9-4 Inscribed Angles

Find each measure. 3. 1.

SOLUTION: SOLUTION: If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its Here, is a semi-circle. 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 ANSWER: intercepted arc. So, 30 Therefore,

2. ANSWER: 66

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,

ANSWER: 126 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,

ANSWER: 72

eSolutions Manual - Powered by Cognero Page 1 9-4 Inscribed Angles

ALGEBRA Find each measure. 5.

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

Therefore, ANSWER: Proof: 54 Statements (Reasons) 1. (Given) 2. (If minor arcs are , their 6. corresponding chords are .) 3. intercepts . intercepts . intercepts . intercepts . (Def. of intercepted arc) 4. (Inscribed of same arc are .) 5. (ASA)

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

Therefore,

ANSWER:

36 Proof: 7. PROOF Write a two-column proof. Statements (Reasons) Given: 1. (Given) Prove: 2. (If minor arcs are , their corresponding chords are .) 3. intercepts . intercepts . intercepts . intercepts . (Def. of eSolutions Manual - Powered by Cognero Page 2 9-4 Inscribed Angles

intercepted arc) 9. x 4. (Inscribed of same arc are .) 5. (ASA)

STRUCTURE Find each value. 8.

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 SOLUTION: 180. 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 is 180. So, ANSWER: 25

10. and

Therefore,

ANSWER: 62

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 .

ANSWER: 122; 80

eSolutions Manual - Powered by Cognero Page 3 9-4 Inscribed Angles

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,

ANSWER: If an angle is inscribed in a circle, then the measure 162 of the angle equals one half the measure of its intercepted arc. Therefore, 12. ANSWER: 70

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: If an angle is inscribed in a circle, then the measure ANSWER: of the angle equals one half the measure of its 46 intercepted arc. Therefore,

ANSWER: 48

eSolutions Manual - Powered by Cognero Page 4 9-4 Inscribed Angles

15. ALGEBRA Find each measure. 17.

SOLUTION:

Here, SOLUTION: The arc is a semicircle. So, 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, ANSWER: Therefore, 32 ANSWER: 140 18.

16.

SOLUTION: If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. SOLUTION: So, Here, The arc is a semicircle. So, Therefore, Then, If an angle is inscribed in a circle, then the measure ANSWER: of the angle equals one half the measure of its 34 intercepted arc. Therefore,

ANSWER: 66

eSolutions Manual - Powered by Cognero Page 5 9-4 Inscribed Angles

19. PROOF Write the specified type of proof. 21. paragraph proof Given:

Prove:

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 = SOLUTION: m∠B. Proof: Given means that

Therefore, m∠A = 5(4) or 20. . Since and

ANSWER: , the equation becomes 20 . Multiplying each side of the 20. equation by 2 results in .

ANSWER:

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 ∠C and ∠D both intercept arc AB, m∠C = m∠D. equation by 2 results in .

Therefore, m∠C = 5(10) – 3 or 47.

ANSWER: 47

eSolutions Manual - Powered by Cognero Page 6 9-4 Inscribed Angles

22. two-column proof ALGEBRA Find each value. Given: Prove: 23. x

SOLUTION: 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 is 4. (AA Similarity) 180. So,

ANSWER: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc ANSWER: are .) 30 3. (Vertical are .) 4. (AA Similarity) 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 is 180. So,

Therefore,

ANSWER: 60

eSolutions Manual - Powered by Cognero Page 7 9-4 Inscribed Angles

25. x STRUCTURE Find each measure. 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 So, opposite angles are supplementary. The sum of the measures of the angles of a triangle is 180. So, Therefore, ANSWER: 135

ANSWER: 28. 12.75

26.

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 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 is ANSWER: 180. So, 80

Therefore,

ANSWER: 51.75

eSolutions Manual - Powered by Cognero Page 8 9-4 Inscribed Angles

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

, so or Therefore, 180. This makes ∠C and ∠A supplementary. ANSWER: Because the sum of the measures of the interior 106 angles of a quadrilateral is 360, . But 30. , so , making them supplementary also.

ANSWER: Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. and are supplementary.

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, ANSWER: . Since by Theorem 10.6 93 and

31. PROOF Write a paragraph proof for Theorem 9.9. , but SOLUTION: , so Given: Quadrilateral ABCD is inscribed in . Prove: and are supplementary. or 180. This makes ∠C and ∠A supplementary. and are supplementary. Because the sum of the measures of the interior angles of a quadrilateral is 360, eSolutions Manual - Powered by Cognero Page 9 9-4 Inscribed Angles

. But 33. , so , making them supplementary also. SOLUTION: Since all the sides of the stop sign are congruent chords of the circle, all the corresponding arcs are SIGNS: A stop sign in the shape of a regular congruent. There are eight adjacent arcs that make octagon is inscribed in a circle. Find each up the circle, so their sum is 360. Thus, the measure 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.

32. Therefore, the measure of ∠RLQ is 22.5. SOLUTION:

Since all the sides of the stop sign are congruent chords of the circle, the corresponding arcs are ANSWER: congruent. There are eight adjacent arcs that make 22.5 up the circle, so their sum is 360. Thus, the measure 34. 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, ANSWER: The sum of the measures of the central angles of a 135 circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc measures Then,

Therefore,

ANSWER: 112.5

eSolutions Manual - Powered by Cognero Page 10 9-4 Inscribed Angles

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 a circle with no interior points in common is 360. Here, circle with no interior points in common is 360. Here, 360 is divided into 8 equal arcs and each arc 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 Therefore, intercepted arc. So, the measure of each inscribed angle is 36. ANSWER: b. Here, 360 is divided into 6 equal arcs and each arc 135 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 of b. 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. ANSWER: a. 36 b. 60

eSolutions Manual - Powered by Cognero Page 11 9-4 Inscribed Angles

measure of the intercepted arc (Case 1)). c. or about 25.7 3. m∠ABC = m(arc AD) + m(arc DC) d. 45 (Substitution) PROOF Write a two-column proof for each case 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) of Theorem 9.6. 37. Case 2 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc Given: P lies inside . is a diameter. Addition Postulate) 6. m∠ABC = m(arc AC) (Substitution) Prove:

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

SOLUTION: Proof: Statements (Reasons) 1. m∠ABC = m∠ABD + m∠DBC ( Addition Postulate) 2. m∠ABD = m(arc AD) SOLUTION: m∠DBC = m(arc DC) (The measure of an Proof: inscribed whose side is a diameter is half the Statements (Reasons) ∠ ∠ ∠ measure of the intercepted arc (Case 1)). 1. m ABC = m DBC – m DBA ( Addition Postulate, Subtraction Property of Equality) 3. m∠ABC = m(arc AD) + m(arc DC) 2. m∠DBC = m(arc DC) (Substitution) ∠ 4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor) m DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the 5. m(arc AD) + m(arc DC) = m(arc AC) (Arc measure of the intercepted arc (Case 1)). Addition Postulate) 3. m∠ABC = m(arc DC) – m(arc DA) 6. m∠ABC = m(arc AC) (Substitution) (Substitution) 4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor) ANSWER: 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 ANSWER: eSolutions Manual - Powered by Cognero Page 12 9-4 Inscribed Angles

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) ANSWER: m∠DBA = m(arc DA) (The measure of an Given: and are inscribed; inscribed whose side is a diameter is half the Prove: measure of the intercepted arc (Case 1)). Proof: 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) Statements (Reasons) 1. and are inscribed; PROOF Write the specified proof for each (Given) theorem. 39. Theorem 9.7, two-column proof 2. ; (Measure

SOLUTION: of an inscribed = half measure of intercepted arc.) 3. (Def. of arcs) Given: and are inscribed; Prove: 4. (Mult. Prop. of Equality) Proof: 5. (Substitution) 6. (Def. of )

40. Theorem 9.8, paragraph proof

SOLUTION:

Statements (Reasons) 1. and are inscribed; (Given) 2. ; (Measure of an inscribed = half measure of intercepted arc.) Part I: Given: is a semicircle. Prove: is a right angle. 3. (Def. of arcs) Proof: Since is a semicircle, then eSolutions Manual - Powered by Cognero Page 13 9-4 Inscribed Angles

. Since is an inscribed angle, 41. MULTIPLE REPRESENTATIONS In this then or 90. So, by definition, problem, you will investigate the relationship between the arcs of a circle that are cut by two parallel is a right angle. chords. a. Geometric Use a compass to draw a circle with Part II: Given: is a right angle. parallel chords and . Connect points A and D Prove: is a semicircle. by drawing segment .

b. Numerical Use a protractor to find and Proof: Since is an inscribed angle, then . Then determine and . What is and by the Multiplication true about these arcs? Explain. c. Verbal Draw another circle and repeat parts a and Property of Equality, m = 2m . Because b. Make a conjecture about arcs of a circle that are is a right angle, m = 90. Then m cut by two parallel chords. = 2(90) or 180. So by definition, is a d. Analytical Use your conjecture to find and semicircle. in the figure at the right. Verify by using ANSWER: 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 m∠A = 15, m∠D = 15, = 30, and = 30. = 2(90) or 180. So by definition, is a The arcs are congruent. semicircle. eSolutions Manual - Powered by Cognero Page 14 9-4 Inscribed Angles

In a circle, two parallel chords cut congruent arcs. 43. rectangle d. In a circle, two parallel chords cut congruent arcs. SOLUTION: So, = . 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. = 4(16) + 6 or 70 ANSWER: = 6(16) – 26 or 70 Therefore, the measure of arcs QS and PR are each Always; rectangles have right angles at each vertex, 70. therefore each pair of opposite angles will be supplementary and inscribed in a circle. ANSWER: 44. parallelogram a. 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. b. Sample answer: , ; , ; The arcs are congruent ANSWER: because they have equal measures. Sometimes; a parallelogram can be inscribed in a c. Sample answer: In a circle, two parallel chords cut circle as long as it is a rectangle. congruent arcs. See students’ work. d. 70; 70 45. rhombus

CONSTRUCT ARGUMENTS Determine SOLUTION: whether the quadrilateral can always, sometimes, The opposite angles of a rhombus are always or never be inscribed in a circle. Explain your congruent. They will only be supplementary when reasoning. they are right angles. So, a rhombus can be inscribed 42. square in a circle as long as it is a square. Since the opposite 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 and inscribed in a circle. Therefore, the statement is ANSWER: always true. Sometimes; a rhombus can be inscribed in a circle as long as it is a square. Since the opposite angles of ANSWER: rhombi that are not squares are not supplementary, Always; squares have right angles at each vertex, they cannot be inscribed in a circle. therefore each pair of opposite angles will be supplementary and inscribed in a circle.

eSolutions Manual - Powered by Cognero Page 15 9-4 Inscribed Angles

46. kite 47. CHALLENGE A square is inscribed in a circle. What is the ratio of the area of the circle to the area SOLUTION: of the square? Exactly one pair of opposite angles of a kite are congruent. To be supplementary, they must each be a SOLUTION: right angle. If one pair of opposite angles for a A square with side s is inscribed in a circle with quadrilateral is supplementary, the other pair must radius r. 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.

ANSWER: Sometimes; as long as the angles that compose the Using the Pythagorean Theorem, pair of congruent opposite angles are right angles.

The area of a square of side s is A = s2 and the area of a circle of radius r is A = .

Therefore, the ratio of the area of the circle to the area of the inscribed square is

ANSWER:

eSolutions Manual - Powered by Cognero Page 16 9-4 Inscribed Angles

48. WRITING IN MATH A right 49. OPEN-ENDED Find and sketch a real-world logo triangle is inscribed in a circle. If the radius of the with an inscribed polygon. circle is given, explain how to find the lengths of the SOLUTION: 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 angle. Therefore, the hypotenuse is a diameter and has a length of 2r. Use trigonometry to find the length of the equal legs. ANSWER: 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?

SOLUTION: Therefore, the length of each leg of the 45-45-90 An inscribed angle has its vertex on the circle. A triangle can be found by multiplying the radius of the central angle has its vertex at the center of the circle. circle in which it is inscribed by . If a central angle intercepts arc AB, then the measure of the central angle is equal to m(arc AB). If an ANSWER: inscribed angle also intercepts arc AB, then the Sample answer: According to theorem 10.8, an measure of the inscribed angle is equal to m(arc inscribed angle of a triangle intercepts a diameter if AB). So, if an inscribed angle and a central angle the angle is a right angle. Therefore, the hypotenuse intercept the same arc, then the measure of the is a diameter and has a length of 2r. Using inscribed angle is one-half the measure of the central trigonometry, each leg = sin or . angle.

ANSWER: An inscribed angle has its vertex on the circle. A central angle has its vertex at the center of the circle. 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.

eSolutions Manual - Powered by Cognero Page 17 9-4 Inscribed Angles

51. Quadrilateral ABCD is inscribed in ⊙K. What is the 52. What is ? measure of ∠A in degrees?

A 96 B 84 SOLUTION: C 48 The sum of the measures of the opposite angles in an D 42 inscribed quadrilateral is 180. E 24

SOLUTION: A triangle inscribed in a semicircle is a right triangle, with the diameter as its hypotenuse. Therefore, ∠P and ∠R are complementary angles. Set the sum of The measure of ∠A = 7(14) + 10 = 108. their measures equal to 90 and solve for x.

ANSWER: 108

The measure of . = 2(m∠P) = 2(42) = 84. The correct choice is B.

ANSWER: B

eSolutions Manual - Powered by Cognero Page 18 9-4 Inscribed Angles

53. Quadrilateral JKLM is inscribed in ⊙C, as shown. a. Which statements are true? Check all that apply. A m∠GFH = 2m∠FEH B m∠GEH = 2m∠GFH C m∠GEH = m∠GFH D m∠GEH = 11.2 E The measure of an inscribed angle is twice the measure of its intercepted arc. Which of the following statements must be true? F The measure of an inscribed angle is half the I. Quadrilateral JKLM contains at least two measure of its intercepted arc. right angles. G The measure of an inscribed angle is equal to the measure of its intercepted arc. II. is a semicircle. H All angles inscribed by the same arc have the III. ∠L and ∠M are supplementary. same measure.

A I only b. The solution for x below contains an error. Choose B II only the step where the error occurs. C II and III only A m∠GEH = m∠GFH D I and II only B 4x + 6 = 5x – 6 SOLUTION: C 9x = 12 Option I is true because JKLM is inscribed in a circle D x = 1.33 and opposite angles are supplementary. Since ∠J is a right angle, then ∠L must also be a right angle. c. What is the measure of ∠GFH? Option II is also true. Triangle JKM would be a right A 22° triangle. Any right triangle inscribed in a circle has B 27° the diameter as its hypotenuse. Therefore, is a C 54° semicircle. D 108° Option III is not necessarily true, as opposite (not adjacent) angles of a quadrilateral inscribed in a circle d. Can you prove △EJG ≅ △FJH? Explain. are supplementary. e. Complete the following proof that shows △EJG ∼ Hence, I and II only are true. The correct choice is △FJH. D.

ANSWER: D (delete anno text) 54. MULTI-STEP Look at the figure. SOLUTION: a. Choices C, F, and H are correct. b. Choice C is incorrect--should be x = 12. c. Using x = 12, 5x – 6 = 5(12) – 6 = 54, so choice C is correct. d. No; you can prove that the triangles are similar because the corresponding angle measures are equal, but the side lengths are not. e.

eSolutions Manual - Powered by Cognero Page 19 9-4 Inscribed Angles

ANSWER: a. C, F, and H b. C c. C d. No; you can prove that the triangles are similar because the corresponding angle measures are equal, but the side lengths are not. e. ∠EGF ≅ ∠FHE

eSolutions Manual - Powered by Cognero Page 20