CHAPTER 6mrjaffe.net/GeomRef/unit6solutions.pdf · Opposite angles are congruent. If one angle...

CHAPTER 6

Lesson 6.1

Think & Discuss (p. 319)

1. As increases, A and D move apart. Therefore ADincreases.

2. The platform of the scissors lift will be raised higher.

Skill Review (p. 320)

1. If two parallel lines are cut by a transversal, then thepairs of consecutive interior angles are supplementary(Theorem 3.5 Consecutive Interior Angles).

2. If two parallel lines are cut by a transversal, then thepairs of alternate interior angles are congruent (Theorem3.4 Alternate Interior Angles).

3. AAS Congruence Theorem

4. SSS Congruence Postulate

5.

Developing Concepts Activity 6.1 (p. 321)

Drawing Conclusions

1. squall, snow, altostratus cloud, hail, showers, lightning

2. squall, hail, showers

� ��12

, �2�

� ��2 � ��3��2

, ��8 � 4�

2 �

� �125

m ��8 � 4

2 � ��3�

� 13

� �169

� �25 � 144

� ��5�2 � �12�2

AB � ��3 � 2�2 � �4 � ��8��2

Q

RP

Y

ZX

Q

RP

Y

ZX

m�B

3. Process, input/output, manual operation, decision, andextract are polygons because they are made up of linesegments and each line segment intersects exactly twoother line segments, one at each endpoint.

4. The fewest number of sides a polygon can have is three,because each segment must intersect exactly two otherline segments. There is no limit other than this on thenumber of sides.

6.1 Guided Practice (p. 325)

1. vertices 2. octagon, 15-gon

3. Yes; no; if the polygon were convex, the string wouldwrap exactly around the polygon with no gaps. If thepolygon were concave, there would be gaps and thelength of the string would be less than the perimeter ofthe polygon.

4. yes 5. No, because one side is not a line segment.

6. No, because two sides intersect only one other side.

7. equilateral 8. none of these 9. regular

10.

11.

6.1 Practice and Applications (pp. 325–328)

12. yes 13. no 14. no 15. no 16. yes 17. no

18. pentagon; convex 19. heptagon; concave

20. heptagon; concave 21. octagon

22. Sample answers: MNPQRSTL, NPQRSTLM,

23. 24. regular 25. equilateral

26. equiangular 27. quadrilateral; regular

28. pentagon; none of these 29. triangle; regular

30. octagon; regular

31. Sample answer: 32. Sample answer:

MP, MQ, MR, MS, MT

� 67�

� 360� � 105� � 113� � 75�

m�A � 360� � m�B � m�C � m�D

� 105�

� 360� � 125� � 60� � 70�

m�A � 360� � m�B � m�C � m�D

108 GeometryChapter 6 Worked-out Solutions Key

Copyright © McDougal Littell Inc. All rights reserved.

MCRBG-0601-SK.qxd 6-14-2001 11:43 AM 08

Geometry 109Chapter 6 Worked-out Solutions Key


Chapter 6 continued


35. Yes; to be concave, a polygon must have an angle withmeasure greater than 180°.

36. is 90° because a regular quadrilateral has fourequal angles whose sum is 360°.

37.

38.

39.

40. The sum of the measures of the interior angles remainsconstant (always 360�).

41.

42.

43.

44.

x � 25

7x� � 175�

7x� � 185� � 360�

4x� � 10� � 108� � 3x� � 67� � 360�

�4x � 10�� 108� � 3x� � 67� � 360�

x � 44

4x� � 176�

4x� � 184� � 360�

2x� � 2x� � 84� � 100� � 360�

x � 20

3x� � 60�

3x� � 300� � 360�

3x� � 150� � 60� � 90� � 360�

x � 67

x� � 293� � 360�

x� � 100� � 87� � 106� � 360�

m�A � 125�

m�A � 235� � 360�

m�A � 85� � 63� � 87� � 360�

m�A � m�B � m�C � m�D � 360�

m�A � 71�

m�A � 289� � 360�

m�A � 55� � 110� � 124� � 360�

m�A � m�B � m�C � m�D � 360�

m�A � 75�

m�A � 285� � 360�

m�A � 95� � 100� � 90� � 360�

m�A � m�B � m�C � m�D � 360�

x � 90

4x� � 360�

x� � x� � x� � x� � 360�

m�ABC

45.

46.

47. Tri–means three; sample answers: triangle is a polygonwith three angles, tricycle is a child’s vehicle with threewheels, tripod is an object with three legs, and triathlon isan athletic competition with three events.

48. hexagon; convex; regular

49. octagon; concave; equilateral

50. pentagon; convex; none of these

51. 17–gon; concave; none of these

52. a. 18–gon; no, it is concave.

b. Step 1: 18–gon, concave; Step 2: decagon; convex;Step 3: heptagon, convex; Step 4: quadrilateral,convex

c. The notches allow the paper to be folded into right orstraight angles without overlapping the paper.

53.

Substituting 60° � y° for x° gives

6.1 Mixed Review (p. 328)

54. 120 55. 63

56.

x � 60

3x� � 180�

x� � 2x� � 180�

x � 60 � 45 � 15

y � 45

�2y� � �90�

480� � 8y� � 6y� � 390� 65°45°

135°

115°

65°45°

135°

115°

8�60� � y�� 6y� � 390�

8x� � 6y� � 390�

8x� � 6y� � 30� � 360�

4x� � 5� � 4x� � 5� � 3y� � 20� � 3y� � 20� � 360�

�4x � 5�� 4x � 5�� 3y � 20�� 3y � 20�� 360�

x� � 60� � y�

x� � y� � 60�

6�x � y�� 360�

6x� � 6y� � 360�

3x� � 3x� � 3y� � 3y� � 360�

x � 9 or �9

�x2�� 81�

�x2�� 279� � 360�

�x2�� 90� � 90� � 99� � 360�

x � 4

70x� � 280�

70x� � 180� � 360�

82� � 25x� � 2� � 20x� � 1� � 25x� � 1� � 360�

82� � �25x � 2�� 20x � 1�� 25x � 1�� 360�

MCRBG-0601-SK.qxd 6-14-2001 11:43 AM Page 109

Chapter 6 continued

57.

58. 59.

60. Plot the midpoints of the triangle on a graph. Connect themidpoints to form the midsegments. will be parallelto the side containing point M. will be parallel to theside containing point N. will be parallel to the sidecontaining point L. Find the slope of and draw a linethrough point M which is parallel to

slope of

Find the slope of and a draw a line through point Nwhich is parallel to

slope of

Find the slope of and draw a line through point Lwhich is parallel to

slope of

The points where the lines intersect are the vertices of thetriangle: (0, 0), (�10, 2) and (�6, 14).

61. Plot the midpoints of the triangle on a graph. Connect themidpoints to form the midsegments. will be parallelto the side containing point M. will be parallel to theside containing point N. will be parallel to the sidecontaining point L. Find the slope of and draw a linethrough point M which is parallel to

slope of

Find the slope of and draw a line through point Nwhich is parallel to LM.

LM

LN ��8 � ��1��2 � ��4� � �

72

LN.LN

MNLM

LN

N

M

L

2 x

y

�2

MN �8 � 1

�8 � ��5� � �73

MN.MN

LM �1 � 7

�5 � ��3� � 3

LM.LM

LN �8 � 7

�8 � ��3� � �15

LN.LN

MNLM

LN

x � 5 x � 90

36x� � 180� 2x� � 180�

25x� � 11x� � 180� x� � x� � 180�

x � 6

29x� � 174�

29x� � 6� � 180�

20x� � 2� � 9x� � 4� � 180�

�20x � 2�� 9x � 4�� 180�slope of


slope of

The points where the lines intersect are the vertices of thetriangle: (1, 13), (5, �1) and (�9, �15).

62. Plot the midpoints of the triangle on a graph. Connect themidpoints to form the midsegments. will be parallelto the side containing point M. will be parallel to theside containing point N. will be parallel to the sidecontaining point L. Find the slope and draw a linethrough point M which is parallel to

slope of

Find the slope of and draw a line through point Nwhich is parallel to

slope


slope of

The points where the lines intersect are the vertices of thetriangle: (3, 9), (1, �1) and (�3, 5).

N

2 x

y

�1

L

M

MN �7 � 2

0 � ��1� � 5

MN.MN

LM �2 � 4

�1 � 2�

23

LM.LM

LN �7 � 40 � 2

� �32

LN.LN

MNLM

LN

L 2 x

y

�2

M

N

MN ��8 � 6�2 � 3

�145

MN.MN

LM �6 � ��1�3 � ��4� � 1






Chapter 6 continued

63. Plot the midpoints of the triangle on a graph. Connect themidpoints to form the midsegments. will be parallelto the side containing point M. will be parallel to theside containing point N. will be parallel to the side containing point L. Find the slope of and draw a linethrough point M which is parallel to

slope of

Find the slope of and draw a line through point Nwhich is parallel to

slope of


slope of

The points where the lines intersect are the vertices of thetriangle: (2, 15), (�4, �9) and (10, �1).

64.

� �25 � 4 � �29

� ��5�2 � ��2�2

BE � ��2 � 3�2 � �1 � 3�2

� �16 � 4

� �02 � ��4�2

AD � ��0 � 0�2 � ��1 � 3�2

� �20 � 2�5

� �4 � 16

� ��2�2 � 42

AC � ��1 � 3�2 � �4 � 0�2

1

x

y

�1

CE

D

BA

L

2 x

y

�2

N

M

MN ��5 � 73 � 6

� 4

MN.MN

LM �7 � 3

6 � ��1� �47

LM.LM

LN ��5 � 3

3 � ��1� � �2

LN.LN

MNLM

LN

Technology Activity 6.2 (p. 329)

1. Yes, ABFC is always a parallelogram because the slopesof the opposite sides are always equal.

2. Opposite sides are congruent.

3. The opposite sides are always congruent.

4. Opposite sides are congruent and have the same slope.

5. Opposite angles are congruent. If one angle increases byx°, the opposite angle also increases by x°. Consecutiveangles are supplementary. If one angle increases by x°, itsconsecutive angle decreases by x°.

6. Opposite angles of a parallelogram are congruent.

Extension

The diagonals of a parallelogram bisect each other.

Lesson 6.2


1. A parallelogram is a quadrilateral with both pairs ofopposite sides parallel.

2. No; only one pair of opposite sides is parallel.

3. Yes; both pairs of opposite sides are parallel.

4. opposite sides of a parallelogram are congruent.

5. diagonals of a parallelogram bisect each other.

6. opposite angles of a parallelogram are congruent.

7. opposite angles of a parallelogram are congruent.

8. diagonals of a parallelogram bisect each other.

9. opposite sides of a parallelogram are congruent.

10. vertical angles are congruent.

11. alternate interior angles are congruent.

12. 13; opposite sides of a parallelogram are congruent.

13. 7; diagonals of a parallelogram bisect each other.


15. 8.2; diagonals of a parallelogram bisect each other.

16. 80°; consecutive angles of a parallelogram are supplementary.


�KMJ;

�KNJ;

JM;

LN;

�LMJ;

�KJM;

KN;

LM;

� �36 � 6

� ��6�2 � 02

CE � ��2 � 4�2 � �1 � 1�2

� �9 � 16 � �25 � 5

� ��3�2 � ��4�2

BD � ��0 � 3�2 � ��1 � 3�2


Chapter 6 continued

18. 100°; opposite angles of a parallelogram are congruent.

19. 29°; opposite sides of a parallelogram are parallel, so � and alternate interior angles LMQ and MQN

are congruent.


20. 10; diagonals of a parallelogram bisect each other.




24. 60°; consecutive angles of a parallelogram are supplementary

25. 120°; opposite angles of a parallelogram are congruent.

26.

27.

30.

32.

34.

35.

36.

37.

t � 25

3t � 15 � 2t � 10

�3t � 15�� 2t � 10��

� 6 4 � f

g � 4 � 2 12 � 3f

g � f � 22f � 5 � 5f � 17

� 100 � 80

d � 90 � 10 c � 90 � 10

d° � �b � 10�� c � � �b � 10�� b � 90

2b � 180

b � 10 � b � 10 � 180

�b � 10�� b � 10�� 180�

w � 1 3 � z

3w � 3 6 � 2z

4w � w � 32z � 1 � 4z � 5

y � 3

3y � 9

x � 2

2x � 4

2x � 4 � 8

n � 110

m � 35

2m� � 70�

b � 101

a � 79

a� � 180� � 101�

x � 14, y � 10

QNLM

38. a. or

b. or

c.

d. SSS

e. corresponding

f.

39. 1. JKLM is a

2. Opposite of a are �.

3. 360°

4. Substitution property of equality

5.

6. Division property of equality

7. Definition of supplementary angles

40. 41. 42.

43.

44. They share a common midpoint. Then, since intersects at its midpoint, bisects by thedefinition of segment bisector. Similarly, bisects .

45. Opposite sides of a parallelogram are parallel. If two parallel lines are cut by a transversal, then alternate interior angles are congruent. Therefore and

. Consecutive angles of a parallelogram are supplementary, so By the substitution property of equality,Therefore is supplementary to

46. and are supplementary by Theorem 6.4, soBy the Alternate Interior Angles

Theorem, so By substitutionthus is supplementary to

47. 48.

49. Corresponding Postulate (If 2 � lines are cut by a transversal, then corresponding are �.)

50. No; the corresponding longer sides of the two are not

51. 52. It increases.

53. It increases. 54. It increases.

m�B � 180� � 120� � 60�

�.�s

�

�

�5; �8�4

�5.�4m�4 � m�5 � 180�,m�3 � m�5.�3 � �5,

m�4 � m�3 � 180�.�3�4

�6.�3m�3 � m�6 � 180�.

m�5 � m�6 � 180�.m�3 � m�5

�3 � �5

PROSOSPROS

PR

�a � c2

, b2�

�a � c2

, b2��a � c, b��c, 0�

m�J; m�K

��

�.

AC

DB � DB

AD � BCAB � DC

AB � DCAD � BC

30 � r s � 40

120 � 4r 3s � 120

3�40� � 4r 60 � 35 � 180

3s � 4r 3s � 3�25� � 15 � 180

3s � 3t � 15 � 180

35� � �3t � 15 �� 180�



28. 29.

q � 9

q � 3 � 6

p � 5

s � 3.5

r � 6

31.

33.

18 � v

6 �v3

4 � u

12 � 3u

2u � 2 � 5u � 10

m � 8

k � 7

k � 4 � 11




Chapter 6 continued

55. Statements Reasons

1. ABCD is a 1. Given.

2. 2. If a quadrilateral is a then itsopposite sides are

3. CEFD is a 3. Given

4. 4. If a quadrilateral is a then itsopposite sides are

5. 5. Transitive Property of


1. PQRS and TUVS 1. Given.are

2. 2. If a quadrilateral is a then its opposite angles are



1. WXYZ is a 1. Given.

2. 2. Definition of a

3. 3. If two lines are cut by a transversal, then alternate interior are

4. 4. ASA Congruence Postulate


1. ABCD, EBGH, 1. Given.and HJKD are s.

2. 2. If a quadrilateral is a then its opposite are


59. Opposite of a are

Consecutive of a are supplementary.

The sum of the measures ofthe with vertex G is

60. B

61. B

62. Sample answer: (See graph in answer to Ex. 64.)

63. Sample answer: Calculate slopes of the segments toshow opposite sides are parallel.

�4, 0�

s � 20

5s � 100

5s � 80 � 180

�2s � 30�� 3s � 50�� 180�

360�.�

��

�.��

120�

120�60�

60�45�

135�45�

105� 75�

B

A

F E

C

DG

75� 105�

135�

��2 � �3

�1 � �4�.��3 � �4

�,�1 � �2

�

�WMZ � �YMX

�.�

�MZW � �MXY��ZWM � �XYM

WZ � XY

�WZ � XY

�.

��1 � �3

�.�2 � �3�,�1 � �2

�s.

�AB � FE

�.�,CD � FE

�.

�.�,AB � CD

�.

64. three; (See graphs below.)

Graphs for Exercises 62 and 64.


65.

66.

67.

68.

70.

71. Yes; if two coplanar lines are to the same line, thenthey are to each other.�

�

�17

��3 � ��4��1 � ��8�

m �y2 � y1

x2 � x1

� 2

�84

�9 � 16 � 2

m �y2 � y1

x2 � x1

� 5�2 � �50

� �49 � 1

� ��7�2 � �1�2

AB � ��1 � ��8��2 � ��3 � ��4��2

� 3�5 � �45

� �36 � 9

� ��6�2 � ��3�2

AB � ��2 � ��4�2 � ��1 � 2�2

� 4�5 � �80

� �16 � 64

� ��4�2 � �8�2

AB � ��6 � 2�2 � �9 � 1�2

1 x

y

1

A

B

C

D (8, 8)

1 x

y

1A

B

C

(�2, 4)

D

1 x

y

1

A

B

C

D (4, 0)

�4, 0�, ��2, 4�, �8, 8�

69.

� �12

��36

��1 � 2

2 � ��4�

m �y2 � y1

x2 � x1


Chapter 6 continued

72. m B � 180� � (65� � 35�) � 80�. If one of a islarger than another then the side opposite the larger is longer than the side opposite the smaller So isthe shortest side and is the longest side.

73. m D � 180� � (90� � 55�) � 35�. If one of a islarger than another then the side opposite the larger is longer than the side opposite the smaller So isthe shortest side and is the longest side.

74. m H � 180� � (45� � 60�) � 75�. If one of a islarger than another then the side opposite the larger is longer than the side opposite the smaller So isthe shortest side and is the longest side.

Lesson 6.3

Activity 6.3 (p. 338)

Yes, it is always a parallelogram.


1. No; a parallelogram is a quadrilateral.

2. Yes; if both pairs of opposite in a quad. are thenthe quad. is a

3. Yes; if an of a quad. is supplementary to both of itsconsecutive , then the quad. is a

4. Yes; if both pairs of opposite sides of a quad. are thenthe quad. is a

5. Prove Then and because parts of s are Since both pairs of oppo-site sides are then ABCD is a

6. Since and and theseare pairs of alt. interior then and Since both pairs of opposite sides of ABCD are ABCDis a

7. Since corresponding angles C and D are , bythe Corresp. Converse. Since alternate interior anglesA and D are , by the Alt. Int. Converse.Then ABCD is by the definition of a

8. Use slopes to show that both pairs of opp. sides are usethe Distance Formula to show that both pairs of opp.sides are ; use slope and the Distance Formula to showthat one pair of opp. sides are both and ; use theMidpoint Formula to show that the diagonals bisect eachother.


9. Yes; if both pairs of opposite sides of a quad. are thenthe quad. is a

10. Yes; if the diagonals of a quad. bisect each other, then thequad. is a

11. No; by the Vertical Thm., this would be true for anyquad.

�

�.

�.�,

��

�;

�.�

�BA � CD��

BC � AD�

�.�,

BC � AD.AB � DC�,�BAC � �DAC,�ACB � �CAD

�.�,�.��

AB � CDBC � AD�BCA � �DAC.

�.�,

�.�

�

�.�,�

GJGH�.

��,��

DFEF�.

��,��

ACAB�.

��,�� 12. Yes; if one of a quad. is supp. to both of its

consecutive then the quad. is a

13. No; one pair of opposite sides which are to each otherand to a diagonal isn’t sufficient to prove that the quad. isa

14. Yes; if one pair of opposite sides of a quad. are both and then the quad. is a

15. Sample answer: Corresp. parts of s are , soand Since both pairs of opposite

sides of ABCD are ABCD is a

16. Sample answer: Since corresp. parts of s are and . By the Alternate Interior

Theorem, . One pair of opposite sides are

both and so ABCD is a

17.

18.

19.

20. Sample answer:

21.

and

22. Midpoint of

Midpoint of

Diagonals and bisect each other.BDAC

� �2, 32�BD � �3 � 1

2,

5 � ��2�2 �

� �2, 32�AC � ��1 � 5

2,

6 � ��3�2 �

BC � ADAB � CD

� 2�17 � �68

� �22 � ��8�2

AD � ��1 � ��1��2 � ��2 � 6�2

� �17

� ��4�2 � 12

CD � ��1 � 5�2 � ��2 � ��3��2

� 2�17 � �68

� �22 � ��8�2

BC � ��5 � 3�2 � ��3 � 5�2

� �17

� �42 � ��1�2

AB � ��3 � ��1�2 � �5 � 6�2

x � 90

2x � 180

�x � 10�� x � 10�� 180°

x � 60

3x � 180

2x� � x� � 180�

x � 70

�.�,�AB � CD�

AB � CD�ABX � �CDX�,��

�.�,AD � CB.AB � CD

��

�.��

�.

�

�.�,�






Chapter 6 continued

23. Slope of

Slope of

Slope of

Slope of

and

24. Slope of

Slope of

Slope of

Slope of

and or and .

25. Sample answer:

Midpoint of

Midpoint of

The diagonals of JKLM bisect each other, so JKLM is a by Thm. 6.9.�

� ��2, �12�

KM � ��1 � ��3�2

, 3 � ��4�

2 �

JL � �2 � ��6�2

, �3 � 2

2 � � ��2, �12�

BC � DABC � DAAB � CDAB � CD

� 2�17

� �68

� �22 � ��8�2

DA � ��1 � ��1��2 � ��2 � 6�2

� �17

� ��4�2 � 12

CD � ��1 � 5�2 � ��2 � ��3�2

� 2�17

� �68

� �22 � ��8�2

BC � ��5 � 3�2 � ��3 � 5�2

� �17

� �42 � ��1�2

AB � ��3 � ��1�2 � �5 � 6�2

DA ��2 � 6

1 � ��1� � �4

CD ��2 � ��3�

1 � 5� �

14

BC ��3 � 55 � 3

� �4

AB �5 � 6

3 � ��1� � �14

BC � ADAB � CD

AD ��2 � 6

1 � ��1� � �4

CD ��2 � ��3�

1 � 5� �

14

BC ��3 � 55 � 3

� �4

AB �5 � 6

3 � ��1� � �14

26. Sample answer:

Slope of

Slope of

Slope of

Slope of

Pairs of opposite sides have the same slope, so they areboth and PQRS is a

27. and Since the opposite pairs ofsides are ABCD is a and

28. Corresp. are so the longer sides are Then bothpairs of opp. sides are and the oblique I is a by definition.

29. If the diagonals of a quad. bisect each other, then thequad. is a

30. Check drawings. Sample answers: By Theorem 6.6: Drawtwo segments, and intersecting at B. Draw twoarcs, one with center A and radius BC and one with center C and radius AB, intersecting at D. Draw and

. By Theorem 6.8: Draw two segments, and intersecting at B. Construct a line through A to anda line through C to , intersecting at D. Draw and . By Theorem 6.10: Draw two segments, and

intersecting at B. Construct a line through A to and construct on the line to . Draw

31. Make and This will form a inwhich case will remain to while the binocularsare being raised or lowered.

32. The sum of the measures of the interior of a quad. is360�, so It isgiven that and so and and, by the substitution property ofequality, . Then

and

By the Consecutive Interior Angles Converse,

Similarly, and RSTU is a by the def. of a

33.

Statements Reasons

1. is supplementary 1. Given.to and

2. and 2. Consecutive Interior Conv.

3. PQRS is a 3. Definition of a

34. the diagonals bisect each other at the origin. Thelength of is a so the length of must also be a,however, it is below the x-axis.

OPOM�0, �a�;

��.

�QP � RSQR � PS

�S.�Q�P

�.�SR � TU

ST � RU.

m�S � m�T � 180�.2�m�S� � 2�m�T� � 360�

m�T � m�S � m�T � m�S � 360�m�S � m�U

m�R � m�T�S � �U,�R � �Tm�T � m�U � 360�.m�R � m�S �

�

AD�BC�AB � DC.AD � BC

DC.BC�ADBC�

BCABCDADAB�

BC�BCABCD

AD

BCAB

�.

��.�,�

AB � CD.��,BC � DA.AB � CD

�.�

PS ��3 � 53 � 2

� �8

RS ��3 � ��4�

3 � 9� �

16

QR ��4 � 49 � 8

� �8

PQ �4 � 58 � 2

� �16


Chapter 6 continued

35. the diagonals of a bisect each other, so

is the midpoint of Let By the

Midpoint Formula, so

and

36. Slope of

Slope of

Slope of

Slope of

Slope of slope of and slope of

slope of

37. a.

b.

c.

d.

EFGH is a parallelogram becauseboth pairs of opposite are .

38. PTRU and PQRS share a common diagonal, PTRUand QTSU share a common diagonal, Since bisects and bisects must also bisect .PTRU is a parallelogram by Thm 6.9.


39. If then If then

40. If then If then

41. If a quadrilateral is a parallelogram, then each pair ofopposite sides are parallel. If each pair of opposite sidesof a quadrilateral are parallel, then the quadrilateral is aparallelogram.

42. A point is on the perpendicular bisector of a segment ifand only if the point is equidistant from the endpoints ofthe segment.

43. A point is on the bisector of an angle if and only if thepoint is equidistant from the two sides of the angle.

4x � 7 � x � 37.x � 10,x � 10.4x � 7 � x � 37,

x2 � 2 � 2.x � 0,x � 0.x2 � 2 � 2,

TUPRTU,SQSQPRTU.

PR.

��

m�H 126�;m�G 54�,m�F 126�,m�E 54�,

m�EHB 27�m�GHC 27�;

m�FGD 63�

m�FGD � 90� � 27� 180�

m�FGD � m�GDF � m�DFG � 180�

m�AFF 27�

63� � 90� � m�AFE 180�

m�AEF � m�EAF � m�AFE � 180�

QP �c � a

b.MN �

NP �c � a

bMQ �

QP ��a � ��c�0 � ��b� �

c � ab

MN �c � ab � 0

�c � a

b

NP �c � ��a�

b � 0�

c � ab

MQ ��c � a�b � 0

�c � a

b

y � �c.

x � �b�0, 0� � �x � b2

, y � c

2 �,

Q � �x, y�.QN.�0, 0�

��b, �c�; 44. The slope of the line through point is

45.

46.

47.

Quiz 1 (p. 346)

1. convex, equilateral, equiangular, regular

2.

The sum of the measures of the interior angles of aquadrilateral is

3.

Statements Reasons

1. ABCG and CDEF 1. Given.are

2. 2. If a quad. is a then the opposite are

3. 3. Vertical Angles Theorem

4. 4. Transitive Property of Congruence.

4. Sample answer: Use slopes to show that both pairs ofopposite sides are parallel, use the Distance Formula toshow that both pairs of opposite sides are congruent, useslope and the Distance Formula to show that one pair ofopposite sides are both parallel and congruent, use theMidpoint Formula to show that the diagonals bisect eachother.

Math and History (p. 346)

1.

� 1,150,000 ft2 � 1

2 �2800 � 1800��500�

A �12 �b1 � b2�h

�A � �E

�BCG � �DCF

�.��DCF � �E�,�A � �BCG,

�s.

360�.

x � 35

4x � 140

4x � 220 � 360

2x� � 2x� � 110� � 110� � 360�

x � 35

�x � �35

x� � 85� � �2x � 50��

x � 48

3x � 144

3x � 36 � 180

x� � �2x � 14�� 50� � 180�

� 60

� 180 � 120

x� � 180� � �52� � 68��

y �14 x �

94

b � �94

�z �14 �1� � b

y � mx � b

14.�1, �2�




Geometry 117Chapter 6 Worked-out Solution Key


Chapter 6 continued


1. rhombus

2. If a parallelogram is a rhombus, then each diagonalbisects a pair of opposite angles. If each diagonal of aparallogram bisects a pair of opposite angles, then theparallelogram is a rhombus. Diagonal bisects and and bisects and

3. always 4. sometimes 5. sometimes 6. always

7. C, D 8. B, D 9. B, D 10. A, B, C, D

11.


12. 13.

Always; a rectangle has Sometimes; if rectangle four right angles and right ABCD is also a rhombus angles are congruent. (a square), then

14. 15.

Always; the diagonals Sometimes; if rectangle of a rectangle are ABCD is also a rhombus congruent. (a square), then the diago-

nals of ABCD are

16. rectangle, square 17. square 18. rhombus, square

19. parallelogram, rectangle, rhombus, square


21. rhombus, square

22.

and bisect each other.

23.

and bisect eachother, bisects and bisects and

24. Pairs of opposite sides are , all foursides are congruent, all four angles are right angles; the diagonals are congruent, bisect each other, and bisectopposite angles.

25. rectangle 26. square

�B C

A D

�SRQ.�SPQPR�PQR.

�PSRQSQS � RP,QSPR�S � �Q,

�P � �R,PQ � QR � RS � SP,PQ � RS, PS � QR,Q R

P S

GIFHGH � FI, �G � �I, �F � �H,FG � HI, FI � GH, FG � HI,G H

F I

�.

B C

A D

B C

A D

AB � BC.

B C

A D

B C

A D

x � 45

2x� � 90�

�PQR.�PSRSQ�SRQ,�SPQPR

27. 28.

Always; if a quad. is a Sometimes; if a rhombus is�, then the opp. are also a rectangle (a square)

then all four angles arecongruent.

29. 30.

Always; each diagonal Always; a rhombus is a of a rhombus bisects a parallelogram with four pair of opposite congruent sides.

31. 32.

Sometimes; if a rhombus Always; a rhombus is a is also a rectangle (square) parallelogram with four then its diagonals are congruent sides.congruent.

33. 34.

35. 36.

37. 38.

39. 40. 41.

42.

43.

44. Sample answer: You would need to know thator is a right angle.

45. Assume temporarily that and thatBy the definition of a �, MNPQ is a �. This

contradicts the given information that It fol-lows that is not parallel to NP.MQ

�1 � �2.MQ � PN.

�1 � �2,MN � PQ,

�D�A, �B, �C,

WY � XZ � 13 � 3 � 10

3 � x

12 � 4x

24 � 5x � 1 � 13 � x

4 � 2�2P � HJ � KJ � KH � 2 � 2 � 2�2 �

45�90�2�2

x � 1

x � 24 2x � 2

8x � 13 � 7x � 11 3x � x � 2

x � 50

3x � 150

x � 5 3x � 30 � 180

2x � 10 �x � 40�� 2x � 10�� 180�

x � 50 x � 18

x� � 180� � 130� 5x� � 90�

B C

A D

B C

A D

�.

B C

A D

B C

A D

� .�

B C

A D

B C

A D


Chapter 6 continued


1. RSTU is a �, 1. Given

2. RSTU is a rhombus. 2. A � is a rhombus if itsdiagonals are

3. 3. Each diagonal of a rhombus bisects a pair of opp.

47. If a � is a rectangle, then its diagonals are congruent. If the diagonals of a � are congruent, then the � is arectangle.

48. If a quadrilateral is a rhombus, then it has four congruentsides. (A rhombus by definition has four congruentsides.)

If a quadrilateral has four congruent sides, then it is arhombus. (Both pairs of opposite sides are congruent, sothe quadrilateral is a �. By definition a � with fourcongruent sides is a rhombus.)

49. If a quadrilateral is a rectangle, then it has four rightangles. (def. of a rectangle)

If a quadrilateral has four right angles, then it is a rectan-gle. (Both pairs of opp. are so the quad. is a �. Bydefinition, a � with 4 right is a rectangle.)

50. If a quadrilateral is a square, then it is a rhombus and arectangle. (A square has four sides, which makes it arhombus and four rt. , which makes it a rectangle.)

If a quadrilateral is a rhombus and a rectangle, then it is asquare. (By definition, a rhombus has four sides and arectangle has four rt. The only quad. that has four sides and four rt. is a square.)

51. Sample answer:

Statements Reasons

1. PQRT is a rhombus. 1. Given

2. 2. A quad. is a rhombus if and only if it has 4 sides.

3. 3. Reflexive Prop. of Congruence

4. 4. SSS Cong. Postulate

5. 5. Corresp. parts of s

are

6. bisects and 6. Def. of bisectorbisects

and �RQP.�PTRQT�QRT,

��TPQPR

�PQT � �RQT�PTQ � �RTQ,

�.�TRP � �QRP;��TPR � �QPR,

�PTQ � �RTQ�PRQ � �PRT;

PR � PR, QT � QT

�PQ � PT � QR � RT

�

��.�

�

�

�

�,�

JL � KM

�.

�STR � �UTR

�.

SU � RT.


1. FGHJ is a �, 1. Givenbisects and

bisects and

2. 2. Def. of bisector

3. 3. Reflexive Prop. ofCongruence

4. 4. ASA Cong. Post.

5. 5. Corresponding parts of congruent triangles arecongruent.

6. 6. Transitive Property ofCongruence

7. FGHJ is a rhombus. 7. Def. of a rhombus

53. Sample answer: Draw a line f and where f is not to f should intersect at H. Construct on f sothat With centers G and J, construct two arcswith radius GH which intersect at K. Draw Since all four sides of GHJK are GHJK is a rhombus.It is not a square or rectangle because and are not

54. Sample answer: Draw a segment Construct linesto at A and B. These lines will be . Choose D onthe at A such that AD � AB. Construct on theat B such that Draw . Since

and ABCD is a �. is supplementary to(Consecutive Int. Thm.) so is a right angle.

Similarly, is a right angle. Then ABCD is a rectan-gle. It is not a square because AD � AB.

55. Rectangle;

If the diagonals of a � are then the � is a rectangle.�,

1 x

y

�1

S (�2, 1)

R (�2, �3) Q (3, �3)

P (3, 1)

� �41

� �52 � ��4�2

QS � ��3 � ��2�2 � ��3 � 1�2

� �41

� �52 � 42

PR � ��3 � ��2�2 � �1 � ��3�2

�C�D��A

�DAD � BC,

AD � BCDCBC � AD.�BC�

�AB�AB.

�.HJGH

�,GK and JK.

HJ � GH.HJGHGH.

�GH

JH � FG � GH � FJ

JH � FJ, FG � FJJH � GH, FG � GH,

�FGJ � �HGJ�FHJ � �FHG,

FH � FH, GJ � GJ

�FJG � �HJG�FGJ � �HGJ,�FHG � �FHJ;

��HFG � �HFJ,

�HGF.�FJHJG

�GHJ,�JFGFH

118 GeometryChapter 6 Worked-out Solution Key


MCRBG-0604-SK.qxd 5-25-2001 11:20 AM 8



Chapter 6 continued

56. Rhombus; slope of

slope of , which is undefined

If the diagonals of a parallelo-gram are perpendicular,then the parallelogram is arhombus.

57. Rectangle;

If the diagonals of a parallel-ogram are then the � is arectangle.

58. Square;

So,

slope of

slope of QR �5 � 2

2 � ��1� � 1

PQ �5 � 22 � 5

� �1

PQ � QR � RS � SP.

� 3�2

� �b��3�2 � ��3�2

SP � ��2 � 5�2 � ��1 � 2�2

� 3�2

� ��3�2 � 32

RS � ��1 � 2�2 � �2 � ��1�2

� 3�2

� �32 � 32

QR � ��2 � ��1�2 � �5 � 2�2

� 3�2

� �32 � ��3�2

PQ � ��5 � 2�2 � �2 � 5�2

�,

1

x

y

1

S (4, �1)

R (2, �3)

P (�1, 4)

Q (�3, 2)

� �58

� ��7�2 � 32

QS � ��3 � 4�2 � �2 � ��1�2

� �58

� ��3�2 � 72

PR � ��1 � 2�2 � �4 � ��3��2

2

x

y

�2

S (1, �5)

P (5, 2)

Q (1, 9)

R (�3, 2)

PR � QS

QS �9 � ��5�

1 � 1�

140

PR �2 � 2

5 � ��3� �08

� 0 slope of

slope of

So,

A quadrilateral is a square if it has 4 congruent sides and 4 right angles.

59. so and so

60.

61. Sample answer: ABDC is a � since cross braces andbisect each other. so ABDC is a rectangle.

Since a rectangle has four right angles,Then, and

By substitution So tabletop is perpendicular to legs

and by the def. of perpendicular.

62. Sample answer: Since ABDC is a rectangle,A, C, and E are collinear as are B, D, and F so

Since ABFD is a � and

63. Rhombus; AECF remains arhombus; AECF remains a rhombus.

64. When A or C is dragged, and remainequal, as do and

65. Each diagonal of a rhombus bisects a pair of opp. (Theorem 6.12)

66. D

67. B

68. In a parallelogram, opposite are and consecutive are supp. Therefore, all four must measure inwhich case, the � would be a rectangle.

69. The diagonals of a � bisect each other. So,and and then by substitution.

and again by substitution, giving and By the Transitive Property of Congruence,OC � OD.

OA � OB �� OB.OCAO � DO

OC � OC � OB � OB,AO � AO � DO � DOAO � OC � DO � OB

DO � OB � DB,AO � OC � ACOD � OB.OA � OC

90��

��

y � 5 x � 9 2�9��y� � 90 9x � 81 2�xy�� 90� �9x � 9�� 90�

x � 4 3x � 12

7x � 3 � 4x � 9

�.

m�CEF.m�AEFm�EACm�FAC

AE � CE � AF � CF;

AB � EF.AE � BF,

AE � BF.AC � BD.

BFAEABm�ABF � 90�.

m�BAE �m�ABF.m�ABD �m�BAC � m�BAEm�DBA � 90�.

m�CAB �AD � BCBC

AD

OM � KN

� �b2 � a2

KN � ��b � 0�2 � �0 � a�2

� �b2 � a2

OM � ��b � 0�2 � �a � 0�2

MN � a.KO � MN,KM � bKM � ON,�b, a�;

�1x

y

1

P (5, 2)

Q (2, 5)

S (2, �1)

R

(�1, 2)

PS � RS.PQ � QR,

RS �2 � 1

�1 � 2�

3�3

� �1

PS �2 � 15 � 2

�33

� 1


Chapter 6 continued

70.

Since and let

(The y-coordinate of A is positive.)

71. Since and lies on the x-axis,

D � (�b, 0). Since OA � OC, OC � b, and C is in quadrant III, the coordinates of C must be negative. So,

72. Slope of

Slope of

Slope of

Slope of

ABCD is a parallelogram with four right angles, that is,ABCD is a rectangle.


73. yes 74. no 75. no 76. no 77. yes 78. yes

79. 80.

EP � 3.3 � 12

EP � 6.6 � 9.9 PD �32 � 1

EP � PC � CE PD � AD � AP

9.9 � CE 32 � AD

6.6 �23CE 1 �

23AD

PC �23CE AP �

23AD

CD � DA

� �1

� ��b � a��b � a�

��a � b��a � b�

��b2 � a2

�a � b ��b2 � a2

a � b � � ��b2 � a2�

��a � b��a � b�

�b2 � a2

a � b�b2 � a2 � 0

a � ��b� �DA:

��b2 � a2

�a � b��b2 � a2 � 0

�a � ��b� �CD:

AB � BC

��a � b�

a � b� �1

��b � a��b � a��a � b��a � b�

��b2 � a2

a � b ��b2 � a2

b � a � �b2 � a2

�a � b��a � b�

0 � ��b2 � a2�b � ��a� �

�b2 � a2

b � aBC:

�b2 � a2 � 0a � b

��b2 � a2

a � bAB:

C � ��a, ��b2 � a2 �.

ODOD � b,OB � OD,

A�a, �b2 � a2 � �b2 � a2 � y

b2 � a2 � y2

b2 � a2 � y2

b � �a2 � y2

b � ��a � 0�2 � �y � 0�2

A � �a, y�.OB � OA, OA � b

� �b2 � bOB � ��0 � b�2 � �0 � 0�281. 82.

83. Assume temporarily that ABCD is a quad. with 4 acutethat is, m�A < 90�, m�B < 90�, m�C < 90�, and

m�D < 90�. Then m�A � m�B � m�C � m�D <360�. This contradicts the Interior Angles of a Quad.Theorem. Therefore, no quad. has 4 acute

Lesson 6.5


1. and

2. A kite is a quadrilateral that has two pairs of consecutivecongruent sides, but opp. sides are not �. In a rhombus,both pairs of opp. sides are �, so a rhombus can’t be akite.

3. isosceles trapezoid 4. kite 5. trapezoid

6. Use coordinates to show that or

7.

8. 9.


Figure for Exs. 10–15:

10. bases 11. legs 12. consecutive sides

13. diagonals 14. opposite angles 15. base angles

16.

17.

18.

19.

20.

21.

22. 23.

24.

25. Yes; X is equidistant from the vertices of the dodecagon,so and by the Base AnglesTheorem. Since trapezoid ABPQ has a pair of congruentbase angles, ABPQ is isosceles.

26. m�AXB �36012

� � 30�

�XAB � �XBAXA � XB

5 � x

16 � 11 � x

8 �12�11 � x�

10 � x 5 � x

14 � x � 4 14 � x � 9

7 �12�x � 4� 7 �

12�x � 9�

MN �12�15 � 9� �

12�24� � 12

MN �12�16 � 14� �

12�30� � 15

MN �12�7 � 9� �

12�16� � 8

m�K � m�L � 98�m�L � 180� � 82� � 98�;m�J � m�M � 82�,

m�L � 180� � 132� � 48�m�J � 180� � 78� � 102�;

m�L � m�K � 136�m�K � 180� � 44� � 136�;m�M � m�J � 44�,

Q R

P S

12�12 � 7� �

12�19� �

192

12�7 � 3� �

12�10� � 5

12�11 � 7� �

12�18� � 9

AD � BC.AC � BD

ABCD

�.

�,

� 13 9 � FB

� 13�39� 6 �

23FB

PD �13AD PB �

23FB






Chapter 6 continued

27. In and so

28. 29.

30.

31.

32.

33.

34. A trapezoid has exactly one pair of parallel oppositesides. Because a � has two pairs of parallel oppositesides, ABCD cannot be a trapezoid.

35. any two positive numbers (except 5 and 5) whose sum is 10

36.

so and are consecutive � sides.

so and are consecutive � sides.

So, the points are vertices of a kite.

BCDCDC � BC,

DAABAB � DA,

� �53

� �22 � ��7�2

DA � ��6 � 4�2 � ��2 � 5�2

� �265

� ��12�2 � ��11�2

CD � ��6 � 6�2 � ��13 � ��2��2

� �265

� �32 � 162

BC � ��3 � ��6��2 � �3 � ��13��2

� �53

� ��7�2 � ��2�2

AB � ��3 � 4�2 � �3 � 5�2

� 90�

m�G � 360� � �100� � 100� � 70��m�F � m�H � 100�

� 70�

m�G � 360� � �110� � 90� � 90�� m�G � 95�

2m�G � 190�

2m�G � 360� � �120� � 50��m�E � m�G

� 9.43

� �89

KL � LM � �82 � 52

� 14.42

� �208

JK � JM � �122 � 82

� 6.40 � 5

� �41 � �25

EF � GF � �52 � 42 BC � DC � �32 � 42

� 8.60 � 3.61

� �74 � �13

EH � GH � �52 � 72 AD � AB � �22 � 32

m�Q � m�P � 105�m�P � 180� � 75� � 105�;m�A � m�B � 75�,

�A � �B,m�AXB � 30�,�ABX,37. Slope of

Slope of

Slope of

Slope of

ABCD is a trapezoid; slope of slope of so slope of and slope of so is not to ABCD is not isosceles;

and CD � 5.

38. Slope of

Slope of

Slope of

Slope of

ABCD is a trapezoid; slope of slope ofso slope of and slope of

so is not to ABCD is isosceles;

39.

40. It is given that ABCD is an isosceles trapezoid withand If is drawn so that ABCE

is a parallelogram, since opposite sides of a parallelogramare congruent, and by the Transitive Propertyof Congruence, By the Base Angles Theorem

Because Thenby the Transitive Property of Congruence. By

the Consecutive Interior Angles Theorem,

—CONTINUED—

�D � �C�AED � �C.AE BC,�D � �AED.

AE � AD.BC � AE

AEAD � BC.AB DC

� 16 in.

� 12�32�

� 12�10 � 22�

BE �12�CD � AF�

EF � GH � �58GH.EF

GH �73,EF � �

73FG EH;

EH � 0,FG �

� �58

� �32 � 72

GH � ��8 � 5�2 � �9 � 2�2

� �58

� �32 � ��7�2

EF � ��4 � 1�2 � �2 � 9�2

EH �9 � 98 � 1

�07

� 0

GH �9 � 28 � 5

�73

FG �2 � 25 � 4

�01

� 0

EF �2 � 94 � 1

� �73

AB � 2�5CD.AB

CD � �43,AB � 2BC AD;

AD � 0,BC �

� �25 � 5

� ��3�2 � 42

CD � ��5 � 8�2 � �4 � 0�2

� 2�5

� �20

� ��2�2 � ��4�2

AB � ��2 � 0�2 � �0 � 4�2

AD �0 � 0

8 � ��2� �0

10� 0

CD �0 � 48 � 5

� �43

BC �4 � 45 � 0

�05

� 0

AB �4 � 0

0 � ��2� �42

� 2


Chapter 6 continued

40. —CONTINUED—and

By the Substitution property of equality,

Finally by the Subtraction property ofequality. So, by definition of congruence.

41. TQRS is an isosceles trapezoid sobecause base of an isosceles trapezoid are �.

and by the Reflexive Prop. of Cong.,by the SAS Congruence Postulate. Then

because corresp. parts of � triangles are �.

42. and by the Midsegment Thm. and

and by the Midsegment Thm.

, so by substitution,

is parallel to both and

because , , and all lie on the same line.

43. If then ACBD is a kite; AC � AD and BC � BD, so the quadrilateral has two pairs of congruentsides, but opposite sides are not congruent. (If AC � BCthen ABDC is a rhombus.) ABDC remains a kite in all 3 cases.

44. The are �; the measures of the change, but the remain �.

45. If a quad. is a kite, then exactly one pair of opp. are �.(Theorem 6.19)


1. 1. Given


3. 3. SSS Congruence Post.

4. 4. Corresp. parts of � s

are �.


6. 6. SAS Congruence Post.

7. 7. Corresp. parts of � s

are �.

8. 8. Def. of linear pairare a linear pair.

9. 9. If two lines intersect toform a linear pair of con-gruent angles, then thelines are

47. Draw (Through any 2 points, there is exactly 1line.) Since and by the SSS Congruence Postulate. becausecorresp. parts of � s are �. Assume temporarily that

then both pairs of opp. of ABCD would be� making ABCD a parallelogram. This contradicts thedef. of a kite. Therefore �B �D.

��B � �D,�

�A � �C�BCD � �BADCD � AD,BC � BA

BD.

�.

AC � BD

�CXB and �AXB

��CXB � �AXB

�CBX � �ABX

BX � BX

��CBX � �ABX

�BCD � �BAD

BD � BD

AD � CDAB � CB,

�

��

AC � BC

BEGEBG

AFCDBE�12�CD � AF�.

BE �12CD �

12AFBE � BG � GE

EG AFEG �12AF

BG CDBG �12CD

TR � SQ�QTS � �RST

TS � TS,QT � RS�

�QTS � �RST

�B � �DABm�B � m�DAB

m�B � m�C � m�DAB � m�C.m�DAB � m�D;m�B � m�C �

m�DAB � m�D � 180�.m�B � m�C � 180�48. Yes; The legs are � and it has one pair of sides.

49. Yes; The diagonals are � and it has one pair of sides.

50. Yes; �A � �B and �D � �C. Then m�A � m�B andm�D � m�C. By the Interior � of a Quad. Theorem,m�A � m�B � m�C � m�D � 360�. By substitutionand the properties of equality, m�A � m�D � 180�. Bythe Consecutive � Converse, . ABCD is not aparallelogram because it is given that �A �C soopposite � are not �. So ABCD is a trapezoid and sinceit has a pair of congruent base angles, then it is an isosce-les trapezoid.

51. 52. C

E

53. In trapezoid PQRS, . Draw a perpendicular seg-ment from Q to and label the point of intersection M.Draw a perpendicular segment from R to and labelthe point of intersection N. becauseperpendicular lines form right angles and all right anglesare congruent. In a plane, if two lines are perpendicularto the same line, then they are parallel to each other, so

lies on , so . Therefore,QRNM is a parallelogram, because both pairs of oppositesides are parallel, and We know that and are right triangles, and it is given that

so by the HL CongruenceTheorem. because corresponding partsof congruent triangles are congruent. by theReflexive Property of Congruence. bythe SAS Congruence Postulate. Therefore,because corresponding parts of congruent triangles arecongruent.


54. If a triangle is scalene, then it has no congruent sides.

55. If a quadrilateral is a kite, then it has perpendicular diag-onals.

56. If a polygon is a pentagon, then it has five sides.

57. 5.6 58. 10 59. 7 60. 11.2 61. 80° 62. 100°

63. Yes; Sample answer:

Slope of

Slope of

Slope of

Slope of

Both pairs of opposite sides are parallel so ABCD is aparallelogram.

AD �8 � 0

�2 � ��5� �83

CD �0 � 02 � 5

�07

� 0

BC �8 � 05 � 2

�83

AB �8 � 85 � 2

�07

� 0

QP � RS�QPS ��RSP

PS � PS�QSP � �RPS

�QMS � �RNPPR � SQ,�RNP

�QMSQM � RN.

QR MNPSMNQM RN.

�QMP � �RNSPS

PSQR PS

6 � x

30 � 5x

15 �12�3x � 2 � 2x � 2�

AB DC






Chapter 6 continued

64. Yes; Sample answer:

Both pairs of opposite sides are congruent, so PQRS is aparallelogram.

Quiz 2 (p. 363)

1. Sample answer: Opposite sides of EBFJ are �, so EBFJis a �. Opposite angles of a � are �, so

by the Congruent Supplements Thm.Since by theSAS Congruence Postulate and since corresponding partsof congruent triangles are congruent,

2.

Opposite sides are congruent and diagonals are congru-ent, so PQSR is a rectangle.

3.

� 3�2

� �18

� ��3�2 � 32

BC � ��0 � 3�2 � �9 � 6�2

� 3�2

� �18

� ��3�2 � ��3�2

AB � ��3 � 0�2 � �6 � 9�2

� 6�5

� �180

� �62 � 122

PS � ��2 � ��4��2 � �5 � ��7��2

� 6�5

� �180

� ��6�2 � 122

QR � ��4 � 2�2 � �5 � ��7��2

QS � ��4 � ��4��2 � ��7 � 5�2 � �144 � 12

PR � ��2 � 2�2 � �5 � ��7��2 � �144 � 12

RS � ��2 � (�4��2 � ��7 � ��7��2 � �36 � 6

� 6PQ � ��2 � ��4��2 � �5 � 5�2 � �36

HJ � JK.

�JFK � �HEJHE � JE � JF � KF,�KFJ � �JEH

�BFJ � �BEJ.

� �12 � 52 � �26

PS � ��4 � 3�2 � ��3 � ��8��2

� �52 � 22 � �29

RS � ��8 � 3�2 � ��6 � ��8��2

� �12 � 52 � �26

QR � ��9 � 8�2 � ��1 � ��6��2

� ��5�2 � ��2�2 � �29

PQ � ��4 � 9�2 � ��3 � ��1��2

Two pairs of consecutive sides are congruent, so ABCD isa kite.

4.

All sides are congruent and diagonals are congruent, soJKLM is a square.

5. Slope of

Slope of

Slope of

Slope of

Exactly one pair of opposite sides are parallel, so PQRSis a trapezoid.

PS RQ

QP ��2 � ��3�1 � ��5� �

16

RQ �3 � ��2�

6 � 1�

55

� 1

SR �9 � 37 � 6

�61

� 6

PS � �3 � 9�5 � 7

��12�12

� 1

� �130

� ��9�2 � 72

JL � ��5 � 4�2 � �6 � ��1��2

� �130

� ��7�2 � ��9�2

MK � ��4 � 3�2 � ��2 � 7�2

� �65

� ��8�2 � ��1�2

MJ � ��5 � 3�2 � �6 � 7�2

� �65

� �12 � ��8�2

LM � ��4 � 3�2 � ��1 � 7�2

� �65

� ��8�2 � ��1�2

KL � ��4 � 4�2 � ��2 � ��1��2

� �65

� ��1�2 � 82

JK � ��5 � ��4��2 � �6 � ��2��2

� �265

� ��3�2 � 162

AD � ��3 � 0�2 � �6 � ��10��2

� �265

� �32 � 162

CD � ��3 � 0�2 � �6 � ��10��2


Chapter 6 continued

6. It is given that ABCD is a trapezoid with andDraw so ABCE is a parallelogram.

by the Corr. Postulate. bythe Trans. Prop. of Cong. by the Converse ofBase Thm. because opp. sides of a parallelo-gram are congruent. By the Trans. Prop. of Congruence,

Lesson 6.6


1. Draw By the Midsegment Theorem for Triangles,and Two lines to the same line are

to each other, so

2.

3.

4.

5.

6.



8.

9.

10.

11.

12.

13.

14. trapezoid 15. isosceles trapezoid 16. trapezoid

17. square 18. kite

19. �, rectangle, square, rhombus, kite 20. rectangle,square, isosceles trapezoid

21. rhombus, square 22. �, rectangle, square, rhombus,isosceles trapezoid

EF � HG.

��HG � DB.EF � DBDB.

AD � BC.

AE � BC�

AD � AE�AED � �D��AED � �C

AE � BC,�D � �C.AB � DC 23. rectangle, square 24. square

25. Show that the quad. has two pairs of consecutive congru-ent sides, but opposite sides are not congruent (def. of akite).

26. Show that the quad. has four rt. with four sides (def.of a square).

27. Show that the quad. has four rt. show that the quad. isa � and that the diagonals are

28. Show that only one pair of opp. sides is (def. of a trape-zoid).

29. Show that exactly 2 sides are parallel and that the nonpar-allel sides are congruent (def. of isosceles trap.); showthat the quad. is a trapezoid and that the pair of base are show that the quad. is a trapezoid and that itsdiagonals are

30. �A and �D or �B and �C; by Consec. Int � Converse, so if �A � �D (or �B � �C),and are not and ABCD is a trapezoid. Sincebase � are �, ABCD is an isosceles trapezoid.

31. and if the diagonals of a quad. bisect eachother, the quad. is a �.

32. and if then one pair of oppo-site sides are both and �, and ABCD is a �. Sincethe diagonals are perpendicular, ABCD is a rhombus.

33. and because the diagonals of ABCD bisecteach other, ABCD is a parallelogram. If the diagonalsof a � are �, then the � is a rectangle.

34. use either SAS orASA to prove that which impliesthat A quad. that has two pairs of consec-utive congruent sides is a kite.

35. Any two consecutive sides; ABCD is a rectangle bythe Rectangle Corollary. If, for example,then since ABCD is a �, and

Then by the Trans. Prop. of �, �� and ABCD is a rhombus. A quad.

that is both a rectangle and a rhombus is a square.

36.

Kite; and but pairs of oppositesides are not necessarily congruent, so it isn’t arhombus.

QR � RS,PQ � PS

� �4 � 2

PS � ��0 � 2�2 � �0 � 0�2

� �34

� �32 � 52

RS � ��5 � 2�2 � �5 � 0�2

� �34

� ��5�2 � ��3�2

QR � ��0 � 5�2 � �2 � 5�2

� �4 � 2

PQ � ��0 � 0�2 � �0 � 2�2

ADCDBCABAD � BC.

AB � CDAB � AD,

AB � AD.�ABC � �ADC,

BC and DC or �BAC and �DAC;

BD;AC

�AB � CD,CD;AB

DE;BE

�DCAB

BC � AD

� .� ;

�

�

� .�;

��



Property � Rectangle Rhombus Square Kite Trapezoid

Both pairs ofopp. sides X X X Xare

Exactly 1 pair of opp. Xsides are

Diagonals are X X X

Diagonals are X X

Diagonals bisect each X X X Xother.

� .

�.

� .

� .

Property � Rectangle Rhombus Square Kite Trapezoid

Both pairs of opp. X X X Xsides are

Exactly 1 pair of opp. sides are

All sides are X X

Both pairs of opp. X X X Xare

Exactly 1 pair of opp. X

are

All are X X� .�

� .�

� .�

� .

� .

� .




Chapter 6 continued

37. Slope of

Slope of

Slope of

Slope of

Isosceles trapezoid; and and are congruent but not parallel.

38.

Slope of

Slope of , which is undefined

Slope of


Trapezoid; and are the only pair of parallel sides,so not a �. No opposite sides are �, so not isosceles.

SPQR

SP �5 � 12 � 2

�40

RS �7 � 57 � 2

�25

QR �7 � 17 � 7

�60

PQ �1 � 17 � 2

�05

� 0

� 4

� �16

SP � ��2 � 2�2 � �5 � 1�2

� �29

� �52 � 22

RS � ��7 � 2�2 � �7 � 5�2

� 6

� �36

QR � ��7 � 7�2 � �1 � 7�2

� 5

� �25

PQ � ��2 � 7�2 � �1 � 1�2

RQPSSR � QP,

� 5�2

� �50

� ��1�2 � ��7�2

PS � ��1 � 2�2 � �1 � 8�2

� �16 � 4

QP � ��1 � 5�2 � �1 � 1�2

� 5�2

� �50

� �12 � ��7�2

RQ � ��5 � 4�2 � �1 � 8�2

� 2

� �4

SR � ��4 � 2�2 � �8 � 8�2

PS �8 � 12 � 1

�71

� 7

QP �1 � 15 � 1

�04

� 0

RQ �8 � 14 � 5

�7

�1� �7

SR �8 � 82 � 4

�0

�2� 0 39. Slope of

Slope of

Slope of

Slope of

Parallelogram; Sample answer: and

40. Slope of

Slope of

Slope of

Slope of

Rectangle; Sample answer: PQRS is a �, and the diago-nals and are congruent.

41.

Rhombus; Sample answer: � SP.PQ � QR � RS

� �41

� �42 � ��5�2

SP � ��5 � 1�2 � �1 � 6�2

� �41

� �42 � 52

RS � ��5 � 1�2 � �11 � 6�2

� �41

� �42 � ��5�2

QR � ��9 � 5�2 � �6 � 11�2

� �41

� ��4�2 � ��5�2

PQ � ��5 � 9�2 � �1 � 6�2

QSPR

� �65

� �12 � 82

QS � ��5 � 4�2 � �9 � 1�2

� �65

� �72 � ��4�2

PR � ��8 � 1�2 � �3 � 7�2

SP �7 � 11 � 4

�6

�3� �2

RS �3 � 18 � 4

�24

�12

QR �3 � 98 � 5

��63

� �2

PQ �9 � 75 � 1

�24

�12

QR � SP.PQ � RS

SP �7 � 10 � 1

�6

�1� �6

RS �2 � 15 � 1

�14

QR �8 � 24 � 5

�6

�1� �6

PQ �8 � 74 � 0

�14


Chapter 6 continued

42. trapezoid 43. isosceles trapezoid

44. APBQ is a kite. Its diagonals are and which are,therefore, .

45. �; if the diagonals of a quad. bisect each other, the quad.is a parallelogram. Since the diagonals are not perpendic-ular, the � is not a rhombus and since the diagonals arenot congruent, the � is not a rectangle.

46. Rhombus; if the diagonals of a quad. bisect each other,the quad is a �. Because the diagonals are perpendicular,the � is a rhombus. Since the � is not a rec-tangle, so it is not a square.

47. Kite; and bisects so you can use con-gruent s to prove that and that does not bisect so ABCD is not a �. Opp. sides arenot so ABCD is a kite.

48. EFLM is a parallelogram; EFGH is a �, so andGHJK is a �, so and

JKLM is a �, so and Thereforeby repeated application of Thm. 3.12 and

by the Trans. Prop. of �.

49. Draw a line through to Draw a line through Eparallel to Label the intersection F. CDEF is a � bydef. of a �. and are rt. because consec.

of a � are supplementary. and are rt.because opp. of a � are and CDEF is a rectan-

gle. The diagonals of a parallelogram bisect each other,so and The diagonals of a rec-tangle are so and By the definition of congruence,

50. ABCD is a quad. with diagonals The diagonalsintersect at N with

by the Base Thm. Let then because the sum of the int. of a quad.is 360°. Therefore x° � 45°. By the Angle Add. Postulate,

or and� 2x� or 90�.

By the same reasoning andThen ABCD is a rectangle.


1. PQRS is a square; 1. GivenE, F, G, and H aremidpoints of the sidesof the square.

2. 2. Definition of squareand

are right angles.

3. 3. Right Angle Congruence Theorem

4. 4. Definition of midpoint and square� EPRH � HS � SE

PF � FQ � QG � GR �

�P � �Q � �R � �S

�S�P, �Q, �R,� SP;PQ � QR � RS

m�CDA � 90�.m�BAD � 90�

� m�NCDm�BCNm�BCD �90�m�ABN � m�BNC � 2x�

�8x� � 360�m�ABN � x��DAN � �NAB

� �CDN � �NDA ��NBC � �BCN � �NCD�ABN � CN � NA.BN � ND �

BD � CA.

DM � CM.DM � CM.1

2DF �12CE,DF � CE,� ,

CM �12CE.DM �

12DF

��

�CFE�DCF�

��DEF�CDECD.

DE.C �

EF � LMEF � LM

KJ � LM.JK � LMGH � JK;HG � JKEF � HG,EF � HG

�,AC,

BDCB � CD.AB � AD�

BD,ACAC � BD

AC � BD,

�

PQAB

5. 5. Definition of are isosceles isosceles triangle

triangles.

6. 6. SAS CongruencePostulate

7. 7. Corresp. parts of s

are

8. EFGH is a rhombus. 8. Rhombus Corollary

9. 9. Base Angles Theorem; corresponding parts of

s are

10. 10. Def of congruent angles and Triangle Sum Theorem(x� � x� � 90� �180�, so x� � 45�)

11. 11. Angle Addition Postulate

12. 12. Subtraction and substitution propertiesof equality

13. EFGH is a square. 13. A rhombus with fourright angles is asquare.

52. Sample answer:

Statements Reasons

1. E, F, G, 1. Givenand H are the midpoints of and

2. 2. Def. of

3. and 3. Def. of midsegmentare midsegments.

4. 4. Midsegment Thm.

5. 5. Substitution prop. ofeq. and trans. prop. ofeq.

6. 6. Def. of

7. EFGH is a rhombus. 7. Rhombus Corollary

�EH � EF � FG � GH

FG � GHEH � EF �

GH �12JKFG �

12LM,

EF �12JK,EH �

12LM,

GHEH, EF, FG,

�JK � LM

JM.JL, KL, KM,

JK � LM,

� 90�m�HGF � m�GHEm�HEF � m�EFG �

m�GHE � 180�m�RHG � m�SHE �m�HGF � 180�,

m�RGH �m�QGF �m�GFE � 180�,

m�GFQ �m�EFP �m�HEF � 180�,m�FEP � m�HES �

m�SHE � m�SEH � 45�m�RGH � m�RHG �m�GFQ � m�FGQ �m�FEP � m�EFP �

�SHE � �SEH� .��RGH � �RHG �

�GFQ � �FGQ ��FEP � �EFP �

�.��EF � FG � GH � HE

�GRH � �HSE�EPF � �FQG �

�HSE�EPF, �FQG, �GRH,






Chapter 6 continued

53. a. Isosceles; by the Angle Addition Post., m�KLM �m�KLJ � m�JLM and m�KJN � m�KJL �m�LJN. Since JKLMN is a regular pentagon,m�KLM � m�KJN and, so, m�KLJ � m�JLM �m�KJL � m�LJN. By the Base � Thm., �KLJ ��KJL, so m�KLJ � m�KJL and by the subtractionprop. of �, m�JLM � m�LJN and �LJN � �JLM.

b. m�LMN � m�JNM and m�LJN � m�JLM; Sincethe sum of measures of the interior � of a quad. is360�, m�LMN � m�JNM � m�LJN � m�JLM �360�. Then, by the substitution prop. of �,2m�JLM � 2m�LMN � 360� and m�JLM �m�LMN � 180�. Since �MLJ and �LMN are supp., by the Consec. Int. � Converse.

c. parallelogram

d. Yes; because JPMN is a parallelogram, opposite sidesmust be congruent. We know that (JKLMNis a regular pentagon), so and Bythe Transitive Property of Congruence,

so JPMN is a rhombus.

54. Isosceles trapezoid; use the SAS Cong. Post. to show that�AND � �BNC, so Then use the Vertical �Thm., the Triangle Sum Thm., and the Base � Thm. toshow that �ACD � �CAB, so that . Finally,since and do not bisect each other, ABCD is not a �, so ABCD is an isosceles trapezoid.


55.

56.

57.

58.

59.

60.

61.

62. m�A � (32x � 15)� � [32(1.75) � 15]� � 71�

63. midsegment

64. midsegment

65. midsegment

Lesson 6.7

Developing Concepts Activity 6.7 (p. 371)

1. The areas, bases and heights are the same.

2. The area of each is the area of the �. The bases andheights are the same for both the and the �.�

12�

�12�AD � BC� �

12�2 � 8� � 5

�12�AD � BC� �

12�4 � 8� � 6

�12�AB � DC� �

12�4 � 10� � 7

x � 1.75

76x � 133

76x � 227 � 360

�32x � 15�� 133� � 80� � �44x � 1�� 360�

A �12bh �

12 � 12 � 8 � 48 sq. units

A �12bh �

12 � 12 � 5 � 30 sq. units

A � lw � �9��6� � 54 sq. units

A � lw � �5��3� � 15 sq. units

A � s2 � 72 � 49 sq. units

A � s2 � 42 � 16 sq. units

BDACAB � CD

AD � BC.

JP � MN,JN � PM �JP � MN.JN � PM

JN � MN

MN � LJ

3. The area of each trapezoid is the area of the parallelo-gram. The base of the parallelogram is equal to the sumof the bases of the trapezoids. The heights are the same.

Extension

parallelogram: A � bh; triangle: ; trapezoid:


1. The midsegment of a trapezoid is the segment thatconnects the midpoints of the legs.

2. 6 3. A 4. E 5. C 6. B 7. D

8.

9.

10.

11.

12.

13.


14.

15.

16.

17.

18.

19.

20. 21.

22. 23.

24. 25.

26. 27.

28. 29.

x � 3 in.

16x � 48

2Ah

� b 12

�2x��16� � 48

A �12

bh 12

d1d2 � A

x � 12 ft

4x � 48 x � 9 cm

12�8��x� � 48 7x � 63

12d1d2 � A bh � A

� 70 sq. units � 168 sq. units

� 12�14��10� � 1

2�8 � 16��14� A �

12d1d2 A �

12�b1 � b2�


� 16 � 15 � 12�24 � 7��24�

A � bh A �12�b1 � b2�h


� 12�38��19� � 1

2�6 � 10��8� A �

12d1d2 A �

12�b1 � b2�h

A �12bh �

12�5��4� � 10 sq. units

A � bh � �22��21� � 462 sq. units

A � bh � �15��8� � 120 sq. units

A � bh � �5��9� � 45 sq. units

A � s2 � 72 � 49 sq. units

A �12bh �

12�5��7� � 17.5 sq. units

A �12h�b1 � b2� �

12�6��8 � 4� � 36 sq. units

A �12d1d2 �

12�12��12� � 72 sq. units

A �12d1d2 �

12�10��8� � 40 sq. units

A � bh � �9��4� � 36 sq. units

A � s2 � 52 � 25 sq. units

A �12bh �

12�7��4� � 14 sq. units

A �12h�b1 � b2�

A �12bh

12


Chapter 6 continued

30. 31.

32. 33.

34.

35. 36.

37. 38.

39. No; although the bases and the heights of two such �sare the same, the angles of the two parallelograms maynot be congruent.

40. Yes. By definition, a rectangle has four right angles. Sothe angles of one rectangle will always be congruent tothe of any other rectangle. Any two rectangles witharea and base 6 ft have a height of 4 ft and are con-gruent.

41.

42.

43.

44. The area would be doubled if the length of one of thediagonals was doubled. The area of the kite would bequadrupled if the lengths of both diagonals were doubled.

45.

1440 in.2

3 in.2 �480 carnations

10 ft2 �144 in.2

1 ft2� 1440 in.2

A � �2��5� � 10 ft2

h � 12

� 192 sq. units h2 � 144

� �16��12� h2 � 400 � 256

A � bh h2 � 202 � 162

5 � b

� 100 sq. units 25 � b2

� 12�20��10� 169 � 144 � b2

A �12d1d2 132 � 122 � b2

8 � b

� 24 sq. units 64 � b2

� 12�8��6� 100 � 36 � b2

A �12bh 102 � 62 � b2

24 ft2�

� 392.5 in.2 � 552 in.2 � �20��16� �

12�9 � 20��5� � �16 � 30�12

A � bh �12�b1 � b2�h A � 2�1

2�b1 � b2�h � 1824 in.2 � 3 ft2 � �48��32� �

12�48��12� � 1

2�3��2� A � bh �

12bh A �

12d1d2

� 5 square units

A � �5�2

S � �1 � 0�2 � �3 � 1�2 � 12 � 22 � 5

A � s2


� 12�2��4� � �3��4�

A �12d1d2 A � bh

2Ah

� b2 � b1

2Ah

� b1 � b2 2Ad2

� d1

A �12

�b1 � b2�h A �12

d1d2 46.

47.

48.

49.

50.

Area of blue

51.

52.

53. Square, square; Sample answer: In quad. EBFJ,and are right angles by the Linear Pair Postulate and

is a right angle by the Interior Angles of aQuadrilateral Theorem. Then EBFJ is a rectangle by theRectangle Corollary. because they are corresp.parts of �s. Then, by the definition of a � and theTransitive Property of Congruence, EBFJ is a rhombusand, therefore, a square. Similarly, HJGD is a square.

�EJ � FJ

�B�F

�E, �J,

� 56 sq. units

� �42 ��72 � Area of blue � bh

� 65 sq. units

� 16 � 49

Area of yellow �12�42 ��42 � �

12�72 ��72 �

� 121 sq. units

ATotal �12�112 ��112 �

� 96 sq. units

� �12��16� � 96

Area of yellow � bh � Area of blue

� 96 sq. units

� �12��8� Area of blue � 2�1

2bh� � 36 sq. units

� 2�3��6� Area of yellow � 4�1

2bh�� s2 � �35�2 � 45 sq. units

� 35

� 45

s � 32 � 62

602,292 in.2

100 in.2� 6023 shakes

� 602,292 in.2 or about 4182.6 ft2 � 561,216 � 41,076

� �1908 � 1644��158� � �252��163� A � 2�1

2�b1 � b2�h � 2�12bh�

1728 in.2

4 in.2 �432 chrysanthemums

12 ft2 �144 in.2

1 ft2� 1728 in.2

A �12

�3��8� � 12 ft2

1152 in.2

2 in.2 �576 daisies

8 ft2 �144 in.2

1 ft2� 1152 in.2

A �12

�5 � 3��2� � 8 ft2






Chapter 6 continued

54. Square; each side has the same length, , so ABCDis a rhombus, and each is a right , so ABCD is a rectangle.

55. length of each side and the area of ABCDwould be

56. The area of The area of

57.

58. Show that the area of Show thatArea of Area of Area of KMNQ.Then, use the distributive and division properties ofequality to show that the Area of

59. Show that the area of Sinceand are Area of Area of

Area of Area of AreaArea of

60.

D

61.

B

62. Area of Area of Area of Areaof The area of So, the area of but PR is thelength of one diagonal, and is the length ofthe other diagonal, Therefore,


63. obtuse; about 64. right; about

65. acute; about


68. 69.

70.

6 � x

2 �13x

x � 6 �43x � 4

x � 6 �23�2x � 6�

1 � x

2 � 2x 7 � x

6 � 4 � 2x 21 � x � 14

4 �23�4 � 2x� 14 �

23�x � 14�

1

1

(�5, 0) (5, 0)

(0, 5) y

x

1

1

(0, 0) (3, 0)

(4, 4)

y

x

15�

90�140�

A �12d1d2.d2.

QT � TSd1,

12�PR��QT � TS�,PQRS �

12�PR��TS�.�PRS ��PQR �

12�PR��QT�.

�PRS.�PQR �PQRS �

A � �5��3� � 15 cm2

A �12�13 � 8��4� � 42 in.2

12h�b1 � b2�.AEGH �GHDF �

AEFD �EBCF �AEFD �ABCD ��,GHDFEBCF

AEGH �12h�b1 � b2�.LPQK �

12�b1 � b2�h.

LMNP �KLPQ �KMNQ � �b1 � b2�h.

bh � A

2bh � 2A

b2 � 2bh � h2 � b2 � h2 � 2A

HJGD � b2.EBFJ � h2.

�b � h�2 � b2 � 2bh � h2.� b � h

��

b � h Quiz 3 (p. 380)

1.

Kite; and but opposite sides arenot congruent.

2. Slope of , which is undefined

Slope of


Slope of

Trapezoid; but and are not parallel.

3.

Slope of

Slope of

—CONTINUED—

XY �1 � 34 � 2

��22

� �1

WX �1 � 04 � 2

�12

� 8 � 22

� ��2�2 � 22

ZW � �0 � 2�2 � �2 � 0�2

� 5

� 22 � 12

YZ � �2 � 0�2 � �3 � 2�2

� 8 � 22

� 22 � ��2�2

XY � �4 � 2�2 � �1 � 3�2

� 5

� ��2�2 � ��1�2

WX � �2 � 4�2 � �0 � 1�2

RSTQQR � ST,

� 8 � 22

� 22 � 22

TQ � �4 � 2�2 � �4 � 2�2

� 5

� ��2�2 � 12

RS � �2 � 4�2 � �1 � 0�2

TQ �4 � 24 � 2

�22

� 1

ST �0 � 44 � 4

� �40

RS �1 � 02 � 4

� �12

QR �2 � 12 � 2

�10

PM � MN,ON � OP

� 13

� ��3�2 � ��2�2

MN � �0 � 3�2 � �1 � 3�2

� 13

� ��2�2 � ��3�2

PM � �1�3�2 � �0 � 3�2

� 1 � 1

OP � �0 � 1�2 � �0 � 0�2

� 1 � 1

ON � �0 � 0�2 � �0 � 1�2


Chapter 6 continued

3. —CONTINUED—

Slope of

Slope of

Parallelogram; Sample answers: and or and

4. 5.

6. 7.

Chapter 6 Review (p. 382)


3.

4.

5.

6.

7.

8.

9. No; you are not given information about opposite sides.

10. Yes; opposite are

11. Yes; you can prove and opposite sidesare

12. Yes; consecutive angles are supplementary.

� .�PQT � �SRT

� .�

� 44 units

� 2�12� � 2�10� P � 2l � 2w

m�DEF � 180� � m�EFG � 180� � 65� � 115�

m�EFG � m�GDE � 65�

DF � DH � FH � 9.5 � 9.5 � 19

FH � DH � 9.5

x � 13

15x � 195

15x � 165 � 360

6x� � 9x� � 75� � 90� � 360�

x � 22.5

8x � 180

8x � 180 � 360

5x� � 3x� � 90� � 90� � 360�

� 115

x � 360 � �67 � 115 � 63� 67� � 115� � 63� � x� � 360�

8 in. � x

� 52.11 cm2 60 �12�15��x�

A �12�8.3 � 11��5.4� A �

12d1d2

x � 12 in. x � 5 in.

12�x��10� � 60 12x � 60

12bh � A bh � A

XY � ZW.WX � YZXY � ZW,WX � YZ

ZW �2 � 00 � 2

�2

�2� �1

YZ �3 � 22 � 0

�12

13. rhombus, square


15. rhombus, square

16.

17.

18. It is given that and by the Reflexive Property of Congruence.

by the SAS Congruence Postulate.by corresponding parts of s are

19.

Square; Sample answer: so PQRSis a rhombus. Diagonals and are �, so PQRS isa rectangle. A quad. that is both a rhombus and a rectan-gle is a square.

20.

—CONTINUED—

� �13

� �22 � ��3�2

QR � ��8 � 6�2 � �5 � 8�2

� 10

� �100

� �36 � 64

PQ � ��6 � 0�2 � �8 � 0�2

QSPRPQ � RS � QR � PS

� 2�17

� �68

� �82 � ��2�2

QS � ��5 � ��3��2 � �6 � 8�2

� �34

� �32 � ��5�2

PS � ��0 � ��3��2 � �3 � 8�2

� �34

� �32 � ��5�2

QR � ��5 � 2�2 � �6 � 11�2

� 2�17

� �68

� �22 � 82

PR � ��2 � 0�2 � �11 � 3�2

� �34

� �52 � 32

RS � ��2 � ��3��2 � �11 � 8�2

� �34

� �52 � 32

PQ � ��5 � 0�2 � �6 � 3�2

� .��ACD � �BDC�ADC � �BCD,DC � DC

�ADC � �BCD.AD � BC

m�BCD � m�ADC � 68�

m�ADC � 180� � m�DAB � 180� � 112� � 68�

m�ABC � m�DAB � 112�

� 11 units

midsegment �12�6 � 16�



MCRBG-060R-SKa.qxd 5-25-2001 11:20 AM Page 130



Chapter 6 continued

20. —CONTINUED—

Slope of

Slope of

Slope of

Slope of

Trapezoid; but and are not parallel.

21.

Slope of

Slope of

Rhombus; and the diagonals arebut not � so PQRS is not a rectangle (square).

22.

� 2�10

� �40

� ��2�2 � ��6�2

PQ � ��5 � ��3��2 � �0 � 6�2

�

PQ � QR � RS � SP

� �72 � 6�2

� �62 � 62

QS � ��4 � ��2��2 � ��5 � 1�2

� �8 � 2�2

� �22 � 22

PR � ��2 � 0�2 � ��1 � ��3��2

QS �1 � ��5��2 � 4

�6

�6� �1

PR ��3 � ��1�

0 � 2�

�2�2

� 1

� �20 � 2�5

� ��4�2 � 22

SP � ��2 � 2�2 � �1 � ��1��2

� �20 � 2�5

� �22 � ��4�2

RS � ��0 � ��2��2 � ��3 � 1�2

� �20 � 2�5

� �42 � ��2�2

QR � ��4 � 0�2 � ��5 � ��3��2

� �20 � 2�5

� �22 � ��4�2

PQ � ��4 � 2�2 � ��5 � ��1��2

RSPQPS � QR,

QR �5 � 88 � 6

� �32

PS ��6 � 04 � 0

��64

� �32

RS ��6 � 54 � 8

��11�4

�114

PQ �8 � 06 � 0

�86

�43

� �52 � 2�13

� �16 � 36

SP � ��4 � 0�2 � ��6 � 0�2

� �137

� �42 � 112

RS � ��8 � 4�2 � �5 � ��6��2

Kite; two pairs of consecutive sides are congruent but opp. sides are not congruent.

23. 24.

25.

Chapter 6 Test (p. 385)

1. Sample answer:

2.

3.

4.

y � 180 � 110 � 70

5. 6. no 7. yes 8. no

9. yes 10. sometimes 11. never 12. always

13. Trapezoid; exactly one pair of sides are parallel.

14. Rhombus; the diagonals bisect each other and are perpendicular.

15. Rectangle; one pair of opposite sides are both � and ,and since one angle is a right angle, the opposite angleand the two consecutive angles are also right angles.

16. Square; the diagonals bisect each other and are congruentso the quad. is a �, a rectangle, and a rhombus. So it is asquare.

�

x � 4 y �72

x � 6 � 10 2y � 7

x � 110

x � 3

y � 8 �2x � �6

12y � 4 3x � 5x � 6

x � 115

x � 245 � 360

x� � 100� � 70� � 75� � 360�

� 12 sq. units

� 12�4��6�

A �12d1d2

� 13.5 ft2 � 2934 in.2

� 12�3 � 6��3� � 1

2�812��7�

A �12�b1 � b2�h A �

12bh

� 2�10

� �40

� �62 � 22

SP � ��1 � ��5��2 � �2 � 0�2

� 4

� �42

RS � ��1 � 1�2 � �6 � 2�2

� 4

� ��4�2

QR � ��3 � 1�2 � �6 � 6�2


Chapter 6 continued

17.

So,

Let O be the origin (where the diagonals meet). OX �OZ � b and OW � OY � a, so the diagonals bisect eachother. One diagonal is vertical and the other is horizontal,so they are perpendicular.

18. 19.

; �ABC � �ADC (SSS Cong. Post.) so �BAC � �DAC. Then �BAP � �DAP (SAS Cong.Post.). Corr. � BPA and DPA are �. Since and form a linear pair of � �, .

20.

Chapter 6 Standardized Test (pp. 386–387)

1. D

2.

C

3. 4. D

C

5.

� �26

� ��5�2 � 12

NP � ��1 � 6�2 � ��6 � ��7��2

� �104 � 2�26

� �22 � 102

MN � ��3 � 1�2 � �4 � ��6��2

q � 18

6q � 108

6q � 72 � 180

8�9� � 17 � 6q � 17 � 180

�8p � 17�� 6q � 17�� 180�

p � 9

�8p � 17�� 7p � 8��

x � 13

20x � 260

20x � 100 � 360

�7x � 1�� 8x � 27�� 108� � �5x � 18�� 360�

� 1218 ft2 � 918 � 300

� �32 � 22��17� � �20��15� A � 2�1

2�b1 � b2�h � 2�12bh�

AC � BDBDAC

AC � BD

� 10.5 in.

x �12�6 � 15�

D

B

CA P

WX � XY � YZ � ZW.

� �a2 � b2 ZW � ��0 � a�2 � ��b � 0�2

� �a2 � b2 YZ � ��0 � ��a��2 � ��b � 0�2

� �a2 � b2 XY � ��0 � ��a��2 � �b � 0�2

� �a2 � b2 WX � ��a � 0�2 � �0 � b�2

Opposite sides are congruent so MNPQ is a �. The diagonals are �, so MNPQ is a rectangle.

C

6. A

7.

Two pairs of consecutive sides are congruent but opposite sides are not congruent.

B

8. The figure in column A is a trapezoid with base lengthsof 8 and 12 and height 9. The area is 90 sq. units. Thefigure in column B is a parallelogram with base 11 andheight 9 so its area is 99 sq. units.

B

9.

D

10. The base units and height units. So,

D

11. Yes; and so FBDH is a parallelogramand the diagonals of a parallelogram bisect each other.Therefore,

12.

� 80� m�BDH � 180� � m�DHF � 180� � 100�

� 100� � 180� � 40� � 40�

m�DHF � m�FBD � 180� � m�ABF � m�CBD

BE � HE.

BD � FHFB � DH,

A �12�10��13� � 65 sq. units.

� 10� 13

30 � x

45 �32x

42 �32x � 3

21 �12�3

2x � 3� 21 �

12��x � 4� � �1

2x � 1�

� �328 � 2�82 � �22 � 182

UR � ��3 � ��5��2 � �11 � ��7��2

� �328 � 2�82 � �22 � ��18�2

TU � ��1 � ��3��2 � ��7 � 11�2

� �8 � 2�2 � ��2�2 � ��2�2

ST � ��3 � ��1��2 � ��9 � ��7��2

� �8 � 2�2 � ��2�2 � 22

RS � ��5 � ��3��2 � ��7 � ��9��2

� �26

� �52 � ��1�2

QM � ��8 � 3�2 � �3 � 4�2

� �104 � 2�26

� �22 � 102

PQ � ��8 � 6�2 � �3 � ��7��2






Chapter 6 continued

13. FBDH is a � (Ex. 11), so � , � , and� so �BEF � �HED (SSS Cong. Post.).

14. FBDH would be a rhombus because a � in which thediagonals are to each other is a rhombus.

15.

16. 17.

18. 19. convex pentagon


1. PSTU is a rectangle, 1. Given

2. and are 2. Def. of rectanglert.

3. 3. All rt. are

4. 4. If a quad is a �, thenopp. sides are

5. 5. SAS Congruence Post.

21. ABCD is an isosceles trapezoid; (1) show that , and and are not ; or (2) show that

and are not , and

22.

Cumulative Practice, Chs. 1–6 (pp. 388–389)

1. 2.

So, would be

3. (�1 � �2)

(Consec. Int. � Thm.)

(Alt. Int. � Thm.)

(Ext. � Thm.)

� 135�

� 45� � 90�

m�URQ � m�RQP � m�QPR

m�RQP � 45� � m�PTS

2m�RQP � 90�

�QS � ST� 2m�RQP � 180� � 90�

2m�RQP � 180� � m�QST

m�RQP � m�SQP

0.89.8999

B C

A

D

499 � 0.04; 18

99 � 0.18; 3599 � 0.35

� 72 sq. units�12�8 � 4��12� A �

12�b1 � b2�h

AC � BD.�BCADAB � CD,�BCADAD � BC

AB � CD,

�PQU � �SRT

� .PU � ST

� .��UPQ � �TSR

�.�TSR�UPQ

PQ � SR.

9 � b

3b � 4 � 4b � 5

UY � TZ

a � 6 y � 2

5a � 30 2y � 4

3a � 2 � 28 � 2a 6y � 1 � 4y � 5

3a � 2 � 2�14 � a� 6y � 1 � 2y � 5 � 2y

PT � 2UW UT � PQ � QR � RS

5 � x

25 � 5x

18 � 5x � 7

9 �12��3x � 10� � �2x � 3�

YZ �12�QR � UT�

�

EDEFHEBEHDBF 4. Sample answers:

a. and b. and

c. and d. and

5. and and so

by the ASA Congruence Postulate. Therefore,since corresponding parts of s are

and by the Alternate InteriorAngles Theorem, so by the AASCongruence Theorem.

6. Since PR � PS (Ex. 5) by the Segment AdditionPostulate, RS � PS � PR. By the Substitution propertyof equality, RS � PS � PS �

7. P is equidistant from and by the Angle BisectorTheorem.

8. concave pentagon

9.

obtuse

10. no 11. no 12. yes; SAS Congruence Postulate

13. yes; HL Congruence Theorem

14. slope of

15.

Since is an isosceles �.�ABCAB � CA,

� �89

� �82 � 52

CA � ��11 � 3�2 � �1 � ��4��2

� 10

� �102

BC � ��3 � 3�2 � �6 � ��4��2

� �89

� �82 � ��5�2

AB � ��11 � 3�2 � �1 � 6�2

y � �58x �

178

�178 � b

�4 � �158 � b

�4 � �58�3� � b

y � mx � b

↔AB �

6 � 13 � 11

� �58

�180 � �21�7� � 1�� 32�

�14�7� � 3� � 95�

�5�7� � 18� � 53�

7 � x

14 � 2x

�2x � �14

19x � 15 � 21x � 1

�5x � 18�� 14x � 3�� 21x � 1��

→QR

→QS

2 � PS.

� �TPS�QPR�QRP � �TSP�2 � �T

PR � PS.�,��QPS

�QPR �m�QPR,m�QPS � 90� �QP � QP,�1 � �2,�QPR � �TPS;�QPR � �QPS

�TSQ�RQS�TSQ�RQS

�QRS�SQT�SPT�QPR


Chapter 6 continued

16. by the Base Angles Theorem.

17. slope of

The slope of the bisector of is .

18. the circumcenter, or the intersection of the three bisec-tors of �ABC

19. The centroid is two thirds of the distance from each ver-tex to the midpoint of the opposite side. Use the medianfrom vertex A to The midpoint of is Callthis D. AD � 8 since 11 � 3 � 8. So the distance from

A to the centroid � � � The x-coordi-

nate of the centroid is then 11 � The

coordinates of the centroid are

20. length of

21. If two angles are supplementary, then they form a linearpair; false; Sample answer: two consecutive angles of aparallelogram are supplementary, but they do not form alinear pair.

22. 4 < XZ < 20

�12�89 �

�892

�12��11 � 3�2 � �1 � ��4��2

midsegment �12AC

1 x

y

1D (3, 1)

A (11, 1)

C (3, �4)

B (3, 6)

( ), 1173

�173 , 1�.163 �

333 �

163 �

173 .

163 .2

3�8�23�AD�

�3, 1�.BCBC.

�

y � �85

x �9710

9710

� b

�32

� �565

� b

�32

� �85

�7� � b

y � mx � b

� �7, �32�

midpoint of AC � �11 � 32

, 1 � 4

2 �

�85

AC�

AC ��4 � 13 � 11

�58

�B � �C 23. the angle opposite the longer side is largerthan the angle opposite the shorter side.

24. Use the converse of the Hinge Theorem in triangles ACDand WYZ. Since

25. rhombus

26.

27. slope of

slope of

slope of

slope of

Sample answer: PQRS is a parallelogram because bothpairs of opposite sides are parallel.

28. Trapezoid; Since and are supple-mentary, but the other two sides are not

29. a. rhombus, kite, square

b. rectangle, isosceles trapezoid, square

30.

31. Sample answer: Show that by theLinear Pair Post. so

by the AAS Congruence Theorem.

32. Rectangle; since and are both perpendicular to so by the Propertyof Perpendicular Lines. One pair of opposite sides ofquad. ADEB are both congruent and parallel, so ADEB isa parallelogram. and are both right angles,so all four angles are right angles and quad. ADEB mustbe a rectangle.

33.

34. The length of the middle shelf is one half the distancebetween the supports on the floor. So, the distancebetween supports on the floor is 60 in. The length of the top shelf is 1

2�30� � 15 in.

138� � 2 � 69�

180� � 42� � 138�

�DEB�ABE

AB � DEBE,DEABAB � DE.�ABC � �DEF,

�DEF�ABC �m�DEF,m�ABC � 90� �

m�C � m�F � 65�

� 51 sq. units

� 12�12 � 5��6�

Area �12�b1 � b2�h

� 172 units

Length of midsegment �12�b1 � b2� �

12�12 � 5�

�.EH � FG,�F�Em�H � 113�.

SP �9 � 41 � 0

� 5

RS �9 � 81 � 9

�18

QR �8 � 39 � 8

� 5

PQ �3 � 48 � 0

� �18

m�Z � 25�; m�W � m�Y � 180� � 25� � 155�

m�D > m�Z.AC > WY,

m�X > m�Z;



MCRBG-060R-SKb.qxd 5-25-2001 11:19 AM Page 134



Chapter 6 continued

35.

Algebra Review (pp. 390–391)

1. 2. 3. 4.

5. 6. 7. 8. 9.

10. 11.

12. 13.

14. 15.

16. 17.

18. 19.

20.

21. 22.

23. 24.

y � 200 q � 9

7y � 1400 4q � 36

7

100�

14y

2q

�4

18

8 � x

x � 4 3x � 4x � 8

5x � 20 3x � 3 � 3 � 4x � 8

x

20�

15

3�x � 1� � 3 � 4�x � 2�

a � �2

3a � �6

15a � 10 � 12a � 16

5�3a � 2� � 2�6a � 8� m �

272

23m � 9 x �79

23m �163 �

113 �9

2x � �72

23m �83 � 8 �

113 5 �

92x �

32

23�m � 4� � 8 �113 12�10 � 9x� �

32

x � 2 x � �1

27x � 54 �4x � 4

�45 � 27x � 9 �4x � 24 � 28

�9�5 � 3x� � 9 �4�x � 6� � 28

76 � x d �134

14 � 12x �16d � �52

14 � 7x � 5x 48 � 16d � �4

7�2 � x� � 5x 16�3 � d� � �4

y � �4 x � 6

�10y � 40 6x � 36

�10y � 80 � �40 6x � 12 � 24

�10�y � 8� � �40 6�x � 2� � 24

x � �3 x � 3

8x � �24 2x � 6

8x � 48 � 24 2x � 14 � 20

8�x � 6� � 24 2�x � 7� � 20

1616 �

11

2016 �

54

82 �

41

220180 �

119

824360 �

10345

228

�114

27 � 230 � 3

�2527

74

2025

�45

� 438.75 in.2 � 1

2�30 � 15��1912�

A �12�b1 � b2�h 25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

x � �864

� �432

y � �29

�4x � 86 �6y � 174

�4x � 4 � 90 �6y � 6 � 168

1

18�

5�4�x � 1�

�38

�21

2�y � 1�

�4 � z

a � 9 �360 � 90z

66a � 594 90 � 90z � 450

3a11

�5422

6

45�

2z � 1015

w � 5

12w � 60 x �959

12w � 24 � 84 9x � 95

3w � 6

28�

34

19x

�95

5 � m d � 4

150 � 30m 6d � 24

6

5m�

625

38

�3

2d

m � 30 y �32

19m � 570 100y � 150

6

19�

m95

y

50�

3100

z �1527

�59

w �4217

27z � 15 17w � 42

275

�3z

w6

�7

17

r �245

t � 12

5r � 24 9t � 108

56

�4r

t

27�

49


Chapter 6 continued

39. 40.

41. 42.

r � �23

w � ±6 �3r � 2

w2 � 36 3r � 6r � 2

w4

�9w

r

3r � 1�

23

p � �3 23

� m

2p � �6 6 � 9m

3p � p � 6 42 � 9m � 36

3

p � 6�

1p

3

m � 4�

914




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