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48 ANSWERS TO EXERCISES

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Answers to Exercisesangles P and D can be drawn at each endpoint

using the protractor.

17. The third angles of the triangles also have the

same measures; are equal in measure

18. You know from the Triangle Sum Conjecture

that m�A � m�B � m�C � 180°, and m�D �m�E � m�F � 180°. By the transitive property,

m�A � m�B � m�C � m�D � m�E � m�F.

You also know that m�A � m�D, and m�B �m�E. You can substitute for m�D and m�E in the

longer equation to get m�A � m�B � m�C �m�A � m�B � m�F. Subtracting equal terms

from both sides, you are left with m�C � m�F.

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

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

triangle is equiangular, each angle has the same

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

20. false

21. false

22. false

23. false

24. true

25. eight; 100

85�

55�

40�P D7 cm

Q

CHAPTER 4 • CHAPTER CHAPTER 4 • CHAPTER

LESSON 4.1

1. The angle measures change, but the sum

remains 180°.

2. 73°

3. 60°

4. 110°

5. 24°

6. 3 � 360° � 180° � 900°

7. 3 � 180° � 180° � 360°

8. 69°; 47°; 116°; 93°; 86°

9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50°

10.

11.

12. First construct �E, using the method used in

Exercise 10.

13.

14. From the Triangle Sum Conjecture

m�A � m�S � m�M � 180°. Because �M is a

right angle, m�M � 90°. By substitution,

m�A � m�S � 90° � 180°. By subtraction,

m�A � m�S � 90°. So two wrongs make a right!

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.

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

Conjecture is used to find that the measure of

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

E

RA�L

�G

Fold

�M

�R

�A

E

RA

�L

�G

�M �A

�R

4

LESSON 4.2

1. 79°

2. 54°

3. 107.5°

4. 44°; 35 cm

5. 76°; 3.5 cm

6. 72°; 36°; 8.6 cm

7. 78°; 93 cm

8. 75 m; 81°

9. 160 in.; 126°

10. a � 124°, b � 56°, c � 56°, d � 38°, e � 38°,

f � 76°, g � 66°, h � 104°, k � 76°, n � 86°,

p � 38°; Possible explanation: The angles with

measures 66° and d form a triangle with the angle

with measure e and its adjacent angle. Because d,

e, and the adjacent angle are all congruent,

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

also the measure of one of the base angles of the

isosceles triangle with vertex angle measure h.

Using the Isosceles Triangle Conjecture, the other

base angle measures d, so 2d � h � 180°, or

76° � h � 180°. Therefore, h � 104°.

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

none

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

be congruent; therefore, each triangle must be

isosceles.

12b. 60°

13. NCA

14. IEC

15.

16.

17. possible answer:

18. perpendicular

19. parallel

20. parallel

21. neither

22. parallelogram

23. 40

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

congruent.

25. New: (3, �3), (�3, �1), (�1, �5). Triangles

are congruent.

0 8 16 24 32 40

Fold 2

Fold 1

Fold 4

Fold 3

105� 60�

45�

M N

G

K

H

P

ED

BA

F

C

ANSWERS TO EXERCISES 49

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USING YOUR ALGEBRA SKILLS 4

1. false 2. true

3. not a solution 4. solution

5. not a solution 6. x � 7

7. y � �4 8. x � �8

9. x � 4.2 10. n � ��12

�

11. x � 2 12. t � 18

13. n � �25

� 14a. x � �49

�

14b. x � �49

�; The two methods produce identical

results. Multiplying by the lowest common

denominator (which is comprised of the factors of

both denominators) and then reducing common

factors (which clears the denominators on either

side) is the same as simply multiplying each numerator

by the opposite denominator (or cross multiplying).

Algebraically you could show that the two methods

are equivalent as follows:

�a�b

� � �dc

�

bd��ab

�� bd��dc

��ab

bd

� � �b

dcd�

ad � bc

The method of “clearing fractions” results in the

method of “cross multiplying.”

15. You get an equation that is always false, so

there is no solution to the equation.

16. Camella is not correct. Because the equation

0 � 0 is always true, the truth of the equation does

not depend on the value of x. Therefore, x can be

any real number. Camella’s answer, x � 0, is only

one of infinitely many solutions.

17.

If x equals the measure of the vertex angle, then

the base angles each measure 2x. Applying the

Triangle Sum Conjecture results in the following

equation:

x � 2x � 2x � 180°

5x � 180°

x � 36°

The measure of the vertex angle is 36° and the

measure of each base angle is 72°.

2x

2x

x

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LESSON 4.3

1. yes

2. no

3. no

4. yes

5. a, b, c

6. c, b, a

7. b, a, c

8. a, c, b

9. a, b, c

10. v, z, y, w, x

11. 6 � length � 102

12. By the Triangle Inequality Conjecture, the

sum 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.

5 6

12

4 5

9

14. 135°

15. 72°

16. 72°

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

from both sides of both equations to get x � 180 � c

and a � b � 180 � c. Substitute a � b for 180 � c

in the first equation to get x � a � b.

18. 45°

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

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

f � 56°, g � 124°, h � 124°

21. By the Triangle Sum Conjecture, the third

angle must measure 36° in the small triangle, but it

measures 32° in the large triangle. These are the

same angle, so they can’t have different measures.

22. ABE

23. FNK

24. cannot be determined


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LESSON 4.4

1. Answers will vary. Possible answer: If three

sides of one triangle are congruent to three sides of

another triangle, then the triangles are congruent

(all corresponding angles are also congruent).

2. Answers will vary. Possible answer: The picture

statement means that if two sides of one triangle

are congruent to two sides of another triangle, and

the angles between those sides are also congruent,

then the two triangles are congruent.

If you know this: then you also know this:

3. Answers will vary. Possible answer:

4. SAS

5. SSS


7. SSS

8. SAS

9. SSS (and the Converse of the Isosceles Triangle

Conjecture)

10. yes, �ABC � �ADE by SAS

11. Possible answer: Boards nailed diagonally in

the corners of the gate form triangles in those

corners. Triangles are rigid, so the triangles in the

gate’s corners will increase the stability of those

corners and keep them from changing shape.

12. FLE by SSS

13. Cannot be determined. SSA is not a congruence

conjecture.

14. AIN by SSS or SAS

15. Cannot be determined. Parts do not correspond.

16. SAO by SAS

17. Cannot be determined. Parts do not correspond.

18. RAY by SAS

19. The midpoint of SD� and PR� is (0, 0).

Therefore, �DRO � �SPO by SAS.

20. Because the LEV is marking out two triangles

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.

23. a � 37°, b � 143°, c � 37°, d � 58°,

e � 37°, f � 53°, g � 48°, h � 84°, k � 96°,

m � 26°, p � 69°, r � 111°, s � 69°; Possible

explanation: The angle with measure h is the vertex

angle of an isosceles triangle with base angles

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

The angle with measure s and the angle with

measure p are corresponding angles formed by

parallel lines, so s � p � 69°.

24. 3 cm � third side � 19 cm

25. See table below.

26a. y � 6 b. y � �133� c. y � ��

34

�x � 2

27. (�5, �3)

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Side length 1 2 3 4 5 … n … 20

Elbows 4 4 4 4 4 4 4

T’s 0 4 8 12 16 76

Crosses 0 1 4 9 16 361

25. (Lesson 4.4)

4n � 4 (n � 1)2

LESSON 4.5

1. If two angles and the included side of one triangle

are congruent to the corresponding side and angles

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.

If you know this: then you also know this:

3. Answers will vary. Possible answer:

4. ASA


6. SAA


8. ASA


10. FED by SSS

11. WTA by ASA or SAA

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

16. Cannot be determined.AAA does not guarantee

congruence.

17. WKL by ASA

18. Yes, �ABC � �ADE by SAA or ASA.

19. Slope AB�� slope CD�� 3 and slope BC��slope DA��

13

�, so AB� � BC�, CD�� DA��, and

BC� � DA��. �ABC � �CDA by SAA.

20.

21. The construction is the same as the

construction using ASA once you find the third

angle, which is used here. (Finding the third angle

is not shown.)

22. Construction will show a similar but larger

(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.

Bisect the right angle. Construct a triangle with

two congruent sides and with a vertex that

measures 135°.

24. 125

25. False. One possible counterexample is a kite.

26. None. One triangle is determined by SAS.

27.

28a. about 100 km southeast of San Francisco

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

to two locations. The third circle narrows it down

to one.

0

0 100 400

50 100 200

Sacramento

CA

LI F

OR

NI A

N E VA D A U T A H

A R I Z O N A

I D A H O

O R E G O N

San Francisco

Eureka

Reno

Las Vegas

Elko

Boise

Los AngelesMiles

Kilometers

K

L

M


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LESSON 4.6

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

(given), and �ABD � �CBD (given), so �DBA ��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.

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.

� �O � �R by CPCTC.

6. Yes. MN�� MA�� and ME�� MR�� (given), and

�M � �M (same angle), so �EMA � �RMN by

SAS. � �E � �R by CPCTC.

7. Yes. BT�� EU� and BU�� ET� (given), and

UT�� UT� (same segment), so �TUB � �UTE by

SSS. � �B � �E by CPCTC.

8. Cannot be determined. �HLF � �LHA by

ASA, but HA�� and HF� are not corresponding sides.

9. Cannot be determined. AAA does not guaran-

tee congruence.

10. Yes. The triangles are congruent by SAS.

11. Yes. The triangles are congruent by SAS, and

the 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.


14. KEI by ASA

15. UTE by SAS

16.

17.

18. a � 112°, b � 68°, c � 44°, d � 44°, e � 136°,

f � 68°, g � 68°, h � 56°, k � 68°, l � 56°, m � 124°;

Possible explanation: f and g are measures of base

angles of an isosceles triangle, so f � g. The vertex

angle measure is 44°, so subtract 44° from 180° and

divide by 2 to get f � 68°. The angle with measure

m is the exterior angle of a triangle. Add the remote

interior angle measures 56° and 68° to get

m � 124°.

19. ASA. The “long segment in the sand” is a

shared side of both triangles

20. (�4, 1)

21. See table below.

22. Value C is always decreasing.

23. x � 3, y � 10

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Number of sides 3 4 5 6 7 … 12 … n

Number of struts needed 0 1 2 3 4 … 9 … n � 3to make polygon rigid

21. (Lesson 4.6)

LESSON 4.7

1. See flowchart below.





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1 SE � SU

�

?

2�E � �U

�

?

4 � �

? � � �

?

ASA CongruenceConjecture

5

3�1 � �2

�

?

MS � SO

�

?

Given

Given CPCTC

Vertical Angles Conjecture

�ESM � �USO

1 I is midpoint of CM

Given

CI � IM

3

Definition of midpoint

2 I is midpoint of BL

�

?�

?

CPCTC

7

5 �1 � �2

�

?

6

�

?

� �

? � � �

?IL � IB 4

�

?

3 NS � NS

WS � SE

WN � NE

4

1 NS is a median

S is a midpoint2 6 7

5

�

?

�

?�

?

Given Same segment

Definitionof median

Definitionof midpoint

�W � � �

?�WSN � � �

?

2

Definition of �?

NS � NS

3�W � �E

�1 � � �

?

�

?

1 NS is anangle bisector

Given

5 � �

? � � �

? 6

4

�

?

�

?�

?

WN � NE

�

?

7 �NEW isisosceles

Given

Vertical Angles Conjecture

Definition of

midpoint�CIL � �MIB

by SAS

CL�� MB��

Given

�ESN

SSS CPCTC

�E

Given CPCTCSAA

�WNS � �ENS

Definition of

isosceles triangle

Same segment

�2

angle bisector

1. (Lesson 4.7)

2. (Lesson 4.7)

3. (Lesson 4.7)

4. (Lesson 4.7)



6. Given: �ABC with BA�� BC�, CD�� AD��

Show: BD� is the angle bisector of �ABC.

7. The angle bisector does not go to the midpoint

of the opposite side in every triangle, only in an

isosceles triangle.

8. NE�, because it is across from the smallest angle

in �NAE. It is shorter than AE�, which is across

from the smallest angle in �LAE.

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.

Given Given

BA � BC CD � AD BD � BD

Same segment

�ABD � �CBD

BD bisects �ABC

SSS

�1 � �2

CPCTC

→

Definition of anglebisector

12B

A

D

C

12. ACK by SSS

13. a � 72°, b � 36°, c � 144°, d � 36°, e � 144°,

f � 18°, g � 162°, h � 144°, j � 36°, k � 54°,

m � 126°

14. The circumcenter is equidistant from all three

vertices because it is on the perpendicular bisector

of each side. Every point on the perpendicular

bisector of a segment is equidistant from the

endpoints. Similarly, the incenter is equidistant

from all three sides because it is on the angle

bisector of each angle, and every point on an angle

bisector is equidistant from the sides of the angle.

15. ASA. The fishing pole forms the side.

“Perpendicular to the ground” forms one angle.

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

angle.

16. �27

�

17.

18. y

x

X

B O

y

4

X'O'

Y'B' 4

Y

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3

SN � SN

4

�3 � �41

�?

�?

�? 6 � �

? � � �? 7

5

�?

�?

�?

SA � NE

2

�?

SE � NA

AIA Conjecture

Same segment

Given

Given AIA Conjecture

1 � �2 SA � NE�

ASA

�ESN � �ANS

This proof shows that in a parallelogram,

opposite sides are congruent.

CPCTC

5. (Lesson 4.7)

LESSON 4.8

1. 6

2. 90°; 18°

3. 45°



6. 1 Isosceles �ABC

with AC � BC

and CD bisects

�C

Given

2 �ADC � �BDC

Conjecture A(Exercise 4)

3 AD � BD

CPCTC

4 CD is a median

Def. of median

7.

8. Yes. First show that the three exterior triangles

are congruent by SAS.

5 �ADC � �BDC

1 AC � BC

Given

4 AD � BD

Def. of midpoint

Def. of angle bisector

9 CD � AB

Def. of altitude

SSS

6 �ACD � �DCB

CPCTC

2 CD � CD

Same segment

3 D is

midpoint

of AB

Given

7 CD is angle

bisector of �ACB

8 CD is altitude

of �ABC

Conjecture B (Exercise 5)


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2

35 �ADC � �BDC

4

�

?

�ABC is isosceleswith AC � BC

1 CD is the bisector of �C

Given Definition of angle bisector

�

? Given

Same segment

CD � CD

2 �ADC � �BDC

1 �ABC is isosceles with AC � BC, and CD is the bisector of �C

Given Conjecture A

Congruent supplementary angles are 90�

4 �1 � �2

�

?

8 �

?

Definition of altitude

7

�

?

3 �1 and �2 form a linear pair

Definition of linear pair

5 �1 and �2 are supplementary

6 �1 and �2 are right angles

Linear Pair Conjecture

CD � AB ��

SAS

�1 � �2

CPCTC

Definition of

perpendicular

CD� is an altitude

4. (Lesson 4.8)

5. (Lesson 4.8)


9.

Drawing the vertex angle bisector as an auxiliary

segment, we have two triangles. We can show them

to be congruent by SAS, as we did in Exercise 4.

Then, �A � �B, by CPCTC. Therefore, base angles

of an isosceles triangle are congruent.

10. The proof is similar to the one on page 245,

but in reverse, and using the Converse of the

Isosceles Triangle Conjecture.

11.

12. a � 128°, b � 128°, c � 52°, d � 76°, e � 104°,

f � 104°, g � 76°, h � 52°, j � 70°, k � 70°, l � 40°,

m � 110°, n � 58°

13. between 16 and 17 minutes

30�

C

A D B

14. y � ��53

� x � 16

15. 120

16. (4, 6) or (4, 0) or any point at which the

x-coordinate is either 1 or 7 and the y-coordinate

does not equal 3

17. Hugo and Duane can locate the site of the

fireworks by creating a diagram using SSS.

18. CnH2n

CC C

CC

C C

H

HH

H

HH

H

HH

HHHH

H

Fireworks

Duane

Hugo

1.5 km

340 m/s • 5 s= 1.7 km

340 m/s • 3 s= 1.02 km

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CHAPTER 4 REVIEW

1. Their rigidity gives strength.

2. The Triangle Sum Conjecture states that the

sum of the measures of the angles in every triangle

is 180°. Possible answers: It applies to all triangles;

many other conjectures rely on it.

3. The angle bisector of the vertex angle is also the

median and the altitude.

4. The distance between A and B is along the

segment connecting them. The distance from A

to C to B can’t be shorter than the distance from A

to B. Therefore, AC � CB � AB. Points A, B, and C

form a triangle. Therefore, the sum of the lengths

of any two sides is greater than the length of the

third side.

5. SSS, SAS, ASA, or SAA

6. In some cases, two different triangles can

be constructed using the same two sides and

non-included angle.


8. ZAP by SAA

9. OSU by SSS


11. APR by SAS

12. NGI by SAS


14. DCE by SAA or ASA

15. RBO or OBR by SAS

16. �AMD � �UMT by SAS, AD�� UT� by

CPCTC



19. �TRI � �ALS by SAA, TR�� AL� by

CPCTC

20. �SVE � �NIK by SSS, EL�� KV�� by

overlapping segments property



23. �LAZ � �IAR by ASA, �LRI � �IZL by

ASA, and �LRD � �IZD by ASA

24. yes. �PTS � �TPO by ASA or SAA

25. �ANG is isosceles, so �A � �G. However,

the sum of m�A � m�N � m�G � 188°. The

measures of the three angles of a triangle must sum

to 180°.

26. �ROW � �NOG by ASA, implying that

OW�� OG��. However, the two segments shown are

not equal in measure.

27. a � g � e � d � b � f � c. Thus, c is the

longest segment, and a and g are the shortest.

28. x � 20°

29. Yes. �TRE � �SAE by SAA, so sides are

congruent by CPCTC.

30. Yes. �FRM � �RFA by SAA. �RFM ��FRA by CPCTC. Because base angles are

congruent, �FRD is isosceles.

31. x � 48°

32. The legs form two triangles that are congruent

by SAS. Because alternate interior angles are

congruent by CPCTC, the seat must be parallel to

the floor.

33. Construct �P and �A to be adjacent. The

angle that forms a linear pair with the conjunction

of �P and �A is �L. Construct �A. Mark off the

length AL on one ray. Construct �L. Extend the

unconnected sides of the angles until they meet.

Label the point of intersection P.

34. Construct �P. Mark off the length PB on one

ray. From point B, mark off the two segments that

intersect the other ray of �P at distance x.

z

xx

P B

S2

S1

A Ly

P

�P�A

�L


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36. Given three sides, only one triangle is possible;

therefore, the shelves on the right hold their shape.

The shelves on the left have no triangles and move

freely as a parallelogram.

37. Possible method: Construct an equilateral

triangle and bisect one angle to obtain 30°.

Adjacent to that angle, construct a right angle

and bisect it to obtain 45°.

38. d, a � b, c, e, f

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2

3 5 � �

? � � �

?

4

6 7� �

? � � �

?

� �

? � � �

?

1�

?

�

?

�

?

�

? �

? �

? �

?

�

?

�

? � �

?

�

? � �

?

�

? � �

? M is midpoint

of TE� and IR�

Given

�TMI � �RME

Vertical angles

TM�� ME��

Definition

of midpoint

�TMI � �EMR

SAS

TI� � RE�

Converse of

AIA Conjecture

�T � �Eor

�R � �I

CPCTC

IM�� MR��

Definition of midpoint

35. (Chapter 4 Review)

answers to exercises - cerritos collegeweb.cerritos.edu/imccance/sitepages/worksheets and...

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