ALGEBRA Quadrilateral ABCD Is a Rhombus. Find Each Value Or Measure. 1. If , Find . SOLUTION

6-5 Special Parallelograms: Rhombi, Squares

ALGEBRA Quadrilateral ABCD is a rhombus. 3. PROOF Write a two-column proof to prove that if Find each value or measure. ABCD is a rhombus with diagonal .

1. If , find .

SOLUTION: A rhombus is a parallelogram with all four sides SOLUTION: congruent. So, Then, is an Given: ABCD is a rhombus with diagonal . isosceles triangle. Therefore, Prove: If a parallelogram is a rhombus, then each diagonal Proof: bisects a pair of opposite angles. So, Statements(Reasons) . 1. ABCD is a rhombus with diagonal . (Given) 2. (Diag. of rhombus bisects ) Therefore, 3. (Refl. Prop.) ANSWER: 4. (Def. of rhombus) 32 5. (SAS) 6. (CPCTC) 2. If AB = 2x + 3 and BC = x + 7, find CD. ANSWER: SOLUTION: Given: ABCD is a rhombus with diagonal . A rhombus is a parallelogram with all four sides Prove: congruent. So, Proof: Statements(Reasons) So, AB = 2(4) + 3 = 11. 1. ABCD is a rhombus with diagonal . (Given) 2. (Diag. of rhombus bisects ) CD is congruent to AB, so CD = 11. 3. (Refl. Prop.) 4. (Def. of rhombus) ANSWER: 5. (SAS) 11 6. (CPCTC)

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4. GAMES The checkerboard below is made up of 64 COORDINATE GEOMETRY Given each set congruent black and red squares. Use this of vertices, determine whether QRST is a information to prove that the board itself is a square. rhombus, a rectangle, or a square. List all that apply. Explain. 5. Q(1, 2), R(–2, –1), S(1, –4), T(4, –1)

SOLUTION: First, graph the quadrilateral.

SOLUTION: Because each side of the board is 8 squares in length and each of the squares is congruent, the lengths of all four sides of the board are equal. Because we know that each of the four quadrilaterals that form the corners of the board are squares, we know that the measure of the angle of each vertex of the board is 90. Therefore, the board

is a square. If the diagonals of the parallelogram are congruent, ANSWER: then it is a rectangle. Use the Distance Formula to Sample answer: Because each side of the board is 8 find the lengths of the diagonals. squares in length and each of the squares is congruent, the lengths of all four sides of the board are equal. Because we know that each of the four quadrilaterals that form the corners of the board are squares, we know that the measure of the angle of each vertex of the board is 90. Therefore, the board So, the parallelogram is a rectangle. Check whether is a square. the two diagonals are perpendicular.

QS has a slope of . RT has a slope of . These slopes are opposite reciprocals.

The diagonals are perpendicular. So, it is a rhombus.

Since the diagonals are both congruent and perpendicular to each other, the parallelogram is a rectangle, a rhombus, and a square.

ANSWER: Rectangle, rhombus, square; consecutive sides are , all sides are .

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6. Q(–2, –1), R(–1, 2), S(4, 1), T(3, –2) ALGEBRA Quadrilateral ABCD is a rhombus. Find each value or measure. SOLUTION:

First, graph the quadrilateral.

7. If AB = 14, find BC.

SOLUTION: A rhombus is a parallelogram with all four sides congruent. So,

Therefore, BC = 14. If the diagonals of the parallelogram are congruent, then it is a rectangle. Use the Distance Formula to ANSWER: find the lengths of the diagonals. 14

8. If , find .

SOLUTION: The diagonals are not congruent. So, the A rhombus is a parallelogram with all four sides parallelogram is not a rectangle. Check whether the congruent. So, Then, is an two diagonals are perpendicular. isosceles triangle. Therefore, If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles. So,

Therefore,

ANSWER: The diagonals are not perpendicular. So, it is not a 27 rhombus either. Since it is neither a rectangle nor a 9. If AP = 3x – 1 and PC = x + 9, find AC. rhombus, it cannot be a square either. SOLUTION: ANSWER: The diagonals of a rhombus bisect each other. None; the diagonals are not congruent or 3x – 1 = x + 9 perpendicular. 2x = 10 x = 5 Therefore, AC = 2(5 + 9) = 28.

ANSWER: 28

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10. If DB = 2x – 4 and PB = 2x – 9, find PD.

SOLUTION: The diagonals of a rhombus bisect each other. So,

SOLUTION: Therefore, PD = PB = 2(7) – 9 = 5. Given:

ANSWER: Prove: WXYZ is a rhombus. 5 11. If , find .

SOLUTION: In a rhombus, consecutive interior angles are supplementary. Proof: Statements(Reasons) 1. (Given) 2. WXYZ is a . (Both pairs of opp. sides are .) Each pair of opposite angles of a rhombus is 3. WXYZ is a rhombus. (If one pair of consecutive congruent. So, sides of a are , the is a rhombus.)

ANSWER: ANSWER: 95 Given: 12. If , find x. Prove: WXYZ is a rhombus. SOLUTION: The diagonals of a rhombus are perpendicular to each other.

ANSWER: Proof: Statements(Reasons) 35 1. (Given) CONSTRUCT ARGUMENTS Write a two- 2. WXYZ is a . (Both pairs of opp. sides are .) column proof. 3. WXYZ is a rhombus. (If one pair of consecutive 13. Given: sides of a are , the is a rhombus.)

Prove: WXYZ is a rhombus. 14. Given: QRST is a parallelogram. eSolutions Manual - Powered by Cognero Page 4 6-5 Special Parallelograms: Rhombi, Squares

Prove: QRST is a square. Proof: Statements(Reasons) 1. QRST is a parallelogram; . (Given) 2. QRST is a rectangle. (If the diags of a are , the is a rectangle.) 3. is a right angle. (Def of rt. ) 4. (Def. of perpendicular) SOLUTION: 5. QRST is a rhombus. (If the diags of a Given: QRST is a parallelogram. are , is a rhombus.) 6. QRST is a square. (Thm. 6.2, if a quadrilateral is a Prove: QRST is a square. rectangle and a rhombus, then it is a square.)

15. Given: ABCD is a parallelogram.

Prove: ABCD is a rhombus.

Proof: Statements(Reasons) 1. QRST is a parallelogram; . (Given) 2. QRST is a rectangle. (If the diags of a are , SOLUTION: the is a rectangle.) Given: ABCD is a parallelogram. 3. is a right angle. (Def of rt. )

4. (Def. of perpendicular) Prove: ABCD is a rhombus. 5. QRST is a rhombus. (If the diags of a are , is a rhombus.) 6. QRST is a square. (Thm. 6.2, if a quadrilateral is a rectangle and a rhombus, then it is a square.)

ANSWER: Given: QRST is a parallelogram.

Prove: QRST is a square. Proof: Statements (Reasons) 1. .ABCD is a parallelogram, (Given) 2. (CPCTC) 3. bisects (Def. of ∠ bisector) eSolutions Manual - Powered by Cognero Page 5 6-5 Special Parallelograms: Rhombi, Squares

4. (CPCTC) Prove: ABHF is a rhombus. 5. bisects (Def. of ∠ bisector) 6. ABCD is a rhombus (If one diagonal of a parallelogram bisects a pair of opp. ∠s, then the parallelogram is a rhombus)

ANSWER: Given: ABCD is a parallelogram.

Proof: Prove: ABCD is a rhombus. Statements (Reasons)

1. ACDH and BCDF are parallelograms; . (Given) 2. (Def. of ) 3. (Trans. Prop) 4. (Def. of ) 5. AC = HD (Def. of segs.) 6. AC = AB + BC, HD = HF + FD (Seg. Add. Proof: Post.) Statements (Reasons) 7. AC – HD = AB + BC – HF – FD (Subt. Prop.) 1. .ABCD is a parallelogram, 8. AB = HF (Subst.) (Given) 9. (Def. of segs.) 2. (CPCTC) 10. (Subst.) 3. bisects (Def. of ∠ bisector) 11. ABFH is a rhombus. (Def. of rhombus) 4. (CPCTC) ANSWER: 5. bisects (Def. of ∠ bisector) Given: ACDH and BCDF are parallelograms; 6. ABCD is a rhombus (If one diagonal of a parallelogram bisects a pair of opp. ∠s, then the . parallelogram is a rhombus) Prove: ABHF is a rhombus.

16. Given: ACDH and BCDF are parallelograms; . Prove: ABFH is a rhombus.

Proof: Statements (Reasons) 1. ACDH and BCDF are parallelograms; . (Given) SOLUTION: 2. (Def. of ) Given: ACDH and BCDF are parallelograms; 3. (Trans. Prop) . 4. (Def. of ) eSolutions Manual - Powered by Cognero Page 6 6-5 Special Parallelograms: Rhombi, Squares

5. AC = HD (Def. of segs.) 18. MODELING A landscaper has staked out the area 6. AC = AB + BC, HD = HF + FD (Seg. Add. for a square garden as shown. She has confirmed Post.) that each side of the quadrilateral formed by the 7. AC – HD = AB + BC – HF – FD (Subt. Prop.) stakes is congruent and that the diagonals are 8. AB = HF (Subst.) perpendicular. Is this information enough for the 9. (Def. of segs.) landscaper to be sure that the garden is a square? 10. (Subst.) Explain your reasoning. 11. ABFH is a rhombus. (Def. of rhombus)

17. ROADWAYS Main Street and High Street intersect as shown in the diagram. Each of the crosswalks is the same length. Classify the quadrilateral formed by the crosswalks. Explain your reasoning.

SOLUTION: Because the four sides of the quadrilateral are congruent and the diagonals are perpendicular, the figure is either a square or a rhombus. To be sure that the garden is a square, she also needs to confirm that the diagonals are congruent. SOLUTION: ANSWER: The measure of the angle formed between the two streets is 29, and vertical angles are congruent, so the no; Sample answer: Because the four sides of the measure of one angle of the quadrilateral is 29. quadrilateral are congruent and the diagonals are Because the crosswalks are the same length, the perpendicular, the figure is either a square or a sides of the quadrilateral are congruent. Therefore, rhombus. To be sure that the garden is a square, she they form a rhombus. also needs to confirm that the diagonals are congruent. ANSWER: rhombus; Sample answer: The measure of the angle formed between the two streets is 29, and vertical angles are congruent, so the measure of one angle of the quadrilateral is 29. Because the crosswalks are the same length, the sides of the quadrilateral are congruent. Therefore, they form a rhombus.

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COORDINATE GEOMETRY Given each set 20. J(–3, –2), K(2, –2), L(5, 2), M(0, 2) of vertices, determine whether JKLM is a SOLUTION: rhombus, a rectangle, or a square. List all that apply. Explain. First, graph the quadrilateral. 19. J(–4, –1), K(1, –1), L(4, 3), M(–1, 3)

SOLUTION: First, graph the quadrilateral.

If the diagonals of the parallelogram are congruent, then it is a rectangle. Use the Distance Formula to find the lengths of the diagonals.

The diagonals are not congruent. So, the parallelogram is not a rectangle. Check whether the two diagonals are perpendicular.

The diagonals are perpendicular. So, it is a rhombus.

ANSWER: Rhombus; the diagonals are . The diagonals are perpendicular. So, it is a rhombus.

ANSWER: Rhombus; the diagonals are .

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21. J(–2, –1), K(–4, 3), L(1, 5), M(3, 1) 22. J(–1, 1), K(4, 1), L(4, 6), M(–1, 6)

SOLUTION: SOLUTION: First, graph the quadrilateral. First, graph the quadrilateral.

If the diagonals of the parallelogram are congruent, If the diagonals of the parallelogram are congruent, then it is a rectangle. Use the Distance Formula to then it is a rectangle. Use the Distance Formula to find the lengths of the diagonals. find the lengths of the diagonals.

The diagonals are not congruent. So, the The diagonals are congruent. So, the parallelogram is parallelogram is not a rectangle. Check whether the a rectangle. Check whether the two diagonals are two diagonals are perpendicular. perpendicular.

The diagonals are perpendicular. So, it is a rhombus. The diagonals are not perpendicular. So, it is not a rhombus either. Because it is neither a rectangle nor Since the diagonals are both congruent and a rhombus, it is not a square either. perpendicular to each other, the parallelogram is a ANSWER: rectangle, a rhombus, and a square. None; the diagonals are not . ANSWER: Square, rectangle, rhombus; all sides are .

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ABCD is a rhombus. If PB = 12, AB = 15, and 26. , find each measure. SOLUTION:

The diagonals are perpendicular to each other. So, in the right triangle PAB,

All four sides of a rhombus are congruent. So, is an isosceles triangle. Then, 23. AP

SOLUTION: ANSWER: The diagonals of a rhombus are perpendicular to each 66 other. So, by the Pythagorean Theorem, AP2 = AB2 – WXYZ is a square. If WT = 3, find each measure. PB2.

ANSWER: 9

27. ZX 24. CP SOLUTION: SOLUTION: The diagonals of a square are congruent and bisect All four sides of a rhombus are congruent and the each other. diagonals are perpendicular to each other. So, ZX = WY = 2(WT) = 6. So, by the Pythagorean Theorem, CP2 = BC2 – PB2. ANSWER: BC = AB. Substitute AB for BC. 6

28. XY

SOLUTION: ANSWER: The diagonals of a square are congruent and bisect 9 each other at right angles. So, YT = XT = WT = 3. 25. By the Pythagorean Theorem, XY2 = YT2 + XT2. SOLUTION: All four sides of a rhombus are congruent. So, is an isosceles triangle. Then,

ANSWER: ANSWER: 24

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29.

SOLUTION: The diagonals of a square are perpendicular to each other. So,

ANSWER: 90 32.

30. SOLUTION: The two pairs of opposite angles are congruent and SOLUTION: the adjacent sides are congruent. Therefore, the In a square, each diagonal bisects a pair of opposite quadrilateral is a rhombus.

angles. So, ANSWER: rhombus ANSWER: 45

Classify each quadrilateral.

33.

SOLUTION: 31. The two pairs of opposite sides are congruent and SOLUTION: one of the angles is a right angle. Therefore, the The diagonals are congruent and bisect each other at quadrilateral is a rectangle. right angles. Therefore, the quadrilateral is a square. ANSWER: ANSWER: rectangle square PROOF Write a paragraph proof. 34. Theorem 6.16

SOLUTION: Given: ABCD is a rhombus. Prove: Each diagonal bisects a pair of opposite angles.

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Given: ABCD is a parallelogram; . Prove: ABCD is a rhombus.

Proof: We are given that ABCD is a rhombus. By definition of rhombus, ABCD is a parallelogram. Opposite angles of a parallelogram are congruent, so . Proof: We are given that ABCD is a parallelogram. because all sides of a rhombus The diagonals of a parallelogram bisect each other, are congruent. by SAS. so because congruence of by CPCTC. segments is reflexive. We are also given that by SAS. by CPCTC. By . Thus, are right angles definition of angle bisector, each diagonal bisects a by the definition of perpendicular lines. Then pair of opposite angles. because all right angles are congruent. Therefore, by SAS. ANSWER: by CPCTC. Opposite sides of Given: ABCD is a rhombus. parallelograms are congruent, so Prove: Each diagonal bisects a pair of opposite angles. . Then since congruence of segments is transitive, . All four sides of ABCD are congruent, so ABCD is a rhombus by definition.

ANSWER: Given: ABCD is a parallelogram; . Prove: ABCD is a rhombus.

Proof: We are given that ABCD is a rhombus. By definition of rhombus, ABCD is a parallelogram. Opposite angles of a parallelogram are congruent, so . because all sides of a rhombus are congruent. by SAS.

by CPCTC. Proof: We are given that ABCD is a parallelogram. by SAS. by CPCTC. By The diagonals of a parallelogram bisect each other, definition of angle bisector, each diagonal bisects a so because congruence of pair of opposite angles. segments is reflexive. We are also given that . Thus, are right angles 35. Theorem 6.17 by the definition of perpendicular lines. Then SOLUTION: because all right angles are eSolutions Manual - Powered by Cognero Page 12 6-5 Special Parallelograms: Rhombi, Squares

congruent. Therefore, by SAS. Given: ABCD is a parallelogram; diagonal by CPCTC. Opposite sides of bisects . parallelograms are congruent, so Prove: ABCD is a rhombus. . Then since congruence of segments is transitive, . All four sides of ABCD are congruent, so ABCD is a rhombus by definition.

36. Theorem 6.18

SOLUTION:

If a diagonal of a parallelogram bisects an angle of a Proof: It is given that ABCD is a parallelogram. parallelogram, then the parallelogram is a rhombus. Since opposite sides of a parallelogram are parallel,

. By definition, are alternate Given: ABCD is a parallelogram; diagonal interior angles of parallel sides . bisects . Because alternate interior angles are congruent, Prove: ABCD is a rhombus. . Congruence of angles is symmetric,

therefore . It is given that bisects , so by definition. By the Transitive Property, . The sides opposite congruent angles in a triangle are congruent, therefore, . So, because a pair of consecutive sides of the parallelogram are congruent, Proof: It is given that ABCD is a parallelogram. ABCD is a rhombus. Since opposite sides of a parallelogram are parallel, . By definition, are alternate interior angles of parallel sides . Because alternate interior angles are congruent, . Congruence of angles is symmetric, therefore . It is given that bisects , so by definition. By the Transitive Property, . The sides opposite congruent angles in a triangle are congruent, therefore, . So, because a pair of consecutive sides of the parallelogram are congruent, ABCD is a rhombus.

ANSWER: If a diagonal of a parallelogram bisects an angle of a parallelogram, then the parallelogram is a rhombus.

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37. Theorem 6.19 38. Theorem 6.20

SOLUTION: SOLUTION: Given: ABCD is a parallelogram; . Given: ABCD is a rectangle and a rhombus. Prove: ABCD is a rhombus. Prove: ABCD is a square.

Proof: Opposite sides of a parallelogram are Proof: We know that ABCD is a rectangle and a congruent, so . We are given rhombus. ABCD is a parallelogram, since all that . So, by the Transitive Property, rectangles and rhombi are parallelograms. By the . So, . Thus, ABCD definition of a rectangle, are is a rhombus by definition. right angles. By the definition of a rhombus, all of the sides are congruent. Therefore, ABCD is a square ANSWER: since ABCD is a parallelogram with all four sides Given: ABCD is a parallelogram; . congruent and all the angles are right. Prove: ABCD is a rhombus. ANSWER: Given: ABCD is a rectangle and a rhombus. Prove: ABCD is a square.

Proof: Opposite sides of a parallelogram are congruent, so . We are given that . So, by the Transitive Property, . So, . Thus, ABCD Proof: We know that ABCD is a rectangle and a is a rhombus by definition. rhombus. ABCD is a parallelogram, since all rectangles and rhombi are parallelograms. By the definition of a rectangle, are right angles. By the definition of a rhombus, all of the sides are congruent. Therefore, ABCD is a square since ABCD is a parallelogram with all four sides congruent and all the angles are right.

CONSTRUCTION Use diagonals to construct each figure. Justify each construction. eSolutions Manual - Powered by Cognero Page 14 6-5 Special Parallelograms: Rhombi, Squares

39. rhombus 40. square

SOLUTION: SOLUTION: Construct the perpendicular bisector of a line Construct the perpendicular bisector of a segment. segment. Place the compass at the midpoint of the Adjust the compass setting to equal the distance from segment. Use the same compass setting to locate a the midpoint of the segment to one of its endpoints. point on the perpendicular bisector above and below Place the compass at the midpoint of the segment. the segment. Connect the endpoints of the segments Draw arcs that intersect the perpendicular bisector with the two points on the perpendicular bisector to above and below the segment. Connect the two form a quadrilateral. points of intersection with the endpoints of the segment to form a quadrilateral.

The diagonals bisect each other, so the quadrilateral The diagonals bisect each other, so the quadrilateral is a parallelogram. Because the diagonals of the is a parallelogram. Because the diagonals of the parallelogram are perpendicular to each other, the parallelogram are congruent and perpendicular, the parallelogram is a rhombus. parallelogram is a square.

ANSWER: ANSWER:

Sample answer: The diagonals bisect each other, so Sample answer: The diagonals bisect each other, so the quadrilateral is a parallelogram. Because the the quadrilateral is a parallelogram. Because the diagonals of the parallelogram are congruent and diagonals of the parallelogram are perpendicular to perpendicular, the parallelogram is a square. each other, the parallelogram is a rhombus. PROOF Write a coordinate proof of each statement. eSolutions Manual - Powered by Cognero Page 15 6-5 Special Parallelograms: Rhombi, Squares

41. The diagonals of a square are perpendicular. slope of SOLUTION: Begin by positioning square ABCD on a coordinate slope of plane. Place vertex A at the origin. Let the length of The slope of is the negative reciprocal of the the bases be a units. Then the rest of the vertices are B(a, 0), C(a, a), and D(0, a). You need to walk slope of , so they are perpendicular. through the proof step by step. Look over what you are given and what you need to prove. Here, you are 42. The segments joining the midpoints of the sides of a given ABCD is a square and you need to prove that rectangle form a rhombus. . Use the properties that you have learned SOLUTION: about squares to walk through the proof. Begin by positioning rectangle ABCD on a coordinate plane. Place vertex A at the origin. Let the length of the bases be a units and the height be b units. Then the rest of the vertices are B(a, 0), C(a, b), and D(0, b). You need to walk through the proof step by step. Look over what you are given and what you need to prove. Here, you are given ABCD is a rectangle and Q, R, S, and T are midpoints of their respective sides and you need to prove that QRST is a rhombus. Use

the properties that you have learned about rhombi to Given: ABCD is a square. walk through the proof. Prove: Proof: slope of

slope of

The slope of is the negative reciprocal of the slope of , so they are perpendicular.

ANSWER: Given: ABCD is a rectangle. Q, R, S, and T are Given: ABCD is a square. midpoints of their respective sides. Prove: Prove: QRST is a rhombus. Proof:

Midpoint Q is .

Midpoint R is .

Midpoint S is .

Midpoint T is . Proof:

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QR = RS = ST = QT QR = RS = ST = QT

QRST is a rhombus. QRST is a rhombus.

ANSWER: 43. DESIGN The tile pattern below consists of regular Given: ABCD is a rectangle. Q, R, S, and T are octagons and quadrilaterals. Classify the midpoints of their respective sides. quadrilaterals in the pattern and explain your Prove: QRST is a rhombus. reasoning.

SOLUTION: Proof: In order to classify the quadrilaterals we need Midpoint Q is . information about the interior angles and the sides. It's given that each quadrilateral is formed by 4 regular octagons. We can use what we know about Midpoint R is . the exterior angles of a regular octagon as well as the sides of a regular octagon to determine which type of Midpoint S is . quadrilateral is in the pattern.

Midpoint T is . The quadrilaterals in the pattern are squares. Because the octagons are regular each side is congruent, and the quadrilaterals share common sides with the octagon, so the quadrilaterals are either rhombuses or squares. The vertices of the quadrilaterals are formed by the exterior angles of the eSolutions Manual - Powered by Cognero Page 17 6-5 Special Parallelograms: Rhombi, Squares

sides of the octagons adjacent to the vertices. The 44. REPAIR The window pane shown needs to be sum of the measures of the exterior angles of a replaced. What are the dimensions of the polygon is always 360 and since a regular octagon replacement pane? has 8 congruent exterior angles, each one measures 45. As shown in the diagram, each angle of the quadrilaterals in the pattern measures 45 + 45 or 90. Therefore, the quadrilateral is a square.

SOLUTION: The window pane is in the shape of a square. The diagonal of a square is the hypotenuse of a right triangle with two consecutive sides of the square as ANSWER: its legs. Let x be the length of each side of the Squares; sample answer: Because the octagons are square. So, by the Pythagorean Theorem, regular each side is congruent, and the quadrilaterals share common sides with the octagon, so the quadrilaterals are either rhombuses or squares. The vertices of the quadrilaterals are formed by the exterior angles of the sides of the octagons adjacent to the vertices. The sum of the measures of the exterior angles of a polygon is always 360 and since a regular octagon has 8 congruent exterior angles, each Therefore, the length of each side of the square is one measures 45. As shown in the diagram, each about 15 inches. angle of the quadrilaterals in the pattern measures 45 + 45 or 90. Therefore, the quadrilateral is a square. ANSWER: square; 15 in.

45. MULTIPLE REPRESENTATIONS In this problem, you will explore the properties of kites, which are quadrilaterals with exactly two distinct pairs of adjacent congruent sides.

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a. Geometric Draw three kites with varying side lengths. Label one kite ABCD, one PQRS, and one WXYZ. Then draw the diagonals of each kite, labeling the point of intersection N for each kite. c. Sample answer: The shorter diagonal of a kite is b. Tabular Measure the distance from N to each bisected by the longer diagonal. vertex. Record your results in a table like the one 46. ERROR ANALYSIS In parallelogram PQRS, shown. . Lola thinks that the parallelogram is a square, and Xavier thinks that it is a rhombus. Is either of them correct? Explain your reasoning.

c. Verbal Make a conjecture about the diagonals of a kite.

SOLUTION: a. Sample answer: SOLUTION: Because they do not know that the sides of the parallelogram are congruent, only that the diagonals are congruent, they can only conclude that the parallelogram is a rectangle. So, neither of them is correct.

b. Use a ruler to measure each length in the table. ANSWER: Neither; Sample answer: Because they do not know that the sides of the parallelogram are congruent, only that the diagonals are congruent, they can only conclude that the parallelogram is a rectangle.

c. The shorter diagonal of a kite is bisected by the 47. REASONING Determine whether the statement is longer diagonal. true or false. Then write the converse, inverse, and contrapositive of the statement and determine the ANSWER: truth value of each. Explain your reasoning. a. Sample answer: If a quadrilateral is a square, then it is a rectangle.

SOLUTION: The statement "If a quadrilateral is a square, then it is a rectangle." is true. b. A rectangle is a quadrilateral with four right angles and a square is both a rectangle and a rhombus, so a eSolutions Manual - Powered by Cognero Page 19 6-5 Special Parallelograms: Rhombi, Squares

square is always a rectangle. ANSWER:

True; sample answer: A rectangle is a quadrilateral with four right angles and a square is both a rectangle and a rhombus, so a square is always a rectangle.

Converse: If a quadrilateral is a rectangle then it is a Converse: If a quadrilateral is a rectangle, then it is a square. False; sample answer: A rectangle is a square. The converse is false. quadrilateral with four right angles. It is not necessarily a rhombus, so it is not necessarily a A rectangle is a quadrilateral with four right angles. It square. is not necessarily a rhombus, so it is not necessarily a square. Inverse: If a quadrilateral is not a square, then it is not a rectangle. False; sample answer: A quadrilateral that has four right angles and two pairs of congruent sides is not a square, but it is a rectangle.

Contrapositive: If a quadrilateral is not a rectangle, then it is not a square. True; sample answer: If a Inverse: If a quadrilateral is not a square, then it is quadrilateral is not a rectangle, it is also not a square not a rectangle. The inverse is false. by definition.

48. CHALLENGE The area of square ABCD is 36 A quadrilateral that has four right angles and two square units and the area of is 20 square units. pairs of congruent sides is not a square, but it is a If , find the length of . rectangle.

Contrapositive: If a quadrilateral is not a rectangle, then it is not a square. The contrapositive is true. SOLUTION: Since the area of the square is 36 square units, the If a quadrilateral is not a rectangle, it is also not a length of each side of the square is 6 units. And, all square by definition. four angles of a square are right angles. So, by the Pythagorean Theorem,

The area of is 20 square units. So, eSolutions Manual - Powered by Cognero Page 20 6-5 Special Parallelograms: Rhombi, Squares

49. OPEN-ENDED Find the vertices of a square with diagonals that are contained in the lines y = x and y = −x + 6. Justify your reasoning.

SOLUTION: First, graph the lines y = x and y = –x + 6. There are 6 units from the origin to the y-intercept of y = –x + 6 and 6 units from the origin to the x-intercept of y = –x + 6. So, three of the vertices of a square will be at (0, 0), ((0, 6), and (6, 0). The point 6 units above (6, 0) is (6, 6).

Also, we have

So, by the HL postulate, (0, 0), (6, 0), (0, 6), (6, 6); the diagonals are perpendicular, and any four points on the lines AE = CF by CPCTC equidistant from the intersection of the diagonals will be the vertices of a square. Therefore, CF = 2.

ANSWER: ANSWER: 2 Sample answer: (0, 0), (6, 0), (0, 6), (6, 6); the diagonals are perpendicular, and any four points on the lines equidistant from the intersection of the diagonals will be the vertices of a square.

50. WRITING IN MATH Compare all of the properties of the following quadrilaterals: parallelograms, rectangles, rhombi, and squares.

SOLUTION: Parallelogram: Opposite sides of a parallelogram are parallel and congruent. Opposite angles of a parallelogram are congruent. The diagonals of a parallelogram bisect each other and each diagonal separates a parallelogram into two congruent triangles.

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ANSWER: Sample answer: Parallelogram: Opposite sides of a parallelogram are parallel and congruent. Opposite angles of a parallelogram are congruent. The diagonals of a parallelogram bisect each other and each diagonal Rectangle: A rectangle has all the properties of a separates a parallelogram into two congruent parallelogram. A rectangle has four right angles. The triangles. diagonals of a rectangle are congruent. Rectangle: A rectangle has all the properties of a parallelogram. A rectangle has four right angles. The diagonals of a rectangle are congruent.

Rhombus: A rhombus has all of the properties of a parallelogram. All sides of a rhombus are congruent. The diagonals of a rhombus are perpendicular and Rhombus: A rhombus has all of the properties of a bisect the angles of the rhombus. parallelogram. All sides of a rhombus are congruent. The diagonals of a rhombus are perpendicular and Square: A square has all of the properties of a bisect the angles of the rhombus. parallelogram. A square has all of the properties of a rectangle. A square has all of the properties of a rhombus.

Square: A square has all of the properties of a parallelogram. A square has all of the properties of a rectangle. A square has all of the properties of a rhombus.

eSolutions Manual - Powered by Cognero Page 22 6-5 Special Parallelograms: Rhombi, Squares

51. Julia is designing a pair of earrings. The figure shows one of the earrings. Julia knows that quadrilateral JKLM is a parallelogram and that m∠KLN = 54.

ABCD is a rhombus. If PB = 15, AB = 25 and m∠ABD = 33, find each measure. 52. CP = ?

A 16 B 18 C 20 D 24

SOLUTION: What should the measure of ∠LKN be in order for Use the Pythagorean Theorem. the earring to be a rhombus? PD2 + CP2 = DC2 A 36 B 54 Using what we know about rhombuses, we can use C 90 substitution. D 108 2 2 2 SOLUTION: 15 + CP = 25 The sum of the measures of the angles of a triangle is 225 + CP2 = 625 180°, so write an equation and solve. CP2 = 400 CP = 20 Let x represent m∠LKN. So, the correct answer is choice C. 54 + x + 90 = 180 x = 36 ANSWER: C So, the correct answer is choice A. 53. AP = ? ANSWER:

A A 16 B 18 C 20 D 24

SOLUTION: In the rhombus, AP = CP = 20.

So, the correct answer is choice C.

ANSWER: C

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54. m∠ACB = ? I. square A 33 II. rectangle B 57 III. rhombus C 63 D 90 A I only B II only SOLUTION: C III only Because we are given ∠ABD = 33, we know that D II and III only ∠ABC = 66. Since opposite angles are congruent in E I, II, and III a rhombus, we also know that ∠ADC = 66. SOLUTION: Because the sum of the interior angle measures of a One way to prove a parallelogram is a rectangle, a quadrilateral is 360, we can subtract and then divide rhombus, or a square is by determining if the by 2 to find the measure of each of the other opposite diagonals are perpendicular (rhombus) or congruent angles, which are another pair of congruent angles. (rectangle). If they are both congruent and perpendicular, then the parallelogram is a square. 360 – 66 – 66 = 228 228 ÷ 2 = 114 To determine if PQRS is a rectangle, use the Distance Formula and see if the diagonals are So, the other two congruent angles each measure congruent. 114°.

To find m∠ACB, divide 114 by 2.

114 ÷ 2 = 57

So, the correct answer is choice B.

ANSWER: B

55. m∠BDA = ?

A 33 B 57 Since PR = QS, the diagonals are congruent and C 63 PQRS is a rectangle. D 90 To determine if PQRS is a rhombus, determine if the SOLUTION: diagonals are perpendicular using the slope formula. In the rhombus, ∠BDA = ∠ABD = 33.

So, the correct answer is choice A.

ANSWER: A 56. MULTI-STEP The vertices of a parallelogram are P(–1, 7), Q(5, 5), R(3, –1), and S(–3, 1). Which of the following terms apply to the parallelogram? eSolutions Manual - Powered by Cognero Page 24 6-5 Special Parallelograms: Rhombi, Squares

58. A supermarket sells trays of vegetables in the shape of a rhombus. The diagonals of the tray form four Since the slope of PR is and the slope of QS is compartments for the vegetables, as shown. , the slopes are opposite reciprocals and the diagonals are perpendicular. Therefore, PQRS is a rhombus.

Since PQRS is both a rectangle and a rhombus, then it is also a square. The correct answer is choice E.

ANSWER:

E Given that m∠EFG = 130, what is m∠HGJ? 57. The diagonals of rhombus ABCD are and . Which of the following is not necessarily true? A 130 B 65 A C 50 B D 25 C and bisect each other. SOLUTION: D bisects ∠A. Since a rhombus is a parallelogram, then consecutive SOLUTION: angles are supplementary. Therefore, if The diagonals of a rhombus bisect each other, are , then . Another perpendicular to each other, and bisect the opposite property of a rhombus is that both diagonals are angle angles. They are not necessarily congruent. bisectors of the opposite angles. Therefore, Therefore, the correct answer is choice B. . The correct answer is choice D. ANSWER: B ANSWER: D

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59. QRST is a parallelogram and x = 10. What is m∠QRS?

SOLUTION: By substituting x = 10 for the measures of ∠TSQ and ∠TQS, it can be determined that they are 80 and 40, respectively.

Based on the Triangle Angle-Sum Theorem, each triangle has a total of 180 degrees. Therefore, the measure of ∠QTS is 60. Since opposite angles are congruent in a parallelogram, the measure of ∠QRS is also 60.

ANSWER: 60