statement is sometimes true. length of the equal legs. 47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the square? SOLUTION: A square with side s is inscribed in a circle with radius r. Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon.

Using the Pythagorean Theorem, SOLUTION: See students’ work.

50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they

2 intercept the same arc, how are they related? The area of a square of side s is A = s and the area SOLUTION: of a circle of radius r is A = . An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the Therefore, the ratio of the area of the circle to the inscribed angle is one-half the measure of the central area of the inscribed square is angle.

48. WRITING IN MATH A right 51. In the circle below, and . triangle is inscribed in a circle. If the radius of the What is ? circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a A 42 triangle intercepts a diameter if the angle is a right B 61 angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the C 80 length of the equal legs. D 84 SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

But, Therefore, the length of each leg of the 45-45-90 Therefore, triangle can be found by multiplying the radius of the The correct choice is A. circle in which it is inscribed by . 52. ALGEBRA Simplify 49. OPEN ENDED Find and sketch a real-world logo 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. with an inscribed polygon. 2 F 9x + 3x – 14 SOLUTION: G 9x2 + 13x – 14 2 See students’ work. H 27x + 37x – 38 2 50. WRITING IN MATH Compare and contrast J 27x + 27x – 26 inscribed angles and central angles of a circle. If they SOLUTION: intercept the same arc, how are they related? Use the Distributive Property to simplify the first SOLUTION: term. An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc Therefore, the correct choice is H. AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the 53. SHORT RESPONSE In the circle below, is a inscribed angle is one-half the measure of the central diameter, AC = 8 inches, and BC = 15 inches. Find angle. the diameter, the radius, and the circumference of the circle. 51. In the circle below, and . What is ?

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter A 42 of the circle. B 61 C 80 The radius is half the diameter. So, the radius of the D 84 circle is about 8.5 in. SOLUTION: The circumference C of a circle of a circle of If an angle is inscribed in a circle, then the measure diameter d is given by of the angle equals one half the measure of its Therefore, the circumference of the circle is intercepted arc. Therefore,

54. SAT/ACT The sum of three consecutive integers is But, –48. What is the least of the three integers? Therefore, A –15 The correct choice is A. B –16 C –17 52. ALGEBRA Simplify D –18 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. E –19 2 F 9x + 3x – 14 SOLUTION: 2 G 9x + 13x – 14 Let the three consecutive integers be x , x +1,and x + 2 H 27x + 37x – 38 2. Then, J 27x2 + 27x – 26 SOLUTION:

Use the Distributive Property to simplify the first So, the three integers are –17, –16, and –15 and the

term. least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure.

Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle. 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter 56. of the circle. SOLUTION: The radius is half the diameter. So, the radius of the If a diameter (or radius) of a circle is perpendicular circle is about 8.5 in. to a chord, then it bisects the chord and its arc. So, The circumference C of a circle of a circle of bisects Therefore, diameter d is given by Therefore, the circumference of the circle is 57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular 54. SAT/ACT The sum of three consecutive integers is to a chord, then it bisects the chord and its arc. So,

–48. What is the least of the three integers? bisects Therefore, A –15 B –16 C –17 D –18 58. E –19 SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular Let the three consecutive integers be x , x +1,and x + to a chord, then it bisects the chord and its arc. So, 2. Then, bisects Therefore,

So, the three integers are –17, –16, and –15 and the Find x. least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure. 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

60.

56. SOLUTION: SOLUTION: The sum of the measures of the central angles of a If a diameter (or radius) of a circle is perpendicular circle with no interior points in common is 360. So, to a chord, then it bisects the chord and its arc. So,

bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 61. SOLUTION: The sum of the measures of the central angles of a 58. circle with no interior points in common is 360. So, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An Find x. image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed? 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x 60. be the length of the base of the larger triangle. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: 61. Since B is the midpoint of SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

62. PHOTOGRAPHY In one of the first cameras

invented, light entered an opening in the front. An 64. AB = 6y – 14, BC = 10 – 2y, AC = ? image was reflected in the back of the camera, SOLUTION: upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 Since B is the midpoint of inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle. 66. AB = 10s + 2, AC = 40, s = ?

SOLUTION: Since B is the midpoint of

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If Find each measure. , what is ? 1.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure

intercepted arc. Therefore, of the angle equals one half the measure of its

intercepted arc. Therefore,

2. ALGEBRA Find each measure. 5.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same intercepted arc. Therefore, arc or congruent arcs, then the angles are congruent. So, 3.

Therefore,

6.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. So, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Therefore, So,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ? Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, Given: bisects . Prove: ALGEBRA Find each measure. 5.

Proof: Statements (Reasons)

1. bisects . (Given) SOLUTION: 2. (Def. of segment bisector) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 3. intercepts . intercepts . (Def. of intercepted arc) So, 4. (Inscribed of same arc are .) 5. (Vertical are .) 11-4Therefore, Inscribed Angles 6. (AAS)

6. CCSS STRUCTURE Find each value. 8.

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same An inscribed angle of a triangle intercepts a diameter arc or congruent arcs, then the angles are congruent. or semicircle if and only if the angle is a right angle. So, So,

The sum of the measures of the angles of a triangle

is 180. So, Therefore,

7. PROOF Write a two-column proof. Therefore, Given: bisects . Prove: 9. x

SOLUTION:

Given: bisects . SOLUTION: Prove: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So, Proof: Statements (Reasons)

1. bisects . (Given) 10. and 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS) SOLUTION: CCSS STRUCTURE Find each value. If a quadrilateral is inscribed in a circle, then its 8. opposite angles are supplementary. eSolutions Manual - Powered by Cognero Page 2 Solve the two equations to find the values of x and y.

Use the values of the variables to find and . SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, ALGEBRA Find each measure. The sum of the measures of the angles of a triangle 11.

is 180. So,

Therefore,

9. x SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

12.

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

10. and intercepted arc. Therefore,

13.

SOLUTION: If a quadrilateral is inscribed in a circle, then its

opposite angles are supplementary. SOLUTION: Solve the two equations to find the values of x and y. The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Use the values of the variables to find and . If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 14. 11.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 15. 12.

SOLUTION:

SOLUTION: Here,

If an angle is inscribed in a circle, then the measure The arc is a semicircle. So, of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. So,

13. Therefore, 16.

SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure Then, of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 14. 17. ALGEBRA Find each measure.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 15. So,

18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. So, If two inscribed angles of a circle intercept the same Therefore, arc or congruent arcs, then the angles are congruent. So, 16.

Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = ALGEBRA Find each measure. 17. m B. ∠

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47. SOLUTION: If two inscribed angles of a circle intercept the same PROOF Write the specified type of proof. arc or congruent arcs, then the angles are congruent. 21. paragraph proof

So, Given:

Prove: Therefore,

19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

. Since and

SOLUTION: , the equation becomes If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. . Multiplying each side of the Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. equation by 2 results in .

22. two-column proof Therefore, m∠A = 5(4) or 20. Given: Prove: 20. ALGEBRA Find each measure.

SOLUTION: Statements (Reasons): 1. (Given) SOLUTION: 2. (Inscribed intercepting same arc are .) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 3. (Vertical are .)

Since ∠C and ∠D both intercept arc AB, m∠C = 4. (AA Similarity) m∠D. ALGEBRA Find each value.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof 23. x Given: SOLUTION: Prove: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, SOLUTION:

Proof: Given means that 24. SOLUTION: . Since and An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. , the equation becomes So, The sum of the measures of the angles of a triangle . Multiplying each side of the is 180. So, equation by 2 results in . Therefore, 22. two-column proof Given: 25. x Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, Statements (Reasons): 1. (Given) 26. 2. (Inscribed intercepting same arc are .) SOLUTION: 3. (Vertical are .) An inscribed angle of a triangle intercepts a diameter 4. (AA Similarity) or semicircle if and only if the angle is a right angle. So, ALGEBRA Find each value. The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

23. x CCSS STRUCTURE Find each measure.

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 27. is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

24. Therefore, SOLUTION: An inscribed angle of a triangle intercepts a diameter 28. or semicircle if and only if the angle is a right angle. So, SOLUTION: The sum of the measures of the angles of a triangle If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. is 180. So,

Therefore, Therefore, 25. x CCSS STRUCTURE Find each measure. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 29.

26. SOLUTION: If a quadrilateral is inscribed in a circle, then its SOLUTION: opposite angles are supplementary. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle Therefore,

is 180. So, 30.

Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its CCSS STRUCTURE Find each measure. opposite angles are supplementary.

Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9. 27. SOLUTION: SOLUTION: Given: Quadrilateral ABCD is inscribed in . If a quadrilateral is inscribed in a circle, then its Prove: and are supplementary. opposite angles are supplementary. and are supplementary. Therefore,

28. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6 Therefore, and CCSS STRUCTURE Find each measure. , but

, so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making 29. them supplementary also.

SOLUTION: SIGNS: A stop sign in the shape of a regular If a quadrilateral is inscribed in a circle, then its octagon is inscribed in a circle. Find each opposite angles are supplementary.

Therefore,

30. measure. SOLUTION: 32. If a quadrilateral is inscribed in a circle, then its SOLUTION: opposite angles are supplementary. Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure Therefore, of each arc joining consecutive vertices is of 360 or 45. 31. PROOF Write a paragraph proof for Theorem 11.9.

SOLUTION: Therefore, measure of arc NPQ is 135. Given: Quadrilateral ABCD is inscribed in .

Prove: and are supplementary. 33. and are supplementary. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its Proof: By arc addition and the definitions of arc measure equals one half the measure of its measure and the sum of central angles, intercepted arc QR. . Since by Theorem 10.6 and

, but Therefore, the measure of RLQ is 22.5. , so or ∠

180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior 34. angles of a quadrilateral is 360, . But SOLUTION: , so , making If an angle is inscribed in a circle, then the measure them supplementary also. of the angle equals one half the measure of its

SIGNS: A stop sign in the shape of a regular intercepted arc. So, octagon is inscribed in a circle. Find each The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Then,

measure. Therefore, 32. 35. SOLUTION: SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are If an angle is inscribed in a circle, then the measure congruent. There are eight adjacent arcs that make of the angle equals one half the measure of its

up the circle, so their sum is 360. Thus, the measure intercepted arc. So, of each arc joining consecutive vertices is of 360 The sum of the measures of the central angles of a or 45. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

Therefore, measure of arc NPQ is 135. measures

33. Then,

SOLUTION: Therefore, Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are ART congruent. There are eight adjacent arcs that make 36. Four different string art star patterns are up the circle, so their sum is 360. Thus, the measure shown. If all of the inscribed angles of each star shown are congruent, find the measure of each of each arc joining consecutive vertices is of 360 inscribed angle. or 45. Since ∠RLQ is inscribed in the circle, its a. measure equals one half the measure of its intercepted arc QR.

b.

Therefore, the measure of ∠RLQ is 22.5.

34.

SOLUTION: c. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, d. 360 is divided into 8 equal arcs and each arc

measures Then,

Therefore, SOLUTION: a. The sum of the measures of the central angles of 35. a circle with no interior points in common is 360. SOLUTION: Here, 360 is divided into 5 equal arcs and each arc

If an angle is inscribed in a circle, then the measure measures of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc. So, of the angle equals one half the measure of its The sum of the measures of the central angles of a intercepted arc. So, the measure of each inscribed circle with no interior points in common is 360. Here, angle is 36. 360 is divided into 8 equal arcs and each arc b. Here, 360 is divided into 6 equal arcs and each arc

measures measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each Then, intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one Therefore, half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. 36. ART Four different string art star patterns are c. Here, 360 is divided into 7 equal arcs and each arc shown. If all of the inscribed angles of each star measures If an angle is inscribed in a shown are congruent, find the measure of each inscribed angle. circle, then the measure of the angle equals one half a. the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed b. by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

c. PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove:

d.

SOLUTION: a. The sum of the measures of the central angles of SOLUTION: a circle with no interior points in common is 360. Proof: Here, 360 is divided into 5 equal arcs and each arc Statements (Reasons) 1. m∠ABC = m∠ABD + m∠DBC ( Addition measures Postulate) If an angle is inscribed in a circle, then the measure 2. m∠ABD = m(arc AD) of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed m∠DBC = m(arc DC) (The measure of an angle is 36. inscribed whose side is a diameter is half the b. Here, 360 is divided into 6 equal arcs and each arc measure of the intercepted arc (Case 1)). measures Each inscribed angle is formed 3. m∠ABC = m(arc AD) + m(arc DC) (Substitution) by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc measure of each inscribed angle is 60. Addition Postulate) c. Here, 360 is divided into 7 equal arcs and each arc 6. m∠ABC = m(arc AC) (Substitution) measures If an angle is inscribed in a 38. Case 3 circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure Given: P lies outside . is a diameter. of each inscribed angle is about 25.7. Prove: d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. SOLUTION: PROOF Write a two-column proof for each case Proof: of Theorem 11.6. Statements (Reasons) 37. Case 2 1. m∠ABC = m∠DBC – m∠DBA ( Addition Given: P lies inside . is a diameter. Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) Prove: m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

SOLUTION: 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc

Proof: Addition Postulate) Statements (Reasons) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 1. m ABC = m ABD + m DBC ( Addition ∠ ∠ ∠ Postulate) 7. m∠ABC = m(arc AC) (Substitution) 2. m∠ABD = m(arc AD) m∠DBC = m(arc DC) (The measure of an PROOF Write the specified proof for each inscribed whose side is a diameter is half the theorem. measure of the intercepted arc (Case 1)). 39. Theorem 11.7, two-column proof 3. m∠ABC = m(arc AD) + m(arc DC) SOLUTION: (Substitution) Given: and are inscribed; 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) Prove: Proof: 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter.

Prove:

Statements (Reasons) 1. and are inscribed; (Given) 2. ; (Measure of an inscribed = half measure of intercepted arc.)

3. (Def. of arcs)

SOLUTION: 4. (Mult. Prop. of Equality) Proof: 5. (Substitution) Statements (Reasons) 6. (Def. of ) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 40. Theorem 11.8, paragraph proof 2. m∠DBC = m(arc DC) SOLUTION: m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc

Addition Postulate) Part I: Given: is a semicircle. 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Prove: is a right angle. Property of Equality) Proof 7. m∠ABC = m(arc AC) (Substitution) : Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition, PROOF Write the specified proof for each theorem. is a right angle. 39. Theorem 11.7, two-column proof SOLUTION: Part II: Given: is a right angle.

Given: and are inscribed; Prove: is a semicircle. Prove: Proof: Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between

Statements (Reasons) the arcs of a circle that are cut by two parallel 1. and are inscribed; chords. (Given) a. GEOMETRIC Use a compass to draw a circle 2. ; (Measure with parallel chords and . Connect points A and D by drawing segment . of an inscribed = half measure of intercepted arc.) b. NUMERICAL Use a protractor to find

3. (Def. of arcs) and . Then determine and .

4. (Mult. Prop. of Equality) What is true about these arcs? Explain. c. VERBAL Draw another circle and repeat parts a 5. (Substitution) and b. Make a conjecture about arcs of a circle that 6. (Def. of ) are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find 40. Theorem 11.8, paragraph proof and in the figure. Verify by using

SOLUTION: inscribed angles to find the measures of the arcs.

SOLUTION: a.

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, b. Sample answer: , ; then or 90. So, by definition, , ; The arcs are congruent is a right angle. because they have equal measures. c. Sample answer: Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m∠A = 15, m∠D = 15, = 30, and = 30. The arcs are congruent. m = 2(90) or 180. So by definition, is a In a circle, two parallel chords cut congruent arcs. semicircle. d. In a circle, two parallel chords cut congruent arcs. 41. MULTIPLE REPRESENTATIONS In this So, = . problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle

with parallel chords and . Connect points A = 4(16) + 6 or 70 and D by drawing segment . = 6(16) – 26 or 70 b. NUMERICAL Use a protractor to find Therefore, the measure of arcs QS and PR are each 70. and . Then determine and . What is true about these arcs? Explain. CCSS ARGUMENTS Determine whether the c. VERBAL Draw another circle and repeat parts a quadrilateral can always, sometimes, or never and b. Make a conjecture about arcs of a circle that be inscribed in a circle. Explain your reasoning. are cut by two parallel chords. 42. square d. ANALYTICAL Use your conjecture to find SOLUTION: and in the figure. Verify by using Squares have right angles at each vertex, therefore inscribed angles to find the measures of the arcs. each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: SOLUTION: Rectangles have right angles at each vertex, a. therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram SOLUTION: The opposite angles of a parallelogram are always b. Sample answer: , ; congruent. They will only be supplementary when they are right angles. So, a parallelogram can be , ; The arcs are congruent inscribed in a circle as long as it is a rectangle. because they have equal measures. Therefore, the statement is sometimes true. c. Sample answer: 45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not m∠A = 15, m∠D = 15, = 30, and = 30. supplementary, they can not be inscribed in a circle. The arcs are congruent. Therefore, the statement is sometimes true. In a circle, two parallel chords cut congruent arcs. 46. kite d. In a circle, two parallel chords cut congruent arcs. So, = . SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must = 4(16) + 6 or 70 also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are = 6(16) – 26 or 70 right angles, a kite can be inscribed in a circle. Therefore, the measure of arcs QS and PR are each Therefore, the statement is sometimes true. 70. 47. CHALLENGE A square is inscribed in a circle. CCSS ARGUMENTS Determine whether the What is the ratio of the area of the circle to the area quadrilateral can always, sometimes, or never of the square? be inscribed in a circle. Explain your reasoning. 42. square SOLUTION: A square with side s is inscribed in a circle with SOLUTION: radius r. Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Using the Pythagorean Theorem, Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram 2 The area of a square of side s is A = s and the area SOLUTION: of a circle of radius r is A = . The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: Therefore, the ratio of the area of the circle to the The opposite angles of a rhombus are always congruent. They will only be supplementary when area of the inscribed square is they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite 48. WRITING IN MATH A right angles of rhombi that are not squares are not triangle is inscribed in a circle. If the radius of the supplementary, they can not be inscribed in a circle. circle is given, explain how to find the lengths of the Therefore, the statement is sometimes true. right triangle’s legs. 46. kite SOLUTION: Sample answer: A triangle will have SOLUTION: two inscribed angles of 45 and one of 90. The Exactly one pair of opposite angles of a kite are hypotenuse is across from the 90, or right angle. congruent. To be supplementary, they must each be According to theorem 10.8, an inscribed angle of a a right angle. If one pair of opposite angles for a triangle intercepts a diameter if the angle is a right quadrilateral is supplementary, the other pair must angle. Therefore, the hypotenuse is a diameter and also be supplementary. So, as long as the angles that has a length of 2r. Use trigonometry to find the compose the pair of congruent opposite angles are length of the equal legs. right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true.

47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the square?

SOLUTION: A square with side s is inscribed in a circle with Therefore, the length of each leg of the 45-45-90 radius r. triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work. Using the Pythagorean Theorem, 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: 2 The area of a square of side s is A = s and the area An inscribed angle has its vertex on the circle. A of a circle of radius r is A = . central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle. Therefore, the ratio of the area of the circle to the area of the inscribed square is 51. In the circle below, and . What is ? 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The A 42 hypotenuse is across from the 90, or right angle. B 61 According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right C 80 angle. Therefore, the hypotenuse is a diameter and D 84 has a length of 2r. Use trigonometry to find the SOLUTION: length of the equal legs. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

But, Therefore,

Therefore, the length of each leg of the 45-45-90 The correct choice is A. triangle can be found by multiplying the radius of the 52. ALGEBRA Simplify circle in which it is inscribed by . 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. F 2 OPEN ENDED 9x + 3x – 14 49. Find and sketch a real-world logo 2 with an inscribed polygon. G 9x + 13x – 14 H 2 SOLUTION: 27x + 37x – 38 2 See students’ work. J 27x + 27x – 26 SOLUTION: 50. WRITING IN MATH Compare and contrast Use the Distributive Property to simplify the first inscribed angles and central angles of a circle. If they term. intercept the same arc, how are they related?

SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the Therefore, the correct choice is H. measure of the inscribed angle is equal to m(arc 53. SHORT RESPONSE In the circle below, is a AB). So, if an inscribed angle and a central angle diameter, AC = 8 inches, and BC = 15 inches. Find intercept the same arc, then the measure of the the diameter, the radius, and the circumference of inscribed angle is one-half the measure of the central the circle. angle.

51. In the circle below, and . What is ?

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle.

A 42 B 61 The radius is half the diameter. So, the radius of the circle is about 8.5 in. C 80 The circumference C of a circle of a circle of D 84 diameter d is given by SOLUTION: Therefore, the circumference of the circle is If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, 54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? A –15 But, B –16 Therefore, C –17 The correct choice is A. D –18 E –19 52. ALGEBRA Simplify 2 SOLUTION: 4(3x – 2)(2x + 4) + 3x + 5x – 6. 2 Let the three consecutive integers be x , x +1,and x + F 9x + 3x – 14 2. Then, G 9x2 + 13x – 14 2 H 27x + 37x – 38 2 J 27x + 27x – 26 SOLUTION: So, the three integers are –17, –16, and –15 and the least of these is –17. Use the Distributive Property to simplify the first Therefore, the correct choice is C. term.

In , FL = 24, HJ = 48, and . Find each measure.

Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of 55. FG the circle. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

SOLUTION: 56. Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter SOLUTION: of the circle. If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, The radius is half the diameter. So, the radius of the bisects Therefore, circle is about 8.5 in. The circumference C of a circle of a circle of 57. NJ diameter d is given by SOLUTION: Therefore, the circumference of the circle is If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, SAT/ACT 54. The sum of three consecutive integers is –48. What is the least of the three integers? A –15 B –16 C –17 58. D –18 SOLUTION: E –19 If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, SOLUTION: Let the three consecutive integers be x , x +1,and x + bisects Therefore, 2. Then,

Find x.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find 59. each measure. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 60. SOLUTION: The sum of the measures of the central angles of a 56. circle with no interior points in common is 360. So, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 61. bisects Therefore, SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

58.

SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the Find x. image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

60. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of 61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

64. AB = 6y – 14, BC = 10 – 2y, AC = ? 62. PHOTOGRAPHY In one of the first cameras SOLUTION: invented, light entered an opening in the front. An image was reflected in the back of the camera, Since B is the midpoint of upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How

tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, 66. AB = 10s + 2, AC = 40, s = ? the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle. SOLUTION: Since B is the midpoint of

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ?

Find each measure. 1.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

of the angle equals one half the measure of its intercepted arc. Therefore,

intercepted arc. Therefore, ALGEBRA Find each measure. 5. 2.

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same If an angle is inscribed in a circle, then the measure arc or congruent arcs, then the angles are congruent. of the angle equals one half the measure of its So, intercepted arc. Therefore,

3. Therefore,

6.

SOLUTION:

Here, is a semi-circle. So, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. So, intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a Therefore, raindrop to make the colors of the rainbow. If , what is ? 7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Given: bisects . Prove: intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector)

3. intercepts . intercepts . SOLUTION: (Def. of intercepted arc) If two inscribed angles of a circle intercept the same 4. (Inscribed of same arc are arc or congruent arcs, then the angles are congruent. .) So, 5. (Vertical are .) 6. (AAS)

Therefore, CCSS STRUCTURE Find each value. 8. 6.

SOLUTION:

An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. If two inscribed angles of a circle intercept the same So, arc or congruent arcs, then the angles are congruent. The sum of the measures of the angles of a triangle So,

is 180. So,

Therefore, Therefore,

7. PROOF Write a two-column proof. 9. x Given: bisects . Prove:

SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter Given: bisects . or semicircle if and only if the angle is a right angle. Prove: So, Since The sum of the measures of the angles of a triangle

is 180. So,

Proof: 10. and Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) SOLUTION: 5. (Vertical are .) If a quadrilateral is inscribed in a circle, then its 6. (AAS) opposite angles are supplementary.

CCSS STRUCTURE Find each value. Solve the two equations to find the values of x and y. 8.

Use the values of the variables to find and .

SOLUTION: ALGEBRA Find each measure. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 11. So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore, SOLUTION: If an angle is inscribed in a circle, then the measure 9. x of the angle equals one half the measure of its intercepted arc. Therefore,

12.

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since SOLUTION: The sum of the measures of the angles of a triangle If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 11-4is Inscribed 180. So, Angles intercepted arc. Therefore,

10. and 13.

SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its The sum of the measures of the central angles of a opposite angles are supplementary. circle with no interior points in common is 360. So,

Solve the two equations to find the values of x and y.

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Use the values of the variables to find and

. intercepted arc. Therefore,

14.

ALGEBRA Find each measure. 11.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 15. intercepted arc. Therefore,

12.

SOLUTION: Here,

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore, intercepted arc. Therefore, 16. eSolutions13. Manual - Powered by Cognero Page 3

SOLUTION: SOLUTION:

The sum of the measures of the central angles of a Here, circle with no interior points in common is 360. So, The arc is a semicircle. So,

Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure

of the angle equals one half the measure of its intercepted arc. Therefore,

intercepted arc. Therefore, 17. ALGEBRA Find each measure.

14.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. intercepted arc. Therefore, So,

15. 18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. intercepted arc. So, So,

Therefore,

16. Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = intercepted arc. Therefore, m∠B.

17. ALGEBRA Find each measure. Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, SOLUTION: 18. ALGEBRA Find each measure. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof SOLUTION: If two inscribed angles of a circle intercept the same Given: arc or congruent arcs, then the angles are congruent. So, Prove:

Therefore,

19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

. Since and

, the equation becomes

SOLUTION: . Multiplying each side of the If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. equation by 2 results in . Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. 22. two-column proof Given: Prove: Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc

are .) 3. (Vertical are .) SOLUTION: 4. (AA Similarity) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. ALGEBRA Find each value. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47. 23. x PROOF Write the specified type of proof. 21. paragraph proof SOLUTION: Given: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Prove: So, The sum of the measures of the angles of a triangle

is 180. So,

24. SOLUTION: SOLUTION: Proof: Given means that An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. . Since and So, The sum of the measures of the angles of a triangle , the equation becomes is 180. So, . Multiplying each side of the Therefore, equation by 2 results in . 25. x 22. two-column proof SOLUTION: Given: An inscribed angle of a triangle intercepts a diameter Prove: or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

SOLUTION: 26. Statements (Reasons): SOLUTION: 1. (Given) An inscribed angle of a triangle intercepts a diameter 2. (Inscribed intercepting same arc or semicircle if and only if the angle is a right angle. are .) So,

3. (Vertical are .) The sum of the measures of the angles of a triangle 4. (AA Similarity) ALGEBRA Find each value. is 180. So,

Therefore,

CCSS STRUCTURE Find each measure.

23. x

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 27. So, The sum of the measures of the angles of a triangle SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. is 180. So,

Therefore,

24. 28. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter If a quadrilateral is inscribed in a circle, then its or semicircle if and only if the angle is a right angle. opposite angles are supplementary.

So, The sum of the measures of the angles of a triangle

is 180. So, Therefore,

Therefore, CCSS STRUCTURE Find each measure. 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 29. is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Therefore, So, The sum of the measures of the angles of a triangle 30. is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its

opposite angles are supplementary. Therefore, CCSS STRUCTURE Find each measure.

Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: Given: Quadrilateral ABCD is inscribed in . 27. Prove: and are supplementary. SOLUTION: and are supplementary. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

28. SOLUTION:

If a quadrilateral is inscribed in a circle, then its Proof: By arc addition and the definitions of arc

opposite angles are supplementary. measure and the sum of central angles, . Since by Theorem 10.6

and

Therefore, , but

CCSS STRUCTURE Find each measure. , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making them supplementary also.

SIGNS: A stop sign in the shape of a regular

29. octagon is inscribed in a circle. Find each SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, measure. 32. 30. SOLUTION: SOLUTION: Since all the sides of the stop sign are congruent If a quadrilateral is inscribed in a circle, then its chords of the circle, the corresponding arcs are opposite angles are supplementary. congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 or 45. Therefore,

Therefore, measure of arc NPQ is 135. 31. PROOF Write a paragraph proof for Theorem 11.9.

SOLUTION: 33. Given: Quadrilateral ABCD is inscribed in . SOLUTION: Prove: and are supplementary. and are supplementary. Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6

and Therefore, the measure of ∠RLQ is 22.5. , but

, so or 34. 180. This makes ∠C and ∠A supplementary. SOLUTION: Because the sum of the measures of the interior If an angle is inscribed in a circle, then the measure angles of a quadrilateral is 360, of the angle equals one half the measure of its . But , so , making intercepted arc. So, them supplementary also. The sum of the measures of the central angles of a SIGNS: A stop sign in the shape of a regular circle with no interior points in common is 360. Here, octagon is inscribed in a circle. Find each 360 is divided into 8 equal arcs and each arc

measures Then,

Therefore,

measure. 35. 32. SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are intercepted arc. So, congruent. There are eight adjacent arcs that make The sum of the measures of the central angles of a up the circle, so their sum is 360. Thus, the measure circle with no interior points in common is 360. Here, of each arc joining consecutive vertices is of 360 360 is divided into 8 equal arcs and each arc

or 45. measures

Therefore, measure of arc NPQ is 135. Then,

Therefore, 33.

SOLUTION: 36. ART Four different string art star patterns are Since all the sides of the stop sign are congruent shown. If all of the inscribed angles of each star chords of the circle, all the corresponding arcs are shown are congruent, find the measure of each congruent. There are eight adjacent arcs that make inscribed angle. up the circle, so their sum is 360. Thus, the measure a. of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

b.

Therefore, the measure of ∠RLQ is 22.5.

c. 34. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Then, SOLUTION: a. The sum of the measures of the central angles of Therefore, a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc 35. measures SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed

intercepted arc. So, angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, measures Each inscribed angle is formed 360 is divided into 8 equal arcs and each arc by intercepting alternate vertices of the star. So, each

measures intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one Then, half the measure of its intercepted arc. So, the measure of each inscribed angle is 60.

Therefore, c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a 36. ART Four different string art star patterns are circle, then the measure of the angle equals one half shown. If all of the inscribed angles of each star the measure of its intercepted arc. So, the measure shown are congruent, find the measure of each of each inscribed angle is about 25.7. inscribed angle. d. a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed b. in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6.

c. 37. Case 2 Given: P lies inside . is a diameter.

Prove:

d.

SOLUTION: SOLUTION: Proof: Statements (Reasons) a. The sum of the measures of the central angles of a circle with no interior points in common is 360. 1. m∠ABC = m∠ABD + m∠DBC ( Addition

Here, 360 is divided into 5 equal arcs and each arc Postulate) 2. m∠ABD = m(arc AD) measures m∠DBC = m(arc DC) (The measure of an If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its inscribed whose side is a diameter is half the

intercepted arc. So, the measure of each inscribed measure of the intercepted arc (Case 1)). angle is 36. 3. m∠ABC = m(arc AD) + m(arc DC) b. Here, 360 is divided into 6 equal arcs and each arc (Substitution) measures Each inscribed angle is formed 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)

by intercepting alternate vertices of the star. So, each 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc intercepted arc measures 120. If an angle is inscribed Addition Postulate) in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the 6. m∠ABC = m(arc AC) (Substitution) measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc 38. Case 3 measures If an angle is inscribed in a Given: P lies outside . is a diameter.

circle, then the measure of the angle equals one half Prove: the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one SOLUTION: half the measure of its intercepted arc. So, the Proof: measure of each inscribed angle is 45. Statements (Reasons) 1. m ABC = m DBC – m DBA ( Addition PROOF Write a two-column proof for each case ∠ ∠ ∠ Postulate, Subtraction Property of Equality) of Theorem 11.6. 37. Case 2 2. m∠DBC = m(arc DC) Given: P lies inside . is a diameter. m∠DBA = m(arc DA) (The measure of an

Prove: inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction SOLUTION: Property of Equality) Proof: 7. m∠ABC = m(arc AC) (Substitution) Statements (Reasons)

1. m∠ABC = m∠ABD + m∠DBC ( Addition

Postulate) PROOF Write the specified proof for each 2. m∠ABD = m(arc AD) theorem. 39. Theorem 11.7, two-column proof m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the SOLUTION: measure of the intercepted arc (Case 1)). Given: and are inscribed; 3. m∠ABC = m(arc AD) + m(arc DC) Prove: (Substitution) Proof: 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)

5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter. Statements (Reasons)

Prove: 1. and are inscribed;

(Given) 2. ; (Measure of an inscribed = half measure of intercepted arc.) 3. (Def. of arcs)

4. (Mult. Prop. of Equality)

5. (Substitution) SOLUTION: 6. (Def. of ) Proof:

Statements (Reasons) 40. Theorem 11.8, paragraph proof 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) SOLUTION: 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

Part I: Given: is a semicircle. 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Prove: is a right angle. Addition Postulate) Proof: Since is a semicircle, then 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) . Since is an inscribed angle, 7. m∠ABC = m(arc AC) (Substitution) then or 90. So, by definition,

is a right angle. PROOF Write the specified proof for each theorem. Part II: Given: is a right angle. 39. Theorem 11.7, two-column proof Prove: is a semicircle. SOLUTION: Proof: Since is an inscribed angle, then Given: and are inscribed; Prove: and by the Multiplication Proof: Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. Statements (Reasons) a. GEOMETRIC Use a compass to draw a circle 1. and are inscribed; with parallel chords and . Connect points A (Given) and D by drawing segment . b. NUMERICAL Use a protractor to find 2. ; (Measure and . Then determine and . of an inscribed = half measure of intercepted arc.) What is true about these arcs? Explain. 3. (Def. of arcs) c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that 4. (Mult. Prop. of Equality) are cut by two parallel chords. d. ANALYTICAL 5. (Substitution) Use your conjecture to find 6. (Def. of ) and in the figure. Verify by using inscribed angles to find the measures of the arcs. 40. Theorem 11.8, paragraph proof SOLUTION:

SOLUTION: a.

Part I: Given: is a semicircle. Prove: is a right angle.

Proof: Since is a semicircle, then b. Sample answer: , ; . Since is an inscribed angle, , ; The arcs are congruent because they have equal measures. then or 90. So, by definition, c. Sample answer: is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then

and by the Multiplication m∠A = 15, m∠D = 15, = 30, and = 30. The arcs are congruent. . Property of Equality, m = 2m Because In a circle, two parallel chords cut congruent arcs. is a right angle, m = 90. Then d. In a circle, two parallel chords cut congruent arcs. m = 2(90) or 180. So by definition, is a So, = . semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel = 4(16) + 6 or 70 chords. a. GEOMETRIC Use a compass to draw a circle = 6(16) – 26 or 70 with parallel chords and . Connect points A Therefore, the measure of arcs QS and PR are each 70. and D by drawing segment . b. NUMERICAL Use a protractor to find CCSS ARGUMENTS Determine whether the and . Then determine and . quadrilateral can always, sometimes, or never What is true about these arcs? Explain. be inscribed in a circle. Explain your reasoning. 42. square c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that SOLUTION: are cut by two parallel chords. Squares have right angles at each vertex, therefore d. ANALYTICAL Use your conjecture to find each pair of opposite angles will be supplementary and in the figure. Verify by using and inscribed in a circle. Therefore, the statement is

inscribed angles to find the measures of the arcs. always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, SOLUTION: the statement is always true. a. 44. parallelogram SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. b. Sample answer: , ; Therefore, the statement is sometimes true. , ; The arcs are congruent because they have equal measures. 45. rhombus c. Sample answer: SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true.

m∠A = 15, m∠D = 15, = 30, and = 30. 46. kite The arcs are congruent. In a circle, two parallel chords cut congruent arcs. SOLUTION: d. In a circle, two parallel chords cut congruent arcs. Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be So, = . a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle.

= 4(16) + 6 or 70 Therefore, the statement is sometimes true. = 6(16) – 26 or 70 CHALLENGE Therefore, the measure of arcs QS and PR are each 47. A square is inscribed in a circle. 70. What is the ratio of the area of the circle to the area of the square? CCSS ARGUMENTS Determine whether the SOLUTION: quadrilateral can always, sometimes, or never A square with side s is inscribed in a circle with be inscribed in a circle. Explain your reasoning. radius r. 42. square SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

Using the Pythagorean Theorem, 43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, 2 the statement is always true. The area of a square of side s is A = s and the area of a circle of radius r is A = . 44. parallelogram SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus Therefore, the ratio of the area of the circle to the SOLUTION: area of the inscribed square is The opposite angles of a rhombus are always 48. WRITING IN MATH A right congruent. They will only be supplementary when triangle is inscribed in a circle. If the radius of the they are right angles. So, a rhombus can be inscribed circle is given, explain how to find the lengths of the in a circle as long as it is a square. Since the opposite right triangle’s legs. angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. SOLUTION: Therefore, the statement is sometimes true. Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The 46. kite hypotenuse is across from the 90, or right angle. SOLUTION: According to theorem 10.8, an inscribed angle of a Exactly one pair of opposite angles of a kite are triangle intercepts a diameter if the angle is a right congruent. To be supplementary, they must each be angle. Therefore, the hypotenuse is a diameter and a right angle. If one pair of opposite angles for a has a length of 2r. Use trigonometry to find the quadrilateral is supplementary, the other pair must length of the equal legs. also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true.

47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the square? Therefore, the length of each leg of the 45-45-90 SOLUTION: triangle can be found by multiplying the radius of the A square with side s is inscribed in a circle with circle in which it is inscribed by . radius r. 49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work.

WRITING IN MATH 50. Compare and contrast Using the Pythagorean Theorem, inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle.

2 If a central angle intercepts arc AB, then the The area of a square of side s is A = s and the area measure of the central angle is equal to m(arc AB). If of a circle of radius r is A = . an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . Therefore, the ratio of the area of the circle to the What is ? area of the inscribed square is

48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the circle is given, explain how to find the lengths of the right triangle’s legs.

SOLUTION: A 42 Sample answer: A triangle will have B 61 two inscribed angles of 45 and one of 90. The C 80 hypotenuse is across from the 90, or right angle. D 84 According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right SOLUTION: angle. Therefore, the hypotenuse is a diameter and If an angle is inscribed in a circle, then the measure has a length of 2r. Use trigonometry to find the of the angle equals one half the measure of its length of the equal legs. intercepted arc. Therefore,

But, Therefore, The correct choice is A.

52. ALGEBRA Simplify Therefore, the length of each leg of the 45-45-90 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. triangle can be found by multiplying the radius of the 2 F 9x + 3x – 14 circle in which it is inscribed by . G 9x2 + 13x – 14 H 2 49. OPEN ENDED Find and sketch a real-world logo 27x + 37x – 38 2 with an inscribed polygon. J 27x + 27x – 26 SOLUTION: SOLUTION: See students’ work. Use the Distributive Property to simplify the first term. 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. Therefore, the correct choice is H. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If 53. SHORT RESPONSE In the circle below, is a an inscribed angle also intercepts arc AB, then the diameter, AC = 8 inches, and BC = 15 inches. Find measure of the inscribed angle is equal to m(arc the diameter, the radius, and the circumference of AB). So, if an inscribed angle and a central angle the circle. intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . What is ? SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle.

The radius is half the diameter. So, the radius of the circle is about 8.5 in. A 42 The circumference C of a circle of a circle of B 61 diameter d is given by C 80 Therefore, the circumference of the circle is D 84 SOLUTION: 54. SAT/ACT The sum of three consecutive integers is If an angle is inscribed in a circle, then the measure –48. What is the least of the three integers? of the angle equals one half the measure of its intercepted arc. Therefore, A –15 B –16 C –17 D But, –18 E Therefore, –19 The correct choice is A. SOLUTION: Let the three consecutive integers be x , x +1,and x + ALGEBRA 52. Simplify 2. Then, 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 F 9x + 3x – 14 G 2 9x + 13x – 14 2 H 27x + 37x – 38 So, the three integers are –17, –16, and –15 and the 2 J 27x + 27x – 26 least of these is –17. Therefore, the correct choice is C. SOLUTION: Use the Distributive Property to simplify the first In , FL = 24, HJ = 48, and . Find term. each measure.

Therefore, the correct choice is H. 55. FG 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find SOLUTION: the diameter, the radius, and the circumference of If a diameter (or radius) of a circle is perpendicular the circle. to a chord, then it bisects the chord and its arc. So, bisects Therefore,

56.

SOLUTION: SOLUTION: Use the Pythagorean Theorem to find the If a diameter (or radius) of a circle is perpendicular hypotenuse of the triangle which is also the diameter to a chord, then it bisects the chord and its arc. So, of the circle. bisects Therefore,

57. NJ The radius is half the diameter. So, the radius of the circle is about 8.5 in. SOLUTION: The circumference C of a circle of a circle of If a diameter (or radius) of a circle is perpendicular diameter d is given by to a chord, then it bisects the chord and its arc. So, Therefore, the circumference of the circle is bisects Therefore,

54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? 58. A –15 B –16 SOLUTION: C –17 If a diameter (or radius) of a circle is perpendicular D –18 to a chord, then it bisects the chord and its arc. So, E –19 bisects Therefore,

SOLUTION: Let the three consecutive integers be x , x +1,and x + 2. Then, Find x.

So, the three integers are –17, –16, and –15 and the least of these is –17. 59. Therefore, the correct choice is C. SOLUTION: In , FL = 24, HJ = 48, and . Find The sum of the measures of the central angles of a each measure. circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 60. bisects Therefore, SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

56. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: 61. If a diameter (or radius) of a circle is perpendicular SOLUTION: to a chord, then it bisects the chord and its arc. So, The sum of the measures of the central angles of a bisects Therefore, circle with no interior points in common is 360. So,

58. SOLUTION: If a diameter (or radius) of a circle is perpendicular 62. PHOTOGRAPHY In one of the first cameras to a chord, then it bisects the chord and its arc. So, invented, light entered an opening in the front. An image was reflected in the back of the camera, bisects Therefore, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is Find x. 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

59. SOLUTION: The sum of the measures of the central angles of a

circle with no interior points in common is 360. So, SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

60. SOLUTION: Therefore, the person being photographed is 5.6 ft The sum of the measures of the central angles of a tall. circle with no interior points in common is 360. So, ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

61.

SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

SOLUTION: The height of the person being photographed is the 66. AB = 10s + 2, AC = 40, s = ? length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will SOLUTION: be proportional. One foot is equivalent to 12 in. Then, Since B is the midpoint of the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

Find each measure. 1. SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

If an angle is inscribed in a circle, then the measure intercepted arc. So, of the angle equals one half the measure of its Therefore,

intercepted arc. Therefore, 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ? 2.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 3. ALGEBRA Find each measure. 5.

SOLUTION: Here, is a semi-circle. So, SOLUTION: If two inscribed angles of a circle intercept the same If an angle is inscribed in a circle, then the measure arc or congruent arcs, then the angles are congruent. of the angle equals one half the measure of its So, intercepted arc. So,

Therefore, Therefore, 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If 6. , what is ?

SOLUTION: If two inscribed angles of a circle intercept the same SOLUTION: arc or congruent arcs, then the angles are congruent. If an angle is inscribed in a circle, then the measure So, of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. Therefore, 5. 7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION:

So, Given: bisects .

Prove:

Therefore,

6.

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) SOLUTION: 4. (Inscribed of same arc are If two inscribed angles of a circle intercept the same .) arc or congruent arcs, then the angles are congruent. 5. (Vertical are .) So,

6. (AAS) CCSS STRUCTURE Find each value. 8. Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, SOLUTION: The sum of the measures of the angles of a triangle

Given: bisects . Prove: is 180. So,

Therefore,

9. x

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are SOLUTION: .) An inscribed angle of a triangle intercepts a diameter 5. (Vertical are .) or semicircle if and only if the angle is a right angle. 6. (AAS) So, Since The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each value. 8. is 180. So,

10. and

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, SOLUTION: The sum of the measures of the angles of a triangle If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. is 180. So, Solve the two equations to find the values of x and y. Therefore,

9. x Use the values of the variables to find and .

ALGEBRA Find each measure. 11. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

10. and intercepted arc. Therefore,

12.

SOLUTION: If a quadrilateral is inscribed in a circle, then its

opposite angles are supplementary. SOLUTION:

If an angle is inscribed in a circle, then the measure Solve the two equations to find the values of x and y. of the angle equals one half the measure of its

intercepted arc. Therefore, Use the values of the variables to find and . 13.

ALGEBRA Find each measure. 11.

SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore,

12. 14.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 15. 13.

SOLUTION:

Here, SOLUTION: The arc is a semicircle. So, The sum of the measures of the central angles of a If an angle is inscribed in a circle, then the measure circle with no interior points in common is 360. So, of the angle equals one half the measure of its

intercepted arc. So, Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 16.

intercepted arc. Therefore,

14.

SOLUTION: Here, SOLUTION: The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure Then, of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc. Therefore, of the angle equals one half the measure of its

intercepted arc. Therefore, 15. 17. ALGEBRA Find each measure.

SOLUTION: Here,

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: If two inscribed angles of a circle intercept the same intercepted arc. So, 11-4 Inscribed Angles arc or congruent arcs, then the angles are congruent. Therefore, So,

16. 18. ALGEBRA Find each measure.

SOLUTION: Here, SOLUTION: If two inscribed angles of a circle intercept the same The arc is a semicircle. So, arc or congruent arcs, then the angles are congruent.

Then, So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, Therefore,

19. ALGEBRA Find each measure. 17. ALGEBRA Find each measure.

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. arc or congruent arcs, then the angles are congruent. So, Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. 18. ALGEBRA Find each measure.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

SOLUTION: eSolutionsTherefore,Manual - Powered by Cognero If two inscribed angles of a circle intercept the samePage 4 arc or congruent arcs, then the angles are congruent. 19. ALGEBRA Find each measure. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given:

SOLUTION: Prove: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20. SOLUTION:

20. ALGEBRA Find each measure. Proof: Given means that

. Since and

, the equation becomes

. Multiplying each side of the

equation by 2 results in . SOLUTION: 22. two-column proof If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Given: Since ∠C and ∠D both intercept arc AB, m∠C = Prove: m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof SOLUTION: Statements (Reasons): Given: 1. (Given) 2. (Inscribed intercepting same arc Prove: are .) 3. (Vertical are .) 4. (AA Similarity)

ALGEBRA Find each value.

SOLUTION:

Proof: Given means that

23. x . Since and

, the equation becomes SOLUTION: An inscribed angle of a triangle intercepts a diameter . Multiplying each side of the or semicircle if and only if the angle is a right angle. So, equation by 2 results in . The sum of the measures of the angles of a triangle

is 180. So, 22. two-column proof Given: Prove: 24. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So,

The sum of the measures of the angles of a triangle SOLUTION: Statements (Reasons): is 180. So, 1. (Given) 2. (Inscribed intercepting same arc Therefore, are .) 3. (Vertical are .) 25. x 4. (AA Similarity) SOLUTION: ALGEBRA Find each value. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

23. x

26. SOLUTION: An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. An inscribed angle of a triangle intercepts a diameter So, or semicircle if and only if the angle is a right angle. The sum of the measures of the angles of a triangle So, The sum of the measures of the angles of a triangle

is 180. So, is 180. So,

24. Therefore, SOLUTION: CCSS STRUCTURE Find each measure. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 27. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 25. x opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, 28. The sum of the measures of the angles of a triangle SOLUTION: If a quadrilateral is inscribed in a circle, then its is 180. So, opposite angles are supplementary.

26.

SOLUTION: Therefore, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. CCSS STRUCTURE Find each measure. So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

CCSS STRUCTURE Find each measure. 29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

27. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 30. opposite angles are supplementary. SOLUTION: Therefore, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 28.

SOLUTION: If a quadrilateral is inscribed in a circle, then its Therefore, opposite angles are supplementary.

31. PROOF Write a paragraph proof for Theorem 11.9.

SOLUTION: Therefore, Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. CCSS STRUCTURE Find each measure. and are supplementary.

29. Proof: By arc addition and the definitions of arc SOLUTION: measure and the sum of central angles, If a quadrilateral is inscribed in a circle, then its . Since by Theorem 10.6 opposite angles are supplementary. and , but

Therefore, , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior 30. angles of a quadrilateral is 360, SOLUTION: . But If a quadrilateral is inscribed in a circle, then its , so , making opposite angles are supplementary. them supplementary also.

SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. measure. and are supplementary. 32. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360

Proof: By arc addition and the definitions of arc or 45. measure and the sum of central angles,

. Since by Theorem 10.6 Therefore, measure of arc NPQ is 135. and 33. , but SOLUTION: , so or Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior congruent. There are eight adjacent arcs that make angles of a quadrilateral is 360, up the circle, so their sum is 360. Thus, the measure . But of each arc joining consecutive vertices is of 360 , so , making or 45. Since ∠RLQ is inscribed in the circle, its them supplementary also. measure equals one half the measure of its intercepted arc QR. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore, the measure of ∠RLQ is 22.5.

measure. 34. 32. SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are intercepted arc. So, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, of each arc joining consecutive vertices is of 360 360 is divided into 8 equal arcs and each arc or 45. measures

Therefore, measure of arc NPQ is 135. Then,

Therefore, 33. SOLUTION: 35. Since all the sides of the stop sign are congruent SOLUTION: chords of the circle, all the corresponding arcs are If an angle is inscribed in a circle, then the measure congruent. There are eight adjacent arcs that make of the angle equals one half the measure of its up the circle, so their sum is 360. Thus, the measure

of each arc joining consecutive vertices is of 360 intercepted arc. So, or 45. Since ∠RLQ is inscribed in the circle, its The sum of the measures of the central angles of a measure equals one half the measure of its circle with no interior points in common is 360. Here, intercepted arc QR. 360 is divided into 8 equal arcs and each arc

measures

Then,

Therefore, Therefore, the measure of ∠RLQ is 22.5.

36. ART Four different string art star patterns are 34. shown. If all of the inscribed angles of each star shown are congruent, find the measure of each SOLUTION: inscribed angle. If an angle is inscribed in a circle, then the measure a. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here,

360 is divided into 8 equal arcs and each arc b.

measures Then,

Therefore,

c. 35. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

Then, SOLUTION: a. The sum of the measures of the central angles of Therefore, a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc

36. ART Four different string art star patterns are measures shown. If all of the inscribed angles of each star shown are congruent, find the measure of each If an angle is inscribed in a circle, then the measure inscribed angle. of the angle equals one half the measure of its a. intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed

by intercepting alternate vertices of the star. So, each b. intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a c. circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc d. measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. SOLUTION: a. The sum of the measures of the central angles of PROOF Write a two-column proof for each case a circle with no interior points in common is 360. of Theorem 11.6. Here, 360 is divided into 5 equal arcs and each arc 37. Case 2 Given: P lies inside . is a diameter. measures Prove: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed

by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one SOLUTION: half the measure of its intercepted arc. So, the Proof: measure of each inscribed angle is 60. Statements (Reasons) c. Here, 360 is divided into 7 equal arcs and each arc 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) measures If an angle is inscribed in a 2. m∠ABD = m(arc AD) circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure m∠DBC = m(arc DC) (The measure of an of each inscribed angle is about 25.7. inscribed whose side is a diameter is half the d. The sum of the measures of the central angles of measure of the intercepted arc (Case 1)). a circle with no interior points in common is 360. 3. m∠ABC = m(arc AD) + m(arc DC) Here, 360 is divided into 8 equal arcs and each arc (Substitution) measures Each inscribed angle is formed 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc in a circle, then the measure of the angle equals one Addition Postulate) half the measure of its intercepted arc. So, the 6. m∠ABC = m(arc AC) (Substitution) measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case 38. Case 3 of Theorem 11.6. Given: P lies outside . is a diameter. 37. Case 2 Given: P lies inside . is a diameter. Prove:

Prove:

SOLUTION: SOLUTION: Proof: Proof: Statements (Reasons) Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate, Subtraction Property of Equality) Postulate) 2. m∠DBC = m(arc DC)

2. m∠ABD = m(arc AD) m∠DBA = m(arc DA) (The measure of an m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) 3. m∠ABC = m(arc AD) + m(arc DC) (Substitution) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 6. m∠ABC = m(arc AC) (Substitution) 7. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter. PROOF Write the specified proof for each theorem. Prove: 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof:

SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition

Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) Statements (Reasons) 1. and are inscribed; m∠DBA = m(arc DA) (The measure of an (Given) inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 2. ; (Measure 3. m∠ABC = m(arc DC) – m(arc DA) of an inscribed = half measure of intercepted arc.) (Substitution) 3. (Def. of arcs) 4. m ABC = [m(arc DC) – m(arc DA)] (Factor) ∠ 4. (Mult. Prop. of Equality) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc

Addition Postulate) 5. (Substitution) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 6. (Def. of ) Property of Equality) 7. m∠ABC = m(arc AC) (Substitution) 40. Theorem 11.8, paragraph proof SOLUTION:

PROOF Write the specified proof for each theorem. 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof: Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition, is a right angle.

Statements (Reasons) Part II: Given: is a right angle. 1. and are inscribed; Prove : is a semicircle. (Given) 2. ; (Measure Proof: Since is an inscribed angle, then of an inscribed = half measure of intercepted arc.) and by the Multiplication

3. (Def. of arcs) Property of Equality, m = 2m . Because

4. (Mult. Prop. of Equality) is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a

5. (Substitution) semicircle. 6. (Def. of ) 41. MULTIPLE REPRESENTATIONS In this 40. Theorem 11.8, paragraph proof problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel SOLUTION: chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain.

c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that Part I: Given: is a semicircle. are cut by two parallel chords. Prove: is a right angle. d. ANALYTICAL Use your conjecture to find Proof: Since is a semicircle, then and in the figure. Verify by using . Since is an inscribed angle, inscribed angles to find the measures of the arcs. then or 90. So, by definition, is a right angle.

Part II: Given: is a right angle.

Prove: is a semicircle. SOLUTION: a. Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then

m = 2(90) or 180. So by definition, is a b. Sample answer: , ; semicircle. , ; The arcs are congruent 41. MULTIPLE REPRESENTATIONS In this because they have equal measures. problem, you will investigate the relationship between c. Sample answer: the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain. m∠A = 15, m∠D = 15, = 30, and = 30. c. VERBAL Draw another circle and repeat parts a The arcs are congruent. and b. Make a conjecture about arcs of a circle that In a circle, two parallel chords cut congruent arcs. are cut by two parallel chords. d. In a circle, two parallel chords cut congruent arcs. d. ANALYTICAL Use your conjecture to find So, = . and in the figure. Verify by using inscribed angles to find the measures of the arcs.

= 4(16) + 6 or 70 = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each 70. SOLUTION: a. CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary b. Sample answer: , ; and inscribed in a circle. Therefore, the statement is always true. , ; The arcs are congruent because they have equal measures. c. Sample answer: 43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram

m∠A = 15, m∠D = 15, = 30, and = 30. SOLUTION: The arcs are congruent. The opposite angles of a parallelogram are always In a circle, two parallel chords cut congruent arcs. congruent. They will only be supplementary when d. In a circle, two parallel chords cut congruent arcs. they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. So, = . Therefore, the statement is sometimes true.

45. rhombus SOLUTION: = 4(16) + 6 or 70 The opposite angles of a rhombus are always congruent. They will only be supplementary when = 6(16) – 26 or 70 they are right angles. So, a rhombus can be inscribed Therefore, the measure of arcs QS and PR are each in a circle as long as it is a square. Since the opposite 70. angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. CCSS ARGUMENTS Determine whether the Therefore, the statement is sometimes true. quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 46. kite 42. square SOLUTION: SOLUTION: Exactly one pair of opposite angles of a kite are Squares have right angles at each vertex, therefore congruent. To be supplementary, they must each be each pair of opposite angles will be supplementary a right angle. If one pair of opposite angles for a and inscribed in a circle. Therefore, the statement is quadrilateral is supplementary, the other pair must always true. also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are 43. rectangle right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. SOLUTION: Rectangles have right angles at each vertex, 47. CHALLENGE A square is inscribed in a circle. therefore each pair of opposite angles will be What is the ratio of the area of the circle to the area supplementary and inscribed in a circle. Therefore, of the square? the statement is always true. SOLUTION: 44. parallelogram A square with side s is inscribed in a circle with radius r. SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus Using the Pythagorean Theorem, SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite 2 angles of rhombi that are not squares are not The area of a square of side s is A = s and the area supplementary, they can not be inscribed in a circle. of a circle of radius r is A = . Therefore, the statement is sometimes true.

46. kite SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that Therefore, the ratio of the area of the circle to the compose the pair of congruent opposite angles are area of the inscribed square is right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the 47. CHALLENGE A square is inscribed in a circle. circle is given, explain how to find the lengths of the What is the ratio of the area of the circle to the area right triangle’s legs. of the square? SOLUTION: SOLUTION: Sample answer: A triangle will have A square with side s is inscribed in a circle with two inscribed angles of 45 and one of 90. The radius r. hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs.

Using the Pythagorean Theorem,

2 Therefore, the length of each leg of the 45-45-90 The area of a square of side s is A = s and the area triangle can be found by multiplying the radius of the of a circle of radius r is A = . circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work.

Therefore, the ratio of the area of the circle to the 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they area of the inscribed square is intercept the same arc, how are they related?

48. WRITING IN MATH A right SOLUTION: triangle is inscribed in a circle. If the radius of the An inscribed angle has its vertex on the circle. A circle is given, explain how to find the lengths of the central angle has its vertex at the center of the circle. right triangle’s legs. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If SOLUTION: an inscribed angle also intercepts arc AB, then the Sample answer: A triangle will have measure of the inscribed angle is equal to m(arc two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. AB). So, if an inscribed angle and a central angle According to theorem 10.8, an inscribed angle of a intercept the same arc, then the measure of the triangle intercepts a diameter if the angle is a right inscribed angle is one-half the measure of the central

angle. Therefore, the hypotenuse is a diameter and angle. has a length of 2r. Use trigonometry to find the length of the equal legs. 51. In the circle below, and . What is ?

Therefore, the length of each leg of the 45-45-90 A triangle can be found by multiplying the radius of the 42 B 61 circle in which it is inscribed by . C 80 D 84 49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its See students’ work. intercepted arc. Therefore,

50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? But, Therefore, SOLUTION: The correct choice is A. An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. 52. ALGEBRA Simplify If a central angle intercepts arc AB, then the 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. measure of the central angle is equal to m(arc AB). If 2 an inscribed angle also intercepts arc AB, then the F 9x + 3x – 14 2 measure of the inscribed angle is equal to m(arc G 9x + 13x – 14 2 AB). So, if an inscribed angle and a central angle H 27x + 37x – 38 intercept the same arc, then the measure of the J 27x2 + 27x – 26 inscribed angle is one-half the measure of the central angle. SOLUTION: Use the Distributive Property to simplify the first 51. In the circle below, and . term. What is ?

Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a A 42 diameter, AC = 8 inches, and BC = 15 inches. Find B 61 the diameter, the radius, and the circumference of C 80 the circle. D 84 SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

SOLUTION: Use the Pythagorean Theorem to find the But, hypotenuse of the triangle which is also the diameter Therefore, of the circle. The correct choice is A. 52. ALGEBRA Simplify The radius is half the diameter. So, the radius of the 2 4(3x – 2)(2x + 4) + 3x + 5x – 6. circle is about 8.5 in. 2 The circumference C of a circle of a circle of F 9x + 3x – 14 diameter d is given by 2 G 9x + 13x – 14 Therefore, the circumference of the circle is 2 H 27x + 37x – 38 J 27x2 + 27x – 26 54. SAT/ACT The sum of three consecutive integers is SOLUTION: –48. What is the least of the three integers? Use the Distributive Property to simplify the first A –15 term. B –16 C –17 D –18 E –19 SOLUTION: Let the three consecutive integers be x , x +1,and x +

Therefore, the correct choice is H. 2. Then, 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle. So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure.

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle.

The radius is half the diameter. So, the radius of the 55. FG circle is about 8.5 in. The circumference C of a circle of a circle of SOLUTION: diameter d is given by If a diameter (or radius) of a circle is perpendicular Therefore, the circumference of the circle is to a chord, then it bisects the chord and its arc. So, bisects Therefore,

54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? A –15 56. B –16 SOLUTION: C –17 If a diameter (or radius) of a circle is perpendicular D –18 to a chord, then it bisects the chord and its arc. So, E –19 bisects Therefore, SOLUTION: Let the three consecutive integers be x , x +1,and x + 57. NJ 2. Then, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. 58. In , FL = 24, HJ = 48, and . Find each measure. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

Find x. 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 59.

SOLUTION: The sum of the measures of the central angles of a 56. circle with no interior points in common is 360. So,

SOLUTION:

If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 60. bisects Therefore, SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 58.

SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

Find x. 61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

60. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 62. PHOTOGRAPHY In one of the first cameras 63. AB = 4x – 5, BC = 11 + 2x, AC = ? invented, light entered an opening in the front. An image was reflected in the back of the camera, SOLUTION: upside down, forming similar triangles. Suppose the Since B is the midpoint of image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Therefore, the person being photographed is 5.6 ft Since B is the midpoint of tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? 66. AB = 10s + 2, AC = 40, s = ? SOLUTION: SOLUTION: Since B is the midpoint of Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore, Find each measure. 1. 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ?

intercepted arc. Therefore, SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 2. intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

intercepted arc. Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 3. arc or congruent arcs, then the angles are congruent. So,

Therefore,

6. SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 4. SCIENCE The diagram shows how light bends in a arc or congruent arcs, then the angles are congruent. raindrop to make the colors of the rainbow. If So, , what is ?

Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. SOLUTION: 5. Given: bisects . Prove:

Proof: SOLUTION: Statements (Reasons) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 1. bisects . (Given) So, 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) Therefore, 4. (Inscribed of same arc are .) 6. 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Therefore, The sum of the measures of the angles of a triangle

7. PROOF Write a two-column proof. is 180. So, Given: bisects . Prove: Therefore,

9. x

SOLUTION: Given: bisects .

Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since

Proof: The sum of the measures of the angles of a triangle Statements (Reasons) 1. bisects . (Given) is 180. So, 2. (Def. of segment bisector) 3. intercepts . intercepts . 10. and (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y.

SOLUTION: Use the values of the variables to find and An inscribed angle of a triangle intercepts a diameter . or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

ALGEBRA Find each measure. is 180. So, 11.

Therefore,

9. x

SOLUTION:

An inscribed angle of a triangle intercepts a diameter 12. or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So,

SOLUTION: 10. and If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

13.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y. SOLUTION: The sum of the measures of the central angles of a Use the values of the variables to find and circle with no interior points in common is 360. So, .

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its ALGEBRA Find each measure. intercepted arc. Therefore, 11.

14.

SOLUTION:

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, 12. 15.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Here,

intercepted arc. Therefore, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 13. intercepted arc. So, Therefore,

16.

SOLUTION: If an angle is inscribed in a circle, then the measure Here, of the angle equals one half the measure of its The arc is a semicircle. So,

intercepted arc. Therefore, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 14. intercepted arc. Therefore,

17. ALGEBRA Find each measure.

15. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So,

Therefore, SOLUTION: 16. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, SOLUTION: 17. ALGEBRA Find each measure. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

18. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

SOLUTION: If two inscribed angles of a circle intercept the same Therefore, m∠C = 5(10) – 3 or 47. arc or congruent arcs, then the angles are congruent. So, PROOF Write the specified type of proof. 21. paragraph proof

Given: Therefore, Prove: 19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

SOLUTION: . Since and If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. , the equation becomes Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. . Multiplying each side of the 11-4 Inscribed Angles equation by 2 results in . Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure. 22. two-column proof Given: Prove:

SOLUTION: SOLUTION: Statements (Reasons): If two inscribed angles of a circle intercept the same 1. (Given) arc or congruent arcs, then the angles are congruent. 2. (Inscribed intercepting same arc Since ∠C and ∠D both intercept arc AB, m∠C = are .) m∠D. 3. (Vertical are .) 4. (AA Similarity)

ALGEBRA Find each value. Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given:

23. x Prove:

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle SOLUTION: is 180. So, Proof: Given means that

. Since and 24. , the equation becomes SOLUTION: An inscribed angle of a triangle intercepts a diameter . Multiplying each side of the or semicircle if and only if the angle is a right angle. So, equation by 2 results in . The sum of the measures of the angles of a triangle

is 180. So, 22. two-column proof Given: Prove: Therefore, 25. x eSolutions Manual - Powered by Cognero SOLUTION: Page 5 An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, SOLUTION: The sum of the measures of the angles of a triangle Statements (Reasons): 1. (Given) is 180. So, 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) 26. 4. (AA Similarity) SOLUTION: ALGEBRA Find each value. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

23. x

Therefore, SOLUTION: CCSS STRUCTURE Find each measure. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 27. SOLUTION: 24. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION:

An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 28.

SOLUTION: is 180. So, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore,

25. x SOLUTION: Therefore, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. CCSS STRUCTURE Find each measure. So, The sum of the measures of the angles of a triangle

is 180. So,

26. 29. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. If a quadrilateral is inscribed in a circle, then its

So, opposite angles are supplementary. The sum of the measures of the angles of a triangle

is 180. So, Therefore,

Therefore, 30. CCSS STRUCTURE Find each measure. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

27. Therefore,

SOLUTION: 31. PROOF Write a paragraph proof for Theorem 11.9. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Therefore, Prove: and are supplementary. and are supplementary. 28. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Proof: By arc addition and the definitions of arc Therefore, measure and the sum of central angles, CCSS STRUCTURE Find each measure. . Since by Theorem 10.6 and , but

, so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior

29. angles of a quadrilateral is 360, . But SOLUTION: , so , making If a quadrilateral is inscribed in a circle, then its them supplementary also. opposite angles are supplementary. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore,

30. SOLUTION: If a quadrilateral is inscribed in a circle, then its measure.

opposite angles are supplementary. 32.

SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are Therefore, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 31. PROOF Write a paragraph proof for Theorem 11.9. of each arc joining consecutive vertices is of 360 SOLUTION: or 45. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. Therefore, measure of arc NPQ is 135. and are supplementary.

33. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure Proof: By arc addition and the definitions of arc of each arc joining consecutive vertices is of 360 measure and the sum of central angles, or 45. Since ∠RLQ is inscribed in the circle, its . Since by Theorem 10.6 measure equals one half the measure of its intercepted arc QR. and , but

, so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior Therefore, the measure of ∠RLQ is 22.5. angles of a quadrilateral is 360, . But , so , making 34. them supplementary also. SOLUTION: SIGNS: A stop sign in the shape of a regular If an angle is inscribed in a circle, then the measure octagon is inscribed in a circle. Find each of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures measure. Then, 32. Therefore, SOLUTION: Since all the sides of the stop sign are congruent 35. chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make SOLUTION: up the circle, so their sum is 360. Thus, the measure If an angle is inscribed in a circle, then the measure of each arc joining consecutive vertices is of 360 of the angle equals one half the measure of its

or 45. intercepted arc. So, The sum of the measures of the central angles of a Therefore, measure of arc NPQ is 135. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

33. measures SOLUTION: Then, Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make Therefore, up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its shown are congruent, find the measure of each

intercepted arc QR. inscribed angle. a.

Therefore, the measure of ∠RLQ is 22.5. b.

34. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its c.

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures d. Then,

Therefore,

35. SOLUTION: SOLUTION: a. The sum of the measures of the central angles of If an angle is inscribed in a circle, then the measure a circle with no interior points in common is 360. of the angle equals one half the measure of its Here, 360 is divided into 5 equal arcs and each arc

intercepted arc. So, measures The sum of the measures of the central angles of a If an angle is inscribed in a circle, then the measure circle with no interior points in common is 360. Here, of the angle equals one half the measure of its 360 is divided into 8 equal arcs and each arc intercepted arc. So, the measure of each inscribed

measures angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc Then, measures Each inscribed angle is formed

Therefore, by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one 36. ART Four different string art star patterns are half the measure of its intercepted arc. So, the shown. If all of the inscribed angles of each star measure of each inscribed angle is 60. shown are congruent, find the measure of each c. Here, 360 is divided into 7 equal arcs and each arc inscribed angle. a. measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. b. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one c. half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2

d. Given: P lies inside . is a diameter.

Prove:

SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc SOLUTION: measures Proof: Statements (Reasons) If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 1. m∠ABC = m∠ABD + m∠DBC ( Addition

intercepted arc. So, the measure of each inscribed Postulate) angle is 36. 2. m∠ABD = m(arc AD) b. Here, 360 is divided into 6 equal arcs and each arc m∠DBC = m(arc DC) (The measure of an measures Each inscribed angle is formed inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed 3. m∠ABC = m(arc AD) + m(arc DC) in a circle, then the measure of the angle equals one (Substitution) half the measure of its intercepted arc. So, the 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) measure of each inscribed angle is 60.

c. Here, 360 is divided into 7 equal arcs and each arc 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc measures If an angle is inscribed in a Addition Postulate) 6. m ABC = m(arc AC) (Substitution) circle, then the measure of the angle equals one half ∠ the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. 38. Case 3 d. The sum of the measures of the central angles of Given: P lies outside . is a diameter. a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc Prove: measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. SOLUTION: 37. Case 2 Proof: Given: P lies inside . is a diameter. Statements (Reasons)

Prove: 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA)

(Substitution) SOLUTION: 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) Proof: Statements (Reasons) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc 1. m∠ABC = m∠ABD + m∠DBC ( Addition Addition Postulate) Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 2. m∠ABD = m(arc AD) Property of Equality) 7. m ABC = m(arc AC) (Substitution) m∠DBC = m(arc DC) (The measure of an ∠

inscribed whose side is a diameter is half the

measure of the intercepted arc (Case 1)). PROOF Write the specified proof for each 3. m∠ABC = m(arc AD) + m(arc DC) theorem. (Substitution) 39. Theorem 11.7, two-column proof 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) SOLUTION:

Given: and are inscribed; 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Prove: Addition Postulate) Proof: 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter.

Prove:

Statements (Reasons) 1. and are inscribed; (Given) 2. ; (Measure

of an inscribed = half measure of intercepted arc.) SOLUTION:

Proof: 3. (Def. of arcs) Statements (Reasons) 4. (Mult. Prop. of Equality) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 5. (Substitution) 2. m∠DBC = m(arc DC) 6. (Def. of ) m∠DBA = m(arc DA) (The measure of an 40. Theorem 11.8, paragraph proof inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). SOLUTION: 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality)

7. m∠ABC = m(arc AC) (Substitution) Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then PROOF Write the specified proof for each . Since is an inscribed angle, theorem. 39. Theorem 11.7, two-column proof then or 90. So, by definition, SOLUTION: is a right angle. Given: and are inscribed; Prove: Part II: Given: is a right angle. Proof: Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle. Statements (Reasons) 1. and are inscribed; 41. MULTIPLE REPRESENTATIONS In this (Given) problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel 2. ; (Measure chords. a. GEOMETRIC Use a compass to draw a circle of an inscribed = half measure of intercepted arc.) with parallel chords and . Connect points A 3. (Def. of arcs) and D by drawing segment .

4. (Mult. Prop. of Equality) b. NUMERICAL Use a protractor to find 5. (Substitution) and . Then determine and . What is true about these arcs? Explain. 6. (Def. of ) c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that 40. Theorem 11.8, paragraph proof are cut by two parallel chords. SOLUTION: d. ANALYTICAL Use your conjecture to find and in the figure. Verify by using inscribed angles to find the measures of the arcs.

SOLUTION: Given: Part I: is a semicircle. a. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition, is a right angle. b. Sample answer: , ;

Part II: Given: is a right angle. , ; The arcs are congruent because they have equal measures. Prove : is a semicircle. c. Sample answer:

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle. m∠A = 15, m∠D = 15, = 30, and = 30. The arcs are congruent. 41. MULTIPLE REPRESENTATIONS In this In a circle, two parallel chords cut congruent arcs. problem, you will investigate the relationship between d. In a circle, two parallel chords cut congruent arcs. the arcs of a circle that are cut by two parallel chords. So, = . a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find = 4(16) + 6 or 70 and . Then determine and . = 6(16) – 26 or 70 What is true about these arcs? Explain. Therefore, the measure of arcs QS and PR are each c. VERBAL Draw another circle and repeat parts a 70. and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. CCSS ARGUMENTS Determine whether the d. ANALYTICAL Use your conjecture to find quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. and in the figure. Verify by using 42. square inscribed angles to find the measures of the arcs. SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

SOLUTION: 43. rectangle a. SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram b. Sample answer: , ; SOLUTION: , ; The arcs are congruent The opposite angles of a parallelogram are always because they have equal measures. congruent. They will only be supplementary when c. Sample answer: they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed m A = 15, m D = 15, = 30, and = 30. ∠ ∠ in a circle as long as it is a square. Since the opposite The arcs are congruent. angles of rhombi that are not squares are not In a circle, two parallel chords cut congruent arcs. supplementary, they can not be inscribed in a circle. d. In a circle, two parallel chords cut congruent arcs. Therefore, the statement is sometimes true. So, = . 46. kite SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be = 4(16) + 6 or 70 a right angle. If one pair of opposite angles for a = 6(16) – 26 or 70 quadrilateral is supplementary, the other pair must Therefore, the measure of arcs QS and PR are each also be supplementary. So, as long as the angles that 70. compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. CCSS ARGUMENTS Determine whether the Therefore, the statement is sometimes true. quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 47. CHALLENGE A square is inscribed in a circle. 42. square What is the ratio of the area of the circle to the area of the square? SOLUTION: Squares have right angles at each vertex, therefore SOLUTION: each pair of opposite angles will be supplementary A square with side s is inscribed in a circle with and inscribed in a circle. Therefore, the statement is radius r. always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, Using the Pythagorean Theorem, the statement is always true.

44. parallelogram SOLUTION:

The opposite angles of a parallelogram are always 2 congruent. They will only be supplementary when The area of a square of side s is A = s and the area they are right angles. So, a parallelogram can be of a circle of radius r is A = . inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed Therefore, the ratio of the area of the circle to the in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not area of the inscribed square is supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true. 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the 46. kite circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: Exactly one pair of opposite angles of a kite are SOLUTION: congruent. To be supplementary, they must each be Sample answer: A triangle will have a right angle. If one pair of opposite angles for a two inscribed angles of 45 and one of 90. The quadrilateral is supplementary, the other pair must hypotenuse is across from the 90, or right angle. also be supplementary. So, as long as the angles that According to theorem 10.8, an inscribed angle of a compose the pair of congruent opposite angles are triangle intercepts a diameter if the angle is a right right angles, a kite can be inscribed in a circle. angle. Therefore, the hypotenuse is a diameter and Therefore, the statement is sometimes true. has a length of 2r. Use trigonometry to find the length of the equal legs. 47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the square? SOLUTION: A square with side s is inscribed in a circle with radius r.

Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. Using the Pythagorean Theorem, SOLUTION: See students’ work.

50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they 2 The area of a square of side s is A = s and the area intercept the same arc, how are they related? of a circle of radius r is A = . SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle Therefore, the ratio of the area of the circle to the intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central area of the inscribed square is angle. 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the 51. In the circle below, and . circle is given, explain how to find the lengths of the What is ? right triangle’s legs. SOLUTION: Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right A 42 angle. Therefore, the hypotenuse is a diameter and B 61 has a length of 2r. Use trigonometry to find the C 80 length of the equal legs. D 84 SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

Therefore, the length of each leg of the 45-45-90 But, triangle can be found by multiplying the radius of the Therefore, circle in which it is inscribed by . The correct choice is A. ALGEBRA 49. OPEN ENDED Find and sketch a real-world logo 52. Simplify 2 with an inscribed polygon. 4(3x – 2)(2x + 4) + 3x + 5x – 6. 2 SOLUTION: F 9x + 3x – 14 2 See students’ work. G 9x + 13x – 14 2 H 27x + 37x – 38 50. WRITING IN MATH Compare and contrast 2 J 27x + 27x – 26 inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: SOLUTION: Use the Distributive Property to simplify the first term. An inscribed angle has its vertex on the circle. A

central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle Therefore, the correct choice is H. intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central 53. SHORT RESPONSE In the circle below, is a angle. diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of 51. In the circle below, and . the circle. What is ?

SOLUTION: Use the Pythagorean Theorem to find the A 42 hypotenuse of the triangle which is also the diameter B 61 of the circle. C 80 D 84 The radius is half the diameter. So, the radius of the circle is about 8.5 in. SOLUTION: The circumference C of a circle of a circle of If an angle is inscribed in a circle, then the measure diameter d is given by of the angle equals one half the measure of its Therefore, the circumference of the circle is intercepted arc. Therefore,

54. SAT/ACT The sum of three consecutive integers is

But, –48. What is the least of the three integers?

Therefore, A –15 The correct choice is A. B –16 52. ALGEBRA Simplify C –17 2 D –18 4(3x – 2)(2x + 4) + 3x + 5x – 6. 2 E –19 F 9x + 3x – 14 G 9x2 + 13x – 14 SOLUTION: Let the three consecutive integers be x , x +1,and x + H 2 27x + 37x – 38 2. Then, J 27x2 + 27x – 26 SOLUTION: Use the Distributive Property to simplify the first term. So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure. Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of

the circle. 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So,

bisects Therefore, SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle. 56.

SOLUTION: The radius is half the diameter. So, the radius of the If a diameter (or radius) of a circle is perpendicular circle is about 8.5 in. to a chord, then it bisects the chord and its arc. So, The circumference C of a circle of a circle of diameter d is given by bisects Therefore, Therefore, the circumference of the circle is 57. NJ

SOLUTION: 54. SAT/ACT The sum of three consecutive integers is If a diameter (or radius) of a circle is perpendicular –48. What is the least of the three integers? to a chord, then it bisects the chord and its arc. So, A –15 bisects Therefore, B –16 C –17 D –18 E –19 58. SOLUTION: SOLUTION: Let the three consecutive integers be x , x +1,and x + If a diameter (or radius) of a circle is perpendicular 2. Then, to a chord, then it bisects the chord and its arc. So, bisects Therefore,

So, the three integers are –17, –16, and –15 and the Find x. least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure. 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

60. 56. SOLUTION: SOLUTION: The sum of the measures of the central angles of a If a diameter (or radius) of a circle is perpendicular circle with no interior points in common is 360. So, to a chord, then it bisects the chord and its arc. So,

bisects Therefore, 57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

61. SOLUTION:

58. The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So,

If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

62. PHOTOGRAPHY In one of the first cameras Find x. invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed? 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, 60. the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? 61. SOLUTION: SOLUTION: Since B is the midpoint of The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An 64. AB = 6y – 14, BC = 10 – 2y, AC = ? image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the SOLUTION: image of the person on the back of the camera is 12 Since B is the midpoint of inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ?

SOLUTION: SOLUTION: Since B is the midpoint of The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle. 66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Therefore, the person being photographed is 5.6 ft tall.

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If

Find each measure. , what is ? 1.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its

intercepted arc. Therefore, intercepted arc. Therefore,

ALGEBRA Find each measure. 2. 5.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. intercepted arc. Therefore, So,

3. Therefore,

6.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same intercepted arc. So, arc or congruent arcs, then the angles are congruent. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ? Therefore, 7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. Therefore, Given: bisects . Prove: ALGEBRA Find each measure. 5.

Proof: Statements (Reasons)

1. bisects . (Given)

SOLUTION: 2. (Def. of segment bisector) If two inscribed angles of a circle intercept the same 3. intercepts . intercepts . arc or congruent arcs, then the angles are congruent. (Def. of intercepted arc) So, 4. (Inscribed of same arc are .) 5. (Vertical are .) Therefore, 6. (AAS) CCSS STRUCTURE Find each value. 6. 8.

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same An inscribed angle of a triangle intercepts a diameter arc or congruent arcs, then the angles are congruent. or semicircle if and only if the angle is a right angle. So, So, The sum of the measures of the angles of a triangle

is 180. So, Therefore, Therefore, 7. PROOF Write a two-column proof. Given: bisects . 9. x Prove:

SOLUTION: SOLUTION: Given: bisects . An inscribed angle of a triangle intercepts a diameter Prove: or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So,

Proof: Statements (Reasons) 10. and 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS) SOLUTION: If a quadrilateral is inscribed in a circle, then its CCSS STRUCTURE Find each value. opposite angles are supplementary. 8. Solve the two equations to find the values of x and y.

Use the values of the variables to find and .

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. ALGEBRA Find each measure. So, 11. The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

9. x SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

12.

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle SOLUTION: is 180. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, 10. and

13.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION: The sum of the measures of the central angles of a Solve the two equations to find the values of x and y. circle with no interior points in common is 360. So,

Use the values of the variables to find and If an angle is inscribed in a circle, then the measure . of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 14. 11.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure

of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 15.

12.

SOLUTION: Here, SOLUTION: The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its

intercepted arc. Therefore, intercepted arc. So, Therefore, 13. 16.

SOLUTION: The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, Here,

The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its

intercepted arc. Therefore, intercepted arc. Therefore,

14. 17. ALGEBRA Find each measure.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 15. So, 18. ALGEBRA Find each measure.

SOLUTION: Here,

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. So, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent.

Therefore, So,

16. Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same intercepted arc. Therefore, arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A =

17. ALGEBRA Find each measure. m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

18. ALGEBRA Find each measure. SOLUTION:

If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47. SOLUTION: PROOF Write the specified type of proof. If two inscribed angles of a circle intercept the same 21. paragraph proof

arc or congruent arcs, then the angles are congruent. Given: So,

Prove:

Therefore,

19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

. Since and

, the equation becomes SOLUTION: If two inscribed angles of a circle intercept the same . Multiplying each side of the arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = equation by 2 results in . m∠B.

22. two-column proof Given: Therefore, m∠A = 5(4) or 20. Prove: 20. ALGEBRA Find each measure.

SOLUTION: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc SOLUTION: are .) If two inscribed angles of a circle intercept the same 3. (Vertical are .) arc or congruent arcs, then the angles are congruent. 4. (AA Similarity) Since ∠C and ∠D both intercept arc AB, m∠C = ALGEBRA Find each value. m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof 23. x

Given: SOLUTION: Prove: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

SOLUTION: 24. Proof: Given means that SOLUTION: . Since and An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.

, the equation becomes So, The sum of the measures of the angles of a triangle

. Multiplying each side of the is 180. So, equation by 2 results in . Therefore, 22. two-column proof 25. x Given: Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, SOLUTION: Statements (Reasons): 1. (Given) 26. 2. (Inscribed intercepting same arc SOLUTION: are .) An inscribed angle of a triangle intercepts a diameter 3. (Vertical are .) or semicircle if and only if the angle is a right angle. 4. (AA Similarity) So, The sum of the measures of the angles of a triangle ALGEBRA Find each value. is 180. So,

Therefore,

CCSS STRUCTURE Find each measure. 23. x

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 27. SOLUTION: is 180. So, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

24. Therefore, SOLUTION: 28. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. SOLUTION: So, If a quadrilateral is inscribed in a circle, then its The sum of the measures of the angles of a triangle opposite angles are supplementary.

is 180. So, 11-4 Inscribed Angles Therefore, Therefore,

25. x CCSS STRUCTURE Find each measure. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 29. SOLUTION: 26. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So,

The sum of the measures of the angles of a triangle Therefore,

is 180. So, 30. SOLUTION: Therefore, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. CCSS STRUCTURE Find each measure.

Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9.

27. SOLUTION: SOLUTION: Given: Quadrilateral ABCD is inscribed in . If a quadrilateral is inscribed in a circle, then its Prove: and are supplementary. opposite angles are supplementary. and are supplementary.

Therefore,

28. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Proof: By arc addition and the definitions of arc measure and the sum of central angles,

. Since by Theorem 10.6

Therefore, and , but CCSS STRUCTURE Find each measure. eSolutions Manual - Powered by Cognero , so Page or6 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making them supplementary also. 29. SOLUTION: SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, measure. 30. 32. SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 Therefore, or 45.

31. PROOF Write a paragraph proof for Theorem 11.9. Therefore, measure of arc NPQ is 135. SOLUTION: Given: Quadrilateral ABCD is inscribed in .

Prove: and are supplementary. 33. and are supplementary. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its Proof: By arc addition and the definitions of arc intercepted arc QR. measure and the sum of central angles, . Since by Theorem 10.6 and

, but Therefore, the measure of ∠RLQ is 22.5. , so or 180. This makes ∠C and ∠A supplementary. 34. Because the sum of the measures of the interior angles of a quadrilateral is 360, SOLUTION: . But If an angle is inscribed in a circle, then the measure , so , making of the angle equals one half the measure of its them supplementary also. intercepted arc. So, SIGNS: A stop sign in the shape of a regular The sum of the measures of the central angles of a octagon is inscribed in a circle. Find each circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Then,

Therefore, measure. 32. 35. SOLUTION: SOLUTION: Since all the sides of the stop sign are congruent If an angle is inscribed in a circle, then the measure chords of the circle, the corresponding arcs are of the angle equals one half the measure of its congruent. There are eight adjacent arcs that make intercepted arc. So, up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, or 45. 360 is divided into 8 equal arcs and each arc

measures Therefore, measure of arc NPQ is 135. Then,

33. Therefore, SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are 36. ART Four different string art star patterns are congruent. There are eight adjacent arcs that make shown. If all of the inscribed angles of each star up the circle, so their sum is 360. Thus, the measure shown are congruent, find the measure of each inscribed angle. of each arc joining consecutive vertices is of 360 a. or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

b.

Therefore, the measure of ∠RLQ is 22.5.

34. c. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So,

The sum of the measures of the central angles of a d. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

Then,

Therefore, SOLUTION: a. The sum of the measures of the central angles of 35. a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc SOLUTION: If an angle is inscribed in a circle, then the measure measures of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, intercepted arc. So, the measure of each inscribed The sum of the measures of the central angles of a angle is 36. circle with no interior points in common is 360. Here, b. Here, 360 is divided into 6 equal arcs and each arc 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed measures by intercepting alternate vertices of the star. So, each Then, intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the Therefore, measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star measures If an angle is inscribed in a shown are congruent, find the measure of each circle, then the measure of the angle equals one half inscribed angle. the measure of its intercepted arc. So, the measure a. of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed

b. by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case c. of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove:

d.

SOLUTION: SOLUTION: a. The sum of the measures of the central angles of Proof: a circle with no interior points in common is 360. Statements (Reasons) Here, 360 is divided into 5 equal arcs and each arc 1. m∠ABC = m∠ABD + m∠DBC ( Addition

measures Postulate) 2. m∠ABD = m(arc AD) If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its m∠DBC = m(arc DC) (The measure of an intercepted arc. So, the measure of each inscribed inscribed whose side is a diameter is half the angle is 36. measure of the intercepted arc (Case 1)). b. Here, 360 is divided into 6 equal arcs and each arc 3. m∠ABC = m(arc AD) + m(arc DC) measures Each inscribed angle is formed (Substitution) by intercepting alternate vertices of the star. So, each 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc half the measure of its intercepted arc. So, the Addition Postulate) measure of each inscribed angle is 60. 6. m∠ABC = m(arc AC) (Substitution) c. Here, 360 is divided into 7 equal arcs and each arc

measures If an angle is inscribed in a 38. Case 3 circle, then the measure of the angle equals one half Given: P lies outside . is a diameter. the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. Prove: d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. SOLUTION: Proof: PROOF Write a two-column proof for each case Statements (Reasons) of Theorem 11.6. 1. m∠ABC = m∠DBC – m∠DBA ( Addition 37. Case 2 Postulate, Subtraction Property of Equality)

Given: P lies inside . is a diameter. 2. m∠DBC = m(arc DC)

Prove: m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc SOLUTION: Addition Postulate) Proof: 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Statements (Reasons) Property of Equality) 1. m∠ABC = m∠ABD + m∠DBC ( Addition 7. m∠ABC = m(arc AC) (Substitution) Postulate) 2. m∠ABD = m(arc AD) PROOF Write the specified proof for each m∠DBC = m(arc DC) (The measure of an theorem. inscribed whose side is a diameter is half the 39. Theorem 11.7, two-column proof measure of the intercepted arc (Case 1)). SOLUTION: 3. m∠ABC = m(arc AD) + m(arc DC) Given: and are inscribed; (Substitution) Prove: 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) Proof:

5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter.

Prove: Statements (Reasons) 1. and are inscribed; (Given) 2. ; (Measure of an inscribed = half measure of intercepted arc.) 3. (Def. of arcs)

4. (Mult. Prop. of Equality) SOLUTION: Proof: 5. (Substitution) Statements (Reasons) 6. (Def. of ) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 40. Theorem 11.8, paragraph proof

2. m∠DBC = m(arc DC) SOLUTION: m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m ABC = [m(arc DC) – m(arc DA)] (Factor) ∠

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Given: Addition Postulate) Part I: is a semicircle. 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Prove: is a right angle. Property of Equality) Proof: Since is a semicircle, then

7. m∠ABC = m(arc AC) (Substitution) . Since is an inscribed angle,

then or 90. So, by definition, PROOF Write the specified proof for each is a right angle. theorem.

39. Theorem 11.7, two-column proof Part II: Given: is a right angle. SOLUTION: Prove: is a semicircle. Given: and are inscribed; Prove: Proof: Since is an inscribed angle, then Proof: and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between

Statements (Reasons) the arcs of a circle that are cut by two parallel chords. 1. and are inscribed; a. GEOMETRIC Use a compass to draw a circle (Given) with parallel chords and . Connect points A 2. ; (Measure and D by drawing segment . b. NUMERICAL Use a protractor to find of an inscribed = half measure of intercepted arc.) and . Then determine and . 3. (Def. of arcs) What is true about these arcs? Explain.

4. (Mult. Prop. of Equality) c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that 5. (Substitution) are cut by two parallel chords. 6. (Def. of ) d. ANALYTICAL Use your conjecture to find and in the figure. Verify by using 40. Theorem 11.8, paragraph proof inscribed angles to find the measures of the arcs. SOLUTION:

SOLUTION: a.

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then

. Since is an inscribed angle, b. Sample answer: , ; then or 90. So, by definition, , ; The arcs are congruent because they have equal measures. is a right angle. c. Sample answer:

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because m∠A = 15, m∠D = 15, = 30, and = 30. is a right angle, m = 90. Then The arcs are congruent. m = 2(90) or 180. So by definition, is a In a circle, two parallel chords cut congruent arcs. semicircle. d. In a circle, two parallel chords cut congruent arcs. So, = . 41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle = 4(16) + 6 or 70 with parallel chords and . Connect points A = 6(16) – 26 or 70 and D by drawing segment . Therefore, the measure of arcs QS and PR are each b. NUMERICAL Use a protractor to find 70.

and . Then determine and . CCSS ARGUMENTS Determine whether the

What is true about these arcs? Explain. quadrilateral can always, sometimes, or never c. VERBAL Draw another circle and repeat parts a be inscribed in a circle. Explain your reasoning. and b. Make a conjecture about arcs of a circle that 42. square are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find SOLUTION: Squares have right angles at each vertex, therefore and in the figure. Verify by using each pair of opposite angles will be supplementary inscribed angles to find the measures of the arcs. and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION:

Rectangles have right angles at each vertex, SOLUTION: therefore each pair of opposite angles will be a. supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when b. Sample answer: , ; they are right angles. So, a parallelogram can be , ; The arcs are congruent inscribed in a circle as long as it is a rectangle. because they have equal measures. Therefore, the statement is sometimes true. c. Sample answer: 45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. m∠A = 15, m∠D = 15, = 30, and = 30. Therefore, the statement is sometimes true. The arcs are congruent. In a circle, two parallel chords cut congruent arcs. 46. kite d. In a circle, two parallel chords cut congruent arcs. SOLUTION: So, = . Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that = 4(16) + 6 or 70 compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. = 6(16) – 26 or 70 Therefore, the statement is sometimes true. Therefore, the measure of arcs QS and PR are each 70. 47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area CCSS ARGUMENTS Determine whether the of the square? quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. SOLUTION: 42. square A square with side s is inscribed in a circle with SOLUTION: radius r. Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Using the Pythagorean Theorem, Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

2 44. parallelogram The area of a square of side s is A = s and the area

SOLUTION: of a circle of radius r is A = . The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus

SOLUTION: Therefore, the ratio of the area of the circle to the The opposite angles of a rhombus are always area of the inscribed square is congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed 48. WRITING IN MATH A right in a circle as long as it is a square. Since the opposite triangle is inscribed in a circle. If the radius of the angles of rhombi that are not squares are not circle is given, explain how to find the lengths of the supplementary, they can not be inscribed in a circle. right triangle’s legs. Therefore, the statement is sometimes true. SOLUTION: 46. kite Sample answer: A triangle will have SOLUTION: two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. Exactly one pair of opposite angles of a kite are According to theorem 10.8, an inscribed angle of a congruent. To be supplementary, they must each be triangle intercepts a diameter if the angle is a right a right angle. If one pair of opposite angles for a angle. Therefore, the hypotenuse is a diameter and quadrilateral is supplementary, the other pair must has a length of 2r. Use trigonometry to find the also be supplementary. So, as long as the angles that length of the equal legs. compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true.

47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the square?

SOLUTION: Therefore, the length of each leg of the 45-45-90 A square with side s is inscribed in a circle with triangle can be found by multiplying the radius of the radius r. circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work.

Using the Pythagorean Theorem, 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A 2 The area of a square of side s is A = s and the area central angle has its vertex at the center of the circle. of a circle of radius r is A = . If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle.

Therefore, the ratio of the area of the circle to the 51. In the circle below, and . area of the inscribed square is What is ? 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: Sample answer: A triangle will have A two inscribed angles of 45 and one of 90. The 42 hypotenuse is across from the 90, or right angle. B 61 According to theorem 10.8, an inscribed angle of a C 80 triangle intercepts a diameter if the angle is a right D 84 angle. Therefore, the hypotenuse is a diameter and SOLUTION: has a length of 2r. Use trigonometry to find the length of the equal legs. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

But, Therefore, The correct choice is A. Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the 52. ALGEBRA Simplify 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. circle in which it is inscribed by . 2 F 9x + 3x – 14 49. OPEN ENDED Find and sketch a real-world logo G 9x2 + 13x – 14 2 with an inscribed polygon. H 27x + 37x – 38 SOLUTION: J 27x2 + 27x – 26 See students’ work. SOLUTION: 50. WRITING IN MATH Compare and contrast Use the Distributive Property to simplify the first inscribed angles and central angles of a circle. If they term. intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If Therefore, the correct choice is H. an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find AB). So, if an inscribed angle and a central angle the diameter, the radius, and the circumference of intercept the same arc, then the measure of the the circle. inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . What is ?

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle.

A 42 The radius is half the diameter. So, the radius of the B 61 circle is about 8.5 in. C 80 The circumference C of a circle of a circle of D 84 diameter d is given by Therefore, the circumference of the circle is SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 54. SAT/ACT The sum of three consecutive integers is intercepted arc. Therefore, –48. What is the least of the three integers? A –15 B –16 But, C –17 Therefore, D –18 The correct choice is A. E –19 52. ALGEBRA Simplify SOLUTION: 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. Let the three consecutive integers be x , x +1,and x + 2 2. Then, F 9x + 3x – 14 G 9x2 + 13x – 14 2 H 27x + 37x – 38 2 J 27x + 27x – 26 So, the three integers are –17, –16, and –15 and the SOLUTION: least of these is –17.

Use the Distributive Property to simplify the first Therefore, the correct choice is C. term. In , FL = 24, HJ = 48, and . Find each measure.

Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find 55. FG the diameter, the radius, and the circumference of SOLUTION: the circle. If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

SOLUTION: 56. Use the Pythagorean Theorem to find the SOLUTION: hypotenuse of the triangle which is also the diameter If a diameter (or radius) of a circle is perpendicular of the circle. to a chord, then it bisects the chord and its arc. So,

bisects Therefore, The radius is half the diameter. So, the radius of the circle is about 8.5 in. 57. NJ The circumference C of a circle of a circle of SOLUTION: diameter d is given by Therefore, the circumference of the circle is If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? A –15

B –16 58. C –17 SOLUTION: D –18 If a diameter (or radius) of a circle is perpendicular E –19 to a chord, then it bisects the chord and its arc. So, SOLUTION: bisects Therefore, Let the three consecutive integers be x , x +1,and x + 2. Then, Find x.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. 59. In , FL = 24, HJ = 48, and . Find each measure. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 60.

SOLUTION: The sum of the measures of the central angles of a 56. circle with no interior points in common is 360. So,

SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 61. bisects Therefore, SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

58. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 62. PHOTOGRAPHY In one of the first cameras bisects Therefore, invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 Find x. inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

60. SOLUTION: The sum of the measures of the central angles of a Therefore, the person being photographed is 5.6 ft circle with no interior points in common is 360. So, tall.

61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An Since B is the midpoint of image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will 66. AB = 10s + 2, AC = 40, s = ? be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x SOLUTION: be the length of the base of the larger triangle. Since B is the midpoint of

Therefore, the person being photographed is 5.6 ft tall.

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3. Find each measure. 1.

SOLUTION: SOLUTION:

If an angle is inscribed in a circle, then the measure Here, is a semi-circle. So, of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc. Therefore, of the angle equals one half the measure of its intercepted arc. So,

2. Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ?

3. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 5. SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, SOLUTION: Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 4. SCIENCE The diagram shows how light bends in a So, raindrop to make the colors of the rainbow. If , what is ? Therefore,

6.

SOLUTION: If an angle is inscribed in a circle, then the measure

of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. ALGEBRA Find each measure. So, 5.

Therefore,

7. PROOF Write a two-column proof. Given: bisects . SOLUTION: Prove: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore, SOLUTION: Given: bisects . 6. Prove:

Proof: Statements (Reasons) SOLUTION: 1. bisects . (Given) If two inscribed angles of a circle intercept the same

arc or congruent arcs, then the angles are congruent. 2. (Def. of segment bisector) So, 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) Therefore, 5. (Vertical are .) 6. (AAS) 7. PROOF Write a two-column proof. CCSS STRUCTURE Find each value. Given: bisects . 8. Prove:

SOLUTION: SOLUTION: Given: bisects . An inscribed angle of a triangle intercepts a diameter Prove: or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

Proof: Therefore, Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 9. x 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS) CCSS STRUCTURE Find each value. SOLUTION: 8. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So,

SOLUTION: 10. and An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, SOLUTION: Therefore, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 9. x Solve the two equations to find the values of x and y.

Use the values of the variables to find and .

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. ALGEBRA Find each measure. So, Since 11. The sum of the measures of the angles of a triangle

is 180. So,

10. and SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

12. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y.

SOLUTION: Use the values of the variables to find and If an angle is inscribed in a circle, then the measure

. of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 13. 11.

SOLUTION: SOLUTION: The sum of the measures of the central angles of a If an angle is inscribed in a circle, then the measure circle with no interior points in common is 360. So, of the angle equals one half the measure of its intercepted arc. Therefore,

If an angle is inscribed in a circle, then the measure 12. of the angle equals one half the measure of its

intercepted arc. Therefore,

14.

intercepted arc. Therefore, SOLUTION: If an angle is inscribed in a circle, then the measure 13. of the angle equals one half the measure of its intercepted arc. Therefore,

15.

SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION:

Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its

intercepted arc. Therefore, intercepted arc. So, Therefore, 14. 16.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. Therefore, Here, The arc is a semicircle. So,

15. Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure. SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

16. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So,

Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: If two inscribed angles of a circle intercept the same intercepted arc. Therefore, arc or congruent arcs, then the angles are congruent. So, 17. ALGEBRA Find each measure.

Therefore,

19. ALGEBRA Find each measure.

18. ALGEBRA Find each measure. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore,

19. ALGEBRA Find each measure.

SOLUTION:

If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47. SOLUTION: PROOF Write the specified type of proof. If two inscribed angles of a circle intercept the same 21. paragraph proof arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = Given: m∠B. Prove:

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

. Since and

, the equation becomes SOLUTION: If two inscribed angles of a circle intercept the same . Multiplying each side of the arc or congruent arcs, then the angles are congruent.

Since ∠C and ∠D both intercept arc AB, m∠C = equation by 2 results in . m∠D. 22. two-column proof Given: Therefore, m∠C = 5(10) – 3 or 47. Prove:

PROOF Write the specified type of proof. 21. paragraph proof

Given:

Prove: SOLUTION: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) SOLUTION: 4. (AA Similarity)

Proof: Given means that ALGEBRA Find each value.

. Since and

, the equation becomes

. Multiplying each side of the 23. x

equation by 2 results in . SOLUTION: An inscribed angle of a triangle intercepts a diameter 22. two-column proof or semicircle if and only if the angle is a right angle. Given: So, Prove: The sum of the measures of the angles of a triangle

is 180. So,

24. SOLUTION: SOLUTION: Statements (Reasons): An inscribed angle of a triangle intercepts a diameter 1. (Given) or semicircle if and only if the angle is a right angle. So, 2. (Inscribed intercepting same arc The sum of the measures of the angles of a triangle

are .) 3. (Vertical are .) is 180. So, 4. (AA Similarity)

ALGEBRA Find each value. Therefore,

25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 23. x So, The sum of the measures of the angles of a triangle

SOLUTION: An inscribed angle of a triangle intercepts a diameter is 180. So, or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 26. SOLUTION: is 180. So, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, 24. The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Therefore, The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each measure. is 180. So,

Therefore,

25. x

SOLUTION: 27. An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. If a quadrilateral is inscribed in a circle, then its

So, opposite angles are supplementary. The sum of the measures of the angles of a triangle

Therefore, is 180. So, 28. 26. SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.

So, The sum of the measures of the angles of a triangle Therefore,

is 180. So, CCSS STRUCTURE Find each measure.

Therefore,

CCSS STRUCTURE Find each measure.

29.

SOLUTION: 27. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, Therefore,

28. 30. SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. opposite angles are supplementary.

Therefore, Therefore,

CCSS STRUCTURE Find each measure. 31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary.

29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6 Therefore, and 30. , but SOLUTION: , so or If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360,

. But 11-4 Inscribed Angles , so , making Therefore, them supplementary also.

31. PROOF Write a paragraph proof for Theorem 11.9. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary.

measure. 32. SOLUTION: Since all the sides of the stop sign are congruent Proof: By arc addition and the definitions of arc chords of the circle, the corresponding arcs are measure and the sum of central angles, congruent. There are eight adjacent arcs that make . Since by Theorem 10.6 up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 and or 45. , but

Therefore, measure of arc NPQ is 135. , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior 33. angles of a quadrilateral is 360, SOLUTION: . But Since all the sides of the stop sign are congruent , so , making chords of the circle, all the corresponding arcs are them supplementary also. congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

measure.

32. Therefore, the measure of ∠RLQ is 22.5. SOLUTION: Since all the sides of the stop sign are congruent 34. chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make SOLUTION: up the circle, so their sum is 360. Thus, the measure If an angle is inscribed in a circle, then the measure of each arc joining consecutive vertices is of 360 of the angle equals one half the measure of its

or 45. intercepted arc. So, The sum of the measures of the central angles of a Therefore, measure of arc NPQ is 135. eSolutions Manual - Powered by Cognero circle with no interior points in common is 360. Here,Page 7 360 is divided into 8 equal arcs and each arc

33. measures SOLUTION: Then, Since all the sides of the stop sign are congruent

chords of the circle, all the corresponding arcs are Therefore, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 35. of each arc joining consecutive vertices is of 360 SOLUTION: or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc QR. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

Therefore, the measure of ∠RLQ is 22.5. measures

Then, 34.

SOLUTION: Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star intercepted arc. So, shown are congruent, find the measure of each The sum of the measures of the central angles of a inscribed angle. circle with no interior points in common is 360. Here, a. 360 is divided into 8 equal arcs and each arc

measures Then,

Therefore, b.

35. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its c.

measures d.

Then,

Therefore,

36. ART Four different string art star patterns are SOLUTION: shown. If all of the inscribed angles of each star a. shown are congruent, find the measure of each The sum of the measures of the central angles of inscribed angle. a circle with no interior points in common is 360. a. Here, 360 is divided into 5 equal arcs and each arc

measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 36. b. b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one c. half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a circle, then the measure of the angle equals one half d. the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed SOLUTION: by intercepting alternate vertices of the star. So, each a. The sum of the measures of the central angles of intercepted arc measures 90. If an angle is inscribed a circle with no interior points in common is 360. in a circle, then the measure of the angle equals one Here, 360 is divided into 5 equal arcs and each arc half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. measures PROOF Write a two-column proof for each case If an angle is inscribed in a circle, then the measure of Theorem 11.6. of the angle equals one half the measure of its 37. Case 2 intercepted arc. So, the measure of each inscribed angle is 36. Given: P lies inside . is a diameter.

b. Here, 360 is divided into 6 equal arcs and each arc Prove: measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a SOLUTION: circle, then the measure of the angle equals one half Proof:

the measure of its intercepted arc. So, the measure Statements (Reasons) of each inscribed angle is about 25.7. 1. m∠ABC = m∠ABD + m∠DBC ( Addition

d. The sum of the measures of the central angles of Postulate) a circle with no interior points in common is 360. 2. m∠ABD = m(arc AD) Here, 360 is divided into 8 equal arcs and each arc m∠DBC = m(arc DC) (The measure of an measures Each inscribed angle is formed inscribed whose side is a diameter is half the

by intercepting alternate vertices of the star. So, each measure of the intercepted arc (Case 1)). intercepted arc measures 90. If an angle is inscribed 3. m∠ABC = m(arc AD) + m(arc DC) in a circle, then the measure of the angle equals one (Substitution) half the measure of its intercepted arc. So, the 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc of Theorem 11.6. Addition Postulate) 37. Case 2 6. m∠ABC = m(arc AC) (Substitution) Given: P lies inside . is a diameter.

Prove: 38. Case 3 Given: P lies outside . is a diameter.

Prove:

SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠ABD + m∠DBC ( Addition SOLUTION: Postulate) Proof:

2. m∠ABD = m(arc AD) Statements (Reasons) m∠DBC = m(arc DC) (The measure of an 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 2. m∠DBC = m(arc DC) 3. m∠ABC = m(arc AD) + m(arc DC) m∠DBA = m(arc DA) (The measure of an (Substitution) inscribed whose side is a diameter is half the 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc (Substitution)

Addition Postulate) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

6. m∠ABC = m(arc AC) (Substitution) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc 38. Case 3 Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Given: P lies outside . is a diameter. Property of Equality)

Prove: 7. m∠ABC = m(arc AC) (Substitution)

PROOF Write the specified proof for each theorem. 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: SOLUTION: Proof: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)).

3. m∠ABC = m(arc DC) – m(arc DA) Statements (Reasons) 1. and are inscribed; (Substitution) (Given) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) 2. ; (Measure 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) of an inscribed = half measure of intercepted arc.) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 3. (Def. of arcs) Property of Equality) 4. (Mult. Prop. of Equality) 7. m∠ABC = m(arc AC) (Substitution) 5. (Substitution) 6. (Def. of ) PROOF Write the specified proof for each theorem. 40. Theorem 11.8, paragraph proof 39. Theorem 11.7, two-column proof SOLUTION: SOLUTION: Given: and are inscribed; Prove: Proof:

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then

Statements (Reasons) . Since is an inscribed angle, 1. and are inscribed; then or 90. So, by definition, (Given) is a right angle. 2. ; (Measure Part II: Given: is a right angle. of an inscribed = half measure of intercepted arc.) Prove: is a semicircle. 3. (Def. of arcs)

4. (Mult. Prop. of Equality) Proof: Since is an inscribed angle, then 5. (Substitution) and by the Multiplication 6. (Def. of ) Property of Equality, m = 2m . Because is a right angle, m = 90. Then 40. Theorem 11.8, paragraph proof m = 2(90) or 180. So by definition, is a SOLUTION: semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Part I: Given: is a semicircle. Use a protractor to find Prove: is a right angle. and . Then determine and . What is true about these arcs? Explain. Proof: Since is a semicircle, then c. VERBAL Draw another circle and repeat parts a . Since is an inscribed angle, and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. then or 90. So, by definition, d. ANALYTICAL Use your conjecture to find is a right angle. and in the figure. Verify by using inscribed angles to find the measures of the arcs. Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication SOLUTION: . Property of Equality, m = 2m Because a. is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel b. Sample answer: , ; chords. a. GEOMETRIC Use a compass to draw a circle , ; The arcs are congruent because they have equal measures. with parallel chords and . Connect points A c. Sample answer: and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain. c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that are cut by two parallel chords.

d. ANALYTICAL Use your conjecture to find m∠A = 15, m∠D = 15, = 30, and = 30. and in the figure. Verify by using The arcs are congruent. inscribed angles to find the measures of the arcs. In a circle, two parallel chords cut congruent arcs. d. In a circle, two parallel chords cut congruent arcs. So, = .

SOLUTION: = 4(16) + 6 or 70 a. = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each 70.

CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. b. Sample answer: , ; 42. square , ; The arcs are congruent SOLUTION:

because they have equal measures. Squares have right angles at each vertex, therefore c. Sample answer: each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex,

therefore each pair of opposite angles will be m∠A = 15, m∠D = 15, = 30, and = 30. supplementary and inscribed in a circle. Therefore, The arcs are congruent. the statement is always true. In a circle, two parallel chords cut congruent arcs. d. In a circle, two parallel chords cut congruent arcs. 44. parallelogram

So, = . SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. = 4(16) + 6 or 70 Therefore, the statement is sometimes true. = 6(16) – 26 or 70 45. rhombus Therefore, the measure of arcs QS and PR are each 70. SOLUTION: The opposite angles of a rhombus are always CCSS ARGUMENTS Determine whether the congruent. They will only be supplementary when quadrilateral can always, sometimes, or never they are right angles. So, a rhombus can be inscribed be inscribed in a circle. Explain your reasoning. in a circle as long as it is a square. Since the opposite 42. square angles of rhombi that are not squares are not SOLUTION: supplementary, they can not be inscribed in a circle. Squares have right angles at each vertex, therefore Therefore, the statement is sometimes true. each pair of opposite angles will be supplementary 46. kite and inscribed in a circle. Therefore, the statement is always true. SOLUTION: Exactly one pair of opposite angles of a kite are 43. rectangle congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a SOLUTION: quadrilateral is supplementary, the other pair must Rectangles have right angles at each vertex, also be supplementary. So, as long as the angles that therefore each pair of opposite angles will be compose the pair of congruent opposite angles are supplementary and inscribed in a circle. Therefore, right angles, a kite can be inscribed in a circle. the statement is always true. Therefore, the statement is sometimes true.

44. parallelogram 47. CHALLENGE A square is inscribed in a circle. SOLUTION: What is the ratio of the area of the circle to the area of the square? The opposite angles of a parallelogram are always congruent. They will only be supplementary when SOLUTION: they are right angles. So, a parallelogram can be A square with side s is inscribed in a circle with inscribed in a circle as long as it is a rectangle. radius r. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite Using the Pythagorean Theorem, angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true.

46. kite 2 SOLUTION: The area of a square of side s is A = s and the area Exactly one pair of opposite angles of a kite are of a circle of radius r is A = . congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true.

47. CHALLENGE A square is inscribed in a circle. Therefore, the ratio of the area of the circle to the What is the ratio of the area of the circle to the area area of the inscribed square is of the square? WRITING IN MATH SOLUTION: 48. A right triangle is inscribed in a circle. If the radius of the A square with side s is inscribed in a circle with circle is given, explain how to find the lengths of the

radius r. right triangle’s legs. SOLUTION: Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right Using the Pythagorean Theorem, angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs.

2 The area of a square of side s is A = s and the area of a circle of radius r is A = .

Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo Therefore, the ratio of the area of the circle to the with an inscribed polygon. area of the inscribed square is SOLUTION: See students’ work. 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the 50. WRITING IN MATH Compare and contrast circle is given, explain how to find the lengths of the inscribed angles and central angles of a circle. If they right triangle’s legs. intercept the same arc, how are they related? SOLUTION: SOLUTION: Sample answer: A triangle will have An inscribed angle has its vertex on the circle. A two inscribed angles of 45 and one of 90. The central angle has its vertex at the center of the circle. hypotenuse is across from the 90, or right angle. If a central angle intercepts arc AB, then the According to theorem 10.8, an inscribed angle of a measure of the central angle is equal to m(arc AB). If triangle intercepts a diameter if the angle is a right an inscribed angle also intercepts arc AB, then the angle. Therefore, the hypotenuse is a diameter and measure of the inscribed angle is equal to m(arc has a length of 2r. Use trigonometry to find the AB). So, if an inscribed angle and a central angle length of the equal legs. intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . What is ?

Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. A 42 B 61 SOLUTION: C 80 See students’ work. D 84 50. WRITING IN MATH Compare and contrast SOLUTION: inscribed angles and central angles of a circle. If they If an angle is inscribed in a circle, then the measure intercept the same arc, how are they related? of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the But, measure of the central angle is equal to m(arc AB). If Therefore, an inscribed angle also intercepts arc AB, then the The correct choice is A. measure of the inscribed angle is equal to m(arc 52. ALGEBRA Simplify AB). So, if an inscribed angle and a central angle 2 intercept the same arc, then the measure of the 4(3x – 2)(2x + 4) + 3x + 5x – 6. 2 inscribed angle is one-half the measure of the central F 9x + 3x – 14 2 angle. G 9x + 13x – 14 2 H 27x + 37x – 38 51. In the circle below, and . 2 What is ? J 27x + 27x – 26 SOLUTION: Use the Distributive Property to simplify the first term.

A 42 B 61 C 80 Therefore, the correct choice is H. D 84 53. SHORT RESPONSE In the circle below, is a SOLUTION: diameter, AC = 8 inches, and BC = 15 inches. Find If an angle is inscribed in a circle, then the measure the diameter, the radius, and the circumference of of the angle equals one half the measure of its the circle. intercepted arc. Therefore,

But, Therefore, The correct choice is A. SOLUTION: 52. ALGEBRA Simplify Use the Pythagorean Theorem to find the 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. hypotenuse of the triangle which is also the diameter 2 F 9x + 3x – 14 of the circle.

G 9x2 + 13x – 14 2 The radius is half the diameter. So, the radius of the H 27x + 37x – 38 circle is about 8.5 in. 2 J 27x + 27x – 26 The circumference C of a circle of a circle of SOLUTION: diameter d is given by Use the Distributive Property to simplify the first Therefore, the circumference of the circle is term.

54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? A –15 B –16 C –17 Therefore, the correct choice is H. D –18 E –19 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find SOLUTION: the diameter, the radius, and the circumference of Let the three consecutive integers be x , x +1,and x + the circle. 2. Then,

So, the three integers are –17, –16, and –15 and the least of these is –17. SOLUTION: Therefore, the correct choice is C. Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter In , FL = 24, HJ = 48, and . Find of the circle. each measure.

The radius is half the diameter. So, the radius of the circle is about 8.5 in. The circumference C of a circle of a circle of diameter d is given by Therefore, the circumference of the circle is 55. FG SOLUTION: 54. SAT/ACT The sum of three consecutive integers is If a diameter (or radius) of a circle is perpendicular –48. What is the least of the three integers? to a chord, then it bisects the chord and its arc. So, A –15 bisects Therefore, B –16

C –17 D –18 E –19 56. SOLUTION: SOLUTION: Let the three consecutive integers be x , x +1,and x + If a diameter (or radius) of a circle is perpendicular

2. Then, to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ So, the three integers are –17, –16, and –15 and the SOLUTION: least of these is –17. Therefore, the correct choice is C. If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, In , FL = 24, HJ = 48, and . Find bisects Therefore, each measure.

58. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 55. FG bisects Therefore, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, Find x. bisects Therefore,

56. SOLUTION: 59. If a diameter (or radius) of a circle is perpendicular SOLUTION: to a chord, then it bisects the chord and its arc. So, The sum of the measures of the central angles of a bisects Therefore, circle with no interior points in common is 360. So,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

58. 60. SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular The sum of the measures of the central angles of a to a chord, then it bisects the chord and its arc. So, circle with no interior points in common is 360. So,

bisects Therefore,

Find x.

61. 59. SOLUTION: SOLUTION: The sum of the measures of the central angles of a The sum of the measures of the central angles of a circle with no interior points in common is 360. So, circle with no interior points in common is 360. So,

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 60. 7 feet, and the camera itself is 15 inches long. How SOLUTION: tall is the person being photographed? The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two

61. triangles are similar. So, their corresponding sides will SOLUTION: be proportional. One foot is equivalent to 12 in. Then, The sum of the measures of the central angles of a the height of the larger triangle is 7(12) = 84 in. Let x circle with no interior points in common is 360. So, be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft 62. PHOTOGRAPHY In one of the first cameras tall. invented, light entered an opening in the front. An image was reflected in the back of the camera, ALGEBRA Suppose B is the midpoint of . upside down, forming similar triangles. Suppose the Use the given information to find the missing image of the person on the back of the camera is 12 measure. inches, the distance from the opening to the person is 63. AB = 4x – 5, BC = 11 + 2x, AC = ? 7 feet, and the camera itself is 15 inches long. How SOLUTION: tall is the person being photographed? Since B is the midpoint of

SOLUTION:

The height of the person being photographed is the 64. AB = 6y – 14, BC = 10 – 2y, AC = ? length of the base of the larger triangle. The two SOLUTION: triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, Since B is the midpoint of the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft tall. 65. BC = 6 – 4m, AC = 8, m = ? ALGEBRA Suppose B is the midpoint of . SOLUTION: Use the given information to find the missing measure. Since B is the midpoint of 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ?

SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ?

Find each measure. 1.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. Therefore,

intercepted arc. Therefore, ALGEBRA Find each measure. 5. 2.

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same If an angle is inscribed in a circle, then the measure arc or congruent arcs, then the angles are congruent. of the angle equals one half the measure of its So, intercepted arc. Therefore,

3. Therefore,

6.

SOLUTION:

Here, is a semi-circle. So, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. So, intercepted arc. So,

Therefore,

4. SCIENCE The diagram shows how light bends in a

raindrop to make the colors of the rainbow. If Therefore,

, what is ? 7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Given: bisects . Prove: intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector)

3. intercepts . intercepts . SOLUTION: (Def. of intercepted arc) If two inscribed angles of a circle intercept the same 4. (Inscribed of same arc are arc or congruent arcs, then the angles are congruent. .)

So, 5. (Vertical are .)

6. (AAS)

Therefore, CCSS STRUCTURE Find each value. 8. 6.

SOLUTION:

An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. If two inscribed angles of a circle intercept the same So, arc or congruent arcs, then the angles are congruent. The sum of the measures of the angles of a triangle So,

is 180. So,

Therefore, Therefore,

7. PROOF Write a two-column proof. 9. x Given: bisects . Prove:

SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter Given: bisects . or semicircle if and only if the angle is a right angle. Prove: So, Since The sum of the measures of the angles of a triangle

is 180. So,

Proof: 10. and Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) SOLUTION: 5. (Vertical are .) If a quadrilateral is inscribed in a circle, then its 6. (AAS) opposite angles are supplementary.

CCSS STRUCTURE Find each value. 8. Solve the two equations to find the values of x and y.

Use the values of the variables to find and .

SOLUTION: An inscribed angle of a triangle intercepts a diameter ALGEBRA Find each measure. or semicircle if and only if the angle is a right angle. 11. So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore, SOLUTION: If an angle is inscribed in a circle, then the measure 9. x of the angle equals one half the measure of its intercepted arc. Therefore,

12.

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since SOLUTION: The sum of the measures of the angles of a triangle If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its is 180. So, intercepted arc. Therefore,

10. and 13.

SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Solve the two equations to find the values of x and y.

If an angle is inscribed in a circle, then the measure Use the values of the variables to find and of the angle equals one half the measure of its

. intercepted arc. Therefore,

14. ALGEBRA Find each measure. 11.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 15. intercepted arc. Therefore,

12.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So,

intercepted arc. Therefore, Therefore, 16. 13.

SOLUTION: SOLUTION:

The sum of the measures of the central angles of a Here, circle with no interior points in common is 360. So, The arc is a semicircle. So,

Then, If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

of the angle equals one half the measure of its intercepted arc. Therefore,

intercepted arc. Therefore, 17. ALGEBRA Find each measure.

14.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. intercepted arc. Therefore, So,

15. 18. ALGEBRA Find each measure.

SOLUTION:

Here,

The arc is a semicircle. So, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. intercepted arc. So, So,

Therefore,

16. Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = intercepted arc. Therefore, m∠B.

17. ALGEBRA Find each measure. Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, SOLUTION: 18. ALGEBRA Find each measure. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof SOLUTION: If two inscribed angles of a circle intercept the same Given: arc or congruent arcs, then the angles are congruent. So, Prove:

Therefore,

19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

. Since and

, the equation becomes

SOLUTION: . Multiplying each side of the If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. equation by 2 results in . Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. 22. two-column proof Given: Prove: Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) SOLUTION: 4. (AA Similarity) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. ALGEBRA Find each value. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47. 23. x PROOF Write the specified type of proof. 21. paragraph proof SOLUTION: Given: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Prove: So, The sum of the measures of the angles of a triangle

is 180. So,

24. SOLUTION: SOLUTION:

Proof: Given means that An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. . Since and So, The sum of the measures of the angles of a triangle , the equation becomes is 180. So, . Multiplying each side of the Therefore, equation by 2 results in . 25. x 22. two-column proof SOLUTION: Given: An inscribed angle of a triangle intercepts a diameter Prove: or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

SOLUTION: 26. Statements (Reasons): 1. (Given) SOLUTION: 2. (Inscribed intercepting same arc An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. are .) So, 3. (Vertical are .) The sum of the measures of the angles of a triangle

4. (AA Similarity) ALGEBRA Find each value. is 180. So,

Therefore,

CCSS STRUCTURE Find each measure.

23. x

SOLUTION: An inscribed angle of a triangle intercepts a diameter

or semicircle if and only if the angle is a right angle. 27. So, The sum of the measures of the angles of a triangle SOLUTION: If a quadrilateral is inscribed in a circle, then its is 180. So, opposite angles are supplementary.

Therefore,

24. 28. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. So, The sum of the measures of the angles of a triangle

is 180. So, Therefore,

Therefore, CCSS STRUCTURE Find each measure. 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 29. is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Therefore, So, The sum of the measures of the angles of a triangle 30. is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore,

CCSS STRUCTURE Find each measure.

Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION:

27. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. SOLUTION: and are supplementary. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

28. SOLUTION: If a quadrilateral is inscribed in a circle, then its Proof: By arc addition and the definitions of arc opposite angles are supplementary. measure and the sum of central angles, . Since by Theorem 10.6

and

Therefore, , but

CCSS STRUCTURE Find each measure. , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making them supplementary also.

SIGNS: A stop sign in the shape of a regular 29. octagon is inscribed in a circle. Find each SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, measure. 32. 30. SOLUTION: SOLUTION: Since all the sides of the stop sign are congruent If a quadrilateral is inscribed in a circle, then its chords of the circle, the corresponding arcs are opposite angles are supplementary. congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 or 45. Therefore,

Therefore, measure of arc NPQ is 135. 31. PROOF Write a paragraph proof for Theorem 11.9.

SOLUTION: 33. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. SOLUTION: and are supplementary. Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6 and Therefore, the measure of ∠RLQ is 22.5. , but

, so or 34. 180. This makes ∠C and ∠A supplementary. SOLUTION: Because the sum of the measures of the interior angles of a quadrilateral is 360, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its . But

, so , making intercepted arc. So, them supplementary also. The sum of the measures of the central angles of a SIGNS: A stop sign in the shape of a regular circle with no interior points in common is 360. Here, octagon is inscribed in a circle. Find each 360 is divided into 8 equal arcs and each arc

measures Then,

Therefore,

measure. 35. 32. SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are intercepted arc. So, congruent. There are eight adjacent arcs that make The sum of the measures of the central angles of a up the circle, so their sum is 360. Thus, the measure circle with no interior points in common is 360. Here, of each arc joining consecutive vertices is of 360 360 is divided into 8 equal arcs and each arc or 45. measures

Therefore, measure of arc NPQ is 135. Then,

Therefore, 33.

SOLUTION: 36. ART Four different string art star patterns are Since all the sides of the stop sign are congruent shown. If all of the inscribed angles of each star chords of the circle, all the corresponding arcs are shown are congruent, find the measure of each congruent. There are eight adjacent arcs that make inscribed angle. up the circle, so their sum is 360. Thus, the measure a. of each arc joining consecutive vertices is of 360 or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

b.

11-4Therefore, Inscribed the Angles measure of ∠RLQ is 22.5.

c. 34. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Then, SOLUTION: a. The sum of the measures of the central angles of Therefore, a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc 35. measures SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed

intercepted arc. So, angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, measures Each inscribed angle is formed 360 is divided into 8 equal arcs and each arc by intercepting alternate vertices of the star. So, each measures intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one Then, half the measure of its intercepted arc. So, the measure of each inscribed angle is 60.

Therefore, c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star circle, then the measure of the angle equals one half shown are congruent, find the measure of each the measure of its intercepted arc. So, the measure

inscribed angle. of each inscribed angle is about 25.7. a. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed b. in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. c. 37. Case 2 eSolutions Manual - Powered by Cognero Given: P lies inside . is a diameter. Page 8

Prove:

d.

SOLUTION: SOLUTION: Proof:

a. The sum of the measures of the central angles of Statements (Reasons) a circle with no interior points in common is 360. 1. m∠ABC = m∠ABD + m∠DBC ( Addition Here, 360 is divided into 5 equal arcs and each arc Postulate) 2. m∠ABD = m(arc AD) measures m∠DBC = m(arc DC) (The measure of an If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its inscribed whose side is a diameter is half the intercepted arc. So, the measure of each inscribed measure of the intercepted arc (Case 1)). angle is 36. 3. m∠ABC = m(arc AD) + m(arc DC) b. Here, 360 is divided into 6 equal arcs and each arc (Substitution) measures Each inscribed angle is formed 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)

by intercepting alternate vertices of the star. So, each 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc intercepted arc measures 120. If an angle is inscribed Addition Postulate) in a circle, then the measure of the angle equals one

half the measure of its intercepted arc. So, the 6. m∠ABC = m(arc AC) (Substitution) measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc 38. Case 3 measures If an angle is inscribed in a Given: P lies outside . is a diameter.

circle, then the measure of the angle equals one half Prove: the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one SOLUTION: half the measure of its intercepted arc. So, the Proof: measure of each inscribed angle is 45. Statements (Reasons) PROOF Write a two-column proof for each case 1. m∠ABC = m∠DBC – m∠DBA ( Addition of Theorem 11.6. Postulate, Subtraction Property of Equality) 37. Case 2 2. m∠DBC = m(arc DC)

Given: P lies inside . is a diameter. m∠DBA = m(arc DA) (The measure of an

Prove: inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction SOLUTION: Property of Equality) Proof: 7. m∠ABC = m(arc AC) (Substitution) Statements (Reasons) 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) PROOF Write the specified proof for each

2. m∠ABD = m(arc AD) theorem. m∠DBC = m(arc DC) (The measure of an 39. Theorem 11.7, two-column proof inscribed whose side is a diameter is half the SOLUTION:

measure of the intercepted arc (Case 1)). Given: and are inscribed; 3. m∠ABC = m(arc AD) + m(arc DC) Prove: (Substitution) Proof: 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)

5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter. Statements (Reasons) Prove: 1. and are inscribed;

(Given) 2. ; (Measure of an inscribed = half measure of intercepted arc.) 3. (Def. of arcs)

4. (Mult. Prop. of Equality)

5. (Substitution) SOLUTION: 6. (Def. of ) Proof: Statements (Reasons) 40. Theorem 11.8, paragraph proof 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) SOLUTION: 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) Part I: Given: is a semicircle. 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Prove: is a right angle. Addition Postulate) Proof: Since is a semicircle, then 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) . Since is an inscribed angle,

7. m∠ABC = m(arc AC) (Substitution) then or 90. So, by definition,

is a right angle. PROOF Write the specified proof for each theorem. Part II: Given: is a right angle. 39. Theorem 11.7, two-column proof Prove: is a semicircle. SOLUTION: Proof: Since is an inscribed angle, then Given: and are inscribed; Prove: and by the Multiplication Proof: Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. Statements (Reasons) a. GEOMETRIC Use a compass to draw a circle 1. and are inscribed; with parallel chords and . Connect points A

(Given) and D by drawing segment . b. NUMERICAL Use a protractor to find 2. ; (Measure and . Then determine and . of an inscribed = half measure of intercepted arc.) What is true about these arcs? Explain. 3. (Def. of arcs) c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that 4. (Mult. Prop. of Equality) are cut by two parallel chords. 5. (Substitution) d. ANALYTICAL Use your conjecture to find 6. (Def. of ) and in the figure. Verify by using inscribed angles to find the measures of the arcs. 40. Theorem 11.8, paragraph proof SOLUTION:

SOLUTION: a.

Part I: Given: is a semicircle. Prove: is a right angle.

Proof: Since is a semicircle, then b. Sample answer: , ; . Since is an inscribed angle, , ; The arcs are congruent because they have equal measures. then or 90. So, by definition, c. Sample answer: is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then

and by the Multiplication m∠A = 15, m∠D = 15, = 30, and = 30. The arcs are congruent. Property of Equality, m = 2m . Because In a circle, two parallel chords cut congruent arcs. is a right angle, m = 90. Then d. In a circle, two parallel chords cut congruent arcs. m = 2(90) or 180. So by definition, is a So, = . semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. = 4(16) + 6 or 70 a. GEOMETRIC Use a compass to draw a circle = 6(16) – 26 or 70 with parallel chords and . Connect points A Therefore, the measure of arcs QS and PR are each 70. and D by drawing segment . b. NUMERICAL Use a protractor to find CCSS ARGUMENTS Determine whether the and . Then determine and . quadrilateral can always, sometimes, or never What is true about these arcs? Explain. be inscribed in a circle. Explain your reasoning. c. VERBAL Draw another circle and repeat parts a 42. square and b. Make a conjecture about arcs of a circle that SOLUTION: are cut by two parallel chords. Squares have right angles at each vertex, therefore d. ANALYTICAL Use your conjecture to find each pair of opposite angles will be supplementary and in the figure. Verify by using and inscribed in a circle. Therefore, the statement is inscribed angles to find the measures of the arcs. always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, SOLUTION: the statement is always true. a. 44. parallelogram SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. b. Sample answer: , ; Therefore, the statement is sometimes true. , ; The arcs are congruent because they have equal measures. 45. rhombus c. Sample answer: SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle.

Therefore, the statement is sometimes true. m∠A = 15, m∠D = 15, = 30, and = 30. 46. kite The arcs are congruent. In a circle, two parallel chords cut congruent arcs. SOLUTION: d. In a circle, two parallel chords cut congruent arcs. Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be So, = . a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. = 4(16) + 6 or 70 Therefore, the statement is sometimes true. = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each 47. CHALLENGE A square is inscribed in a circle. 70. What is the ratio of the area of the circle to the area of the square? CCSS ARGUMENTS Determine whether the SOLUTION: quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. A square with side s is inscribed in a circle with 42. square radius r. SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle Using the Pythagorean Theorem, SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, 2 the statement is always true. The area of a square of side s is A = s and the area of a circle of radius r is A = . 44. parallelogram SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus Therefore, the ratio of the area of the circle to the SOLUTION: area of the inscribed square is The opposite angles of a rhombus are always WRITING IN MATH congruent. They will only be supplementary when 48. A right they are right angles. So, a rhombus can be inscribed triangle is inscribed in a circle. If the radius of the in a circle as long as it is a square. Since the opposite circle is given, explain how to find the lengths of the

angles of rhombi that are not squares are not right triangle’s legs. supplementary, they can not be inscribed in a circle. SOLUTION: Therefore, the statement is sometimes true. Sample answer: A triangle will have 46. kite two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. SOLUTION: According to theorem 10.8, an inscribed angle of a Exactly one pair of opposite angles of a kite are triangle intercepts a diameter if the angle is a right congruent. To be supplementary, they must each be angle. Therefore, the hypotenuse is a diameter and a right angle. If one pair of opposite angles for a has a length of 2r. Use trigonometry to find the quadrilateral is supplementary, the other pair must length of the equal legs. also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true.

47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area

of the square? Therefore, the length of each leg of the 45-45-90 SOLUTION: triangle can be found by multiplying the radius of the A square with side s is inscribed in a circle with circle in which it is inscribed by . radius r. 49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work.

50. WRITING IN MATH Compare and contrast Using the Pythagorean Theorem, inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. 2 If a central angle intercepts arc AB, then the The area of a square of side s is A = s and the area measure of the central angle is equal to m(arc AB). If of a circle of radius r is A = . an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . Therefore, the ratio of the area of the circle to the What is ? area of the inscribed square is

48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the circle is given, explain how to find the lengths of the right triangle’s legs.

SOLUTION: A 42 Sample answer: A triangle will have B 61 two inscribed angles of 45 and one of 90. The C 80 hypotenuse is across from the 90, or right angle. D 84 According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right SOLUTION: angle. Therefore, the hypotenuse is a diameter and If an angle is inscribed in a circle, then the measure has a length of 2r. Use trigonometry to find the of the angle equals one half the measure of its length of the equal legs. intercepted arc. Therefore,

But, Therefore, The correct choice is A.

52. ALGEBRA Simplify Therefore, the length of each leg of the 45-45-90 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. triangle can be found by multiplying the radius of the 2 F 9x + 3x – 14 circle in which it is inscribed by . G 9x2 + 13x – 14 2 49. OPEN ENDED Find and sketch a real-world logo H 27x + 37x – 38 with an inscribed polygon. J 27x2 + 27x – 26 SOLUTION: SOLUTION: See students’ work. Use the Distributive Property to simplify the first term. 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. Therefore, the correct choice is H. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If 53. SHORT RESPONSE In the circle below, is a an inscribed angle also intercepts arc AB, then the diameter, AC = 8 inches, and BC = 15 inches. Find measure of the inscribed angle is equal to m(arc the diameter, the radius, and the circumference of AB). So, if an inscribed angle and a central angle the circle. intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . What is ? SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle.

The radius is half the diameter. So, the radius of the circle is about 8.5 in. A 42 The circumference C of a circle of a circle of B 61 diameter d is given by C 80 Therefore, the circumference of the circle is D 84 SOLUTION: If an angle is inscribed in a circle, then the measure 54. SAT/ACT The sum of three consecutive integers is of the angle equals one half the measure of its –48. What is the least of the three integers? intercepted arc. Therefore, A –15 B –16 C –17 But, D –18 Therefore, E –19 The correct choice is A. SOLUTION: Let the three consecutive integers be x , x +1,and x + 52. ALGEBRA Simplify 2. Then, 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 F 9x + 3x – 14 G 9x2 + 13x – 14 2 H 27x + 37x – 38 So, the three integers are –17, –16, and –15 and the J 27x2 + 27x – 26 least of these is –17. Therefore, the correct choice is C. SOLUTION: Use the Distributive Property to simplify the first In , FL = 24, HJ = 48, and . Find term. each measure.

Therefore, the correct choice is H. 55. FG 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find SOLUTION: the diameter, the radius, and the circumference of If a diameter (or radius) of a circle is perpendicular the circle. to a chord, then it bisects the chord and its arc. So, bisects Therefore,

56. SOLUTION: SOLUTION: Use the Pythagorean Theorem to find the If a diameter (or radius) of a circle is perpendicular hypotenuse of the triangle which is also the diameter to a chord, then it bisects the chord and its arc. So, of the circle. bisects Therefore,

57. NJ The radius is half the diameter. So, the radius of the circle is about 8.5 in. SOLUTION: The circumference C of a circle of a circle of If a diameter (or radius) of a circle is perpendicular diameter d is given by to a chord, then it bisects the chord and its arc. So, Therefore, the circumference of the circle is bisects Therefore,

54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? 58. A –15 B –16 SOLUTION: C –17 If a diameter (or radius) of a circle is perpendicular D –18 to a chord, then it bisects the chord and its arc. So, E –19 bisects Therefore,

SOLUTION: Let the three consecutive integers be x , x +1,and x + 2. Then, Find x.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. 59. SOLUTION: In , FL = 24, HJ = 48, and . Find The sum of the measures of the central angles of a each measure. circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 60. bisects Therefore, SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

56. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: 61. If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, SOLUTION: The sum of the measures of the central angles of a bisects Therefore, circle with no interior points in common is 360. So,

58. SOLUTION: If a diameter (or radius) of a circle is perpendicular 62. PHOTOGRAPHY In one of the first cameras to a chord, then it bisects the chord and its arc. So, invented, light entered an opening in the front. An image was reflected in the back of the camera, bisects Therefore, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is Find x. 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

60. SOLUTION: Therefore, the person being photographed is 5.6 ft The sum of the measures of the central angles of a tall. circle with no interior points in common is 360. So, ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

61.

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two 66. AB = 10s + 2, AC = 40, s = ? triangles are similar. So, their corresponding sides will SOLUTION: be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x Since B is the midpoint of be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft tall.

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

Find each measure. 1. SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

If an angle is inscribed in a circle, then the measure intercepted arc. So, of the angle equals one half the measure of its Therefore,

intercepted arc. Therefore, 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ? 2.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 3. ALGEBRA Find each measure. 5.

SOLUTION: Here, is a semi-circle. So, SOLUTION: If two inscribed angles of a circle intercept the same If an angle is inscribed in a circle, then the measure arc or congruent arcs, then the angles are congruent. of the angle equals one half the measure of its So, intercepted arc. So,

Therefore, Therefore, 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If 6. , what is ?

SOLUTION: If two inscribed angles of a circle intercept the same SOLUTION: arc or congruent arcs, then the angles are congruent. If an angle is inscribed in a circle, then the measure So, of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. Therefore, 5. 7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION:

So, Given: bisects .

Prove:

Therefore,

6.

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) SOLUTION: 4. (Inscribed of same arc are If two inscribed angles of a circle intercept the same .) arc or congruent arcs, then the angles are congruent. 5. (Vertical are .) So,

6. (AAS) CCSS STRUCTURE Find each value. 8. Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, SOLUTION: The sum of the measures of the angles of a triangle

Given: bisects . Prove: is 180. So,

Therefore,

9. x

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are SOLUTION: .) An inscribed angle of a triangle intercepts a diameter 5. (Vertical are .) or semicircle if and only if the angle is a right angle. 6. (AAS) So, Since The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each value. 8. is 180. So,

10. and

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, SOLUTION: The sum of the measures of the angles of a triangle If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. is 180. So, Solve the two equations to find the values of x and y. Therefore,

9. x Use the values of the variables to find and .

ALGEBRA Find each measure. 11. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

10. and intercepted arc. Therefore,

12.

SOLUTION: If a quadrilateral is inscribed in a circle, then its

opposite angles are supplementary. SOLUTION:

If an angle is inscribed in a circle, then the measure Solve the two equations to find the values of x and y. of the angle equals one half the measure of its

intercepted arc. Therefore, Use the values of the variables to find and . 13.

ALGEBRA Find each measure. 11.

SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore,

12. 14.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 15. 13.

SOLUTION:

Here, SOLUTION: The arc is a semicircle. So, The sum of the measures of the central angles of a If an angle is inscribed in a circle, then the measure circle with no interior points in common is 360. So, of the angle equals one half the measure of its

intercepted arc. So, Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 16.

intercepted arc. Therefore,

14.

SOLUTION: Here, SOLUTION: The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure Then, of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc. Therefore, of the angle equals one half the measure of its

intercepted arc. Therefore, 15. 17. ALGEBRA Find each measure.

SOLUTION: Here,

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: If two inscribed angles of a circle intercept the same intercepted arc. So, arc or congruent arcs, then the angles are congruent. Therefore, So,

16. 18. ALGEBRA Find each measure.

SOLUTION: Here, SOLUTION: If two inscribed angles of a circle intercept the same The arc is a semicircle. So, arc or congruent arcs, then the angles are congruent.

Then, So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, Therefore,

19. ALGEBRA Find each measure. 17. ALGEBRA Find each measure.

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. arc or congruent arcs, then the angles are congruent. So, Since ∠A and ∠B both intercept arc CD , m∠A = m∠B. 18. ALGEBRA Find each measure.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

SOLUTION: Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 19. ALGEBRA Find each measure. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given:

SOLUTION: Prove: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20. SOLUTION:

20. ALGEBRA Find each measure. Proof: Given means that

. Since and

, the equation becomes

. Multiplying each side of the

equation by 2 results in . SOLUTION: 22. two-column proof If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Given: Since ∠C and ∠D both intercept arc AB, m∠C = Prove: m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof SOLUTION: Statements (Reasons): Given: 1. (Given) 2. (Inscribed intercepting same arc Prove: are .) 3. (Vertical are .) 4. (AA Similarity)

ALGEBRA Find each value.

SOLUTION:

Proof: Given means that

23. x . Since and

, the equation becomes SOLUTION: An inscribed angle of a triangle intercepts a diameter . Multiplying each side of the or semicircle if and only if the angle is a right angle. So, equation by 2 results in . The sum of the measures of the angles of a triangle

is 180. So, 22. two-column proof Given: Prove: 24. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So,

The sum of the measures of the angles of a triangle SOLUTION: Statements (Reasons): is 180. So, 1. (Given) 2. (Inscribed intercepting same arc Therefore, are .) 3. (Vertical are .) 25. x 4. (AA Similarity) SOLUTION: ALGEBRA Find each value. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

23. x

26. SOLUTION: An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. An inscribed angle of a triangle intercepts a diameter So, or semicircle if and only if the angle is a right angle. The sum of the measures of the angles of a triangle So, The sum of the measures of the angles of a triangle

is 180. So, is 180. So,

24. Therefore, SOLUTION: CCSS STRUCTURE Find each measure. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 27. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 25. x opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, 28. The sum of the measures of the angles of a triangle SOLUTION: If a quadrilateral is inscribed in a circle, then its is 180. So, opposite angles are supplementary.

26.

SOLUTION: Therefore, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. CCSS STRUCTURE Find each measure. So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

CCSS STRUCTURE Find each measure. 29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

27. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 30. opposite angles are supplementary. SOLUTION: Therefore, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 28.

SOLUTION: If a quadrilateral is inscribed in a circle, then its Therefore, opposite angles are supplementary.

31. PROOF Write a paragraph proof for Theorem 11.9.

SOLUTION: Therefore, Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. CCSS STRUCTURE Find each measure. and are supplementary.

29. Proof: By arc addition and the definitions of arc SOLUTION: measure and the sum of central angles, If a quadrilateral is inscribed in a circle, then its . Since by Theorem 10.6 opposite angles are supplementary. and , but

Therefore, , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior 30. angles of a quadrilateral is 360, SOLUTION: . But If a quadrilateral is inscribed in a circle, then its , so , making opposite angles are supplementary. them supplementary also.

SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. measure. and are supplementary. 32. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360

Proof: By arc addition and the definitions of arc or 45. measure and the sum of central angles,

. Since by Theorem 10.6 Therefore, measure of arc NPQ is 135. and 33. , but SOLUTION: , so or Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior congruent. There are eight adjacent arcs that make angles of a quadrilateral is 360, up the circle, so their sum is 360. Thus, the measure . But of each arc joining consecutive vertices is of 360 , so , making or 45. Since ∠RLQ is inscribed in the circle, its them supplementary also. measure equals one half the measure of its intercepted arc QR. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore, the measure of ∠RLQ is 22.5.

measure. 34. 32. SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are intercepted arc. So, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, of each arc joining consecutive vertices is of 360 360 is divided into 8 equal arcs and each arc or 45. measures

Therefore, measure of arc NPQ is 135. Then,

Therefore, 33. SOLUTION: 35. Since all the sides of the stop sign are congruent SOLUTION: chords of the circle, all the corresponding arcs are If an angle is inscribed in a circle, then the measure congruent. There are eight adjacent arcs that make of the angle equals one half the measure of its up the circle, so their sum is 360. Thus, the measure

of each arc joining consecutive vertices is of 360 intercepted arc. So, or 45. Since ∠RLQ is inscribed in the circle, its The sum of the measures of the central angles of a measure equals one half the measure of its circle with no interior points in common is 360. Here, intercepted arc QR. 360 is divided into 8 equal arcs and each arc

measures

Then,

Therefore, Therefore, the measure of ∠RLQ is 22.5.

36. ART Four different string art star patterns are 34. shown. If all of the inscribed angles of each star shown are congruent, find the measure of each SOLUTION: inscribed angle. If an angle is inscribed in a circle, then the measure a. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here,

360 is divided into 8 equal arcs and each arc b.

measures Then,

Therefore,

c. 35. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

measures

Then, SOLUTION: a. The sum of the measures of the central angles of Therefore, a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc

36. ART Four different string art star patterns are measures shown. If all of the inscribed angles of each star shown are congruent, find the measure of each If an angle is inscribed in a circle, then the measure inscribed angle. of the angle equals one half the measure of its a. intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed

by intercepting alternate vertices of the star. So, each b. intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a c. circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc d. measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. SOLUTION: a. The sum of the measures of the central angles of PROOF Write a two-column proof for each case a circle with no interior points in common is 360. of Theorem 11.6. Here, 360 is divided into 5 equal arcs and each arc 37. Case 2 Given: P lies inside . is a diameter. measures Prove: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed

by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one SOLUTION: half the measure of its intercepted arc. So, the Proof: measure of each inscribed angle is 60. Statements (Reasons) c. Here, 360 is divided into 7 equal arcs and each arc 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) measures If an angle is inscribed in a 2. m∠ABD = m(arc AD) circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure m∠DBC = m(arc DC) (The measure of an of each inscribed angle is about 25.7. inscribed whose side is a diameter is half the d. The sum of the measures of the central angles of measure of the intercepted arc (Case 1)). a circle with no interior points in common is 360. 3. m∠ABC = m(arc AD) + m(arc DC) Here, 360 is divided into 8 equal arcs and each arc (Substitution) measures Each inscribed angle is formed 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc in a circle, then the measure of the angle equals one Addition Postulate) 11-4half Inscribed the measure Angles of its intercepted arc. So, the 6. m∠ABC = m(arc AC) (Substitution) measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case 38. Case 3 of Theorem 11.6. Given: P lies outside . is a diameter. 37. Case 2 Given: P lies inside . is a diameter. Prove:

Prove:

SOLUTION: SOLUTION: Proof: Proof: Statements (Reasons) Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate, Subtraction Property of Equality) Postulate) 2. m∠DBC = m(arc DC)

2. m∠ABD = m(arc AD) m∠DBA = m(arc DA) (The measure of an m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) 3. m∠ABC = m(arc AD) + m(arc DC) (Substitution) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 6. m∠ABC = m(arc AC) (Substitution) 7. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter. PROOF Write the specified proof for each theorem. Prove: 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof:

SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition eSolutions Manual - Powered by Cognero Page 9 Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) Statements (Reasons) 1. and are inscribed; m∠DBA = m(arc DA) (The measure of an (Given) inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 2. ; (Measure 3. m∠ABC = m(arc DC) – m(arc DA) of an inscribed = half measure of intercepted arc.) (Substitution) 3. (Def. of arcs) 4. m ABC = [m(arc DC) – m(arc DA)] (Factor) ∠ 4. (Mult. Prop. of Equality) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc

Addition Postulate) 5. (Substitution) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 6. (Def. of ) Property of Equality) 7. m∠ABC = m(arc AC) (Substitution) 40. Theorem 11.8, paragraph proof SOLUTION:

PROOF Write the specified proof for each theorem. 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof: Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition, is a right angle.

Statements (Reasons) Part II: Given: is a right angle. 1. and are inscribed; Prove : is a semicircle. (Given) 2. ; (Measure Proof: Since is an inscribed angle, then of an inscribed = half measure of intercepted arc.) and by the Multiplication

3. (Def. of arcs) Property of Equality, m = 2m . Because

4. (Mult. Prop. of Equality) is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a

5. (Substitution) semicircle. 6. (Def. of ) 41. MULTIPLE REPRESENTATIONS In this 40. Theorem 11.8, paragraph proof problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel SOLUTION: chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain.

c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that Part I: Given: is a semicircle. are cut by two parallel chords. Prove: is a right angle. d. ANALYTICAL Use your conjecture to find Proof: Since is a semicircle, then and in the figure. Verify by using . Since is an inscribed angle, inscribed angles to find the measures of the arcs. then or 90. So, by definition, is a right angle.

Part II: Given: is a right angle.

Prove: is a semicircle. SOLUTION: a. Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then

m = 2(90) or 180. So by definition, is a b. Sample answer: , ; semicircle. , ; The arcs are congruent 41. MULTIPLE REPRESENTATIONS In this because they have equal measures. problem, you will investigate the relationship between c. Sample answer: the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain. m∠A = 15, m∠D = 15, = 30, and = 30. c. VERBAL Draw another circle and repeat parts a The arcs are congruent. and b. Make a conjecture about arcs of a circle that In a circle, two parallel chords cut congruent arcs. are cut by two parallel chords. d. In a circle, two parallel chords cut congruent arcs. d. ANALYTICAL Use your conjecture to find So, = . and in the figure. Verify by using inscribed angles to find the measures of the arcs.

= 4(16) + 6 or 70 = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each 70. SOLUTION: a. CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary b. Sample answer: , ; and inscribed in a circle. Therefore, the statement is always true. , ; The arcs are congruent because they have equal measures. c. Sample answer: 43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram

m∠A = 15, m∠D = 15, = 30, and = 30. SOLUTION: The arcs are congruent. The opposite angles of a parallelogram are always In a circle, two parallel chords cut congruent arcs. congruent. They will only be supplementary when d. In a circle, two parallel chords cut congruent arcs. they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. So, = . Therefore, the statement is sometimes true.

45. rhombus SOLUTION: = 4(16) + 6 or 70 The opposite angles of a rhombus are always congruent. They will only be supplementary when = 6(16) – 26 or 70 they are right angles. So, a rhombus can be inscribed Therefore, the measure of arcs QS and PR are each in a circle as long as it is a square. Since the opposite 70. angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. CCSS ARGUMENTS Determine whether the Therefore, the statement is sometimes true. quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 46. kite 42. square SOLUTION: SOLUTION: Exactly one pair of opposite angles of a kite are Squares have right angles at each vertex, therefore congruent. To be supplementary, they must each be each pair of opposite angles will be supplementary a right angle. If one pair of opposite angles for a and inscribed in a circle. Therefore, the statement is quadrilateral is supplementary, the other pair must always true. also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are 43. rectangle right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. SOLUTION: Rectangles have right angles at each vertex, 47. CHALLENGE A square is inscribed in a circle. therefore each pair of opposite angles will be What is the ratio of the area of the circle to the area supplementary and inscribed in a circle. Therefore, of the square? the statement is always true. SOLUTION: 44. parallelogram A square with side s is inscribed in a circle with radius r. SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus Using the Pythagorean Theorem, SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite 2 angles of rhombi that are not squares are not The area of a square of side s is A = s and the area supplementary, they can not be inscribed in a circle. of a circle of radius r is A = . Therefore, the statement is sometimes true.

46. kite SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that Therefore, the ratio of the area of the circle to the compose the pair of congruent opposite angles are area of the inscribed square is right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the 47. CHALLENGE A square is inscribed in a circle. circle is given, explain how to find the lengths of the What is the ratio of the area of the circle to the area right triangle’s legs. of the square? SOLUTION: SOLUTION: Sample answer: A triangle will have A square with side s is inscribed in a circle with two inscribed angles of 45 and one of 90. The radius r. hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs.

Using the Pythagorean Theorem,

2 Therefore, the length of each leg of the 45-45-90 The area of a square of side s is A = s and the area triangle can be found by multiplying the radius of the of a circle of radius r is A = . circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work.

Therefore, the ratio of the area of the circle to the 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they area of the inscribed square is intercept the same arc, how are they related?

48. WRITING IN MATH A right SOLUTION: triangle is inscribed in a circle. If the radius of the An inscribed angle has its vertex on the circle. A circle is given, explain how to find the lengths of the central angle has its vertex at the center of the circle. right triangle’s legs. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If SOLUTION: an inscribed angle also intercepts arc AB, then the Sample answer: A triangle will have measure of the inscribed angle is equal to m(arc two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. AB). So, if an inscribed angle and a central angle According to theorem 10.8, an inscribed angle of a intercept the same arc, then the measure of the triangle intercepts a diameter if the angle is a right inscribed angle is one-half the measure of the central

angle. Therefore, the hypotenuse is a diameter and angle. has a length of 2r. Use trigonometry to find the length of the equal legs. 51. In the circle below, and . What is ?

Therefore, the length of each leg of the 45-45-90 A triangle can be found by multiplying the radius of the 42 B 61 circle in which it is inscribed by . C 80 D 84 49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its See students’ work. intercepted arc. Therefore,

50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? But, Therefore, SOLUTION: The correct choice is A. An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. 52. ALGEBRA Simplify If a central angle intercepts arc AB, then the 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. measure of the central angle is equal to m(arc AB). If 2 an inscribed angle also intercepts arc AB, then the F 9x + 3x – 14 2 measure of the inscribed angle is equal to m(arc G 9x + 13x – 14 2 AB). So, if an inscribed angle and a central angle H 27x + 37x – 38 intercept the same arc, then the measure of the J 27x2 + 27x – 26 inscribed angle is one-half the measure of the central angle. SOLUTION: Use the Distributive Property to simplify the first 51. In the circle below, and . term. What is ?

Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a A 42 diameter, AC = 8 inches, and BC = 15 inches. Find B 61 the diameter, the radius, and the circumference of C 80 the circle. D 84 SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

SOLUTION: Use the Pythagorean Theorem to find the But, hypotenuse of the triangle which is also the diameter Therefore, of the circle. The correct choice is A. 52. ALGEBRA Simplify The radius is half the diameter. So, the radius of the 2 4(3x – 2)(2x + 4) + 3x + 5x – 6. circle is about 8.5 in. 2 The circumference C of a circle of a circle of F 9x + 3x – 14 diameter d is given by 2 G 9x + 13x – 14 Therefore, the circumference of the circle is 2 H 27x + 37x – 38 J 27x2 + 27x – 26 54. SAT/ACT The sum of three consecutive integers is SOLUTION: –48. What is the least of the three integers? Use the Distributive Property to simplify the first A –15 term. B –16 C –17 D –18 E –19 SOLUTION: Let the three consecutive integers be x , x +1,and x +

Therefore, the correct choice is H. 2. Then, 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle. So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure.

The radius is half the diameter. So, the radius of the 55. FG circle is about 8.5 in. The circumference C of a circle of a circle of SOLUTION: diameter d is given by If a diameter (or radius) of a circle is perpendicular Therefore, the circumference of the circle is to a chord, then it bisects the chord and its arc. So, bisects Therefore,

54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? A –15 56. B –16 SOLUTION: C –17 If a diameter (or radius) of a circle is perpendicular D –18 to a chord, then it bisects the chord and its arc. So, E –19 bisects Therefore, SOLUTION: Let the three consecutive integers be x , x +1,and x + 57. NJ 2. Then, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. 58. In , FL = 24, HJ = 48, and . Find each measure. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

Find x. 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 59.

SOLUTION: The sum of the measures of the central angles of a 56. circle with no interior points in common is 360. So,

SOLUTION:

If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 60. bisects Therefore, SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 58.

SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

Find x. 61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

60. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 62. PHOTOGRAPHY In one of the first cameras 63. AB = 4x – 5, BC = 11 + 2x, AC = ? invented, light entered an opening in the front. An image was reflected in the back of the camera, SOLUTION: upside down, forming similar triangles. Suppose the Since B is the midpoint of image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Therefore, the person being photographed is 5.6 ft Since B is the midpoint of tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? 66. AB = 10s + 2, AC = 40, s = ? SOLUTION: SOLUTION: Since B is the midpoint of Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

Find each measure. 1.

SOLUTION:

Here, is a semi-circle. So, SOLUTION:

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. So, Therefore, 2. 4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If , what is ?

intercepted arc. Therefore, SOLUTION: If an angle is inscribed in a circle, then the measure 3. of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. So, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Therefore, So,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If Therefore, , what is ?

6.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If two inscribed angles of a circle intercept the same intercepted arc. Therefore, arc or congruent arcs, then the angles are congruent. So, ALGEBRA Find each measure. 5.

Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove: SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

SOLUTION:

Therefore, Given: bisects . Prove: 6.

Proof: Statements (Reasons)

1. bisects . (Given) SOLUTION: 2. (Def. of segment bisector) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 3. intercepts . intercepts . (Def. of intercepted arc) So, 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS) Therefore, CCSS STRUCTURE Find each value. 7. PROOF Write a two-column proof. 8. Given: bisects . Prove:

SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter Given: bisects . or semicircle if and only if the angle is a right angle.

Prove: So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore, Proof: Statements (Reasons) 9. x 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS) SOLUTION: An inscribed angle of a triangle intercepts a diameter CCSS STRUCTURE Find each value. or semicircle if and only if the angle is a right angle. 8. So, Since The sum of the measures of the angles of a triangle

is 180. So,

10. and SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its Therefore, opposite angles are supplementary.

9. x Solve the two equations to find the values of x and y.

Use the values of the variables to find and .

SOLUTION: ALGEBRA Find each measure. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 11. So, Since The sum of the measures of the angles of a triangle

is 180. So,

10. and SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

12.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y. SOLUTION: If an angle is inscribed in a circle, then the measure Use the values of the variables to find and of the angle equals one half the measure of its . intercepted arc. Therefore,

13. ALGEBRA Find each measure. 11.

SOLUTION: The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

12. intercepted arc. Therefore,

14.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 13. intercepted arc. Therefore,

15.

SOLUTION: The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, Here,

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. So,

intercepted arc. Therefore, Therefore,

16. 14.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Here, intercepted arc. Therefore, The arc is a semicircle. So, Then, 15. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 16. arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So,

Then, SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent.

intercepted arc. Therefore, So,

17. ALGEBRA Find each measure. Therefore,

19. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent.

So, SOLUTION: If two inscribed angles of a circle intercept the same 18. ALGEBRA Find each measure. arc or congruent arcs, then the angles are congruent.

Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

Therefore,

19. ALGEBRA Find each measure. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. SOLUTION: 21. paragraph proof

If two inscribed angles of a circle intercept the same Given: arc or congruent arcs, then the angles are congruent. Since A and B both intercept arc CD , m A = ∠ ∠ ∠ Prove: m∠B.

Therefore, m∠A = 5(4) or 20. ALGEBRA Find each measure. 20.

SOLUTION: Proof: Given means that

. Since and

, the equation becomes

. Multiplying each side of the SOLUTION: If two inscribed angles of a circle intercept the same equation by 2 results in . arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D. 22. two-column proof Given: Prove:

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given: SOLUTION: Prove: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) 4. (AA Similarity) SOLUTION: ALGEBRA Find each value.

Proof: Given means that

. Since and

, the equation becomes 23. x

. Multiplying each side of the SOLUTION: equation by 2 results in . An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.

22. two-column proof So, The sum of the measures of the angles of a triangle Given:

Prove: is 180. So,

24.

SOLUTION: An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. Statements (Reasons): So, 1. (Given) The sum of the measures of the angles of a triangle 2. (Inscribed intercepting same arc are .) is 180. So, 3. (Vertical are .)

4. (AA Similarity) Therefore, ALGEBRA Find each value. 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 23. x

is 180. So, SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 26. So, The sum of the measures of the angles of a triangle SOLUTION: An inscribed angle of a triangle intercepts a diameter is 180. So, or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

24. is 180. So, SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, CCSS STRUCTURE Find each measure. The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

25. x 27. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter If a quadrilateral is inscribed in a circle, then its or semicircle if and only if the angle is a right angle. opposite angles are supplementary. So, The sum of the measures of the angles of a triangle Therefore,

is 180. So, 28. SOLUTION: 26. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Therefore, The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each measure. is 180. So,

Therefore,

CCSS STRUCTURE Find each measure.

29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 27. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore,

Therefore, 30.

28. SOLUTION: If a quadrilateral is inscribed in a circle, then its SOLUTION: opposite angles are supplementary. If a quadrilateral is inscribed in a circle, then its

opposite angles are supplementary.

Therefore, Therefore, 31. PROOF Write a paragraph proof for Theorem 11.9. CCSS STRUCTURE Find each measure. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary.

29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6

Therefore, and , but 30. , so or SOLUTION: 180. This makes ∠C and ∠A supplementary. If a quadrilateral is inscribed in a circle, then its Because the sum of the measures of the interior opposite angles are supplementary. angles of a quadrilateral is 360, . But , so , making them supplementary also. Therefore, SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each 31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary.

measure. 32. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make Proof: By arc addition and the definitions of arc up the circle, so their sum is 360. Thus, the measure measure and the sum of central angles, of each arc joining consecutive vertices is of 360 . Since by Theorem 10.6 or 45. and

, but Therefore, measure of arc NPQ is 135.

, so or 33. 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior SOLUTION: angles of a quadrilateral is 360, Since all the sides of the stop sign are congruent . But chords of the circle, all the corresponding arcs are , so , making congruent. There are eight adjacent arcs that make them supplementary also. up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR.

measure. Therefore, the measure of ∠RLQ is 22.5. 32. SOLUTION: 34. Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are SOLUTION: congruent. There are eight adjacent arcs that make If an angle is inscribed in a circle, then the measure up the circle, so their sum is 360. Thus, the measure of the angle equals one half the measure of its

of each arc joining consecutive vertices is of 360 intercepted arc. So, or 45. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, Therefore, measure of arc NPQ is 135. 360 is divided into 8 equal arcs and each arc

measures 33. Then, SOLUTION: Since all the sides of the stop sign are congruent Therefore, chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make 35. up the circle, so their sum is 360. Thus, the measure SOLUTION: of each arc joining consecutive vertices is of 360 If an angle is inscribed in a circle, then the measure or 45. Since ∠RLQ is inscribed in the circle, its of the angle equals one half the measure of its measure equals one half the measure of its intercepted arc QR. intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Therefore, the measure of ∠RLQ is 22.5. Then,

34. Therefore, SOLUTION: 36. ART Four different string art star patterns are If an angle is inscribed in a circle, then the measure shown. If all of the inscribed angles of each star of the angle equals one half the measure of its shown are congruent, find the measure of each

intercepted arc. So, inscribed angle. a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

Then, b.

Therefore,

35. SOLUTION: If an angle is inscribed in a circle, then the measure c. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc d.

measures

Then,

Therefore, SOLUTION: 36. ART Four different string art star patterns are a. The sum of the measures of the central angles of shown. If all of the inscribed angles of each star a circle with no interior points in common is 360. shown are congruent, find the measure of each Here, 360 is divided into 5 equal arcs and each arc inscribed angle. a. measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc b. measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the c. measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each SOLUTION: intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one a. The sum of the measures of the central angles of half the measure of its intercepted arc. So, the a circle with no interior points in common is 360. measure of each inscribed angle is 45. Here, 360 is divided into 5 equal arcs and each arc PROOF Write a two-column proof for each case measures of Theorem 11.6. If an angle is inscribed in a circle, then the measure 37. Case 2 of the angle equals one half the measure of its Given: P lies inside . is a diameter. intercepted arc. So, the measure of each inscribed angle is 36. Prove: b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc SOLUTION: measures If an angle is inscribed in a Proof: Statements (Reasons) circle, then the measure of the angle equals one half 1. m∠ABC = m∠ABD + m∠DBC ( Addition the measure of its intercepted arc. So, the measure Postulate) of each inscribed angle is about 25.7. 2. m∠ABD = m(arc AD) d. The sum of the measures of the central angles of a circle with no interior points in common is 360. m∠DBC = m(arc DC) (The measure of an Here, 360 is divided into 8 equal arcs and each arc inscribed whose side is a diameter is half the measures Each inscribed angle is formed measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc AD) + m(arc DC) by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed (Substitution) in a circle, then the measure of the angle equals one 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) PROOF Write a two-column proof for each case of Theorem 11.6. 6. m∠ABC = m(arc AC) (Substitution) 37. Case 2 Given: P lies inside . is a diameter. 38. Case 3

Prove: Given: P lies outside . is a diameter.

Prove:

SOLUTION:

Proof: Statements (Reasons) SOLUTION: 1. m∠ABC = m∠ABD + m∠DBC ( Addition Proof: Postulate) Statements (Reasons) 2. m∠ABD = m(arc AD) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) m∠DBC = m(arc DC) (The measure of an 2. m∠DBC = m(arc DC) inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). m∠DBA = m(arc DA) (The measure of an 3. m∠ABC = m(arc AD) + m(arc DC) inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). (Substitution) 3. m∠ABC = m(arc DC) – m(arc DA) 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) (Substitution)

5. m(arc AD) + m(arc DC) = m(arc AC) (Arc 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 38. Case 3 Property of Equality) Given: P lies outside . is a diameter. 7. m∠ABC = m(arc AC) (Substitution)

Prove: PROOF Write the specified proof for each theorem. 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof: SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an

inscribed whose side is a diameter is half the Statements (Reasons) measure of the intercepted arc (Case 1)). 1. and are inscribed; 3. m∠ABC = m(arc DC) – m(arc DA) (Given) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) 2. ; (Measure of an inscribed = half measure of intercepted arc.) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 3. (Def. of arcs)

6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 4. (Mult. Prop. of Equality) Property of Equality) 7. m∠ABC = m(arc AC) (Substitution) 5. (Substitution) 11-4 Inscribed Angles 6. (Def. of )

PROOF Write the specified proof for each 40. Theorem 11.8, paragraph proof theorem. 39. Theorem 11.7, two-column proof SOLUTION: SOLUTION: Given: and are inscribed; Prove: Proof:

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle,

Statements (Reasons) then or 90. So, by definition, 1. and are inscribed; is a right angle. (Given) Part II: Given: is a right angle. 2. ; (Measure Prove: is a semicircle. of an inscribed = half measure of intercepted arc.) 3. (Def. of arcs) Proof: Since is an inscribed angle, then

4. (Mult. Prop. of Equality) and by the Multiplication

5. (Substitution) Property of Equality, m = 2m . Because 6. (Def. of ) is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a 40. Theorem 11.8, paragraph proof semicircle. SOLUTION: 41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . Part I: Given: is a semicircle. What is true about these arcs? Explain. Prove: is a right angle. c. VERBAL Draw another circle and repeat parts a Proof: Since is a semicircle, then and b. Make a conjecture about arcs of a circle that . Since is an inscribed angle, are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find then or 90. So, by definition, and in the figure. Verify by using is a right angle. inscribed angles to find the measures of the arcs. eSolutions Manual - Powered by Cognero Page 10 Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication SOLUTION: a. Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between b. Sample answer: , ; the arcs of a circle that are cut by two parallel , ; The arcs are congruent chords. because they have equal measures. a. GEOMETRIC Use a compass to draw a circle c. Sample answer: with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain. c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. m∠A = 15, m∠D = 15, = 30, and = 30. d. ANALYTICAL Use your conjecture to find The arcs are congruent. and in the figure. Verify by using In a circle, two parallel chords cut congruent arcs. inscribed angles to find the measures of the arcs. d. In a circle, two parallel chords cut congruent arcs. So, = .

= 4(16) + 6 or 70 SOLUTION: = 6(16) – 26 or 70 a. Therefore, the measure of arcs QS and PR are each 70.

CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square

b. Sample answer: , ; SOLUTION: , ; The arcs are congruent Squares have right angles at each vertex, therefore because they have equal measures. each pair of opposite angles will be supplementary c. Sample answer: and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, m∠A = 15, m∠D = 15, = 30, and = 30. the statement is always true. The arcs are congruent. 44. parallelogram In a circle, two parallel chords cut congruent arcs. d. In a circle, two parallel chords cut congruent arcs. SOLUTION: So, = . The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true. = 4(16) + 6 or 70 45. rhombus = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each SOLUTION: 70. The opposite angles of a rhombus are always congruent. They will only be supplementary when CCSS ARGUMENTS Determine whether the they are right angles. So, a rhombus can be inscribed quadrilateral can always, sometimes, or never in a circle as long as it is a square. Since the opposite be inscribed in a circle. Explain your reasoning. angles of rhombi that are not squares are not 42. square supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true. SOLUTION: Squares have right angles at each vertex, therefore 46. kite each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is SOLUTION: always true. Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a 43. rectangle quadrilateral is supplementary, the other pair must SOLUTION: also be supplementary. So, as long as the angles that Rectangles have right angles at each vertex, compose the pair of congruent opposite angles are therefore each pair of opposite angles will be right angles, a kite can be inscribed in a circle.

supplementary and inscribed in a circle. Therefore, Therefore, the statement is sometimes true. the statement is always true. 47. CHALLENGE A square is inscribed in a circle. 44. parallelogram What is the ratio of the area of the circle to the area of the square? SOLUTION: SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when A square with side s is inscribed in a circle with they are right angles. So, a parallelogram can be radius r. inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when Using the Pythagorean Theorem, they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true. 2 46. kite The area of a square of side s is A = s and the area of a circle of radius r is A = . SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. Therefore, the ratio of the area of the circle to the

47. CHALLENGE A square is inscribed in a circle. area of the inscribed square is What is the ratio of the area of the circle to the area of the square? 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the SOLUTION: circle is given, explain how to find the lengths of the A square with side s is inscribed in a circle with right triangle’s legs. radius r. SOLUTION: Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and

Using the Pythagorean Theorem, has a length of 2r. Use trigonometry to find the length of the equal legs.

2 The area of a square of side s is A = s and the area of a circle of radius r is A = .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon.

Therefore, the ratio of the area of the circle to the SOLUTION: area of the inscribed square is See students’ work. 50. WRITING IN MATH Compare and contrast 48. WRITING IN MATH A right inscribed angles and central angles of a circle. If they triangle is inscribed in a circle. If the radius of the intercept the same arc, how are they related? circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. Sample answer: A triangle will have If a central angle intercepts arc AB, then the two inscribed angles of 45 and one of 90. The measure of the central angle is equal to m(arc AB). If hypotenuse is across from the 90, or right angle. an inscribed angle also intercepts arc AB, then the According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right measure of the inscribed angle is equal to m(arc angle. Therefore, the hypotenuse is a diameter and AB). So, if an inscribed angle and a central angle has a length of 2r. Use trigonometry to find the intercept the same arc, then the measure of the length of the equal legs. inscribed angle is one-half the measure of the central angle.

51. In the circle below, and . What is ?

Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by . A 42 49. OPEN ENDED Find and sketch a real-world logo B 61 with an inscribed polygon. C 80 SOLUTION: D 84 See students’ work. SOLUTION: 50. WRITING IN MATH Compare and contrast If an angle is inscribed in a circle, then the measure inscribed angles and central angles of a circle. If they of the angle equals one half the measure of its intercept the same arc, how are they related? intercepted arc. Therefore,

SOLUTION: An inscribed angle has its vertex on the circle. A But, central angle has its vertex at the center of the circle.

If a central angle intercepts arc AB, then the Therefore,

measure of the central angle is equal to m(arc AB). If The correct choice is A. an inscribed angle also intercepts arc AB, then the 52. ALGEBRA Simplify measure of the inscribed angle is equal to m(arc 2 4(3x – 2)(2x + 4) + 3x + 5x – 6. AB). So, if an inscribed angle and a central angle F 2 intercept the same arc, then the measure of the 9x + 3x – 14 2 inscribed angle is one-half the measure of the central G 9x + 13x – 14 2 angle. H 27x + 37x – 38 J 27x2 + 27x – 26 51. In the circle below, and . What is ? SOLUTION: Use the Distributive Property to simplify the first term.

A 42

B 61 Therefore, the correct choice is H. C 80 53. SHORT RESPONSE In the circle below, is a D 84 diameter, AC = 8 inches, and BC = 15 inches. Find SOLUTION: the diameter, the radius, and the circumference of If an angle is inscribed in a circle, then the measure the circle. of the angle equals one half the measure of its intercepted arc. Therefore,

But, Therefore, SOLUTION: The correct choice is A. Use the Pythagorean Theorem to find the 52. ALGEBRA Simplify hypotenuse of the triangle which is also the diameter of the circle. 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 F 9x + 3x – 14 2 The radius is half the diameter. So, the radius of the G 9x + 13x – 14 circle is about 8.5 in. 2 H 27x + 37x – 38 The circumference C of a circle of a circle of J 27x2 + 27x – 26 diameter d is given by Therefore, the circumference of the circle is SOLUTION: Use the Distributive Property to simplify the first

term. 54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? A –15 B –16 C –17 D –18 Therefore, the correct choice is H. E –19 SOLUTION: 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find Let the three consecutive integers be x , x +1,and x + the diameter, the radius, and the circumference of 2. Then, the circle.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. SOLUTION: In , FL = 24, HJ = 48, and . Find Use the Pythagorean Theorem to find the each measure. hypotenuse of the triangle which is also the diameter of the circle.

The radius is half the diameter. So, the radius of the circle is about 8.5 in. The circumference C of a circle of a circle of diameter d is given by 55. FG Therefore, the circumference of the circle is SOLUTION: If a diameter (or radius) of a circle is perpendicular 54. SAT/ACT The sum of three consecutive integers is to a chord, then it bisects the chord and its arc. So, –48. What is the least of the three integers? bisects Therefore, A –15 B –16 C –17 D –18 56. E –19 SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular Let the three consecutive integers be x , x +1,and x + to a chord, then it bisects the chord and its arc. So, 2. Then, bisects Therefore,

57. NJ SOLUTION: So, the three integers are –17, –16, and –15 and the If a diameter (or radius) of a circle is perpendicular least of these is –17. to a chord, then it bisects the chord and its arc. So,

Therefore, the correct choice is C. bisects Therefore,

In , FL = 24, HJ = 48, and . Find each measure. 58. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular Find x. to a chord, then it bisects the chord and its arc. So, bisects Therefore,

56. 59. SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular The sum of the measures of the central angles of a to a chord, then it bisects the chord and its arc. So, circle with no interior points in common is 360. So,

bisects Therefore, 57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

60. 58. SOLUTION: The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

Find x.

61. SOLUTION: 59. The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How 60. tall is the person being photographed? SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will 61. be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x SOLUTION: be the length of the base of the larger triangle. The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Therefore, the person being photographed is 5.6 ft tall.

62. PHOTOGRAPHY In one of the first cameras ALGEBRA Suppose B is the midpoint of . invented, light entered an opening in the front. An Use the given information to find the missing image was reflected in the back of the camera, measure. upside down, forming similar triangles. Suppose the 63. AB = 4x – 5, BC = 11 + 2x, AC = ? image of the person on the back of the camera is 12 inches, the distance from the opening to the person is SOLUTION: 7 feet, and the camera itself is 15 inches long. How Since B is the midpoint of tall is the person being photographed?

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: The height of the person being photographed is the SOLUTION: length of the base of the larger triangle. The two Since B is the midpoint of triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft 65. BC = 6 – 4m, AC = 8, m = ? tall. SOLUTION: ALGEBRA Suppose B is the midpoint of . Since B is the midpoint of Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of 66. AB = 10s + 2, AC = 40, s = ? SOLUTION:

Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

Find each measure. If an angle is inscribed in a circle, then the measure 1. of the angle equals one half the measure of its intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If

SOLUTION: , what is ? If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

2. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

SOLUTION: ALGEBRA Find each measure. If an angle is inscribed in a circle, then the measure 5. of the angle equals one half the measure of its intercepted arc. Therefore,

3.

SOLUTION:

Here, is a semi-circle. So, Therefore,

If an angle is inscribed in a circle, then the measure 6. of the angle equals one half the measure of its intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If

, what is ? SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore, SOLUTION: If an angle is inscribed in a circle, then the measure 7. PROOF Write a two-column proof. of the angle equals one half the measure of its Given: bisects . Prove: intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

SOLUTION: Given: bisects . Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, Proof: Statements (Reasons) Therefore, 1. bisects . (Given) 2. (Def. of segment bisector) 6. 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. SOLUTION: 8. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore, SOLUTION: An inscribed angle of a triangle intercepts a diameter 7. PROOF Write a two-column proof. or semicircle if and only if the angle is a right angle. Given: bisects . So, Prove: The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

SOLUTION: 9. x Given: bisects . Prove:

Proof: SOLUTION: Statements (Reasons) An inscribed angle of a triangle intercepts a diameter 1. bisects . (Given) or semicircle if and only if the angle is a right angle. 2. (Def. of segment bisector) So, Since The sum of the measures of the angles of a triangle 3. intercepts . intercepts .

(Def. of intercepted arc) 4. (Inscribed of same arc are is 180. So, .) 5. (Vertical are .) 6. (AAS) 10. and

CCSS STRUCTURE Find each value. 8.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter Solve the two equations to find the values of x and y. or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle Use the values of the variables to find and . is 180. So,

Therefore, ALGEBRA Find each measure. 9. x 11.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure An inscribed angle of a triangle intercepts a diameter of the angle equals one half the measure of its or semicircle if and only if the angle is a right angle. So, Since intercepted arc. Therefore, The sum of the measures of the angles of a triangle 12. is 180. So,

10. and

intercepted arc. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 13.

Solve the two equations to find the values of x and y.

Use the values of the variables to find and . SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, ALGEBRA Find each measure. 11. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

14. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

12. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

15. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, SOLUTION: 13. Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So,

Therefore, SOLUTION: 16. The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

intercepted arc. Therefore, SOLUTION: Here, 14. The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

SOLUTION: 17. ALGEBRA Find each measure. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

15.

SOLUTION: If two inscribed angles of a circle intercept the same SOLUTION: arc or congruent arcs, then the angles are congruent. So, Here, The arc is a semicircle. So, 18. ALGEBRA Find each measure. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

16.

SOLUTION: Therefore, Here, The arc is a semicircle. So, 19. ALGEBRA Find each measure. Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

SOLUTION: If two inscribed angles of a circle intercept the same Therefore, m∠A = 5(4) or 20. arc or congruent arcs, then the angles are congruent. ALGEBRA Find each measure. So, 20.

18. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same SOLUTION: arc or congruent arcs, then the angles are congruent. If two inscribed angles of a circle intercept the same Since ∠C and ∠D both intercept arc AB, m∠C = arc or congruent arcs, then the angles are congruent. m D. ∠ So,

Therefore, Therefore, m∠C = 5(10) – 3 or 47. PROOF Write the specified type of proof. ALGEBRA Find each measure. 19. 21. paragraph proof

Given:

Prove:

SOLUTION: If two inscribed angles of a circle intercept the same SOLUTION: arc or congruent arcs, then the angles are congruent.

Since ∠A and ∠B both intercept arc CD , m∠A = Proof: Given means that m∠B. . Since and

Therefore, m∠A = 5(4) or 20. , the equation becomes

20. ALGEBRA Find each measure. . Multiplying each side of the

equation by 2 results in .

22. two-column proof Given: Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = SOLUTION: m D. ∠ Statements (Reasons):

1. (Given) 2. (Inscribed intercepting same arc Therefore, m∠C = 5(10) – 3 or 47. are .) 3. (Vertical are .) PROOF Write the specified type of proof. 4. (AA Similarity) 21. paragraph proof

Given: ALGEBRA Find each value.

Prove:

23. x

SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Proof: Given means that So, The sum of the measures of the angles of a triangle . Since and is 180. So, , the equation becomes

. Multiplying each side of the 24.

equation by 2 results in . SOLUTION: An inscribed angle of a triangle intercepts a diameter 22. two-column proof or semicircle if and only if the angle is a right angle. Given: So, Prove: The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

25. x SOLUTION: SOLUTION: Statements (Reasons): An inscribed angle of a triangle intercepts a diameter 1. (Given) or semicircle if and only if the angle is a right angle. 2. (Inscribed intercepting same arc So, are .) The sum of the measures of the angles of a triangle 3. (Vertical are .)

4. (AA Similarity) is 180. So, ALGEBRA Find each value. 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, 23. x The sum of the measures of the angles of a triangle

SOLUTION: is 180. So, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.

So, Therefore, The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each measure. is 180. So,

24. SOLUTION: 27. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. SOLUTION: So, If a quadrilateral is inscribed in a circle, then its The sum of the measures of the angles of a triangle opposite angles are supplementary.

is 180. So, Therefore,

Therefore, 28.

25. x SOLUTION: If a quadrilateral is inscribed in a circle, then its SOLUTION: opposite angles are supplementary. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle Therefore,

is 180. So, CCSS STRUCTURE Find each measure.

26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 29. SOLUTION: is 180. So, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

CCSS STRUCTURE Find each measure. Therefore,

30. SOLUTION: If a quadrilateral is inscribed in a circle, then its 27. opposite angles are supplementary.

SOLUTION:

If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore, Therefore, 31. PROOF Write a paragraph proof for Theorem 11.9. 28. SOLUTION:

SOLUTION: Given: Quadrilateral ABCD is inscribed in .

If a quadrilateral is inscribed in a circle, then its Prove: and are supplementary. opposite angles are supplementary. and are supplementary.

Therefore,

CCSS STRUCTURE Find each measure.

Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6 and , but 29. , so or SOLUTION: If a quadrilateral is inscribed in a circle, then its 180. This makes ∠C and ∠A supplementary. opposite angles are supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making them supplementary also. Therefore, SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each 30. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

measure. Therefore, 32. 31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are Given: Quadrilateral ABCD is inscribed in . congruent. There are eight adjacent arcs that make Prove: and are supplementary. up the circle, so their sum is 360. Thus, the measure and are supplementary. of each arc joining consecutive vertices is of 360 or 45.

Therefore, measure of arc NPQ is 135.

33.

SOLUTION: Proof: By arc addition and the definitions of arc measure and the sum of central angles, Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are . Since by Theorem 10.6 congruent. There are eight adjacent arcs that make and up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 , but or 45. Since ∠RLQ is inscribed in the circle, its , so or measure equals one half the measure of its intercepted arc QR. 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making them supplementary also. Therefore, the measure of ∠RLQ is 22.5. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each 34. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, measure. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 32. 360 is divided into 8 equal arcs and each arc

SOLUTION: measures Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are Then, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure Therefore, of each arc joining consecutive vertices is of 360 or 45. 35. SOLUTION:

Therefore, measure of arc NPQ is 135. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

33. intercepted arc. So, SOLUTION: The sum of the measures of the central angles of a Since all the sides of the stop sign are congruent circle with no interior points in common is 360. Here, chords of the circle, all the corresponding arcs are 360 is divided into 8 equal arcs and each arc

congruent. There are eight adjacent arcs that make measures up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 Then, or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its Therefore, intercepted arc QR. 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star shown are congruent, find the measure of each inscribed angle. a. Therefore, the measure of ∠RLQ is 22.5.

34. SOLUTION: If an angle is inscribed in a circle, then the measure b. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc c.

measures Then,

Therefore, d. 35. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, SOLUTION: The sum of the measures of the central angles of a a. The sum of the measures of the central angles of circle with no interior points in common is 360. Here, a circle with no interior points in common is 360. 360 is divided into 8 equal arcs and each arc Here, 360 is divided into 5 equal arcs and each arc

measures measures

Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed Therefore, angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star measures Each inscribed angle is formed shown are congruent, find the measure of each by intercepting alternate vertices of the star. So, each inscribed angle. intercepted arc measures 120. If an angle is inscribed a. in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a

b. circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc c. measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45. d. PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove: SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc

measures If an angle is inscribed in a circle, then the measure

of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed SOLUTION: angle is 36. Proof: b. Here, 360 is divided into 6 equal arcs and each arc Statements (Reasons) measures Each inscribed angle is formed 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) by intercepting alternate vertices of the star. So, each 2. m∠ABD = m(arc AD) intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one m∠DBC = m(arc DC) (The measure of an half the measure of its intercepted arc. So, the inscribed whose side is a diameter is half the measure of each inscribed angle is 60. measure of the intercepted arc (Case 1)). c. Here, 360 is divided into 7 equal arcs and each arc 3. m∠ABC = m(arc AD) + m(arc DC) measures If an angle is inscribed in a (Substitution) circle, then the measure of the angle equals one half 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc d. The sum of the measures of the central angles of Addition Postulate) a circle with no interior points in common is 360. 6. m∠ABC = m(arc AC) (Substitution) Here, 360 is divided into 8 equal arcs and each arc

measures Each inscribed angle is formed 38. Case 3 by intercepting alternate vertices of the star. So, each Given: P lies outside . is a diameter. intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one Prove: half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove: SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC)

m∠DBA = m(arc DA) (The measure of an SOLUTION: inscribed whose side is a diameter is half the Proof: measure of the intercepted arc (Case 1)). Statements (Reasons) 3. m∠ABC = m(arc DC) – m(arc DA) 1. m ABC = m ABD + m DBC ( Addition ∠ ∠ ∠ (Substitution) Postulate) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) 2. m∠ABD = m(arc AD)

m∠DBC = m(arc DC) (The measure of an 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc inscribed whose side is a diameter is half the Addition Postulate) measure of the intercepted arc (Case 1)). 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 3. m∠ABC = m(arc AD) + m(arc DC) 7. m∠ABC = m(arc AC) (Substitution) (Substitution)

4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) PROOF Write the specified proof for each 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc theorem. Addition Postulate) 39. Theorem 11.7, two-column proof

6. m∠ABC = m(arc AC) (Substitution) SOLUTION: Given: and are inscribed; 38. Case 3 Prove: Given: P lies outside . is a diameter. Proof:

Prove:

Statements (Reasons) SOLUTION: 1. and are inscribed;

Proof: (Given) Statements (Reasons) 2. ; (Measure 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) of an inscribed = half measure of intercepted arc.)

2. m∠DBC = m(arc DC) 3. (Def. of arcs)

m∠DBA = m(arc DA) (The measure of an 4. (Mult. Prop. of Equality) inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 5. (Substitution) 6. (Def. of ) 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 40. Theorem 11.8, paragraph proof 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) SOLUTION: 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 7. m∠ABC = m(arc AC) (Substitution)

PROOF Write the specified proof for each theorem. Part I: Given: is a semicircle. 39. Theorem 11.7, two-column proof Prove: is a right angle. SOLUTION: Proof: Since is a semicircle, then Given: and are inscribed; . Since is an inscribed angle, Prove: Proof: then or 90. So, by definition, is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Statements (Reasons) Property of Equality, m = 2m . Because 1. and are inscribed; is a right angle, m = 90. Then (Given) m = 2(90) or 180. So by definition, is a 2. ; (Measure semicircle. of an inscribed = half measure of intercepted arc.) 41. MULTIPLE REPRESENTATIONS In this 3. (Def. of arcs) problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel 4. (Mult. Prop. of Equality) chords. a. GEOMETRIC Use a compass to draw a circle 5. (Substitution) with parallel chords and . Connect points A 6. (Def. of ) and D by drawing segment . b. NUMERICAL Use a protractor to find 40. Theorem 11.8, paragraph proof and . Then determine and . SOLUTION: What is true about these arcs? Explain. c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find and in the figure. Verify by using inscribed angles to find the measures of the arcs.

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, SOLUTION: then or 90. So, by definition, a. is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

b. Sample answer: , ; Proof: Since is an inscribed angle, then , ; The arcs are congruent and by the Multiplication because they have equal measures. c. Sample answer: Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle.

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. m∠A = 15, m∠D = 15, = 30, and = 30. a. GEOMETRIC Use a compass to draw a circle The arcs are congruent. with parallel chords and . Connect points A In a circle, two parallel chords cut congruent arcs. d. In a circle, two parallel chords cut congruent arcs. and D by drawing segment .

b. NUMERICAL Use a protractor to find So, = . and . Then determine and . What is true about these arcs? Explain. c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that = 4(16) + 6 or 70

are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find = 6(16) – 26 or 70 11-4 Inscribed Angles Therefore, the measure of arcs QS and PR are each and in the figure. Verify by using 70. inscribed angles to find the measures of the arcs. CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square SOLUTION: Squares have right angles at each vertex, therefore SOLUTION: each pair of opposite angles will be supplementary a. and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, b. Sample answer: , ; therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, , ; The arcs are congruent the statement is always true. because they have equal measures. c. Sample answer: 44. parallelogram SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus m∠A = 15, m∠D = 15, = 30, and = 30. The arcs are congruent. SOLUTION: In a circle, two parallel chords cut congruent arcs. The opposite angles of a rhombus are always d. In a circle, two parallel chords cut congruent arcs. congruent. They will only be supplementary when So, = . they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true. = 4(16) + 6 or 70 46. kite = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each SOLUTION: 70. Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be CCSS ARGUMENTS Determine whether the a right angle. If one pair of opposite angles for a quadrilateral can always, sometimes, or never quadrilateral is supplementary, the other pair must be inscribed in a circle. Explain your reasoning. also be supplementary. So, as long as the angles that 42. square compose the pair of congruent opposite angles are right angles, a kite can be inscribed in a circle. SOLUTION: Therefore, the statement is sometimes true. Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary 47. CHALLENGE A square is inscribed in a circle. and inscribed in a circle. Therefore, the statement is What is the ratio of the area of the circle to the area always true. of the square? eSolutions Manual - Powered by Cognero SOLUTION: Page 11 43. rectangle A square with side s is inscribed in a circle with SOLUTION: radius r. Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram SOLUTION: Using the Pythagorean Theorem, The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true. 2 45. rhombus The area of a square of side s is A = s and the area of a circle of radius r is A = . SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true. Therefore, the ratio of the area of the circle to the 46. kite area of the inscribed square is SOLUTION: Exactly one pair of opposite angles of a kite are 48. WRITING IN MATH A right congruent. To be supplementary, they must each be triangle is inscribed in a circle. If the radius of the a right angle. If one pair of opposite angles for a circle is given, explain how to find the lengths of the quadrilateral is supplementary, the other pair must right triangle’s legs. also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are SOLUTION: right angles, a kite can be inscribed in a circle. Sample answer: A triangle will have Therefore, the statement is sometimes true. two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. 47. CHALLENGE A square is inscribed in a circle. According to theorem 10.8, an inscribed angle of a What is the ratio of the area of the circle to the area triangle intercepts a diameter if the angle is a right of the square? angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the SOLUTION: length of the equal legs. A square with side s is inscribed in a circle with radius r.

Therefore, the length of each leg of the 45-45-90

Using the Pythagorean Theorem, triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon.

2 The area of a square of side s is A = s and the area SOLUTION: See students’ work. of a circle of radius r is A = . 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the Therefore, the ratio of the area of the circle to the measure of the central angle is equal to m(arc AB). If area of the inscribed square is an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc 48. WRITING IN MATH A right AB). So, if an inscribed angle and a central angle triangle is inscribed in a circle. If the radius of the intercept the same arc, then the measure of the circle is given, explain how to find the lengths of the inscribed angle is one-half the measure of the central right triangle’s legs. angle. SOLUTION: Sample answer: A triangle will have 51. In the circle below, and . two inscribed angles of 45 and one of 90. The What is ? hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs.

A 42 B 61 C 80 D 84 SOLUTION:

Therefore, the length of each leg of the 45-45-90 If an angle is inscribed in a circle, then the measure triangle can be found by multiplying the radius of the of the angle equals one half the measure of its intercepted arc. Therefore, circle in which it is inscribed by .

OPEN ENDED 49. Find and sketch a real-world logo But, with an inscribed polygon. Therefore, SOLUTION: The correct choice is A. See students’ work. 52. ALGEBRA Simplify 50. WRITING IN MATH Compare and contrast 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 inscribed angles and central angles of a circle. If they F 9x + 3x – 14 intercept the same arc, how are they related? G 9x2 + 13x – 14 SOLUTION: 2 H 27x + 37x – 38 An inscribed angle has its vertex on the circle. A J 2 central angle has its vertex at the center of the circle. 27x + 27x – 26 If a central angle intercepts arc AB, then the SOLUTION: measure of the central angle is equal to m(arc AB). If Use the Distributive Property to simplify the first an inscribed angle also intercepts arc AB, then the term. measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle. Therefore, the correct choice is H. 51. In the circle below, and . What is ? 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle.

A 42 B 61 C 80 SOLUTION: D 84 Use the Pythagorean Theorem to find the SOLUTION: hypotenuse of the triangle which is also the diameter If an angle is inscribed in a circle, then the measure of the circle. of the angle equals one half the measure of its intercepted arc. Therefore, The radius is half the diameter. So, the radius of the circle is about 8.5 in. The circumference C of a circle of a circle of But, diameter d is given by Therefore, Therefore, the circumference of the circle is The correct choice is A.

52. ALGEBRA Simplify 54. SAT/ACT The sum of three consecutive integers is 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. –48. What is the least of the three integers? 2 F 9x + 3x – 14 A –15 2 B –16 G 9x + 13x – 14 C –17 H 2 27x + 37x – 38 D –18 2 J 27x + 27x – 26 E –19 SOLUTION: SOLUTION: Use the Distributive Property to simplify the first Let the three consecutive integers be x , x +1,and x + term. 2. Then,

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is H. Therefore, the correct choice is C.

53. SHORT RESPONSE In the circle below, is a In , FL = 24, HJ = 48, and . Find diameter, AC = 8 inches, and BC = 15 inches. Find each measure. the diameter, the radius, and the circumference of the circle.

55. FG SOLUTION: SOLUTION: Use the Pythagorean Theorem to find the If a diameter (or radius) of a circle is perpendicular hypotenuse of the triangle which is also the diameter to a chord, then it bisects the chord and its arc. So, of the circle. bisects Therefore,

The radius is half the diameter. So, the radius of the circle is about 8.5 in. The circumference C of a circle of a circle of 56. diameter d is given by Therefore, the circumference of the circle is SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 54. SAT/ACT The sum of three consecutive integers is bisects Therefore, –48. What is the least of the three integers? A –15 57. NJ B –16 SOLUTION: C –17 If a diameter (or radius) of a circle is perpendicular D –18 to a chord, then it bisects the chord and its arc. So, E –19 bisects Therefore, SOLUTION: Let the three consecutive integers be x , x +1,and x + 2. Then, 58. SOLUTION: If a diameter (or radius) of a circle is perpendicular So, the three integers are –17, –16, and –15 and the to a chord, then it bisects the chord and its arc. So, least of these is –17. Therefore, the correct choice is C. bisects Therefore,

In , FL = 24, HJ = 48, and . Find each measure. Find x.

59.

55. FG SOLUTION: SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, If a diameter (or radius) of a circle is perpendicular

to a chord, then it bisects the chord and its arc. So,

bisects Therefore,

56. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 60. bisects Therefore, SOLUTION: 57. NJ The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION:

If a diameter (or radius) of a circle is perpendicular

to a chord, then it bisects the chord and its arc. So, bisects Therefore,

58. SOLUTION: 61. If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, SOLUTION: bisects Therefore, The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Find x.

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera,

59. upside down, forming similar triangles. Suppose the SOLUTION: image of the person on the back of the camera is 12 The sum of the measures of the central angles of a inches, the distance from the opening to the person is circle with no interior points in common is 360. So, 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

60. SOLUTION: The height of the person being photographed is the SOLUTION: length of the base of the larger triangle. The two The sum of the measures of the central angles of a triangles are similar. So, their corresponding sides will circle with no interior points in common is 360. So, be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft tall.

61. ALGEBRA Suppose B is the midpoint of . SOLUTION: Use the given information to find the missing measure. The sum of the measures of the central angles of a 63. AB = 4x – 5, BC = 11 + 2x, AC = ? circle with no interior points in common is 360. So, SOLUTION: Since B is the midpoint of

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How 64. AB = 6y – 14, BC = 10 – 2y, AC = ? tall is the person being photographed? SOLUTION: Since B is the midpoint of

SOLUTION: The height of the person being photographed is the 65. BC = 6 – 4m, AC = 8, m = ? length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will SOLUTION: be proportional. One foot is equivalent to 12 in. Then, Since B is the midpoint of the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft 66. AB = 10s + 2, AC = 40, s = ? tall. SOLUTION: ALGEBRA Suppose B is the midpoint of . Since B is the midpoint of Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

Find each measure. 1.

SOLUTION: Here, is a semi-circle. So,

SOLUTION: If an angle is inscribed in a circle, then the measure If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its of the angle equals one half the measure of its intercepted arc. So,

intercepted arc. Therefore, Therefore,

4. SCIENCE The diagram shows how light bends in a 2. raindrop to make the colors of the rainbow. If , what is ?

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

3. intercepted arc. Therefore,

ALGEBRA Find each measure. 5.

SOLUTION:

Here, is a semi-circle. So,

SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent.

intercepted arc. So, So, Therefore,

4. SCIENCE The diagram shows how light bends in a Therefore, raindrop to make the colors of the rainbow. If

, what is ? 6.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure If two inscribed angles of a circle intercept the same of the angle equals one half the measure of its arc or congruent arcs, then the angles are congruent. So, intercepted arc. Therefore,

ALGEBRA Find each measure. 5. Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, SOLUTION: Given: bisects . Therefore, Prove:

6.

Proof: Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) SOLUTION: 3. intercepts . intercepts . If two inscribed angles of a circle intercept the same (Def. of intercepted arc) arc or congruent arcs, then the angles are congruent. 4. (Inscribed of same arc are So, .) 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. Therefore, 8. 7. PROOF Write a two-column proof. Given: bisects . Prove:

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. SOLUTION: So, Given: bisects . The sum of the measures of the angles of a triangle Prove: is 180. So,

Therefore,

Proof: 9. x Statements (Reasons) 1. bisects . (Given) 2. (Def. of segment bisector) 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) SOLUTION: 6. (AAS) An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. CCSS STRUCTURE Find each value. So, Since 8. The sum of the measures of the angles of a triangle

is 180. So,

10. and

SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle SOLUTION: is 180. So, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore, Solve the two equations to find the values of x and y.

9. x Use the values of the variables to find and .

ALGEBRA Find each measure. SOLUTION: 11. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So, SOLUTION: If an angle is inscribed in a circle, then the measure 10. and of the angle equals one half the measure of its intercepted arc. Therefore,

12.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

Use the values of the variables to find and intercepted arc. Therefore, . 13.

ALGEBRA Find each measure. 11.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure intercepted arc. Therefore, of the angle equals one half the measure of its

intercepted arc. Therefore, 12. 14.

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore, 13. 15.

SOLUTION: SOLUTION: The sum of the measures of the central angles of a Here, circle with no interior points in common is 360. So, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Therefore,

intercepted arc. Therefore, 16.

14.

SOLUTION: SOLUTION: Here, If an angle is inscribed in a circle, then the measure The arc is a semicircle. So, of the angle equals one half the measure of its Then, intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 15. intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, SOLUTION: If two inscribed angles of a circle intercept the same Therefore, arc or congruent arcs, then the angles are congruent. 16. So, 18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, SOLUTION: Then, If two inscribed angles of a circle intercept the same If an angle is inscribed in a circle, then the measure arc or congruent arcs, then the angles are congruent. of the angle equals one half the measure of its So,

intercepted arc. Therefore, Therefore, 17. ALGEBRA Find each measure. 19. ALGEBRA Find each measure.

SOLUTION:

If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: So, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 18. ALGEBRA Find each measure. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

Therefore, SOLUTION: If two inscribed angles of a circle intercept the same ALGEBRA Find each measure. 19. arc or congruent arcs, then the angles are congruent.

Since C and D both intercept arc AB, m C = ∠ ∠ ∠ m∠D.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

SOLUTION: Given: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Prove: Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure. SOLUTION:

Proof: Given means that

. Since and

, the equation becomes

. Multiplying each side of the

SOLUTION: equation by 2 results in . If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 22. two-column proof Since ∠C and ∠D both intercept arc AB, m∠C = Given: m∠D. Prove:

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given: SOLUTION: Statements (Reasons): Prove: 1. (Given) 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) 4. (AA Similarity)

ALGEBRA Find each value. SOLUTION:

Proof: Given means that

. Since and 23. x , the equation becomes SOLUTION: . Multiplying each side of the An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. equation by 2 results in . So, The sum of the measures of the angles of a triangle 22. two-column proof Given: is 180. So, Prove:

24. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. SOLUTION: So, Statements (Reasons): The sum of the measures of the angles of a triangle 1. (Given) 2. (Inscribed intercepting same arc is 180. So, are .) 3. (Vertical are .) Therefore, 4. (AA Similarity) 25. x ALGEBRA Find each value. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

23. x is 180. So,

SOLUTION: An inscribed angle of a triangle intercepts a diameter 26. or semicircle if and only if the angle is a right angle. SOLUTION:

So, An inscribed angle of a triangle intercepts a diameter The sum of the measures of the angles of a triangle or semicircle if and only if the angle is a right angle.

So, is 180. So, The sum of the measures of the angles of a triangle

is 180. So, 24. SOLUTION: Therefore, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. CCSS STRUCTURE Find each measure. So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore, 27. 25. x SOLUTION: If a quadrilateral is inscribed in a circle, then its SOLUTION: opposite angles are supplementary. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Therefore, So, The sum of the measures of the angles of a triangle 28. is 180. So, SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. Therefore, So, The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each measure.

is 180. So,

Therefore,

CCSS STRUCTURE Find each measure.

29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

27. SOLUTION: Therefore, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. 30. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 28. opposite angles are supplementary. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore,

31. PROOF Write a paragraph proof for Theorem 11.9.

Therefore, SOLUTION: CCSS STRUCTURE Find each measure. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary.

29. SOLUTION: Proof: By arc addition and the definitions of arc If a quadrilateral is inscribed in a circle, then its measure and the sum of central angles, opposite angles are supplementary. . Since by Theorem 10.6 and

Therefore, , but , so or 30. 180. This makes ∠C and ∠A supplementary. SOLUTION: Because the sum of the measures of the interior angles of a quadrilateral is 360, If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. . But , so , making them supplementary also.

SIGNS: A stop sign in the shape of a regular Therefore, octagon is inscribed in a circle. Find each

31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary. measure. 32. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure Proof: By arc addition and the definitions of arc of each arc joining consecutive vertices is of 360

measure and the sum of central angles, or 45. . Since by Theorem 10.6

and Therefore, measure of arc NPQ is 135. , but 33. , so or SOLUTION: 180. This makes ∠C and ∠A supplementary. Since all the sides of the stop sign are congruent Because the sum of the measures of the interior chords of the circle, all the corresponding arcs are angles of a quadrilateral is 360, congruent. There are eight adjacent arcs that make . But up the circle, so their sum is 360. Thus, the measure , so , making of each arc joining consecutive vertices is of 360 them supplementary also. or 45. Since ∠RLQ is inscribed in the circle, its SIGNS: A stop sign in the shape of a regular measure equals one half the measure of its octagon is inscribed in a circle. Find each intercepted arc QR.

Therefore, the measure of ∠RLQ is 22.5. measure. 32. 34. SOLUTION: SOLUTION: Since all the sides of the stop sign are congruent If an angle is inscribed in a circle, then the measure chords of the circle, the corresponding arcs are of the angle equals one half the measure of its congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure intercepted arc. So, of each arc joining consecutive vertices is of 360 The sum of the measures of the central angles of a or 45. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

Therefore, measure of arc NPQ is 135. measures Then, 33. Therefore, SOLUTION: Since all the sides of the stop sign are congruent 35. chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make SOLUTION: up the circle, so their sum is 360. Thus, the measure If an angle is inscribed in a circle, then the measure of each arc joining consecutive vertices is of 360 of the angle equals one half the measure of its

or 45. Since ∠RLQ is inscribed in the circle, its intercepted arc. So, measure equals one half the measure of its intercepted arc QR. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

Then, Therefore, the measure of ∠RLQ is 22.5.

Therefore, 34. 36. ART Four different string art star patterns are SOLUTION: shown. If all of the inscribed angles of each star If an angle is inscribed in a circle, then the measure shown are congruent, find the measure of each of the angle equals one half the measure of its inscribed angle. a. intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures b. Then,

Therefore,

35. c. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So,

The sum of the measures of the central angles of a d. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

Then,

Therefore, SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. ART 36. Four different string art star patterns are Here, 360 is divided into 5 equal arcs and each arc shown. If all of the inscribed angles of each star shown are congruent, find the measure of each measures inscribed angle. a. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc

measures Each inscribed angle is formed b. by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc c. measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of d. a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one SOLUTION: half the measure of its intercepted arc. So, the a. The sum of the measures of the central angles of measure of each inscribed angle is 45. a circle with no interior points in common is 360. PROOF Write a two-column proof for each case Here, 360 is divided into 5 equal arcs and each arc of Theorem 11.6.

measures 37. Case 2 Given: P lies inside . is a diameter. If an angle is inscribed in a circle, then the measure

of the angle equals one half the measure of its Prove: intercepted arc. So, the measure of each inscribed angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. SOLUTION: c. Here, 360 is divided into 7 equal arcs and each arc Proof: Statements (Reasons) measures If an angle is inscribed in a 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure 2. m∠ABD = m(arc AD) of each inscribed angle is about 25.7. m∠DBC = m(arc DC) (The measure of an d. The sum of the measures of the central angles of a circle with no interior points in common is 360. inscribed whose side is a diameter is half the Here, 360 is divided into 8 equal arcs and each arc measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc AD) + m(arc DC) measures Each inscribed angle is formed (Substitution) by intercepting alternate vertices of the star. So, each 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) intercepted arc measures 90. If an angle is inscribed

in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc

measure of each inscribed angle is 45. Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution) PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 38. Case 3

Given: P lies inside . is a diameter. Given: P lies outside . is a diameter.

Prove: Prove:

SOLUTION: SOLUTION: Proof: Proof: Statements (Reasons) Statements (Reasons) 1. m∠ABC = m∠ABD + m∠DBC ( Addition 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate) Postulate, Subtraction Property of Equality) 2. m∠ABD = m(arc AD) 2. m∠DBC = m(arc DC) m∠DBC = m(arc DC) (The measure of an m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc AD) + m(arc DC) 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) (Substitution) 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc AD) + m(arc DC) = m(arc AC) (Arc 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality)

38. Case 3 7. m∠ABC = m(arc AC) (Substitution) Given: P lies outside . is a diameter.

Prove: PROOF Write the specified proof for each theorem. 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof:

SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC)

m∠DBA = m(arc DA) (The measure of an Statements (Reasons) inscribed whose side is a diameter is half the 1. and are inscribed; measure of the intercepted arc (Case 1)). (Given) 3. m∠ABC = m(arc DC) – m(arc DA) 2. ; (Measure (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) of an inscribed = half measure of intercepted arc.) 3. (Def. of arcs) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 4. (Mult. Prop. of Equality) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction 5. (Substitution) Property of Equality) 6. (Def. of ) 7. m∠ABC = m(arc AC) (Substitution) 40. Theorem 11.8, paragraph proof PROOF Write the specified proof for each SOLUTION: theorem. 39. Theorem 11.7, two-column proof SOLUTION: Given: and are inscribed; Prove: Proof:

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition,

Statements (Reasons) is a right angle. 1. and are inscribed; Part II: Given: is a right angle. (Given) Prove: is a semicircle. 2. ; (Measure of an inscribed = half measure of intercepted arc.) Proof: Since is an inscribed angle, then 3. (Def. of arcs) and by the Multiplication

4. (Mult. Prop. of Equality) Property of Equality, m = 2m . Because 5. (Substitution) is a right angle, m = 90. Then 6. (Def. of ) m = 2(90) or 180. So by definition, is a semicircle. 40. Theorem 11.8, paragraph proof 41. MULTIPLE REPRESENTATIONS In this SOLUTION: problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain. Part I: Given: is a semicircle. c. VERBAL Draw another circle and repeat parts a Prove: is a right angle. and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. Proof: Since is a semicircle, then d. ANALYTICAL Use your conjecture to find . Since is an inscribed angle, and in the figure. Verify by using then or 90. So, by definition, inscribed angles to find the measures of the arcs. is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then SOLUTION: a. and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a semicircle. b. Sample answer: , ; 41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between , ; The arcs are congruent the arcs of a circle that are cut by two parallel because they have equal measures. chords. c. Sample answer: a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain.

c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that m∠A = 15, m∠D = 15, = 30, and = 30.

are cut by two parallel chords. The arcs are congruent. In a circle, two parallel chords cut congruent arcs. d. ANALYTICAL Use your conjecture to find d. In a circle, two parallel chords cut congruent arcs. and in the figure. Verify by using So, = . inscribed angles to find the measures of the arcs.

= 4(16) + 6 or 70 = 6(16) – 26 or 70 SOLUTION: Therefore, the measure of arcs QS and PR are each 70. a. CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square SOLUTION:

b. Sample answer: , ; Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary , ; The arcs are congruent and inscribed in a circle. Therefore, the statement is because they have equal measures. always true. c. Sample answer: 43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

m∠A = 15, m∠D = 15, = 30, and = 30. 44. parallelogram The arcs are congruent. SOLUTION: In a circle, two parallel chords cut congruent arcs. The opposite angles of a parallelogram are always d. In a circle, two parallel chords cut congruent arcs. congruent. They will only be supplementary when

So, = . they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus = 4(16) + 6 or 70 SOLUTION:

= 6(16) – 26 or 70 The opposite angles of a rhombus are always Therefore, the measure of arcs QS and PR are each congruent. They will only be supplementary when 70. they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite CCSS ARGUMENTS Determine whether the angles of rhombi that are not squares are not quadrilateral can always, sometimes, or never supplementary, they can not be inscribed in a circle. be inscribed in a circle. Explain your reasoning. Therefore, the statement is sometimes true. 42. square 46. kite SOLUTION: Squares have right angles at each vertex, therefore SOLUTION: each pair of opposite angles will be supplementary Exactly one pair of opposite angles of a kite are and inscribed in a circle. Therefore, the statement is congruent. To be supplementary, they must each be always true. a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must 43. rectangle also be supplementary. So, as long as the angles that compose the pair of congruent opposite angles are SOLUTION: right angles, a kite can be inscribed in a circle. Rectangles have right angles at each vertex, Therefore, the statement is sometimes true. therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, 47. CHALLENGE A square is inscribed in a circle. the statement is always true. What is the ratio of the area of the circle to the area of the square? 44. parallelogram SOLUTION: SOLUTION: A square with side s is inscribed in a circle with The opposite angles of a parallelogram are always radius r. congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always Using the Pythagorean Theorem, congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not

supplementary, they can not be inscribed in a circle. 2 Therefore, the statement is sometimes true. The area of a square of side s is A = s and the area of a circle of radius r is A = . 46. kite SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must also be supplementary. So, as long as the angles that

compose the pair of congruent opposite angles are Therefore, the ratio of the area of the circle to the 11-4right Inscribed angles, Anglesa kite can be inscribed in a circle. Therefore, the statement is sometimes true. area of the inscribed square is

47. CHALLENGE A square is inscribed in a circle. 48. WRITING IN MATH A right What is the ratio of the area of the circle to the area triangle is inscribed in a circle. If the radius of the of the square? circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: A square with side s is inscribed in a circle with SOLUTION: radius r. Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs. Using the Pythagorean Theorem,

2 The area of a square of side s is A = s and the area of a circle of radius r is A = . Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work. Therefore, the ratio of the area of the circle to the area of the inscribed square is 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they 48. WRITING IN MATH A right intercept the same arc, how are they related? triangle is inscribed in a circle. If the radius of the SOLUTION: circle is given, explain how to find the lengths of the right triangle’s legs. An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. SOLUTION: If a central angle intercepts arc AB, then the Sample answer: A triangle will have measure of the central angle is equal to m(arc AB). If two inscribed angles of 45 and one of 90. The an inscribed angle also intercepts arc AB, then the hypotenuse is across from the 90, or right angle. measure of the inscribed angle is equal to m(arc According to theorem 10.8, an inscribed angle of a AB). So, if an inscribed angle and a central angle triangle intercepts a diameter if the angle is a right intercept the same arc, then the measure of the angle. Therefore, the hypotenuse is a diameter and inscribed angle is one-half the measure of the central has a length of 2r. Use trigonometry to find the angle. length of the equal legs. 51. In the circle below, and . What is ?

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Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by . A 42 B 61 49. OPEN ENDED Find and sketch a real-world logo C 80 with an inscribed polygon. D 84 SOLUTION: SOLUTION: See students’ work. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its 50. WRITING IN MATH Compare and contrast intercepted arc. Therefore, inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: But, An inscribed angle has its vertex on the circle. A Therefore, central angle has its vertex at the center of the circle. The correct choice is A. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If 52. ALGEBRA Simplify an inscribed angle also intercepts arc AB, then the 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 measure of the inscribed angle is equal to m(arc F 9x + 3x – 14 AB). So, if an inscribed angle and a central angle G 9x2 + 13x – 14 intercept the same arc, then the measure of the H 2 inscribed angle is one-half the measure of the central 27x + 37x – 38 2 angle. J 27x + 27x – 26 SOLUTION: 51. In the circle below, and . Use the Distributive Property to simplify the first What is ? term.

Therefore, the correct choice is H. A 42 B 61 53. SHORT RESPONSE In the circle below, is a C 80 diameter, AC = 8 inches, and BC = 15 inches. Find D 84 the diameter, the radius, and the circumference of the circle. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

But, SOLUTION: Therefore, Use the Pythagorean Theorem to find the The correct choice is A. hypotenuse of the triangle which is also the diameter of the circle. 52. ALGEBRA Simplify 2 4(3x – 2)(2x + 4) + 3x + 5x – 6. The radius is half the diameter. So, the radius of the 2 F 9x + 3x – 14 circle is about 8.5 in. G 9x2 + 13x – 14 The circumference C of a circle of a circle of 2 diameter d is given by H 27x + 37x – 38 2 Therefore, the circumference of the circle is J 27x + 27x – 26 SOLUTION: Use the Distributive Property to simplify the first 54. SAT/ACT The sum of three consecutive integers is term. –48. What is the least of the three integers? A –15 B –16 C –17 D –18 E –19 Therefore, the correct choice is H. SOLUTION: Let the three consecutive integers be x , x +1,and x + 53. SHORT RESPONSE In the circle below, is a 2. Then, diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find SOLUTION: each measure. Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter of the circle.

The radius is half the diameter. So, the radius of the circle is about 8.5 in. The circumference C of a circle of a circle of 55. FG diameter d is given by SOLUTION: Therefore, the circumference of the circle is If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 54. SAT/ACT The sum of three consecutive integers is bisects Therefore, –48. What is the least of the three integers? A –15 B –16 56. C –17 D –18 SOLUTION: E –19 If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, SOLUTION: Let the three consecutive integers be x , x +1,and x + bisects Therefore, 2. Then, 57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, So, the three integers are –17, –16, and –15 and the least of these is –17. bisects Therefore, Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure. 58. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

55. FG Find x. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

59.

56. SOLUTION: The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 60. SOLUTION: The sum of the measures of the central angles of a 58. circle with no interior points in common is 360. So, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

Find x.

61. SOLUTION: The sum of the measures of the central angles of a 59. circle with no interior points in common is 360. So,

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

60. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x 61. be the length of the base of the larger triangle. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . 62. PHOTOGRAPHY In one of the first cameras Use the given information to find the missing invented, light entered an opening in the front. An measure. image was reflected in the back of the camera, 63. AB = 4x – 5, BC = 11 + 2x, AC = ? upside down, forming similar triangles. Suppose the SOLUTION: image of the person on the back of the camera is 12 inches, the distance from the opening to the person is Since B is the midpoint of 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

64. AB = 6y – 14, BC = 10 – 2y, AC = ?

SOLUTION: SOLUTION: Since B is the midpoint of The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

65. BC = 6 – 4m, AC = 8, m = ? Therefore, the person being photographed is 5.6 ft SOLUTION: tall. Since B is the midpoint of ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION:

Since B is the midpoint of 66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Find each measure. intercepted arc. So, 1. Therefore,

intercepted arc. Therefore,

2. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 5. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

3. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore, SOLUTION:

Here, is a semi-circle. So, 6.

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If SOLUTION: , what is ? If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore,

7. PROOF Write a two-column proof. SOLUTION: Given: bisects . If an angle is inscribed in a circle, then the measure Prove: of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 5. SOLUTION: Given: bisects . Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Proof: So, Statements (Reasons)

1. bisects . (Given) 2. (Def. of segment bisector) Therefore, 3. intercepts . intercepts . (Def. of intercepted arc) 6. 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8.

SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. 7. PROOF Write a two-column proof. So, The sum of the measures of the angles of a triangle Given: bisects .

Prove: is 180. So,

Therefore,

9. x SOLUTION: Given: bisects . Prove:

SOLUTION: An inscribed angle of a triangle intercepts a diameter Proof: or semicircle if and only if the angle is a right angle. Statements (Reasons)

So, Since 1. bisects . (Given) The sum of the measures of the angles of a triangle 2. (Def. of segment bisector) 3. intercepts . intercepts . is 180. So, (Def. of intercepted arc) 4. (Inscribed of same arc are .) 10. and 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

SOLUTION: Solve the two equations to find the values of x and y.

An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Use the values of the variables to find and The sum of the measures of the angles of a triangle .

is 180. So,

Therefore, ALGEBRA Find each measure. 11. 9. x

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its An inscribed angle of a triangle intercepts a diameter intercepted arc. Therefore, or semicircle if and only if the angle is a right angle. So, Since 12. The sum of the measures of the angles of a triangle

is 180. So,

10. and

intercepted arc. Therefore,

SOLUTION: 13. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y.

Use the values of the variables to find and . SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

ALGEBRA Find each measure. 11. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

14.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, SOLUTION: 12. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

15.

intercepted arc. Therefore, SOLUTION: Here,

13. The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

16. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. Therefore, Here, The arc is a semicircle. So,

14. Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure. SOLUTION:

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

15.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, SOLUTION: 18. ALGEBRA Find each measure. Here,

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

16. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore, SOLUTION: Here, 19. ALGEBRA Find each measure.

The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20. SOLUTION: If two inscribed angles of a circle intercept the same 20. ALGEBRA Find each measure. arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: Since ∠C and ∠D both intercept arc AB, m∠C = If two inscribed angles of a circle intercept the same m∠D. arc or congruent arcs, then the angles are congruent. So,

Therefore, m∠C = 5(10) – 3 or 47.

Therefore, PROOF Write the specified type of proof. 21. paragraph proof

19. ALGEBRA Find each measure. Given:

Prove:

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same Proof: Given means that arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = . Since and m∠B. , the equation becomes

Therefore, m∠A = 5(4) or 20. . Multiplying each side of the 20. ALGEBRA Find each measure. equation by 2 results in .

22. two-column proof Given: Prove:

SOLUTION:

If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: Since ∠C and ∠D both intercept arc AB, m∠C = Statements (Reasons): m∠D. 1. (Given) 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) Therefore, m∠C = 5(10) – 3 or 47. 4. (AA Similarity) PROOF Write the specified type of proof. ALGEBRA Find each value. 21. paragraph proof

Given:

Prove:

23. x

SOLUTION: An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. So, Proof: Given means that The sum of the measures of the angles of a triangle

. Since and is 180. So,

, the equation becomes 24. . Multiplying each side of the SOLUTION: equation by 2 results in . An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, 22. two-column proof The sum of the measures of the angles of a triangle Given: Prove: is 180. So,

Therefore,

25. x SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter Statements (Reasons): or semicircle if and only if the angle is a right angle.

1. (Given) So, The sum of the measures of the angles of a triangle 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) is 180. So, 4. (AA Similarity)

ALGEBRA Find each value. 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 23. x is 180. So, SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, CCSS STRUCTURE Find each measure. The sum of the measures of the angles of a triangle

is 180. So,

24. 27. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter If a quadrilateral is inscribed in a circle, then its or semicircle if and only if the angle is a right angle. opposite angles are supplementary. So, The sum of the measures of the angles of a triangle Therefore,

is 180. So, 28. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 25. x opposite angles are supplementary.

SOLUTION:

An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Therefore, The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each measure. is 180. So,

26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 29. So, SOLUTION: The sum of the measures of the angles of a triangle If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. is 180. So,

Therefore, Therefore, CCSS STRUCTURE Find each measure.

30. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

27. SOLUTION: If a quadrilateral is inscribed in a circle, then its Therefore, opposite angles are supplementary. 31. PROOF Write a paragraph proof for Theorem 11.9. Therefore, SOLUTION: Given: Quadrilateral ABCD is inscribed in . 28. Prove: and are supplementary. SOLUTION: and are supplementary. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

CCSS STRUCTURE Find each measure. Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6 and , but

, so or 29. 180. This makes ∠C and ∠A supplementary. SOLUTION: Because the sum of the measures of the interior If a quadrilateral is inscribed in a circle, then its angles of a quadrilateral is 360, opposite angles are supplementary. . But , so , making them supplementary also.

SIGNS: A stop sign in the shape of a regular Therefore, octagon is inscribed in a circle. Find each

30. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

measure. 32.

Therefore, SOLUTION: Since all the sides of the stop sign are congruent PROOF 31. Write a paragraph proof for Theorem 11.9. chords of the circle, the corresponding arcs are SOLUTION: congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. of each arc joining consecutive vertices is of 360 and are supplementary. or 45.

Therefore, measure of arc NPQ is 135.

33. SOLUTION: Since all the sides of the stop sign are congruent Proof: By arc addition and the definitions of arc chords of the circle, all the corresponding arcs are measure and the sum of central angles, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure . Since by Theorem 10.6 of each arc joining consecutive vertices is of 360 and or 45. Since ∠RLQ is inscribed in the circle, its , but measure equals one half the measure of its intercepted arc QR. , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making Therefore, the measure of ∠RLQ is 22.5. them supplementary also.

SIGNS: A stop sign in the shape of a regular 34. octagon is inscribed in a circle. Find each SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, measure. 360 is divided into 8 equal arcs and each arc

32. measures SOLUTION: Then, Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are Therefore, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 35. of each arc joining consecutive vertices is of 360 or 45. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Therefore, measure of arc NPQ is 135. intercepted arc. So, 33. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, SOLUTION: 360 is divided into 8 equal arcs and each arc Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are measures congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure Then, of each arc joining consecutive vertices is of 360 Therefore, or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR. 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star shown are congruent, find the measure of each inscribed angle. a.

Therefore, the measure of ∠RLQ is 22.5.

34. b. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So,

The sum of the measures of the central angles of a c. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures

Then, d. Therefore,

35. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. So, a. The sum of the measures of the central angles of a circle with no interior points in common is 360. The sum of the measures of the central angles of a Here, 360 is divided into 5 equal arcs and each arc circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures

measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Then, intercepted arc. So, the measure of each inscribed angle is 36.

Therefore, b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed 36. ART Four different string art star patterns are by intercepting alternate vertices of the star. So, each shown. If all of the inscribed angles of each star intercepted arc measures 120. If an angle is inscribed shown are congruent, find the measure of each in a circle, then the measure of the angle equals one inscribed angle. half the measure of its intercepted arc. So, the a. measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure b. of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed

c. by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case d. of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove:

SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc

measures If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed Proof:

angle is 36. Statements (Reasons) b. Here, 360 is divided into 6 equal arcs and each arc 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) measures Each inscribed angle is formed 2. m∠ABD = m(arc AD) by intercepting alternate vertices of the star. So, each m∠DBC = m(arc DC) (The measure of an intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one inscribed whose side is a diameter is half the

half the measure of its intercepted arc. So, the measure of the intercepted arc (Case 1)). measure of each inscribed angle is 60. 3. m∠ABC = m(arc AD) + m(arc DC) c. Here, 360 is divided into 7 equal arcs and each arc (Substitution) measures If an angle is inscribed in a 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)

circle, then the measure of the angle equals one half 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc the measure of its intercepted arc. So, the measure Addition Postulate) of each inscribed angle is about 25.7. 6. m ABC = m(arc AC) (Substitution) d. The sum of the measures of the central angles of ∠ a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc 38. Case 3 measures Each inscribed angle is formed Given: P lies outside . is a diameter.

by intercepting alternate vertices of the star. So, each Prove: intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove: SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). SOLUTION: 3. m ABC = m(arc DC) – m(arc DA) Proof: ∠ Statements (Reasons) (Substitution) 1. m∠ABC = m∠ABD + m∠DBC ( Addition 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) Postulate) 2. m∠ABD = m(arc AD) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) m∠DBC = m(arc DC) (The measure of an 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction inscribed whose side is a diameter is half the Property of Equality) measure of the intercepted arc (Case 1)). 7. m∠ABC = m(arc AC) (Substitution) 3. m∠ABC = m(arc AD) + m(arc DC) (Substitution) 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) PROOF Write the specified proof for each theorem. 39. Theorem 11.7, two-column proof 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Addition Postulate) SOLUTION:

6. m∠ABC = m(arc AC) (Substitution) Given: and are inscribed; Prove: 38. Case 3 Proof: Given: P lies outside . is a diameter.

Prove:

Statements (Reasons) 1. and are inscribed;

(Given) SOLUTION: Proof: 2. ; (Measure Statements (Reasons) of an inscribed = half measure of intercepted arc.) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 3. (Def. of arcs)

2. m∠DBC = m(arc DC) 4. (Mult. Prop. of Equality) m∠DBA = m(arc DA) (The measure of an 5. (Substitution) inscribed whose side is a diameter is half the 6. (Def. of ) measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) 40. Theorem 11.8, paragraph proof

(Substitution) SOLUTION: 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 7. m∠ABC = m(arc AC) (Substitution)

PROOF Write the specified proof for each Part I: Given: is a semicircle. theorem. Prove: is a right angle. 39. Theorem 11.7, two-column proof Proof: Since is a semicircle, then SOLUTION: . Since is an inscribed angle, Given: and are inscribed; then or 90. So, by definition, Prove: Proof: is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because Statements (Reasons) is a right angle, m = 90. Then 1. and are inscribed; m = 2(90) or 180. So by definition, is a (Given) semicircle. 2. ; (Measure 41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between of an inscribed = half measure of intercepted arc.) the arcs of a circle that are cut by two parallel 3. (Def. of arcs) chords. a. GEOMETRIC Use a compass to draw a circle 4. (Mult. Prop. of Equality) with parallel chords and . Connect points A 5. (Substitution) and D by drawing segment . 6. (Def. of ) b. NUMERICAL Use a protractor to find and . Then determine and . 40. Theorem 11.8, paragraph proof What is true about these arcs? Explain. SOLUTION: c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find and in the figure. Verify by using inscribed angles to find the measures of the arcs.

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then SOLUTION: . Since is an inscribed angle, a. then or 90. So, by definition, is a right angle.

Given Part II: : is a right angle. Prove: is a semicircle. b. Sample answer: , ; , ; The arcs are congruent Proof: Since is an inscribed angle, then because they have equal measures. c. and by the Multiplication Sample answer:

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between m∠A = 15, m∠D = 15, = 30, and = 30. the arcs of a circle that are cut by two parallel The arcs are congruent. chords. In a circle, two parallel chords cut congruent arcs. a. GEOMETRIC Use a compass to draw a circle d. In a circle, two parallel chords cut congruent arcs. with parallel chords and . Connect points A So, = .

and D by drawing segment . b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain. c. VERBAL Draw another circle and repeat parts a = 4(16) + 6 or 70 and b. Make a conjecture about arcs of a circle that = 6(16) – 26 or 70 are cut by two parallel chords. Therefore, the measure of arcs QS and PR are each d. ANALYTICAL Use your conjecture to find 70. and in the figure. Verify by using CCSS ARGUMENTS Determine whether the inscribed angles to find the measures of the arcs. quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary SOLUTION: and inscribed in a circle. Therefore, the statement is always true. a. 43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, b. Sample answer: , ; the statement is always true. , ; The arcs are congruent because they have equal measures. 44. parallelogram c. Sample answer: SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus

SOLUTION: m∠A = 15, m∠D = 15, = 30, and = 30. The arcs are congruent. The opposite angles of a rhombus are always In a circle, two parallel chords cut congruent arcs. congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed d. In a circle, two parallel chords cut congruent arcs. in a circle as long as it is a square. Since the opposite So, = . angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true.

46. kite

= 4(16) + 6 or 70 SOLUTION: = 6(16) – 26 or 70 Exactly one pair of opposite angles of a kite are Therefore, the measure of arcs QS and PR are each congruent. To be supplementary, they must each be 70. a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must CCSS ARGUMENTS Determine whether the also be supplementary. So, as long as the angles that quadrilateral can always, sometimes, or never compose the pair of congruent opposite angles are be inscribed in a circle. Explain your reasoning. right angles, a kite can be inscribed in a circle. 42. square Therefore, the statement is sometimes true. SOLUTION: 47. CHALLENGE A square is inscribed in a circle. Squares have right angles at each vertex, therefore What is the ratio of the area of the circle to the area each pair of opposite angles will be supplementary of the square? and inscribed in a circle. Therefore, the statement is always true. SOLUTION: A square with side s is inscribed in a circle with 43. rectangle radius r. SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram Using the Pythagorean Theorem, SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be

inscribed in a circle as long as it is a rectangle. 2 Therefore, the statement is sometimes true. The area of a square of side s is A = s and the area of a circle of radius r is A = . 45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the ratio of the area of the circle to the Therefore, the statement is sometimes true. area of the inscribed square is 46. kite 48. WRITING IN MATH A right SOLUTION: triangle is inscribed in a circle. If the radius of the Exactly one pair of opposite angles of a kite are circle is given, explain how to find the lengths of the congruent. To be supplementary, they must each be right triangle’s legs. a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must SOLUTION: also be supplementary. So, as long as the angles that Sample answer: A triangle will have compose the pair of congruent opposite angles are two inscribed angles of 45 and one of 90. The right angles, a kite can be inscribed in a circle. hypotenuse is across from the 90, or right angle. Therefore, the statement is sometimes true. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right 47. CHALLENGE A square is inscribed in a circle. angle. Therefore, the hypotenuse is a diameter and What is the ratio of the area of the circle to the area has a length of 2r. Use trigonometry to find the of the square? length of the equal legs. SOLUTION: A square with side s is inscribed in a circle with radius r.

Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by . Using the Pythagorean Theorem, 49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION: See students’ work. 2 The area of a square of side s is A = s and the area of a circle of radius r is A = . 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If

Therefore, the ratio of the area of the circle to the an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc area of the inscribed square is AB). So, if an inscribed angle and a central angle 48. WRITING IN MATH A right intercept the same arc, then the measure of the triangle is inscribed in a circle. If the radius of the inscribed angle is one-half the measure of the central

circle is given, explain how to find the lengths of the angle. right triangle’s legs. 51. In the circle below, and . SOLUTION: What is ? Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs. A 42 B 61 C 80 D 84 SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Therefore, the length of each leg of the 45-45-90 intercepted arc. Therefore, triangle can be found by multiplying the radius of the circle in which it is inscribed by . But, 49. OPEN ENDED Find and sketch a real-world logo Therefore, with an inscribed polygon. The correct choice is A. SOLUTION: 52. ALGEBRA Simplify See students’ work. 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 50. WRITING IN MATH Compare and contrast F 9x + 3x – 14 2 inscribed angles and central angles of a circle. If they G 9x + 13x – 14 2 intercept the same arc, how are they related? H 27x + 37x – 38 SOLUTION: J 27x2 + 27x – 26 An inscribed angle has its vertex on the circle. A SOLUTION: central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the Use the Distributive Property to simplify the first

measure of the central angle is equal to m(arc AB). If term. an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the 11-4inscribed Inscribed angle Angles is one -half the measure of the central angle. Therefore, the correct choice is H.

51. In the circle below, and . 53. SHORT RESPONSE In the circle below, is a What is ? diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle.

A 42 B 61 SOLUTION: C 80 Use the Pythagorean Theorem to find the D 84 hypotenuse of the triangle which is also the diameter of the circle. SOLUTION: If an angle is inscribed in a circle, then the measure The radius is half the diameter. So, the radius of the of the angle equals one half the measure of its circle is about 8.5 in. intercepted arc. Therefore, The circumference C of a circle of a circle of

diameter d is given by Therefore, the circumference of the circle is But, Therefore,

The correct choice is A. 54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? 52. ALGEBRA Simplify A 2 –15 4(3x – 2)(2x + 4) + 3x + 5x – 6. B –16 2 F 9x + 3x – 14 C –17 G 9x2 + 13x – 14 D –18 2 H 27x + 37x – 38 E –19 2 J 27x + 27x – 26 SOLUTION: SOLUTION: Let the three consecutive integers be x , x +1,and x + 2. Then, Use the Distributive Property to simplify the first term.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. Therefore, the correct choice is H. In , FL = 24, HJ = 48, and . Find each measure. 53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle.

55. FG eSolutions Manual - Powered by Cognero Page 13 SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular Use the Pythagorean Theorem to find the to a chord, then it bisects the chord and its arc. So, hypotenuse of the triangle which is also the diameter bisects Therefore, of the circle.

The radius is half the diameter. So, the radius of the circle is about 8.5 in. 56. The circumference C of a circle of a circle of SOLUTION: diameter d is given by If a diameter (or radius) of a circle is perpendicular Therefore, the circumference of the circle is to a chord, then it bisects the chord and its arc. So,

bisects Therefore,

54. SAT/ACT The sum of three consecutive integers is 57. NJ –48. What is the least of the three integers? A –15 SOLUTION: B –16 If a diameter (or radius) of a circle is perpendicular C –17 to a chord, then it bisects the chord and its arc. So, D –18 bisects Therefore, E –19 SOLUTION: Let the three consecutive integers be x , x +1,and x + 58. 2. Then, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So,

So, the three integers are –17, –16, and –15 and the bisects Therefore, least of these is –17. Therefore, the correct choice is C. Find x. In , FL = 24, HJ = 48, and . Find each measure.

59. SOLUTION: The sum of the measures of the central angles of a 55. FG circle with no interior points in common is 360. So, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

56. SOLUTION: If a diameter (or radius) of a circle is perpendicular 60. to a chord, then it bisects the chord and its arc. So, SOLUTION:

bisects Therefore, The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

58. 61. SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular The sum of the measures of the central angles of a to a chord, then it bisects the chord and its arc. So, circle with no interior points in common is 360. So, bisects Therefore,

Find x.

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the 59. image of the person on the back of the camera is 12 inches, the distance from the opening to the person is SOLUTION: 7 feet, and the camera itself is 15 inches long. How The sum of the measures of the central angles of a tall is the person being photographed? circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the 60. length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will SOLUTION: be proportional. One foot is equivalent to 12 in. Then, The sum of the measures of the central angles of a the height of the larger triangle is 7(12) = 84 in. Let x circle with no interior points in common is 360. So, be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing

61. measure. SOLUTION: 63. AB = 4x – 5, BC = 11 + 2x, AC = ? The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, Since B is the midpoint of

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 64. AB = 6y – 14, BC = 10 – 2y, AC = ? inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How SOLUTION: tall is the person being photographed? Since B is the midpoint of

SOLUTION: 65. BC = 6 – 4m, AC = 8, m = ? The height of the person being photographed is the SOLUTION: length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will Since B is the midpoint of be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

66. AB = 10s + 2, AC = 40, s = ? Therefore, the person being photographed is 5.6 ft SOLUTION: tall. Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Find each measure. intercepted arc. So, 1. Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If

, what is ? SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

2. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

ALGEBRA Find each measure. 5. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

3. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

SOLUTION: Therefore, Here, is a semi-circle. So,

6. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

4. SCIENCE The diagram shows how light bends in a raindrop to make the colors of the rainbow. If SOLUTION: , what is ? If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore,

7. PROOF Write a two-column proof. SOLUTION: Given: If an angle is inscribed in a circle, then the measure bisects . of the angle equals one half the measure of its Prove:

intercepted arc. Therefore,

ALGEBRA Find each measure. 5. SOLUTION: Given: bisects . Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent.

So, Proof:

Statements (Reasons) 1. bisects . (Given)

Therefore, 2. (Def. of segment bisector) 3. intercepts . intercepts . 6. (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

SOLUTION: Therefore, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 7. PROOF Write a two-column proof. So, Given: bisects . The sum of the measures of the angles of a triangle Prove: is 180. So,

Therefore,

9. x SOLUTION: Given: bisects . Prove:

SOLUTION:

Proof: An inscribed angle of a triangle intercepts a diameter Statements (Reasons) or semicircle if and only if the angle is a right angle. So, Since 1. bisects . (Given) The sum of the measures of the angles of a triangle 2. (Def. of segment bisector) 3. intercepts . intercepts . is 180. So, (Def. of intercepted arc) 4. (Inscribed of same arc are .) 10. and 5. (Vertical are .) 6. (AAS)

CCSS STRUCTURE Find each value. 8.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

SOLUTION: Solve the two equations to find the values of x and y. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Use the values of the variables to find and The sum of the measures of the angles of a triangle .

is 180. So,

Therefore, ALGEBRA Find each measure. 11. 9. x

SOLUTION: If an angle is inscribed in a circle, then the measure SOLUTION: of the angle equals one half the measure of its An inscribed angle of a triangle intercepts a diameter intercepted arc. Therefore, or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle 12.

is 180. So,

10. and

intercepted arc. Therefore,

SOLUTION: 13. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y.

Use the values of the variables to find and . SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

ALGEBRA Find each measure. 11. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

14.

intercepted arc. Therefore, SOLUTION: 12. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

15.

intercepted arc. Therefore, SOLUTION: Here, 13. The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

16. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, Here, The arc is a semicircle. So, 14. Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

15.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So, SOLUTION: Here, 18. ALGEBRA Find each measure.

The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, Therefore,

16. SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore, SOLUTION: Here, 19. ALGEBRA Find each measure.

The arc is a semicircle. So,

Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20. SOLUTION: If two inscribed angles of a circle intercept the same 20. ALGEBRA Find each measure. arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: Since ∠C and ∠D both intercept arc AB, m∠C = If two inscribed angles of a circle intercept the same m∠D. arc or congruent arcs, then the angles are congruent. So,

Therefore, m∠C = 5(10) – 3 or 47.

Therefore, PROOF Write the specified type of proof. 21. paragraph proof 19. ALGEBRA Find each measure. Given:

Prove:

SOLUTION: SOLUTION: If two inscribed angles of a circle intercept the same Proof: Given means that arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = . Since and m∠B.

, the equation becomes

Therefore, m∠A = 5(4) or 20. . Multiplying each side of the ALGEBRA Find each measure. 20. equation by 2 results in .

22. two-column proof Given: Prove:

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: Since ∠C and ∠D both intercept arc AB, m∠C = Statements (Reasons): m∠D. 1. (Given) 2. (Inscribed intercepting same arc are .)

Therefore, m∠C = 5(10) – 3 or 47. 3. (Vertical are .) 4. (AA Similarity) PROOF Write the specified type of proof. 21. paragraph proof ALGEBRA Find each value.

Given:

Prove:

23. x

SOLUTION: An inscribed angle of a triangle intercepts a diameter SOLUTION: or semicircle if and only if the angle is a right angle. So, Proof: Given means that The sum of the measures of the angles of a triangle

. Since and is 180. So,

, the equation becomes 24. . Multiplying each side of the SOLUTION: equation by 2 results in . An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, 22. two-column proof The sum of the measures of the angles of a triangle Given:

Prove: is 180. So,

Therefore,

25. x SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter Statements (Reasons): or semicircle if and only if the angle is a right angle. 1. (Given) So, 2. (Inscribed intercepting same arc The sum of the measures of the angles of a triangle are .) 3. (Vertical are .) is 180. So, 4. (AA Similarity)

ALGEBRA Find each value. 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle 23. x

is 180. So, SOLUTION: An inscribed angle of a triangle intercepts a diameter Therefore, or semicircle if and only if the angle is a right angle. So, CCSS STRUCTURE Find each measure. The sum of the measures of the angles of a triangle

is 180. So,

24. 27. SOLUTION: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. So, The sum of the measures of the angles of a triangle Therefore, is 180. So, 28. Therefore, SOLUTION: If a quadrilateral is inscribed in a circle, then its 25. x opposite angles are supplementary. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Therefore, The sum of the measures of the angles of a triangle CCSS STRUCTURE Find each measure. is 180. So,

26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. 29. So, The sum of the measures of the angles of a triangle SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. is 180. So,

Therefore, Therefore, CCSS STRUCTURE Find each measure.

30. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

27.

SOLUTION: If a quadrilateral is inscribed in a circle, then its Therefore, opposite angles are supplementary.

31. PROOF Write a paragraph proof for Theorem 11.9. Therefore, SOLUTION: 28. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. SOLUTION: and are supplementary. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

CCSS STRUCTURE Find each measure. Proof: By arc addition and the definitions of arc measure and the sum of central angles, . Since by Theorem 10.6 and , but

29. , so or 180. This makes ∠C and ∠A supplementary. SOLUTION: Because the sum of the measures of the interior If a quadrilateral is inscribed in a circle, then its angles of a quadrilateral is 360, opposite angles are supplementary. . But , so , making them supplementary also.

Therefore, SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

measure. 32. Therefore, SOLUTION: 31. PROOF Write a paragraph proof for Theorem 11.9. Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are SOLUTION: congruent. There are eight adjacent arcs that make Given: Quadrilateral ABCD is inscribed in . up the circle, so their sum is 360. Thus, the measure Prove: and are supplementary. of each arc joining consecutive vertices is of 360 and are supplementary. or 45.

Therefore, measure of arc NPQ is 135.

33. SOLUTION: Since all the sides of the stop sign are congruent Proof: By arc addition and the definitions of arc chords of the circle, all the corresponding arcs are measure and the sum of central angles, congruent. There are eight adjacent arcs that make . Since by Theorem 10.6 up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 and or 45. Since ∠RLQ is inscribed in the circle, its , but measure equals one half the measure of its intercepted arc QR. , so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But , so , making Therefore, the measure of ∠RLQ is 22.5. them supplementary also.

SIGNS: A stop sign in the shape of a regular 34. octagon is inscribed in a circle. Find each SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, measure. 360 is divided into 8 equal arcs and each arc 32. measures SOLUTION: Then, Since all the sides of the stop sign are congruent

chords of the circle, the corresponding arcs are Therefore, congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 35. of each arc joining consecutive vertices is of 360 or 45. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Therefore, measure of arc NPQ is 135. intercepted arc. So, 33. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, SOLUTION: 360 is divided into 8 equal arcs and each arc Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are measures congruent. There are eight adjacent arcs that make

up the circle, so their sum is 360. Thus, the measure Then, of each arc joining consecutive vertices is of 360 Therefore, or 45. Since ∠RLQ is inscribed in the circle, its measure equals one half the measure of its intercepted arc QR. 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star shown are congruent, find the measure of each inscribed angle. a.

Therefore, the measure of ∠RLQ is 22.5.

34. b. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So,

The sum of the measures of the central angles of a c. circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Then,

d. Therefore,

35. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. So, a. The sum of the measures of the central angles of a circle with no interior points in common is 360. The sum of the measures of the central angles of a Here, 360 is divided into 5 equal arcs and each arc circle with no interior points in common is 360. Here,

360 is divided into 8 equal arcs and each arc measures

measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Then, intercepted arc. So, the measure of each inscribed angle is 36.

Therefore, b. Here, 360 is divided into 6 equal arcs and each arc measures Each inscribed angle is formed 36. ART Four different string art star patterns are shown. If all of the inscribed angles of each star by intercepting alternate vertices of the star. So, each shown are congruent, find the measure of each intercepted arc measures 120. If an angle is inscribed inscribed angle. in a circle, then the measure of the angle equals one a. half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. c. Here, 360 is divided into 7 equal arcs and each arc measures If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure b. of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Each inscribed angle is formed c. by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

d. PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove:

SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 5 equal arcs and each arc

measures

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. So, the measure of each inscribed Proof: angle is 36. Statements (Reasons) b. Here, 360 is divided into 6 equal arcs and each arc 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) measures Each inscribed angle is formed 2. m∠ABD = m(arc AD) by intercepting alternate vertices of the star. So, each m∠DBC = m(arc DC) (The measure of an intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one inscribed whose side is a diameter is half the half the measure of its intercepted arc. So, the measure of the intercepted arc (Case 1)). measure of each inscribed angle is 60. 3. m∠ABC = m(arc AD) + m(arc DC) c. Here, 360 is divided into 7 equal arcs and each arc (Substitution) measures If an angle is inscribed in a 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)

circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc

of each inscribed angle is about 25.7. Addition Postulate)

d. The sum of the measures of the central angles of 6. m∠ABC = m(arc AC) (Substitution) a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc 38. Case 3 measures Each inscribed angle is formed Given: P lies outside . is a diameter.

by intercepting alternate vertices of the star. So, each Prove: intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

Prove: SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the SOLUTION: measure of the intercepted arc (Case 1)). Proof: 3. m∠ABC = m(arc DC) – m(arc DA) Statements (Reasons) (Substitution) 1. m∠ABC = m∠ABD + m∠DBC ( Addition 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) Postulate)

2. m∠ABD = m(arc AD) 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc

m∠DBC = m(arc DC) (The measure of an Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction inscribed whose side is a diameter is half the Property of Equality) measure of the intercepted arc (Case 1)). 7. m∠ABC = m(arc AC) (Substitution) 3. m∠ABC = m(arc AD) + m(arc DC)

(Substitution) 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) PROOF Write the specified proof for each theorem. 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc 39. Theorem 11.7, two-column proof Addition Postulate) SOLUTION: 6. m ABC = m(arc AC) (Substitution) ∠ Given: and are inscribed; Prove: 38. Case 3 Proof: Given: P lies outside . is a diameter.

Prove:

Statements (Reasons) 1. and are inscribed;

(Given) SOLUTION: Proof: 2. ; (Measure Statements (Reasons) of an inscribed = half measure of intercepted arc.) 1. m∠ABC = m∠DBC – m∠DBA ( Addition Postulate, Subtraction Property of Equality) 3. (Def. of arcs)

2. m∠DBC = m(arc DC) 4. (Mult. Prop. of Equality) m∠DBA = m(arc DA) (The measure of an 5. (Substitution) inscribed whose side is a diameter is half the 6. (Def. of ) measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) 40. Theorem 11.8, paragraph proof (Substitution) SOLUTION: 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) 7. m∠ABC = m(arc AC) (Substitution)

PROOF Write the specified proof for each Part I: Given: is a semicircle. theorem. Prove: is a right angle. 39. Theorem 11.7, two-column proof Proof: Since is a semicircle, then SOLUTION: . Since is an inscribed angle, Given: and are inscribed; then or 90. So, by definition, Prove: Proof: is a right angle.

Part II: Given: is a right angle. Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because Statements (Reasons) is a right angle, m = 90. Then 1. and are inscribed; m = 2(90) or 180. So by definition, is a (Given) semicircle.

2. ; (Measure 41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between of an inscribed = half measure of intercepted arc.) the arcs of a circle that are cut by two parallel

3. (Def. of arcs) chords. a. GEOMETRIC Use a compass to draw a circle 4. (Mult. Prop. of Equality) with parallel chords and . Connect points A

5. (Substitution) and D by drawing segment .

6. (Def. of ) b. NUMERICAL Use a protractor to find and . Then determine and . 40. Theorem 11.8, paragraph proof What is true about these arcs? Explain. SOLUTION: c. VERBAL Draw another circle and repeat parts a and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. d. ANALYTICAL Use your conjecture to find and in the figure. Verify by using inscribed angles to find the measures of the arcs.

Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then SOLUTION: . Since is an inscribed angle, a. then or 90. So, by definition, is a right angle.

Part II: Given: is a right angle.

Prove: is a semicircle. b. Sample answer: , ;

, ; The arcs are congruent Proof: Since is an inscribed angle, then because they have equal measures. and by the Multiplication c. Sample answer:

41. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel m∠A = 15, m∠D = 15, = 30, and = 30.

chords. The arcs are congruent. In a circle, two parallel chords cut congruent arcs. a. GEOMETRIC Use a compass to draw a circle d. In a circle, two parallel chords cut congruent arcs. with parallel chords and . Connect points A So, = . and D by drawing segment .

b. NUMERICAL Use a protractor to find and . Then determine and . What is true about these arcs? Explain.

c. VERBAL Draw another circle and repeat parts a = 4(16) + 6 or 70 and b. Make a conjecture about arcs of a circle that = 6(16) – 26 or 70 are cut by two parallel chords. Therefore, the measure of arcs QS and PR are each d. ANALYTICAL Use your conjecture to find 70. and in the figure. Verify by using CCSS ARGUMENTS Determine whether the inscribed angles to find the measures of the arcs. quadrilateral can always, sometimes, or never be inscribed in a circle. Explain your reasoning. 42. square SOLUTION: Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary SOLUTION: and inscribed in a circle. Therefore, the statement is a. always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, b. Sample answer: , ; the statement is always true. , ; The arcs are congruent because they have equal measures. 44. parallelogram c. Sample answer: SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus

m∠A = 15, m∠D = 15, = 30, and = 30. SOLUTION: The arcs are congruent. The opposite angles of a rhombus are always In a circle, two parallel chords cut congruent arcs. congruent. They will only be supplementary when d. In a circle, two parallel chords cut congruent arcs. they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite So, = . angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true.

46. kite = 4(16) + 6 or 70 SOLUTION:

= 6(16) – 26 or 70 Exactly one pair of opposite angles of a kite are Therefore, the measure of arcs QS and PR are each congruent. To be supplementary, they must each be 70. a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must CCSS ARGUMENTS Determine whether the also be supplementary. So, as long as the angles that quadrilateral can always, sometimes, or never compose the pair of congruent opposite angles are be inscribed in a circle. Explain your reasoning. right angles, a kite can be inscribed in a circle. 42. square Therefore, the statement is sometimes true. SOLUTION: Squares have right angles at each vertex, therefore 47. CHALLENGE A square is inscribed in a circle. each pair of opposite angles will be supplementary What is the ratio of the area of the circle to the area and inscribed in a circle. Therefore, the statement is of the square? always true. SOLUTION: A square with side s is inscribed in a circle with 43. rectangle radius r. SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram Using the Pythagorean Theorem, SOLUTION: The opposite angles of a parallelogram are always congruent. They will only be supplementary when they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. 2 Therefore, the statement is sometimes true. The area of a square of side s is A = s and the area of a circle of radius r is A = . 45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not supplementary, they can not be inscribed in a circle. Therefore, the statement is sometimes true. Therefore, the ratio of the area of the circle to the area of the inscribed square is 46. kite SOLUTION: 48. WRITING IN MATH A right triangle is inscribed in a circle. If the radius of the Exactly one pair of opposite angles of a kite are circle is given, explain how to find the lengths of the congruent. To be supplementary, they must each be right triangle’s legs. a right angle. If one pair of opposite angles for a quadrilateral is supplementary, the other pair must SOLUTION: also be supplementary. So, as long as the angles that Sample answer: A triangle will have compose the pair of congruent opposite angles are two inscribed angles of 45 and one of 90. The right angles, a kite can be inscribed in a circle. hypotenuse is across from the 90, or right angle.

Therefore, the statement is sometimes true. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right 47. CHALLENGE A square is inscribed in a circle. angle. Therefore, the hypotenuse is a diameter and What is the ratio of the area of the circle to the area has a length of 2r. Use trigonometry to find the of the square? length of the equal legs. SOLUTION: A square with side s is inscribed in a circle with radius r.

Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the

Using the Pythagorean Theorem, circle in which it is inscribed by . 49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon. SOLUTION:

2 See students’ work. The area of a square of side s is A = s and the area of a circle of radius r is A = . 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If Therefore, the ratio of the area of the circle to the an inscribed angle also intercepts arc AB, then the area of the inscribed square is measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle 48. WRITING IN MATH A right intercept the same arc, then the measure of the triangle is inscribed in a circle. If the radius of the inscribed angle is one-half the measure of the central circle is given, explain how to find the lengths of the angle. right triangle’s legs. 51. In the circle below, and . SOLUTION: What is ? Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a triangle intercepts a diameter if the angle is a right angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the

length of the equal legs. A 42 B 61 C 80 D 84 SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Therefore, the length of each leg of the 45-45-90 intercepted arc. Therefore, triangle can be found by multiplying the radius of the circle in which it is inscribed by . But, 49. OPEN ENDED Find and sketch a real-world logo Therefore, with an inscribed polygon. The correct choice is A.

SOLUTION: 52. ALGEBRA Simplify See students’ work. 2 4(3x – 2)(2x + 4) + 3x + 5x – 6. 2 50. WRITING IN MATH Compare and contrast F 9x + 3x – 14 inscribed angles and central angles of a circle. If they G 9x2 + 13x – 14 intercept the same arc, how are they related? 2 H 27x + 37x – 38 SOLUTION: J 27x2 + 27x – 26 An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. SOLUTION: If a central angle intercepts arc AB, then the Use the Distributive Property to simplify the first measure of the central angle is equal to m(arc AB). If term. an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the inscribed angle is one-half the measure of the central angle. Therefore, the correct choice is H.

51. In the circle below, and . 53. SHORT RESPONSE In the circle below, is a What is ? diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle.

A 42 B 61 SOLUTION: C 80 Use the Pythagorean Theorem to find the D 84 hypotenuse of the triangle which is also the diameter of the circle. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its The radius is half the diameter. So, the radius of the intercepted arc. Therefore, circle is about 8.5 in. The circumference C of a circle of a circle of diameter d is given by Therefore, the circumference of the circle is

But, Therefore, The correct choice is A. 54. SAT/ACT The sum of three consecutive integers is 52. ALGEBRA Simplify –48. What is the least of the three integers? 2 A –15 4(3x – 2)(2x + 4) + 3x + 5x – 6. B –16 F 2 9x + 3x – 14 C –17 2 G 9x + 13x – 14 D –18 2 H 27x + 37x – 38 E –19 2 J 27x + 27x – 26 SOLUTION: SOLUTION: Let the three consecutive integers be x , x +1,and x + Use the Distributive Property to simplify the first 2. Then, term.

So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C. Therefore, the correct choice is H. In , FL = 24, HJ = 48, and . Find 53. SHORT RESPONSE In the circle below, is a each measure. diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle.

55. FG SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular Use the Pythagorean Theorem to find the to a chord, then it bisects the chord and its arc. So, hypotenuse of the triangle which is also the diameter bisects Therefore, of the circle.

The radius is half the diameter. So, the radius of the

circle is about 8.5 in. 56. The circumference C of a circle of a circle of SOLUTION: diameter d is given by If a diameter (or radius) of a circle is perpendicular Therefore, the circumference of the circle is to a chord, then it bisects the chord and its arc. So,

bisects Therefore, SAT/ACT 54. The sum of three consecutive integers is 57. NJ –48. What is the least of the three integers? A –15 SOLUTION: B –16 If a diameter (or radius) of a circle is perpendicular C –17 to a chord, then it bisects the chord and its arc. So, D –18 bisects Therefore, E –19 SOLUTION: Let the three consecutive integers be x , x +1,and x + 2. Then, 58. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So,

So, the three integers are –17, –16, and –15 and the bisects Therefore, 11-4least Inscribed of these Angles is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find Find x. each measure.

59. SOLUTION:

55. FG The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

56. SOLUTION: If a diameter (or radius) of a circle is perpendicular 60. to a chord, then it bisects the chord and its arc. So, SOLUTION: bisects Therefore, The sum of the measures of the central angles of a 57. NJ circle with no interior points in common is 360. So,

58. 61. SOLUTION: If a diameter (or radius) of a circle is perpendicular SOLUTION: to a chord, then it bisects the chord and its arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. So, bisects Therefore,

Find x.

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the 59. image of the person on the back of the camera is 12 inches, the distance from the opening to the person is SOLUTION: eSolutions Manual - Powered by Cognero 7 feet, and the camera itself is 15 inches long. HowPage 14 The sum of the measures of the central angles of a tall is the person being photographed? circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the 60. length of the base of the larger triangle. The two SOLUTION: triangles are similar. So, their corresponding sides will The sum of the measures of the central angles of a be proportional. One foot is equivalent to 12 in. Then, circle with no interior points in common is 360. So, the height of the larger triangle is 7(12) = 84 in. Let x

be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft tall.

ALGEBRA Suppose B is the midpoint of . 61. Use the given information to find the missing measure. SOLUTION: 63. AB = 4x – 5, BC = 11 + 2x, AC = ? The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION: Since B is the midpoint of

62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 64. AB = 6y – 14, BC = 10 – 2y, AC = ? 7 feet, and the camera itself is 15 inches long. How SOLUTION: tall is the person being photographed? Since B is the midpoint of

SOLUTION: 65. BC = 6 – 4m, AC = 8, m = ? The height of the person being photographed is the SOLUTION: length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will Since B is the midpoint of be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x be the length of the base of the larger triangle.

66. AB = 10s + 2, AC = 40, s = ? Therefore, the person being photographed is 5.6 ft SOLUTION: tall. Since B is the midpoint of ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: Since B is the midpoint of

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of

Find each measure. 1.

intercepted arc. Therefore,

2.

3.

SOLUTION: Here, is a semi-circle. So,

4. SCIENCE The diagram shows how light bends in a Find each measure. raindrop to make the colors of the rainbow. If 1. , what is ?

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION:

intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, 2. ALGEBRA Find each measure. 5.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SOLUTION: intercepted arc. Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 3. So,

Therefore,

6.

SOLUTION: Here, is a semi-circle. So,

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, SOLUTION: Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 4. SCIENCE The diagram shows how light bends in a So, raindrop to make the colors of the rainbow. If , what is ?

Therefore,

7. PROOF Write a two-column proof. Given: bisects . Prove:

intercepted arc. Therefore, SOLUTION:

ALGEBRA Find each measure. Given: bisects . 5. Prove:

Proof: Statements (Reasons) SOLUTION: 1. bisects . (Given) If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. 2. (Def. of segment bisector) So, 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are Therefore, .) 5. (Vertical are .) 6. (AAS) 6. CCSS STRUCTURE Find each value. 8.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: So, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

Therefore, is 180. So, 7. PROOF Write a two-column proof. Given: bisects . Therefore, Prove: 9. x

SOLUTION:

Given: bisects . Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

Proof: Statements (Reasons) is 180. So, 1. bisects . (Given) 2. (Def. of segment bisector) 10. and 3. intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (Vertical are .) 6. (AAS) CCSS STRUCTURE Find each value. SOLUTION: 8. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y.

Use the values of the variables to find and SOLUTION: . An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle ALGEBRA Find each measure.

11. is 180. So,

Therefore,

9. x

12. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, Since The sum of the measures of the angles of a triangle

is 180. So, SOLUTION: If an angle is inscribed in a circle, then the measure 10. and of the angle equals one half the measure of its

intercepted arc. Therefore,

13.

SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Solve the two equations to find the values of x and y. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, Use the values of the variables to find and . If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, ALGEBRA Find each measure. 11. 14.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, intercepted arc. Therefore,

12. 15.

SOLUTION: SOLUTION:

If an angle is inscribed in a circle, then the measure Here, of the angle equals one half the measure of its The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure intercepted arc. Therefore, of the angle equals one half the measure of its intercepted arc. So, 13. Therefore,

16.

SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, SOLUTION: Here, If an angle is inscribed in a circle, then the measure The arc is a semicircle. So, of the angle equals one half the measure of its Then,

intercepted arc. Therefore, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

14. intercepted arc. Therefore,

17. ALGEBRA Find each measure.

SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore, SOLUTION: 15. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

18. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. So, SOLUTION:

Therefore, If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent.

16. So,

Therefore,

19. ALGEBRA Find each measure.

SOLUTION: Here, The arc is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore, SOLUTION: If two inscribed angles of a circle intercept the same 17. ALGEBRA Find each measure. arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = m∠B.

Therefore, m∠A = 5(4) or 20.

20. ALGEBRA Find each measure.

18. ALGEBRA Find each measure.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since ∠C and ∠D both intercept arc AB, m∠C = m∠D.

SOLUTION: Therefore, m∠C = 5(10) – 3 or 47. If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. PROOF Write the specified type of proof. So, 21. paragraph proof

Given:

Therefore, Prove: 19. ALGEBRA Find each measure.

SOLUTION:

Proof: Given means that

. Since and SOLUTION: If two inscribed angles of a circle intercept the same , the equation becomes arc or congruent arcs, then the angles are congruent. Since ∠A and ∠B both intercept arc CD , m∠A = . Multiplying each side of the m∠B. equation by 2 results in .

Therefore, m∠A = 5(4) or 20. 22. two-column proof 20. ALGEBRA Find each measure. Given: Prove:

SOLUTION: Statements (Reasons): SOLUTION: 1. (Given) If two inscribed angles of a circle intercept the same 2. (Inscribed intercepting same arc arc or congruent arcs, then the angles are congruent. are .) Since ∠C and ∠D both intercept arc AB, m∠C = 3. (Vertical are .) m∠D. 4. (AA Similarity)

ALGEBRA Find each value.

Therefore, m∠C = 5(10) – 3 or 47.

PROOF Write the specified type of proof. 21. paragraph proof

Given: 23. x

Prove: SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

SOLUTION: is 180. So,

Proof: Given means that 24. . Since and SOLUTION: , the equation becomes An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, . Multiplying each side of the The sum of the measures of the angles of a triangle

equation by 2 results in . is 180. So, 22. two-column proof

Given: Therefore, Prove: 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

SOLUTION: Statements (Reasons): is 180. So, 1. (Given) 2. (Inscribed intercepting same arc are .) 26. 3. (Vertical are .) SOLUTION: 4. (AA Similarity) An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. ALGEBRA Find each value. So, The sum of the measures of the angles of a triangle

is 180. So,

Therefore, 23. x

CCSS STRUCTURE Find each measure. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So, 27. SOLUTION: If a quadrilateral is inscribed in a circle, then its 24. opposite angles are supplementary.

SOLUTION: Therefore, An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, 28. The sum of the measures of the angles of a triangle SOLUTION: If a quadrilateral is inscribed in a circle, then its is 180. So, opposite angles are supplementary.

Therefore,

25. x Therefore, SOLUTION: CCSS STRUCTURE Find each measure. An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle

is 180. So,

29. 26. SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its An inscribed angle of a triangle intercepts a diameter opposite angles are supplementary. or semicircle if and only if the angle is a right angle.

So, The sum of the measures of the angles of a triangle

Therefore, is 180. So,

Therefore, 30. SOLUTION: CCSS STRUCTURE Find each measure. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

27. 31. PROOF Write a paragraph proof for Theorem 11.9. SOLUTION: SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. Therefore, and are supplementary.

28. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Proof: By arc addition and the definitions of arc measure and the sum of central angles,

Therefore, . Since by Theorem 10.6 CCSS STRUCTURE Find each measure. and , but

, so or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, . But 29. , so , making SOLUTION: them supplementary also. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each

Therefore,

30. SOLUTION: measure. If a quadrilateral is inscribed in a circle, then its 32. opposite angles are supplementary. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are congruent. There are eight adjacent arcs that make Therefore, up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 31. PROOF Write a paragraph proof for Theorem 11.9. or 45. SOLUTION: Given: Quadrilateral ABCD is inscribed in . Therefore, measure of arc NPQ is 135. Prove: and are supplementary. and are supplementary. 33. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure of each arc joining consecutive vertices is of 360 Proof: By arc addition and the definitions of arc or 45. Since ∠RLQ is inscribed in the circle, its measure and the sum of central angles, measure equals one half the measure of its . Since by Theorem 10.6 intercepted arc QR. and , but

, so or Therefore, the measure of ∠RLQ is 22.5. 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, 34. . But , so , making SOLUTION: them supplementary also. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its SIGNS: A stop sign in the shape of a regular octagon is inscribed in a circle. Find each intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc

measures Then, measure. Therefore, 32.

SOLUTION: 35. Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are SOLUTION: congruent. There are eight adjacent arcs that make If an angle is inscribed in a circle, then the measure up the circle, so their sum is 360. Thus, the measure of the angle equals one half the measure of its

of each arc joining consecutive vertices is of 360 intercepted arc. So, or 45. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, Therefore, measure of arc NPQ is 135. 360 is divided into 8 equal arcs and each arc

measures 33. Then, SOLUTION: Since all the sides of the stop sign are congruent Therefore, chords of the circle, all the corresponding arcs are congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure 36. ART Four different string art star patterns are of each arc joining consecutive vertices is of 360 shown. If all of the inscribed angles of each star shown are congruent, find the measure of each or 45. Since ∠RLQ is inscribed in the circle, its inscribed angle. measure equals one half the measure of its a. intercepted arc QR.

b. Therefore, the measure of ∠RLQ is 22.5.

34.

SOLUTION: If an angle is inscribed in a circle, then the measure c. of the angle equals one half the measure of its

intercepted arc. So, The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc d.

measures Then,

Therefore,

35. SOLUTION: a. The sum of the measures of the central angles of SOLUTION: a circle with no interior points in common is 360. If an angle is inscribed in a circle, then the measure Here, 360 is divided into 5 equal arcs and each arc of the angle equals one half the measure of its measures intercepted arc. So, If an angle is inscribed in a circle, then the measure The sum of the measures of the central angles of a of the angle equals one half the measure of its circle with no interior points in common is 360. Here, intercepted arc. So, the measure of each inscribed 360 is divided into 8 equal arcs and each arc angle is 36. b. Here, 360 is divided into 6 equal arcs and each arc measures measures Each inscribed angle is formed Then, by intercepting alternate vertices of the star. So, each

Therefore, intercepted arc measures 120. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the 36. ART Four different string art star patterns are measure of each inscribed angle is 60. shown. If all of the inscribed angles of each star c. Here, 360 is divided into 7 equal arcs and each arc shown are congruent, find the measure of each inscribed angle. measures If an angle is inscribed in a a. circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is about 25.7. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc b. measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the c. measure of each inscribed angle is 45. PROOF Write a two-column proof for each case of Theorem 11.6. 37. Case 2 Given: P lies inside . is a diameter.

d. Prove:

SOLUTION: a. The sum of the measures of the central angles of a circle with no interior points in common is 360. SOLUTION: Here, 360 is divided into 5 equal arcs and each arc Proof:

measures Statements (Reasons) 1. m∠ABC = m∠ABD + m∠DBC ( Addition If an angle is inscribed in a circle, then the measure Postulate) of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed 2. m∠ABD = m(arc AD) angle is 36. m∠DBC = m(arc DC) (The measure of an b. Here, 360 is divided into 6 equal arcs and each arc inscribed whose side is a diameter is half the measures Each inscribed angle is formed measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc AD) + m(arc DC) by intercepting alternate vertices of the star. So, each intercepted arc measures 120. If an angle is inscribed (Substitution) in a circle, then the measure of the angle equals one 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) half the measure of its intercepted arc. So, the measure of each inscribed angle is 60. 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc c. Here, 360 is divided into 7 equal arcs and each arc Addition Postulate) measures If an angle is inscribed in a 6. m∠ABC = m(arc AC) (Substitution) circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure 38. Case 3 of each inscribed angle is about 25.7. Given: P lies outside . is a diameter. d. The sum of the measures of the central angles of a circle with no interior points in common is 360. Prove: Here, 360 is divided into 8 equal arcs and each arc measures Each inscribed angle is formed by intercepting alternate vertices of the star. So, each intercepted arc measures 90. If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. So, the measure of each inscribed angle is 45.

PROOF Write a two-column proof for each case SOLUTION: of Theorem 11.6. Proof: 37. Case 2 Statements (Reasons)

Given: P lies inside . is a diameter. 1. m∠ABC = m∠DBC – m∠DBA ( Addition

Prove: Postulate, Subtraction Property of Equality) 2. m∠DBC = m(arc DC) m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution)

4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) SOLUTION: Proof: 5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Statements (Reasons) Addition Postulate) 1. m∠ABC = m∠ABD + m∠DBC ( Addition 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Postulate) Property of Equality) 2. m∠ABD = m(arc AD) 7. m∠ABC = m(arc AC) (Substitution) m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the PROOF Write the specified proof for each measure of the intercepted arc (Case 1)). theorem. 3. m∠ABC = m(arc AD) + m(arc DC) 39. Theorem 11.7, two-column proof (Substitution) SOLUTION: 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) Given: and are inscribed; Prove: 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Proof: Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution)

38. Case 3 Given: P lies outside . is a diameter.

Prove:

Statements (Reasons) 1. and are inscribed; (Given) 2. ; (Measure

of an inscribed = half measure of intercepted arc.)

SOLUTION: 3. (Def. of arcs)

Proof: 4. (Mult. Prop. of Equality) Statements (Reasons) 1. m∠ABC = m∠DBC – m∠DBA ( Addition 5. (Substitution) Postulate, Subtraction Property of Equality) 6. (Def. of ) 2. m∠DBC = m(arc DC) 40. Theorem 11.8, paragraph proof m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the SOLUTION: measure of the intercepted arc (Case 1)). 3. m∠ABC = m(arc DC) – m(arc DA) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)

5. m(arc DA) + m(arc AC) = m(arc DC) (Arc Addition Postulate) 6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction Property of Equality) Part I: Given: is a semicircle. Prove 7. m∠ABC = m(arc AC) (Substitution) : is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, PROOF Write the specified proof for each theorem. then or 90. So, by definition, 39. Theorem 11.7, two-column proof is a right angle. SOLUTION: Given: and are inscribed; Part II: Given: is a right angle. Prove: Prove: is a semicircle. Proof: Proof: Since is an inscribed angle, then and by the Multiplication

Statements (Reasons) 41. MULTIPLE REPRESENTATIONS In this 1. and are inscribed; problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel (Given) chords. 2. ; (Measure a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A of an inscribed = half measure of intercepted arc.) and D by drawing segment . 3. (Def. of arcs) b. NUMERICAL Use a protractor to find

4. (Mult. Prop. of Equality) and . Then determine and . What is true about these arcs? Explain.

5. (Substitution) c. VERBAL Draw another circle and repeat parts a 6. (Def. of ) and b. Make a conjecture about arcs of a circle that are cut by two parallel chords. 40. Theorem 11.8, paragraph proof d. ANALYTICAL Use your conjecture to find SOLUTION: and in the figure. Verify by using inscribed angles to find the measures of the arcs.

SOLUTION:

a. Part I: Given: is a semicircle. Prove: is a right angle. Proof: Since is a semicircle, then . Since is an inscribed angle, then or 90. So, by definition, b. Sample answer: , ; is a right angle. , ; The arcs are congruent because they have equal measures. Part II: Given: is a right angle. c. Sample answer: Prove: is a semicircle.

Proof: Since is an inscribed angle, then and by the Multiplication

Property of Equality, m = 2m . Because is a right angle, m = 90. Then m = 2(90) or 180. So by definition, is a m∠A = 15, m∠D = 15, = 30, and = 30. semicircle. The arcs are congruent. In a circle, two parallel chords cut congruent arcs. 41. MULTIPLE REPRESENTATIONS In this d. In a circle, two parallel chords cut congruent arcs. problem, you will investigate the relationship between So, = . the arcs of a circle that are cut by two parallel chords. a. GEOMETRIC Use a compass to draw a circle with parallel chords and . Connect points A

and D by drawing segment . = 4(16) + 6 or 70 b. NUMERICAL Use a protractor to find = 6(16) – 26 or 70 and . Then determine and . Therefore, the measure of arcs QS and PR are each What is true about these arcs? Explain. 70. c. VERBAL Draw another circle and repeat parts a CCSS ARGUMENTS Determine whether the and b. Make a conjecture about arcs of a circle that quadrilateral can always, sometimes, or never are cut by two parallel chords. be inscribed in a circle. Explain your reasoning. d. ANALYTICAL Use your conjecture to find 42. square and in the figure. Verify by using SOLUTION: inscribed angles to find the measures of the arcs. Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: SOLUTION: a. Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram SOLUTION: b. Sample answer: , ; The opposite angles of a parallelogram are always , ; The arcs are congruent congruent. They will only be supplementary when because they have equal measures. they are right angles. So, a parallelogram can be c. Sample answer: inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when they are right angles. So, a rhombus can be inscribed in a circle as long as it is a square. Since the opposite m∠A = 15, m∠D = 15, = 30, and = 30. angles of rhombi that are not squares are not The arcs are congruent. supplementary, they can not be inscribed in a circle. In a circle, two parallel chords cut congruent arcs. Therefore, the statement is sometimes true. d. In a circle, two parallel chords cut congruent arcs. 46. kite So, = . SOLUTION: Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a right angle. If one pair of opposite angles for a

= 4(16) + 6 or 70 quadrilateral is supplementary, the other pair must = 6(16) – 26 or 70 also be supplementary. So, as long as the angles that Therefore, the measure of arcs QS and PR are each compose the pair of congruent opposite angles are 70. right angles, a kite can be inscribed in a circle. Therefore, the statement is sometimes true. CCSS ARGUMENTS Determine whether the quadrilateral can always, sometimes, or never 47. CHALLENGE A square is inscribed in a circle. be inscribed in a circle. Explain your reasoning. What is the ratio of the area of the circle to the area 42. square of the square? SOLUTION: SOLUTION: Squares have right angles at each vertex, therefore A square with side s is inscribed in a circle with each pair of opposite angles will be supplementary radius r. and inscribed in a circle. Therefore, the statement is always true.

43. rectangle SOLUTION: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be Using the Pythagorean Theorem, supplementary and inscribed in a circle. Therefore, the statement is always true.

44. parallelogram

SOLUTION: 2 The area of a square of side s is A = s and the area The opposite angles of a parallelogram are always congruent. They will only be supplementary when of a circle of radius r is A = . they are right angles. So, a parallelogram can be inscribed in a circle as long as it is a rectangle. Therefore, the statement is sometimes true.

45. rhombus SOLUTION: The opposite angles of a rhombus are always congruent. They will only be supplementary when Therefore, the ratio of the area of the circle to the they are right angles. So, a rhombus can be inscribed area of the inscribed square is in a circle as long as it is a square. Since the opposite angles of rhombi that are not squares are not 48. WRITING IN MATH A right supplementary, they can not be inscribed in a circle. triangle is inscribed in a circle. If the radius of the Therefore, the statement is sometimes true. circle is given, explain how to find the lengths of the 46. kite right triangle’s legs. SOLUTION: SOLUTION: Exactly one pair of opposite angles of a kite are Sample answer: A triangle will have congruent. To be supplementary, they must each be two inscribed angles of 45 and one of 90. The a right angle. If one pair of opposite angles for a hypotenuse is across from the 90, or right angle. quadrilateral is supplementary, the other pair must According to theorem 10.8, an inscribed angle of a also be supplementary. So, as long as the angles that triangle intercepts a diameter if the angle is a right compose the pair of congruent opposite angles are angle. Therefore, the hypotenuse is a diameter and right angles, a kite can be inscribed in a circle. has a length of 2r. Use trigonometry to find the Therefore, the statement is sometimes true. length of the equal legs.

47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the square? SOLUTION: A square with side s is inscribed in a circle with radius r. Therefore, the length of each leg of the 45-45-90 triangle can be found by multiplying the radius of the circle in which it is inscribed by .

49. OPEN ENDED Find and sketch a real-world logo with an inscribed polygon.

Using the Pythagorean Theorem, SOLUTION: See students’ work.

50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they intercept the same arc, how are they related? 2 The area of a square of side s is A = s and the area SOLUTION: of a circle of radius r is A = . An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the Therefore, the ratio of the area of the circle to the inscribed angle is one-half the measure of the central angle. area of the inscribed square is 51. In the circle below, and . 48. WRITING IN MATH A right What is ? triangle is inscribed in a circle. If the radius of the circle is given, explain how to find the lengths of the right triangle’s legs. SOLUTION: Sample answer: A triangle will have two inscribed angles of 45 and one of 90. The hypotenuse is across from the 90, or right angle. According to theorem 10.8, an inscribed angle of a A 42 triangle intercepts a diameter if the angle is a right B 61 angle. Therefore, the hypotenuse is a diameter and C 80 has a length of 2r. Use trigonometry to find the D 84 length of the equal legs. SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

But, Therefore, the length of each leg of the 45-45-90 Therefore, triangle can be found by multiplying the radius of the The correct choice is A.

circle in which it is inscribed by . 52. ALGEBRA Simplify 2 49. OPEN ENDED Find and sketch a real-world logo 4(3x – 2)(2x + 4) + 3x + 5x – 6. 2 with an inscribed polygon. F 9x + 3x – 14 2 SOLUTION: G 9x + 13x – 14 2 See students’ work. H 27x + 37x – 38 J 27x2 + 27x – 26 50. WRITING IN MATH Compare and contrast inscribed angles and central angles of a circle. If they SOLUTION: intercept the same arc, how are they related? Use the Distributive Property to simplify the first term. SOLUTION: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an inscribed angle also intercepts arc AB, then the measure of the inscribed angle is equal to m(arc Therefore, the correct choice is H. AB). So, if an inscribed angle and a central angle intercept the same arc, then the measure of the 53. SHORT RESPONSE In the circle below, is a inscribed angle is one-half the measure of the central diameter, AC = 8 inches, and BC = 15 inches. Find angle. the diameter, the radius, and the circumference of the circle. 51. In the circle below, and . What is ?

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter

A 42 of the circle.

B 61 C 80 The radius is half the diameter. So, the radius of the D 84 circle is about 8.5 in. The circumference C of a circle of a circle of SOLUTION: diameter d is given by If an angle is inscribed in a circle, then the measure Therefore, the circumference of the circle is of the angle equals one half the measure of its intercepted arc. Therefore,

54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? But, A –15 Therefore, B –16 The correct choice is A. C –17 D –18 52. ALGEBRA Simplify E –19 4(3x – 2)(2x + 4) + 3x2 + 5x – 6. 2 F 9x + 3x – 14 SOLUTION: 2 Let the three consecutive integers be x , x +1,and x + G 9x + 13x – 14 2 2. Then, H 27x + 37x – 38 J 27x2 + 27x – 26 SOLUTION: Use the Distributive Property to simplify the first So, the three integers are –17, –16, and –15 and the term. least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure.

Therefore, the correct choice is H.

53. SHORT RESPONSE In the circle below, is a diameter, AC = 8 inches, and BC = 15 inches. Find the diameter, the radius, and the circumference of the circle. 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

SOLUTION: Use the Pythagorean Theorem to find the hypotenuse of the triangle which is also the diameter 56.

of the circle. SOLUTION:

If a diameter (or radius) of a circle is perpendicular The radius is half the diameter. So, the radius of the to a chord, then it bisects the chord and its arc. So,

circle is about 8.5 in. bisects Therefore, The circumference C of a circle of a circle of diameter d is given by 57. NJ Therefore, the circumference of the circle is SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 54. SAT/ACT The sum of three consecutive integers is –48. What is the least of the three integers? bisects Therefore, A –15 B –16 C –17

D –18 58. E –19 SOLUTION: SOLUTION: If a diameter (or radius) of a circle is perpendicular Let the three consecutive integers be x , x +1,and x + to a chord, then it bisects the chord and its arc. So, 2. Then, bisects Therefore,

Find x. So, the three integers are –17, –16, and –15 and the least of these is –17. Therefore, the correct choice is C.

In , FL = 24, HJ = 48, and . Find each measure. 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

60.

56. SOLUTION: The sum of the measures of the central angles of a SOLUTION: circle with no interior points in common is 360. So, If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore,

57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 61. SOLUTION: The sum of the measures of the central angles of a 58. circle with no interior points in common is 360. So,

SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 62. PHOTOGRAPHY In one of the first cameras invented, light entered an opening in the front. An Find x. image was reflected in the back of the camera, upside down, forming similar triangles. Suppose the image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed? 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x 60. be the length of the base of the larger triangle.

Therefore, the person being photographed is 5.6 ft

tall.

ALGEBRA Suppose B is the midpoint of . Use the given information to find the missing measure. 63. AB = 4x – 5, BC = 11 + 2x, AC = ? SOLUTION: 61. Since B is the midpoint of SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So,

11-4 Inscribed Angles

62. PHOTOGRAPHY In one of the first cameras 64. AB = 6y – 14, BC = 10 – 2y, AC = ? invented, light entered an opening in the front. An image was reflected in the back of the camera, SOLUTION: upside down, forming similar triangles. Suppose the Since B is the midpoint of image of the person on the back of the camera is 12 inches, the distance from the opening to the person is 7 feet, and the camera itself is 15 inches long. How tall is the person being photographed?

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of SOLUTION: The height of the person being photographed is the length of the base of the larger triangle. The two triangles are similar. So, their corresponding sides will be proportional. One foot is equivalent to 12 in. Then, the height of the larger triangle is 7(12) = 84 in. Let x 66. AB = 10s + 2, AC = 40, s = ? be the length of the base of the larger triangle. SOLUTION: Since B is the midpoint of

Therefore, the person being photographed is 5.6 ft tall.

64. AB = 6y – 14, BC = 10 – 2y, AC = ? SOLUTION: Since B is the midpoint of eSolutions Manual - Powered by Cognero Page 15

65. BC = 6 – 4m, AC = 8, m = ? SOLUTION: Since B is the midpoint of

66. AB = 10s + 2, AC = 40, s = ? SOLUTION: Since B is the midpoint of