Answers to Exercises

CHAPTER 4 • CHAPTER CHAPTER 4 • CHAPTER angles P and D can be drawn at each endpoint4

using the protractor.

LESSON 4.1 Q

551. The angle measures change, but the sum

remains 180°.

2. 73°

40 85

P D7 cm3. 60°

4. 110° 17. The third angles of the triangles also have the

same measures; are equal in measure 5. 24°

18. You know from the Triangle Sum Conjecture6. 3 360° 180° 900°

that mA mB mC 180°, and mD 7. 3 180° 180° 360°

mE mF 180°. By the transitive property,

8. 69°; 47°; 116°; 93°; 86°

mA mB mC mD mE mF.

9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50° You also know that mA mD, and mB

10. mE. You can substitute for mD and mE in the

R longer equation to get mA mB mC

mA mB mF. Subtracting equal terms

from both sides, you are left with mC mF.

M A

19. For any triangle, the sum of the angle measures

11. is 180°, by the Triangle Sum Conjecture. Since the

G

triangle is equiangular, each angle has the same

L

measure, say x. So x x x 180°, and x 60°.

20. false12. First construct E, using the method used in

Exercise 10.

E

A R

Fold13. 21. false

ER

M A G RAL

22. false

14. From the Triangle Sum Conjecture

mA mS mM 180°. Because M is a

right angle, mM 90°. By substitution,

mA mS 90° 180°. By subtraction,

mA mS 90°. So two wrongs make a right!

23. false

15. Answers will vary. See the proof on page 202.

To prove the Triangle Sum Conjecture, the Linear

Pair Conjecture and the Alternate Interior Angles

Conjecture must be accepted as true.

24. true

16. It is easier to draw PDQ if the Triangle Sum

25. eight; 100

Conjecture is used to find that the measure of

D is 85°. Then PD can be drawn to be 7 cm, and

48 ANSWERS TO EXERCISES

Answers to ExercisesAnswers to Exercises

F15.LESSON 4.2

1. 79° D E

2. 54° C

3. 107.5°

A B4. 44°; 35 cm

P5. 76°; 3.5 cm 16.

6. 72°; 36°; 8.6 cm

M N7. 78°; 93 cm

K

8. 75 m; 81°

G H9. 160 in.; 126°

10. a 124°, b 56°, c 56°, d 38°, e 38°, 17. possible answer:

f 76°, g 66°, h 104°, k 76°, n 86°, Fold 1 Fold 3

p 38°; Possible explanation: The angles with

Fold 2

Fold 4measures 66° and d form a triangle with the angle

60105with measure e and its adjacent angle. Because d,

45

e, and the adjacent angle are all congruent,

3d 66° 180°. Solve to get d 38°. This is

18. perpendicular

also the measure of one of the base angles of the

19. parallelisosceles triangle with vertex angle measure h.

20. parallelUsing the Isosceles Triangle Conjecture, the other

21. neitherbase angle measures d, so 2d h 180°, or

76° h 180°. Therefore, h 104°. 22. parallelogram

11. a 36°, b 36°, c 72°, d 108°, e 36°; 23. 40

none

12a. Yes. Two sides are radii of a circle. Radii must

0 8 16 24 32 40be congruent; therefore, each triangle must be

24. New: (6, 3), (2, 5), (3, 0). Triangles areisosceles.

congruent.12b. 60°

25. New: (3, 3), (3, 1), (1, 5). Triangles13. NCA

are congruent.

14. IEC

ANSWERS TO EXERCISES 49USING YOUR ALGEBRA SKILLS 4 15. You get an equation that is always false, so

there is no solution to the equation.

1. false 2. true

16. Camella is not correct. Because the equation

3. not a solution 4. solution 0 0 is always true, the truth of the equation does

5. not a solution 6. x 7 not depend on the value of x. Therefore, x can be

7. y 4 8. x 8 any real number. Camella’s answer, x 0, is only

one of infinitely many solutions.19. x 4.2 10. n 2 17.

2x

11. x 2 12. t 18

x

2x2 9 13. n 14a. x 5 4 If x equals the measure of the vertex angle, then

9

14b. x ; The two methods produce identical the base angles each measure 2x. Applying the4

results. Multiplying by the lowest common Triangle Sum Conjecture results in the following

denominator (which is comprised of the factors of equation:

both denominators) and then reducing common x 2x 2x 180°

factors (which clears the denominators on either 5x 180°

side) is the same as simply multiplying each numerator

x 36°

by the opposite denominator (or cross multiplying).

The measure of the vertex angle is 36° and theAlgebraically you could show that the two methods

measure of each base angle is 72°.are equivalent as follows:

a c b d

a c bd bd b d

abd bcd b d

ad bc

The method of “clearing fractions” results in the

method of “cross multiplying.”

50 ANSWERS TO EXERCISES

Answers to ExercisesAnswers to Exercises

14. 135°LESSON 4.3

15. 72°

1. yes

16. 72°

2. no

17. a b c 180° and x c 180°. Subtract c

45

from both sides of both equations to get x 180 c9

and a b 180 c. Substitute a b for 180 c3. no

in the first equation to get x a b.56

12 18. 45°

4. yes 19. a 52°, b 38°, c 110°, d 35°

5. a, b, c 20. a 90°, b 68°, c 112°, d 112°, e 68°,

6. c, b, a f 56°, g 124°, h 124°

7. b, a, c 21. By the Triangle Sum Conjecture, the third

8. a, c, b angle must measure 36° in the small triangle, but it

measures 32° in the large triangle. These are the9. a, b, c

same angle, so they can’t have different measures.10. v, z, y, w, x

22. ABE11. 6 length 102

23. FNK12. By the Triangle Inequality Conjecture, the

24. cannot be determinedsum of 11 cm and 25 cm should be greater than

48 cm.

13. b 55°, but 55° 130° 180°, which is

impossible by the Triangle Sum Conjecture.

ANSWERS TO EXERCISES 51LESSON 4.4 13. Cannot be determined. SSA is not a congruence

conjecture.

1. Answers will vary. Possible answer: If three

14. AIN by SSS or SAS

sides of one triangle are congruent to three sides of

15. Cannot be determined. Parts do not correspond.another triangle, then the triangles are congruent

16. SAO by SAS(all corresponding angles are also congruent).

17. Cannot be determined. Parts do not correspond.2. Answers will vary. Possible answer: The picture

18. RAY by SASstatement means that if two sides of one triangle

are congruent to two sides of another triangle, and 19. The midpoint of SD and PR is (0, 0).

the angles between those sides are also congruent, Therefore, DRO SPO by SAS.

then the two triangles are congruent. 20. Because the LEV is marking out two triangles

If you know this: then you also know this: that are congruent by SAS, measuring the length

of the segment leading to the finish will also

approximate the distance across the crater.

21. 22.

3. Answers will vary. Possible answer:

23. a 37°, b 143°, c 37°, d 58°,4. SAS

e 37°, f 53°, g 48°, h 84°, k 96°,5. SSS

m 26°, p 69°, r 111°, s 69°; Possible 6. cannot be determined

explanation: The angle with measure h is the vertex

7. SSS

angle of an isosceles triangle with base angles

8. SAS measuring 48°, so h 2(48) 180°, and h 84°.

9. SSS (and the Converse of the Isosceles Triangle The angle with measure s and the angle with

Conjecture) measure p are corresponding angles formed by

10. yes, ABC ADE by SAS parallel lines, so s p 69°.

11. Possible answer: Boards nailed diagonally in 24. 3 cm third side 19 cm

the corners of the gate form triangles in those 25. See table below.

corners. Triangles are rigid, so the triangles in the 13 3 26a. y 6 b. y c. y x 24gate’s corners will increase the stability of those 3

corners and keep them from changing shape. 27. (5, 3)

12. FLE by SSS

25. (Lesson 4.4)

Side length 12345 … n …20

Elbows 4444444

T’s 04 812 16 76

Crosses 01 49 16 361

24n 4 (n 1)

52 ANSWERS TO EXERCISES

Answers to ExercisesAnswers to Exercises

CALIFORNIA

O REG O N

N E VA D A

I DAHO

ARIZ ONA

U TA H

21. The construction is the same as the LESSON 4.5

construction using ASA once you find the third

1. If two angles and the included side of one triangle angle, which is used here. (Finding the third angle

are congruent to the corresponding side and angles is not shown.)

of another triangle, then the triangles are congruent.

2. If two angles and a non-included side of one

triangle are congruent to the corresponding side

and angles of another triangle, then the triangles

are congruent.

22. Construction will show a similar but larger

If you know this: then you also know this: (or smaller) triangle constructed from a drawn

triangle by duplicating two angles on either end of a

new side that is not congruent to the corresponding

side.

23. Draw a line segment. Construct a perpendicular.3. Answers will vary. Possible answer:

Bisect the right angle. Construct a triangle with

two congruent sides and with a vertex that

measures 135°.

24. 125

4. ASA 25. False. One possible counterexample is a kite.

5. cannot be determined 26. None. One triangle is determined by SAS.

6. SAA L

7. cannot be determined

8. ASA

9. cannot be determined K M

10. FED by SSS

11. WTA by ASA or SAA 27.

12. SAT by SAS

13. PRN by ASA or SAS; SRE by ASA

14. Cannot be determined. Parts do not

correspond.

15. MRA by SAS

28a. about 100 km southeast of San Francisco

16. Cannot be determined.AAA does not guarantee

28b. Yes. No, two towns would narrow it downcongruence.

to two locations. The third circle narrows it down

17. WKL by ASA

to one.

18. Yes, ABC ADE by SAA or ASA.

19. Slope AB slope CD 3 and slope BC Boise

1 Eureka slope DA ,so AB BC,CD DA, and

3

BC DA. ABC CDA by SAA.

ElkoReno

Sacramento

San 20. Francisco

Las

Vegas

Los Angeles

Miles

0 50 100 200

0 100 400

Kilometers

ANSWERS TO EXERCISES 53

13. cannot be determinedLESSON 4.6

14. KEI by ASA

1. Yes. BD BD (same segment), A C

15. UTE by SAS(given), and ABD CBD (given), so DBA

16. DBC by SAA. AB CB by CPCTC.

2. Yes. CN WN and C W (given), and

RNC ONW (vertical angles), CNR

WON by ASA. RN ON by CPCTC.

3. Cannot be determined. The congruent parts

lead to the ambiguous case SSA. 17.

4. Yes. S I, G A (given), and TS IT

(definition of midpoint), so TIA TSG by

SAA. SG IA by CPCTC.

5. Yes. FO FR and UO UR (given), and UF

UF (same segment), so FOU FRU by SSS. 18. a 112°, b 68°, c 44°, d 44°, e 136°,

O R by CPCTC. f 68°, g 68°, h 56°, k 68°, l 56°, m 124°;

Possible explanation: f and g are measures of base6. Yes. MN MA and ME MR (given), and

angles of an isosceles triangle, so f g.The vertexM M (same angle), so EMA RMN by

angle measure is 44°, so subtract 44° from 180° andSAS. E R by CPCTC.

divide by 2 to get f 68°. The angle with measure 7. Yes. BT EU and BU ET (given), and

m is the exterior angle of a triangle. Add the remote UT UT (same segment), so TUB UTE by

interior angle measures 56° and 68° to get SSS. B E by CPCTC.

m 124°.

8. Cannot be determined. HLF LHA by

19. ASA. The “long segment in the sand” is aASA, but HA and HF are not corresponding sides.

shared side of both triangles

9. Cannot be determined. AAA does not guaran-

20. (4, 1)tee congruence.

21. See table below.10. Yes. The triangles are congruent by SAS.

22. Value C is always decreasing.11. Yes. The triangles are congruent by SAS, and

23. x 3, y 10the angles are congruent by CPCTC.

12. Draw AC and DF to form ABC and DEF.

AB CB DE FE because all were drawn with

the same radius. AC DF for the same reason.

ABC DEF by SSS. Therefore, B E by

CPCTC.

21. (Lesson 4.6)

Number of sides 34 5 6 7 … 12 … n

01 2 3 4 … 9 … n 3Number of struts needed

to make polygon rigid

54 ANSWERS TO EXERCISES

Answers to Exercises

Answers to Exercises

3. See flowchart below.LESSON 4.7

4. See flowchart below.

1. See flowchart below.

2. See flowchart below.

1. (Lesson 4.7)

1 SE SU

? Given

ESM USO

2 4 5? ?E U MS SO

? ?Given ASA Congruence CPCTC

Conjecture

3 1 2

? Vertical Angles Conjecture

2. (Lesson 4.7)

1 3 I is midpoint of CM CI IM

Given Definition

of midpoint

2 4 6 7 ? ? ?I is midpoint of BL IL IB CL MB

? ? ?Definition ofGiven CPCTC CIL MIB

midpoint by SAS

5 1 2

? Vertical Angles Conjecture

3. (Lesson 4.7)

1 3NS is a median NS NS

Given Same segment

EESN

2 4 6 7? ?S is a midpoint WS SE WSN W

? ?Definition Definition SSS CPCTC

of median of midpoint

5 WN NE

? Given

4. (Lesson 4.7)

1 2NS is an ?1 2

angle bisector

?Definition of Given angle bisector WNS ENS

7 NEW is3 5 6? ?W E WN NE isosceles

? ? ?Given SAA CPCTC ? Definition of

isosceles triangle

4 NS NS

? Same segment

ANSWERS TO EXERCISES 555. See flowchart below. 12. ACK by SSS

6. Given: ABC with BA BC,CD AD 13. a 72°, b 36°, c 144°, d 36°, e 144°,

f 18°, g 162°, h 144°, j 36°, k 54°,Show: BD is the angle bisector of ABC.

m 126°

14. The circumcenter is equidistant from all threeA

vertices because it is on the perpendicular bisector

1 D

B of each side. Every point on the perpendicular2

bisector of a segment is equidistant from the

C endpoints. Similarly, the incenter is equidistant

from all three sides because it is on the angle

BA BC CD AD BD BD bisector of each angle, and every point on an angle

Same segmentGiven Given bisector is equidistant from the sides of the angle.

15. ASA. The fishing pole forms the side.

“Perpendicular to the ground” forms one angle.ABD CBD

SSS “Same angle on her line of sight” forms the other

angle.

1 2 216.

CPCTC 7

17.

→

BD bisects ABC

Definition of angle

bisector

7. The angle bisector does not go to the midpoint

of the opposite side in every triangle, only in an

18. y

isosceles triangle.

4 Y X8. NE, because it is across from the smallest angle

BOin NAE. It is shorter than AE, which is across

x

B' Y'4from the smallest angle in LAE.

O' X'9. The triangles are congruent by SSS, so the two

central angles cannot have different measures.

10. PRN by ASA; SRE by ASA

11. Cannot be determined. Parts do not

correspond.

5. (Lesson 4.7)

1 3SA NE 3 4

? AIA ConjectureGiven

ESN ANS

42 6 71 2 ? ? ? ?SE NA SA NE

AIA Conjecture? ? ? ASA ? CPCTCGiven

This proof shows that in a parallelogram,

5 opposite sides are congruent.SN SN

Same segment

56 ANSWERS TO EXERCISES

Answers to Exercises

Answers to Exercises

1 2 37.LESSON 4.8 AC BC CD CD D is

midpointGiven Same segment

of AB1. 6

Given2. 90°; 18°

5 4ADC BDC AD BD3. 45°

SSS Def. of midpoint

4. See flowchart below.

6 7ACD DCB CD is angle5. See flowchart below.

bisector of ACBCPCTC

6. 1 2 Def. of angle bisectorIsosceles ABC ADC BDC

8with AC BC CD is altitudeConjecture A 9

CD ABand CD bisects of ABC(Exercise 4)

C Conjecture B Def. of altitude

(Exercise 5)Given

3 4 8. Yes. First show that the three exterior trianglesAD BD CD is a median

are congruent by SAS.CPCTC Def. of median

4. (Lesson 4.8)

1 2 ? 1 2CD is the bisector of C

Given Definition of

angle bisector

3 ABC is isosceles 5 ADC BDC

with AC BC

? SAS

Given

4 CD CD

Same segment

5. (Lesson 4.8)

1 ABC is isosceles with

2 4 AC BC, and CD is ADC BDC 1 2

the bisector of C ? CPCTCConjecture A

6 1 and 2 are Given

right angles

Congruent

3 5 supplementary 1 and 2 form 1 and 2 are

angles are 90 a linear pair supplementary

Definition of Linear Pair

7 linear pair Conjecture CD AB

Definition of ?

perpendicular

8 ?CD is an altitude

Definition of

altitude

ANSWERS TO EXERCISES 57