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Answers to Exercises13. 1714. 1515. the twelfth century 16. The angles of the trapezoid measure 67.5 and112.5; 67.5 is half the value of each angle of a regularoctagon, and 112.5 is half the value of 360 135.

17. Answers will vary; see the answer forDeveloping Proof on page 259. Using the TriangleSum Conjecture, a b j c d k e f l g h i 4(180), or 720. The four angles inthe center sum to 360, so j k l i 360.Subtract to get a b c d e f g h 360.18. x 12019. The segments joining the opposite midpointsof a quadrilateral always bisect each other.20. D21. Counterexample: The base angles of anisosceles right triangle measure 45; thus they arecomplementary.

67.5

135

ANSWERS TO EXERCISES 61

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Exe

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CHAPTER 5 CHAPTER CHAPTER 5 CHAPTER

LESSON 5.1

1. See table below.2. See table below.3. 1224. 1365. 108; 366. 108; 1067. 105; 828. 120; 389. The sum of the interior angle measures of thequadrilateral is 358. It should be 360.10. The measures of the interior angles shown sumto 554. However, the figure is a pentagon, so themeasures of its interior angles should sum to 540.11. 1812. a 116, b 64, c 90, d 82, e 99,f 88, g 150, h 56, j 106, k 74,m 136, n 118, p 99; Possible explanation:The sum of the angles of a quadrilateral is 360, soa b 98 d 360. Substituting 116 for aand 64 for b gives d 82. Using the largerquadrilateral, e p 64 98 360.Substituting e for p, the equation simplifies to 2e 198, so e 99. The sum of the angles of apentagon is 540, so e p f 138 116 540. Substituting 99 for e and p gives f 88.

5

Number of sides of polygon 7 8 9 10 11 20 55 100

Sum of measures of angles 900 1080 1260 1440 1620 3240 9540 17640

Number of sides of 5 6 7 8 9 10 12 16 100equiangular polygon

Measure of each angle 108 120 12847 135 140 144 150 15712

17625

of equiangular polygon

1. (Lesson 5.1)

2. (Lesson 5.1)

62 ANSWERS TO EXERCISES

LESSON 5.2

1. 3602. 72; 603. 154. 435. a 108

6. b 4513

7. c 5137, d 11557

8. e 72, f 45, g 117, h 1269. a 30, b 30, c 106, d 13610. a 162, b 83, c 102, d 39, e 129,f 51, g 55, h 97, k 8311. See flowchart below.12. Yes. The maximum is three. The minimum iszero. A polygon might have no acute interiorangles.

13. Answers will vary. Possible proof using the diagram on the left: a b i 180, c d h 180, and e f g 180 by the TriangleSum Conjecture. a b c d e f g h i 540 by the addition property of equality.Therefore, the sum of the measures of the angles ofa pentagon is 540. To use the other diagram,students must remember to subtract 360 toaccount for angle measures k through o.14. regular polygons: equilateral triangle andregular dodecagon; angle measures: 60, 150,and 15015. regular polygons: square, regular hexagon, andregular dodecagon; angle measures: 90, 120,and 15016. Yes. RAC DCA by SAS. AD CR byCPCTC.17. Yes. DAT RAT by SSS. D R byCPCTC.

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1 a b 180

2 c d 180

4 a + b + c + d + e + f =

3 e f 180Subtraction propertyof equality

Addition property of equality

6 b + d + f =

?

?

?

?

?

?

?

5 a + c + e =

Linear Pair Conjecture

Linear Pair Conjecture

Linear Pair Conjecture

540

180

360

Triangle Sum Conjecture

11. (Lesson 5.2)

LESSON 5.3

1. 64 cm 2. 21; 146 3. 52; 1284. 15 cm 5. 72; 61 6. 99; 38 cm 7. w 120, x 45, y 308. w 1.6 cm, x 48, y 429. See flowchart below.10. Answers may vary. This proof uses the KiteAngle Bisector Conjecture.

Given: Kite BENY with vertex angles B and NShow: Diagonal BN is the perpendicular bisectorof diagonal YE.

From the definition of kite, BE BY. From theKite Angle Bisector Conjecture, 1 2. BXBX because they are the same segment. By SAS,BXY BXE. So by CPCTC, XY XE.Because YXB and EXB form a linear pair, theyare supplementary, so mYXB mEXB 180. By CPCTC, YXB EXB, or mYXB mEXB, so by substitution, 2mYXB 180, ormYXB 90. So mYXB mEXB 90.Because XY XE and YXB and EXB are rightangles, BN is the perpendicular bisector of YE.11. possible answer: E I

12. possible answer:

The other base is ZI. Q and U are a pair of baseangles. Z and I are a pair of base angles.

Z I

Q U

E

KT

I

B N

Y

E

21 X

13. possible answer:

OW is the other base. S and H are a pair ofbase angles. O and W are a pair of base angles.SW HO.14. Only one kite is possible because three sides determine a triangle.15.

16. infinitely many, possible construction:

17. 80, 80, 100, 10018. Because ABCD is an isosceles trapezoid, A B. AGF BHE by SAA. Thus, AG BHby CPCTC.19. a 80, b 20, c 160, d 20, e 80,f 80, g 110, h 70, m 110,n 100;Possible explanation: Because d forms a linear pairwith e and its congruent adjacent angle,d 2e 180.Substituting d 20 gives 2e 160,so e 80.Using theVerticalAngles Conjecture and d 20, theunlabeled angle in the small right triangle measures20, which means h 70. Because g and h are alinear pair, they are supplementary, so g 110.

B O

NE

S H

WI

F

E

BN

W O

S H

ANSWERS TO EXERCISES 63

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9. (Lesson 5.3)

? and ?

61 and5

3 ? Congruence

shortcutDefinition ofangle bisector

4

?

Given

1 BE BY

Given

2 EN YN

Same segment

3? ?

BEN ? ?

?

BN BN

BYN 2 BN bisects B, BN bisects N

4

CPCTCSSS

64 ANSWERS TO EXERCISES

LESSON 5.4

1. three; one 2. 28 3. 60; 140 4. 655. 23 6. 129; 73; 42 cm7. 35 8. See flowchart below.9. Parallelogram. Draw a diagonal of the originalquadrilateral. The diagonal forms two tri-angles.Each of the two midsegments is parallel to thediagonal, and thus the midsegments are parallel toeach other. Now draw the other diagonal of theoriginal quadrilateral. By the same reasoning, thesecond pair of midsegments is parallel. Therefore,the quadrilateral formed by joining the midpoints isa parallelogram.10. The length of the edge of the top basemeasures 30 m. We know this by the TrapezoidMidsegment Conjecture.11. Ladie drives a stake into the ground to create atriangle for which the trees are the other twovertices. She finds the midpoint from the stake toeach tree. The distance between these midpoints ishalf the distance between the trees.

12. Explanations will vary.

80

40 60

60 cm

Cabin

13. If a quadrilateral is a kite, then exactly onediagonal bisects a pair of opposite angles. Both theoriginal and converse statements are true.14. a 54, b 72, c 108, d 72, e 162,f 18, g 81, h 49.5, i 130.5, k 49.5,m 162, n 99; Possible explanation: The thirdangle of the triangle containing f and g measures81, so using the Vertical Angles Conjecture, thevertex angle of the triangle containing h alsomeasures 81. Subtract 81 from 180 and divide by2 to get h 49.5. The other base angle must alsomeasure 49.5. By the Corresponding AnglesConjecture, k 49.5.15. (3, 8)16. (0, 8)17. coordinates: E(2, 3.5), Z(6, 5); the slope ofEZ 38, and the slope of YT

38

18.

There is only one kite, but more than one way toconstruct it.

K

R

NF

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FOA withmidsegment LN

Given

LN OA3

4

1

Triangle MidsegmentConjecture

IOA withmidsegment RD

Given

2

5

Two lines parallel to the same line are parallel

?

?

?

OA RD

LN RDTriangle Midsegment Conjecture

8. (Lesson 5.4)

LESSON 5.5

1. 34 cm; 27 cm 2. 132; 483. 16 in.; 14 in.4. 63 m 5. 80 6. 63; 787.

8.

9. Vh

Vw

D

P

P

O

R

R

T S

AL

10.

11. (b a, c)12. possible answer:

13. See flowchart below.14. The parallelogram linkage is used for thesewing box so that the drawers remain parallel toeach other (and to the ground) so that the contentscannot fall out.

15. a 135, b 9016. a 120, b 108, c 90, d 42, e 69

a

a

b

b

c

c

d

d

Vb

Vc

ANSWERS TO EXERCISES 65

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ET NT

EA LN

2

LEAN is a parallelogram 7

CPCTC

1

AIA Conjecture

3 AEN LNE

?

?

4

AIA Conjecture

? 6

ASA

?

8

CPCTC

?

9

Definition of segment bisector

?

5

Opposite sides congruent

?

Given EAL NLA AET LNT

AE LNAT LT

EN and LA bisecteach other

Definition ofparallelogram

13. (Lesson 5.5)

66 ANSWERS TO EXERCISES

17. x 104, y 98. The quadrilaterals on theleft and right sides are kites. Nonvertex angles arecongruent. The quadrilateral at the bottom is anisosceles trapezoid. Base angles are congruent, andconsecutive angles between the bases aresupplementary.18. a 84, b 9619. No. The congruent angles and side do notcorrespond.20.

21. Parallelogram. Because the triangles arecongruent by SAS, 1 2. So, the segments are

parallel. Use a similar argument to show that theother pair of opposite sides is parallel.

22. Kite or dart. Radii of the same circle arecongruent. If the circles have equal radii, a rhombusis formed.

1

2

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USING YOUR ALGEBRA SKILLS 5

1.

2.

3. y

x

(2, 9)

(0, 6)

y

x

(3, 8)

(0, 4)

y

x(0, 1)

(1, 1)

ANSWERS TO EXERCISES 67

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4. y x 2

5. y 163 x

71

43

6. y x 17. y 3x 5

8. y 25 x 85

9. y 80 4x10. y 3x 26

11. y 14 x 3

12. y 65 x

13. y x 1

14. y 29 x 493

68 ANSWERS TO EXERCISES

LESSON 5.6

1. Sometimes true; it is true only if theparallelogram is a rectangle.

2. Always true; by the definition of rectangle, allthe angles are congruent. By the Quadrilateral SumConjecture and division, they all measure 90, soany two angles in a rectangle, including consecutiveangles, are supplementary.3. Always true by the Rectangle DiagonalsConjecture.4. Sometimes true; it is true only if the rectangle isa square.

5. Always true by the Square Diagonals Conjecture.6. Sometimes true; it is true only if the rhombus isequiangular.

7. Always true; all squares fit the definition ofrectangle.8. Always true; all sides of a square are congruentand form right angles, so the sides become the legsof the isosceles right triangle and the diagonal isthe hypotenuse.9. Always true by the Parallelogram OppositeAngles Conjecture.10. Sometimes true; it is true only if theparallelogram is a rectangle. Consecutive angles ofa parallelogram are always supplementary, but arecongruent only if they are right angles.11. 20 12. 3713. 45, 9014. DIAM is not a rhombus because it is notequilateral and opposite sides are not parallel.15. BOXY is a rectangle because its adjacent sidesare perpendicular.16. Yes. TILE is a rhombus, and a rhombus is aparallelogram.

false

true

falsetrue

false

true

17.

18. Constructions will vary.

19. one possible construction:

20. Converse: If the diagonals of a quadrilateralare congruent and bisect each other, then thequadrilateral is a rectangle.Given: Quadrilateral ABCD with diagonals AC BD. AC and BD bisect each otherShow: ABCD is a rectangle

Because the diagonals are congruent and bisecteach other, AE BE DE EC. Using theVertical Angles Conjecture, AEB CED andBEC DEA. So AEB CED and AED CEB by SAS. Using the Isosceles TriangleConjecture and CPCTC, 1 2 5 6,and 3 4 7 8. Each angle of thequadrilateral is the sum of two angles, one fromeach set, so for example, mDAB m1 m8.By the addition property of equality, m1 m8 m2 m3 m5 m4 m6 m7. So mDAB mABC mBCD mCDA. So the quadrilateral is equiangular. Using1 5 and the Converse of AIA, AB CD.Using 3 7 and the Converse of AIA,BC AD. Therefore ABCD is an equiangularparallelogram, so it is a rectangle.21. If the diagonals are congruent and bisect eachother, then the room is rectangular (converse of theRectangle Diagonals Conjecture).

A

E

B

CD

81 2

3

456

7

I E

P S

A

B E

K A

B

K

E

E V

L O

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22. The platform stays parallel to the floorbecause opposite sides of a rectangle are parallel (a rectangle is a parallelogram).23. The crosswalks form a parallelogram: Thestreets are of different widths, so the crosswalks areof different lengths. The streets would have to meetat right angles for the crosswalks to form a rectangle.The corners would have to be right angles and thestreets would also have to be of the same width forthe crosswalk to form a square.24. Place one side of the ruler along one side ofthe angle. Draw a line with the other side of theruler. Repeat with the other side of the angle. Drawa line from the vertex of the angle to the pointwhere the two lines meet.25. Rotate your ruler so that each endpoint of thesegment barely shows on each side of the ruler.Draw the parallel lines on each side of your ruler.Now rotate your ruler the other way and repeat theprocess to get a rhombus. The original segment isone diagonal of the rhombus. The other diagonalwill be the perpendicular bisector of the originalsegment.26. See f lowchart below.27. Yes, it is true for rectangles.Given: 1 2 3 4Show: ABCD is a rectangle By the Quadrilateral Sum Conjecture, m1 m2 m3 m4 360. It is given that allfour angles are congruent, so each angle measures

ANSWERS TO EXERCISES 69

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90. Because 4 and 5 form a linear pair,m4 m5 180. Substitute 90 for m4 andsolve to get m5 90. By definition of congruentangles, 5 3, and 5 and 3 are alternate interior angles, so AD BC by the Converse of theParallel Lines Conjecture. Similarly, 1 and 5 arecongruent corresponding angles, so AB CD by theConverse of the Parallel Lines Conjecture. Thus,ABCD is a parallelogram by the definition ofparallelogram. Because it is an equiangular parallelogram, ABCD is a rectangle.28. a 54, b 36, c 72, d 108, e 36,f 144, g 18, h 48, j 48, k 8429. possible answers: (1, 0); (0, 1); (1, 2); (2, 3)

30. y 89x 896 or 8x 9y 86

31. y 170x

152 or 7x 10y 24

32. velocity 1.8 mi/h; angle of path 106.1clockwise from the north

2 mi/h

1.5 mi/h

60

QU AD QD AU

QU AD

1

4

Given

Given

DU DU

3

2

Same segment

QUD ADU 5 1 2 3 4

Converse of the Parallel Lines Conjecture

6 QUAD is a parallelogram

Definition of parallelogram

7

QUAD is a rhombus

9 QU UA AD DQ 8

?

?

?

?

QD AU

Given SSSCPCTC

Definition ofrhombus

26. (Lesson 5.6)

70 ANSWERS TO EXERCISES

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LESSON 5.7

1. See flowchart below. 2. See flowchart below.3. See flowchart below.4.

Given

1 SP OA

Converse of AIA Conjecture

7 PA SO

Definition of parallelogram

8 SOAP is a parallelogram

Given

2 SP OA

SAS

5SOP APO

Same segment

4

PO PO

AIA Conjecture

31 2

CPCTC

63 4

5. parallelogram 6. sample flowchart proof:

Opposite sides ofrectangle are congruent

Definition of rectangle

SAS

Same segment

4

CPCTC

5

31 2 YOG OYI YO OY IY GO

YG IO

YOG OYI

3 4

1 2

SO KA

AIADefinition ofparallelogram

1SOA 7

2

Given

3

4

5SOAK is aparallelogram

6

?

?

OA ?

SA ?

?

?

?

65

GivenConjecture proved in Exercise 1

21BAT THB

CPCTC

3BAT THB

BAH = HBA

?

? ?

?

?

Parallelogram BATH with diagonal BT

4 Parallelogram BATH with diagonal HA

9

8

65

7

1 ?

4

? ? WATR is aparallelogram

4 ?

? ?

? ?

?

? ?

?

?

?

?

?

?

? ?

1

? ?

2

? ?

3

AKS

ASA

Same segment

AIA

SA

Definition ofparallelogram

SK

CPCTC

ATHTHA

Given Conjecture provedin Exercise 1

WA RT

WR AT WRT TAW

WT WT

Given2 RT WA

RW TA

CPCTC Converse of the Parallel Lines Conjecture

Definition ofparallelogram

CPCTC

3Given SSS

Same segment

by Converse of the ParallelLines Conjecture

1. (Lesson 5.7)

2. (Lesson 5.7)

3. (Lesson 5.7)

ANSWERS TO EXERCISES 71

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7.

8. Because AR is parallel to ZT, corresponding3 and 2 are congruent. Opposite sides ofparallelogram ZART are congruent so AR TZ.Because the trapezoid is isosceles, AR PT, andsubstituting gives ZT PT making PTZisosceles and 1 and 2 congruent. Bysubstitution, 1 and 3 are congruent.

Given Parallelogram Opposite Sides Conjecture

Parallelogram Opposite Sides Conjecture

SSS

5

CPCTC

6

4 2 AR EB 3 BR EA

Given

1 BEAR is a parallelogram

RE AB

Definition of rectangle

7 BEAR is a rectangle

EBR ARB RAE BEA

EBR ARB RAE BEA

9. See sample flowchart below.10. If the fabric is pulled along the warp or the weft,nothing happens.However, if the fabric is pulledalong the bias, it can be stretched because therectangles are pulled into parallelograms.11. 30 angles in 4-pointed star, 30 angles in 6-pointed star; yes12. He should measure the alternate interiorangles to see whether theyre congruent.If they are,the edges are parallel.13. ES: y 2x 3; QI: y 12x 214. (12, 7)

15. 13

16.

6 inches

9

12 3

6

9. (Lesson 5.7)

GivenSame segment

2 GR TH

Given

3 5 RGT HTG

SAS CPCTC

6

4RGT HTG

Isosceles TrapezoidConjecture

GH TR 1 Isosceles trapezoid

GTHR GT GT

72 ANSWERS TO EXERCISES

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CHAPTER 5 REVIEW

1. 360 divided by the number of sides 2. Sample answers: Using an interior angle, set theinterior angle measure formula equal to the angleand solve for n. Using an exterior angle, divide into360. Or find the interior angle measure and gofrom there.3. Trace both sides of the ruler as shown at right.

4. Make a rhombus using the double-edgedstraightedge, and draw a diagonal connecting theangle vertex to the opposite vertex.

5. Sample answer: Measure the diagonals withstring to see if they are congruent and bisect eachother.6. Sample answer: Draw a third point and connectit with each of the two points to form two sides of atriangle. Find the midpoints of the two sides andconnect them to construct the midsegment. The

distance between the two points is twice the lengthof the midsegment.7. x 10, y 40 8. x 60 cm9. a 116, c 64 10. 10011. x 38 cm12. y 34 cm, z 51 cm13. See table below.14. a 72, b 10815. a 120, b 60, c 60, d 120, e 60,f 30, g 108, m 24, p 84; Possible explanation: Because c 60, the angle that formsa linear pair with e and its congruent adjacentangle measures 60. So 60 2e 180, and e 60. The triangle containing f has a 60 angle.The other angle is a right angle because it forms alinear pair with a right angle. So f 30 by theTriangle Sum Conjecture. Because g is an interiorangle in an equiangular pentagon, divide 540 by 5to get g 108.16. 15 stones17. (1, 0)18. When the swing is motionless, the seat, the barat the top, and the chains form a rectangle. Whenyou swing left to right, the rectangle changes to aparallelogram. The opposite sides stay equal inlength, so they stay parallel. The seat and the bar at the top are also parallel to the ground.19. a = 60, b = 120

IsoscelesKite trapezoid Parallelogram Rhombus Rectangle Square

Opposite sidesare parallel No No Yes Yes Yes Yes

Opposite sides are congruent No No Yes Yes Yes Yes

Opposite anglesare congruent No No Yes Yes Yes Yes

Diagonals bisecteach other No No Yes Yes Yes Yes

Diagonals are perpendicular Yes No No Yes No Yes

Diagonals are congruent No Yes No No Yes Yes

Exactly one lineYesof symmetry Yes No No No No

Exactly two linesof symmetry No No No Yes Yes No

13. (Chapter 5 Review)

ANSWERS TO EXERCISES 73

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20.

Speed: 901.4 km/h. Direction: slightly west ofnorth. Figure is approximate.21.

22. possible answers:

23.

PL z

yN E

x

FL

Y

R

FLx

Y

R

x12

E

RS

Q

x12

900 km/h

50 km/h

Resultantvector

24. possible answers:

25. 20 sides26. 12 cm27. See flowchart below.28. possible answer:

Given: Parallelogram ABCDShow: AB CD and AD CBSee flowchart below.

12

43

A B

D C

Fx

z

R

D

Y

Fx

z

R

D

Y

CPCTC

8 AB CD

Definition ofparallelogram

2

1

AD CB

Definition ofparallelogram

3 AB CD

ABCD is a parallelogram

Same segment

5 BD BD

AIA

4 1 3

AIA

6 2 4

CPCTC

9 AD CB

ASA

7 ABD CDB

Given

? ?

3 5

2 DE ?

NE ?

4 DN ?

DENI isa rhombus

1

6 1 ? 3 ?

7DN bisectsIDE and INE

?

?

?

?

?

?

?

Given

Definition ofrhombus

Same segment

SSSCPCTC Definition of

angle bisector

DI

NI DEN DIN 2, 4

DN

Definition of rhombus

27. (Chapter 5 Review)

28. (Chapter 5 Review)

Discovering Geometry Teaching ResourcesTeaching and Worksheet MastersAnswers to ExercisesCHAPTER 0CHAPTER 1CHAPTER 2CHAPTER 3CHAPTER 4CHAPTER 5LESSON 5.1LESSON 5.2LESSON 5.3LESSON 5.4LESSON 5.5USING YOUR ALGEBRA SKILLS 5LESSON 5.6LESSON 5.7CHAPTER 5 REVIEW

CHAPTER 6CHAPTER 7CHAPTER 8CHAPTER 9CHAPTER 10CHAPTER 11CHAPTER 12CHAPTER 13

Solutions ManualAssessment ResourcesPractice Your Skills with AnswersCondensed Lessons: A Tool for Parents and TutorsCondensed Lessons in SpanishDiscovering Geometry with The Geometers SketchpadMore Projects and ExplorationsTracing Proof in Discovering GeometryA Guide for ParentsA Guide for Parents in Spanish