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TRANSCRIPT
CHAPTER 5
80 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Think & Discuss (p. 261)
1. Answers may vary.
Sample answer:
Position B may be the best position because he wouldhave less space for the ball to pass him. He would also bemore toward the middle of the area, so he wouldn’t haveas far to move to the left or right to intercept the ball.
2. The is approximately An opponent could movecloser to the goal to increase the shooting angle.
Skill Review (p. 262)
1. Sample answer:
3.
4.
5.
6. The slope of the line perpendicular to is because
Lesson 5.1
Developing Concepts Activity 5.1 (p. 263)
Exploring the Concept
1.–3. Sample answer
5. m�CMA � 90�
BMA
C
�12 � 2 � �1.
�12BC
� 2��4�2
�0 � 4
�2 � 0m �
y2 � y1
x2 � x1
� 5
� �25
� �9 � 16
� ���3�2 � 42
AB � ��0 � 3�2 � �4 � 0�2
� ��1, 2�� ��22
, 42�M � �0 � ��2�
2,
4 � 02 �
A BM
38�.m�x
Drawing Conclusions
1. is the perpendicular bisector of because
with M between A and B and M on line
and is perpendicular to since
2. Sample answers:
3. Sample answer:
The distances from D to the endpoints A and B are equal,the distances from E to the endpoints A and B are equal,the distances from F to the endpoints A and B are equal,and the distances from G to the endpoints A and B areequal.
4. Any point on the perpendicular bisector has equal dis-tances to the endpoints of the segments.
5.1 Guided Practice (p. 267)
1. If D is on the perpendicular bisector of then D isequidistant from A and B.
2. Point G must be on the angle bisector of by Theorem 5.4.
3.
4. and are both right angles and are congruent.
5. because C is on the perpendicular bisector of
6.
7. The distance from M to is equal to the distance fromM to
5.1 Practice and Applications (pp. 268–271)
8. No; C is not on the perpendicular bisector of becausethe measures of and are not equal.BCAC
AB
→PN.
→PL
m�LPM � m�NPM
AB.AC � BC
�BDC�ADC
AD � BD
�HJK
K
J
H
G
AB,
BMA
CD
EF
G
m�CAB � 90�.AB↔CM
↔CDAM � MB
AB↔CM
2. Sample answer:
P
4. Sample answer:
MA � MB � 32 mm
6. Sample answer:
CA � CB � 37 mm
Point D Point E Point F Point G
GB � 67 mmFB � 58 mmEB � 48 mmDB � 43 mm
GA � 67 mmFA � 58 mmEA � 48 mmDA � 43 mm
Geometry 81Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
9. No; the diagram does not show that
10. No; along with the information given, we would alsoneed
11. No; since P is not equidistant from the sides of P isnot on the bisector of
12. No; since we do not know for sure that one of the dis-tances given is a perpendicular distance.
13. No; the diagram does not show that the segments withequal length are perpendicular segments.
14.
16. 17.
18. Point U must be on the perpendicular bisector
19.
20. Point M must be on the angle bisector 21. B
22.
(because triangle TVX is a right triangle)
A
23.
because triangle UVW is a right triangle.
because
C
24. F 25. D 26. E
27. Given: P is on m.
Prove:Statements Reasons
1. P is on line m. 1. Given
2. 2. By construction
3. 3. Reflexive Property of Congruence
4. 4. SSS Congruence Postulate
5. 5. Corresponding parts of congruenttriangles are congruent.
6. 6. Theorem 3.1↔CP � AB
�CPA � �CPB
�CPA � �CPB
CP � CP
CA � CB
PA � PB
↔CP�AB.
m�VWX � 40�
m�VWX.m�VWU �
m�VWX � 50� � 90�
m�VWU � m�UVW � 90�
m�XTV � 60�
m�TV � 30� � 90�
m�XTV � m�TVX � 90�
→JN.
NQ � 2
↔SV.
SR � 17VT � 8
AD � BD � 5 cm
A B
D
4 cm 4 cm
3 cm
�A.�A,
AP � PB.
CA � CB. 28.
Statements Reasons
1. 1. Given
2. 2. Definition of congruent segments
3. and 3. Definition of perpendicular are right angles. lines
4. 4. Reflexive Property forCongruence.
5. 5. SAS Congruence Postulate
6. 6. Corresponding parts of congru-ent triangles are congruent.
7. 7. Definition of congruent segments
29.
Statements Reasons
1. Construct at P 1. Perpendicular Postulate
2. and are 2. Definition of perpendicular right angles. lines
3. and are 3. Definition of right trianglesright triangles.
4. 4. Definition of congruent
5. 5. Reflexive Property ofCongruence
6. 6. HL Congruence Theorem
7. 7. Corresponding parts of con-gruent triangles are congru-ent.
8. is the perpendicular 8. Definition of perpendicularbisector of and C is bisectoron the perpendicular bisector of
30.
Statements Reasons
1. is the perpendicular 1. Givenbisector of
2. 2. Definition of perpencicular bisector of a segment
3. 3. Perpendicular Bisector Theorem
4. 4. Definition of congruent seg-ments
5. 5. Reflexive Property ofCongruence
6. 6. SSS Congruence Theorem
7. 7. Corresponding parts of con-gruent triangles are congru-ent.
8. 8. SAS Congruence TheoremTheorem�GMH � �GMK
�GJH � �GJK
�GJH � �GJK
GJ � GJ, GM � GM
GH � GK, HJ � JK
GH � GK
GJ � HK, HJ � JK
HK.GJ
AB.
AB
↔CP
AP � BP
�CPA � �CPB
CP � CP
CA � CB
�CPB�CPA
�CPB�CPA
↔CP � AB
CA � CB
CA � CB
�APC � �BPC
CP � CP
�CPB�CPA
AP � BP
↔CP � AB
15.
D is about 1.4 inches fromeach side of �A.
A
D3 in.
Chapter 5 continued
82 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
31. The post is the perpendicular bisector between the ends ofthe wires.
32.
Statements Reasons
1. D is in the interior of 1. Given
2. D is equidistant from 2. Given
and
3. 3. Definition of equidistant
4. 4. Definition of distance froma point to a line
5. and are 5. If two lines are then theyright angles form 4 right angles.
6. 6. Definition of right angles
7. 7. Reflexive Property of Congruence.
8. 8. HL Congruence Theorem
9. 9. Corresponding parts of congruent triangles are congruent.
10. bisects and 10. Definition of angle bisectorpoint D is on the bisector of
33. Line l is the perpendicular bisector of
34. should be the angle bisector of to give thegoalie equal distances to travel on both sides.
35. The increases as the puck gets closer to the goal.This change makes it more difficult for the goalie becausethe goalie has a greater area to defend since the distancesfrom goalie to the sides of (the shooting angle)increase.
36. Answers may vary.
Sample answer:
This demonstrates the Perpendicular Bisector Theorembecause D is on the perpendicular bisector of and D isequidistant from A and B.
AB
D2A � D2B � 48 mm
D1A � D1B � 40 mm
BA C
D1
D2
�APB
m�APB
�APB→PG
AB.
�ABC.
�ABC→BD
�ABD � �CBD
�DAB � �DCB
BD � BD
�DAB � �DCB
�,�DCB�DAB
DA � →BA, DC �
→BC
DA � DC
→BC.
→BA
�ABC.
37. a–c.
The perpendicular bisectors meet in one point.
d. The fire station at A should respond because it is closest to the house at x.
38.
because the product of their slopes is
because the product of their slopes is
39.
40. bisects because the perpendicular distances
from W to and are equal. We know these are
perpendicular distances because in problem 38 it was
shown that and
5.1 Mixed Review (p. 271)
41.
r � 6 cm
2r � 12
d � 12 cm
WT�→YZ.WS�
→YX
→YT
→YS
�XYZ→YW
� �10
� �1 � 9
� ���1�2 � ��3�2
WT � ��5 � 6�2 � �1 � 4�2
� �10
� �9 � 1
� ���3�2 � 12
WS � ��3 � 6�2 � �5 � 4�2
�3 � ��13� � �1�.
�1.WT�→YZ
� �13
��26
slope of YZ � m �0 � 28 � 2
� 3��3�1
slope of WT � m �1 � 45 � 6
��13
� 3 � �1�.
�1.WS�→YX
� 3�62
slope of →YX � m �
8 � 24 � 2
�1
�3slope of WS � m �
5 � 43 � 6
A
B
C
X
42.
� 37.68 cm
� 2 � 3.14 � 6
C � 2�r
Geometry 83Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
43.
45.
46. 47.
48.
49.
50.
52.
Lesson 5.2
Developing Concepts Activity (p. 272)
1.–4. Yes, all three bisectors intersect at the same point.
Conjecture: For any acute scalene triangle, the threeperpendicular bisectors of the three sides will intersectat the same point.
The three segments that are formed by connecting thevertices of the triangle to the point of intersection ofthe perpendicular bisectors of the sides have the samelength.
5.2 Guided Practice (p. 275)
1. If three or more points intersect at the same point, thelines are concurrent.
2. The incenter is made up of the words in and center. Theincenter is the “center” for the circle that is “in” the trian-gle. The circumcenter is made up of the parts circum andcenter. Circum can be short for circumference which isthe distance around the circle and can help us to remem-ber it is the circle around the triangle.
3. 4. MK � MJ � 5GC � GA � 7
AP � BP � CP � 38 mm
x � 8
6x � 48
10x � 48 � 4x
�10x � 22�� � 70� � 4x�
x � 59
x� � 31� � 90�
� 0�0
11m �
�8 � ��8�8 � ��3�
� �113
m �12 � 11�10 � 3
�87
��8�7
m �0 � 8
�7 � 0� 0�
05
m �5 � 59 � 4
� �45
�8
�10m �
5 � ��3��6 � 4
� 113.04 cm2
� 3.14 � 36
� 3.14�6�2
A � �r2 5.2 Practice and Applications (pp. 275–278)
5.
The perpendicular bisectors intersect outside the obtuse triangle.
7.
The perpendicular bisectors intersect at a point on the right triangle.
10. always 11. always 12. never 13. sometimes
14. 15.
16. by the Pythagorean Theorem.
17. Let the midpoint of be called point R. Then
by the Pythagorean Theorem.
18. The student’s conclusion is false because D is not thepoint of intersection of the angle bisectors. D is the pointof intersection of the perpendicular bisectors of the sidesof the triangle. So
19. The student’s conclusion is false because the angle bisec-tors of a triangle intersect in a point that is equidistantfrom the sides of the triangle, but MQ and MN are notnecessarily distances to the sides; M is equidistant from
and LJ.JK, LK,
DA � DC � DB.
QN � PQ � 25.
25 � PQ
625 � �PQ�2
576 � 49 � �PQ�2
�24�2 � �7�2 � �PQ�2
�PR�2 � �QR�2 � �PQ�2
PR � RM �PM2
�482
� 24.
PM
KB � CK � 3
CK � 3
�CK�2 � 9
16 � �CK�2 � 25
42 � �CK�2 � 52
�JC�2 � �CK�2 � �KJ�2
WB � WC � 20DR � SD � 9
51.
x � 34
2x � 34 � x
�2x � 6�� � 40� � x�
6.
The perpendicular bisectorsintersect at a point inside anacute triangle.
8. and 9.
Sample answer:
The segments are congruent. This confirmsTheorem 5.6.
A B
C
D
44.
� �5
�5
�1
m �10 � 5
�2 � ��1�
Chapter 5 continued
84 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
20. To find the point that is equidistant from each location,draw the triangle, construct the perpendicular bisectorsfor each side, and the point of intersection is the pointthat is equidistant from each location.
21. Point H is the point of intersection of the perpendicular bisectors. So H is equidistant fromeach location. H would bethe best location for the new home.
22.
Statements Reasons
1. the bisectors of 1. Givenand
2. 2. bisects so D isequidistant from the sidesof
3. 3. bisects so D isequidistant from the sidesof
4. 4. Transitive property ofequality
5. D is on the angle bisector 5. Converse of the Angle of Bisector Theorem
6. The angle bisectors 6. Given and Steps 2, 3, 4 intersect and 5 at a point D that is
equidistant from and
23. She could construct the perpendicular bisectors to findthe point that is equidistant from the vertices of the trian-gle. By doing so, she would see that the perpendicularbisectors do intersect at a point on the hypotenuse. Sincethe point on the hypotenuse would be the point of inter-section of the perpendicular bisectors, then it would beequidistant from the vertices.
24.–25. The radius isapproximately
26. ft inches
30 in. in. per year years
The mycelium is approximately 3.75 years old.
� 3.75� 8
� 30212
212 ft.
1
1
y
x
A (2, 5)
B (6, 3)
C (4, 1)
CA.BC,AB,
�C.
DF � DG
�ABC.
�ABC,→BDDE � DF
�BAC.
�BAC,→ADDE � DG
DG � CA
DF�BC,DE�AB,�C,�B,�A,
ABC,
School
Office
FactoryH
27.
E
28.
by the Pythagorean Theorem
because T is the midpoint of
C
29. The midpoint of is
The midpoint of is
The midpoint of is
The slope of
The perpendicular bisector has slope because
The line is the equation of the perpendicular bisector of
The slope of
—CONTINUED—
BC �0 � 6
18 � 12�
�66
� �1.
AB.y � �2x � 15
y � 3 � �2x � 12
y � 3 � �2�x � 6�
�2 �12
� �1.
�2
AB �6 � 0
12 � 0�
612
�12
.
�0 � 182
, 0 � 0
2 � � �182
, 02� � �9, 0�.
AC
�12 � 182
, 6 � 0
2 � � �302
, 62� � �15, 3�.
BC
�0 � 122
, 0 � 6
2 � � �122
, 62� � �6, 3�.
AB
XY � 10
XY � 2 � 5
XY.
XY � 2�XT�
XT � 5
�XT�2 � 25
�XT�2 � 144 � 169
�XT�2 � 122 � 132
�XT�2 � �TW�2 � �XW�2
XW � WZ � 13
m�ADC � 140�
m�ADC � 40� � 180�
m�ADC � m�DCA � m�CAD � 180�
m�DCA � m�CAD � 40�
12 �m�BCA� �12 �m�CAB� � 40�
12 �m�BCA � m�CAB� �12 � 80�
m�BCA � m�CAB � 80�
100� � m�BCA � m�CAB � 180�
m�ABC � m�BCA � m�CAB � 180�
Geometry 85Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
29. —CONTINUED—
The perpendicular bisector of has slope 1 because
The line is the equation of the perpendicularbisector of
The slope of so is
horizontal.
So the perpendicular bisector is the vertical line
30. The lines and intersect at the pointbecause
To show is also on plug the valuesin to see if it is true.
Since is true, the point is on
31. Let P be the point
Since P is equidistant from A,B, and C.
AP � BP � CP � 3�10,
� 3�10
� �9 � �10
� �90
� �81 � 9
� ���9�2 � ��3�2
CP � ��9 � 18�2 � ��3 � 0�2
� 3�10
� �9 � �10
� �90
� �9 � 81
� ���3�2 � ��9�2
BP � ��9 � 12�2 � ��3 � 6�2
� 3�10
� �9 � �10
� �90
� �81 � 9
� �92 � ��3�2
AP � ��9 � 0�2 � ��3 � 0�2
�9, �3�.
y � x � 12.�9, �3��3 � �3
�3 � �3
�3 � 9 � 12
y � x � 12
y � x � 12,�9, �3�y � �2�9� � 15 � �18 � 15 � �3.�9, �3�
x � 9y � �2x � 15
x � 9.
ACAC �0 � 0
18 � 0�
018
� 0,
BC.y � x � 12
y � 3 � x � 15
y � 3 � 1 � �x � 15�
�1 � 1 � �1.BC
5.2 Mixed Review (p. 278)
32.
34. j has slope because
The equation of j is
35. j has slope because
The equation of is
36. j has slope because
The equation of j is
37. j has slope because
The equation of j is
38. There is enough information to prove byusing the SAS Congruent Postulate.
39. There is not enough information given to proveOne pair of congruent sides, one side
congruent to itself, and one pair of congruent angles aregiven. But the angles must be the included angles and theyare not.
40. There is enough information given to proveOne pair of congruent legs and one
pair of congruent hypotenuses are given. The HLCongruence Theorem can be used to prove
Lesson 5.3
5.3 Guided Practice (p. 282)
1. The centroid of a triangle is the point where the threemedians intersect.
2. The legs, and of right are also altitudesof because is the perpendicular segment fromK to and is the perpendicular segment from L toKM.
LMLMKM�KLM
�KLMLM,KM
�PMN � �KML.
�PMN � �KML.
�GJF � �GJH.
�ABC � �DEC
y � �1110 x �
565 .
y � �1110 x �
565
y � 9 � �1110 x �
115
y � 9 � �1110 �x � 2�
y � ��9� � �1110 �x � ��2��
�1110 � 10
11 � �1.�1110
y �32 x � 5.
y �32 x � 5
y � 8 �32 x � 3
y � 8 �32 �x � 2�
�23 � 3
2 � �1.32
y �12 x �
52.j
y �12 x �
52
y � 6 �12 x �
72
y � 6 �12 �x � 7�
�2 � 12 � �1.1
2
y � �13 x �
133 .
y � �13 x �
133
y � 4 � �13 x �
13
y � 4 � �13 �x � 1�
�13 � 3 � �1.�
13
� 22.5 square units
� 452
� 12 � 9 � 5
A �12 bh 33.
� 77 square units
� 1542
� 12 � 22 � 7
A �12 bh
Chapter 5 continued
86 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
3. indicates that G is the midpoint of Therefore, is a median of
4. indicates that is an altitude of
5. indicates the is bisected. So is an angle bisector of
6. and indicates G is the midpoint ofand is a perpendicular bisector of but it is
also a median and an altitude.
7. indicates and In this case is a
perpendicular bisector, an angle bisector, a median, andan altitude of
5.3 Practice and Applications (pp. 282–284)
8.
9.
11.
12.
13.–14. Sample answer:
is an acute triangle.
15. Yes, they all met at the same point. It is labeled as pointD.
�ABC
A
LD
N
M B
C
PHEP
�48
�12
PHEH
�4
12�
13
� 48 units
� 15 � 15 � 18
� 2 � 7.5 � 15 � 2 � 9
� 2�DG� � 15 � 2�DH�
Perimeter of �DEF � DE � EF � FD
15 � EF
225 � �EF�2
144 � 81 � �EF�2
122 � 92 � �EF�2
�EH�2 � �HF�2 � �EF�2
12 � EH
32 � 8 � EH
8 �23 EH
EP �23 EH
FH � DH � 9
�DEF.
EGDG � FG.�DGE � �FGE,�DEG � �FEG,�DGE � �FGE
�DEF,EGDFDG � FGEG�DF
�DEF.EG�DEF�DEG � �FEG
�DEF.EGEG�DF
�DEF.EGDF.DG � FG 16. Sample answer:
The distance from the centroid to a vertex is two thirds ofthe distance from that vertex to the midpoint of the oppo-site side.
17.
18.
19.
The coordinates of T are
20.
21. � �4, 4�� �82
, 82�M � �5 � 3
2,
2 � 62 �
23
�23
69
�23
NTNR
�23
NR � 11 � 2 � 9
NT � 11 � 5 � 6
� �2, 2�� �42
, 42� R � ��1 � 5
2,
�2 � 62 �
�5, 6 � 4� � �5, 2�.
PT � 4
PT �23 � 6
PT �23 PQ
� 6
� �36
� �0 � 36
� �02 � ��6�2
PQ � ��5 � 5�2 � �0 � 6�2
� �5, 0�� �102
, 02�Q � ��1 � 11
2,
�2 � 22 �
40 � 40
40 �23 � 60
CD �23 �CM�
CM � 60 mm
CD � 40 mm
54 � 54
54 �23 � 81
BD �23 �BL�
BL � 81 mm
BD � 54 mm
64 � 64
64 �23 � 96
AD �23 �AN�
AN � 96
AD � 64 mm
10.
PH � 4
PH � 8 � 12
PH � PE � EH
Geometry 87Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
22.
23.
24. Sample answer:
P is the orthocenter of
25. Sample answer:
The orthocenter of ispoint G.
26. Sample answer:
P is the orthocenter of
27.C
FA B
D
GH
E
�KLM.
P
K
ML
�EFGE
G F
�ABC.
A
P
B
C
2�5 � 2�5
2�5 �23 � 3�5
JP �23 JM
� 3�5
� �9 � �5
� �45
� �9 � 36
� ���3�2 � ��6�2
JM � ��4 � 7�2 � �4 � 10�
� 2�5
� �4 � �5
� �20
� �4 � 16
� ���2�2 � ��4�2
JP � ��5 � 7�2 � �6 � 10�2 28. G and H are the same point.
29. When is measured, it is found that Sincethen G and H must be the same point; therefore
the lines containing the three altitudes intersect at onepoint.
30.–32.
33. The measure of the angle between r and is approxi-mately
34. a.
b.
c. This is in the drawing above.
d.
The length of the altitude is equal to twice the area divid-ed by the base.
12 � BE
90 � 7.5 � BE
90 �12
� 15 � BE
90 �12
� CA � BE
A �12
bh
� 90 square units�12
� 20 � 9A �12
bh
CD � 9
�CD�2 � 81
144 � �CD�2 � 225
122 � �CD�2 � 152
�AD�2 � �CD�2 � �AC�2
20�.MP
R
Sr
B
L
D
C
M
P
GH � 0,GH � 0.GH
e.
2Ab
� h
2A � bh
A �12
bh
Chapter 5 continued
88 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
35.
Statements Reasons
1. is isosceles 1. Givenis a median to base
2. D is the midpoint of 2. Definition of median?
3. 3. Definition of midpoint?
4. 4. Definition of isosceles triangle
5. 5. Reflexive Property ofCongruence
6. 6. SSS Congruence Postulate
7. 7. Corresponding parts ofcongruent triangles arecongruent.
8. and are a 8. Definition of linear pairlinear pair.
9. 9. If two lines intersect toform a linear pair of con-gruent angles, then thelines are perpendicular.
10. is also an altitude. 10. Definition of altitude.
36. No, medians to the legs of an isosceles triangle are notalso altitudes. It was the fact that the legs are congruentthat made it possible to prove Exercise 35. One leg andthe other side would have to be congruent as well for thisto be true.
37. Yes, the medians of an equilateral triangle are also alti-tudes because the proof for the isosceles triangle could beused for the equilateral triangle.
Yes, the medians would be contained in the angle bisec-tors. By looking at the proof in Exercise 35, it can beseen that the median was also the angle bisector since thetwo triangles are congruent.
Yes the medians would be contained in the perpendicularbisectors because it was shown in Exercise 35 that themedian was perpendicular to the side at the midpoint.
38. The median of an equilateral triangle is also a perpendic-ular bisector of a side, an altitude, and an angle bisector.
5.3 Mixed Review (p. 284)
39. The parallel line would also have slope
The equation of the line through P that is parallel tois y � �x � 8.y � �x � 3
y � �x � 8
y � 7 � �x � 1
y � 7 � �1�x � 1�
�1.
BD
BD�AC
�BDA�BDC
�BDC � �BDA
�BDC � �BDA
BD � BD
AB � CB
AD � CD
AC.
AC.BD�ABC
40. The parallel line would also have slope
The equation of the line through P that is parallel tois
41. The parallel line would also have slope 3.
The equation of the line through P that is parallel tois
42. The parallel line would also have slope
The equation of the line through P that is parallel tois
43. because you need the angles which do nothave or as a side.
44. because you need the angles which have oras a side.
45. Sample answer:
Quiz 1 (p. 285)
1.
x � 16
4x � 3x � 16
4x � 9 � 3x � 25
� 5�10
� �25 � �10
� �250
� �169 � 81
� �132 � 92
h � ��13 � 0�2 � �9 � 0�2
1
1
y
x
(0, 9)
(13, 0)
HJEF�F � �J
GJDF�E � �H
y � �12 x.y � �
12 x � 1
y � �12 x
y � 2 � �12 x � 2
y � ��2� � �12 �x � 4�
�12.
y � 3x � 21.y � 3x � 5
y � 3x � 21
y � 9 � 3x � 12
y � ��9� � 3�x � 4�
y � �2x � 14.y � �2x � 3
y � �2x � 14
y � 8 � �2x � 6
y � 8 � �2�x � 3�
y � ��8� � �2�x � ��3��
�2.
2.
y � 12
2y � 24
3y � y � 24
Geometry 89Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
3.
4. because the perpendicular bisectors intersect at a point equidistant from the vertices of the triangle.
5. The balancing point would be at point G because that isthe centroid of the triangle.
5.3 Math and History (p. 285)
1. You need to go to the post office (P), then the market(M), then the library (L) or in reverse order.
2. The goalie’s position on the angle bisector optimizes thechance of blocking a scoring shot because the distancethe goalie would have to travel to either side of the anglewould be the same. The goalie would not have fartherthan that distance to either side of himself to block.
Technology Activity 5.3 (p. 286)
Investigate
1.
So is the angle bisector of
2. F was the point of intersection of the angle bisectors ofangles and by construction. Since
is the angle bisector of So the three angle bisectors are concurrent.
3. and This makes a medi-an also.
4. Since F was the point of intersection of two medians byconstruction. Since G is the midpoint of and is the third median of the triangle. Since F is on
by construction, F is on all three medians and themedians are concurrent.
5.
Yes,
6. No, the quotient does not change because it is the
ratios based on measurements using the centroid.
ADAF
AD �23 AF
ADAF
�72
108�
23
AF � 108 mm
AD � 72 mm
CGCG
ABAG � BG,
CGBG � 57 mm.AG � 57 mm
�BAC.→AFm�BAF � m�CAF,
�BCA�ABC
�BAC.→AF
m�CAF � 42�
m�BAF � 42�
VT � VS � 10
10 � VT
100 � �VT�2
36 � 64 � �VT�2
62 � 82 � �VT�2 Extension
For any triangle in which the angle bisector is containedin the same ine as the median, the line will also containan altitude and perpendicular bisector of the triangle.
In this drawing is the angle bisector of andis the median of the triangle. So Sinceis the angle bisector, or
Then by the HL Congruence Theorem.Therefore, or because corresponding parts of congruent triangles arecongruent. Also by the HL CongruenceTheorem. So, or because corresponding parts of congruent triangles arecongruent. Then
So which are linear pairs.Therefore, and This wouldmean that would be an altitude and a perpendicularbisector, also.
Lesson 5.4
5.4 Guided Practice (p. 290)
1. In if M is the midpoint of N is the midpointof and P is the midpoint of then and
are midsegments of triangle ABC.
2. It is convenient to position one of the sides of the trianglealong the x-axis because some of the coordinates of twoof the points will be zero and the side will be horizontaland have no slope.
3. 4.
5.
7.
9.
� 30.6
� 10.6 � 8 � 12
Perimeter � GJ � JH � GH
� 10.6
� 12 � 21.2
GJ �12 � EF
� 16
� 2 � 8
DF � 2 � DG
� 21.2
� 2 � 10.6
EF � 2 � EH
GH � DEJH � DF
PNNP,MN,BC,AC,
AB,�ABC,
ADAD�CD.�ADC � �ADB
m�ADC � m�ADBm�ADE.m�CDE � m�ADF � m�BDE �
m�CDF � m�BDE�CDF � �BDE�CFD � �BED
m�ADE � m�ADF�ADE � �ADF�ADE � �ADF
AD � AD.DE � DF.DE � DF
→AD
DC � DB.AD�BAC
→AD
A E B
D
CF
6.
8.
� 8
� 12 � 16
JH �12 � DF
� 12
� 12 � 24
GH �12 � DE
Chapter 5 continued
90 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
10.
11.
12. 13.
14.
15.
16.
17.
19. The following pairs of angles are congruent:
and because they are correspondingangles, as are and and because they are alternate interiorangles.
So, by the Transitive Property of Congruence,and are congruent, as are and Then and
all congruent by the Third Angles Theoremand the Transitive Property of Congruence.
20. Sample answer:
Use the construction of the perpendicular bisectors tofind the midpoints D, E, and F.
are the midsegments for �ABC.DE, EF, DF
A B
E
C
F
D
�MNC are�B, �ALB, �LMN,�NLM.�LMA,�C,
�BNL,�LNM�A, �NMC,�BLN,
�CMN � �LNM�AML � �MLN�LMA,�BNL, �C,
�NMC�BLN, �A,
� 31
� 24 � 7
� 3 � 8 � 7
LM � 3x � 7
x � 8
�12 x � �4
3x �72 x � 4
3x � 7 �72 x � 3
3x � 7 �12 �7x � 6�
LM �12 � BC
� 18
� 2 � 9
BC � 2 � NC
� 14� 2 � 7AB � 2 � MN
� 10�12 � 20LN �
12 � AC
AB � MNLM � BC
AB � 5.4 � 10 yds � 54 yd
� �29 � 5.4
� �25 � 4
� �52 � 22
AB � ��6 � 1�2 � �6 � 4�2
� �1, 4�
� �22
, 82
A � �0 � 22
, 0 � 8
2 21.
22. Slope of
Slope of
Since the slopes of and are equal,
So and
23.
24.
—CONTINUED—
�a � c
b
�2�a � c�
2b
�2a � 2c
2b
�2a � 2c2b � 0
�a � c
b
Slope of DF �a � cb � 0
� �c, 0�� �2c2
, 02F � �0 � 2c
2,
0 � 02
y
xA(0, 0) F (c, 0) B(2c, 0)
E(a � c, b)
C (2a, 2b)
D (a, b)
DF �12
BC.BC � DF
DF �12
BC
�12
� �89��89
2�
�89�4
��894
DF � �22.25
� �89
� �25 � 64
� �52 � 82
BC � ��10 � 5�2 � �6 � ��2��2
� �22.25
� �6.25 � 16
� �2.52 � 42
DF ���5 �52
2
� �4 � 0�2
BC � DF.DFBC
�85
� 4 �25
�452
DF �4 � 0
5 �52
�85
BC �6 � ��2�
10 � 5
� �5, 4�� �102
, 82F � �0 � 10
2,
2 � 62
� �152
, 2� �152
, 42E � �5 � 10
2,
�2 � 62
� �52
, 0� �52
, 02D � �0 � 5
2,
2 � ��2�2
� �6, 6�
� �122
, 122
B � �2 � 102
, 8 � 4
2
18.
� 6
� 24 � 18
� 6 � 4 � 18
AB � 6x � 18
x � 4
�2x � �8
x � 3x � 8
x � 1 � 3x � 9
x � 1 �12 �6x � 18�
MN �12 � AB
� 9
� 12 � 18
LM �12 � BC
Geometry 91Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
24. —CONTINUED—
Since the slopes of and are equal,
Since the slopes of and are equal,
25.
26.
Slope of
Draw a line through M with slope
Slope of
Draw a line through L with slope 5.
MN �9 � 45 � 4
�51
� 5
13
.
LN �4 � 34 � 1
�13
2
2
y
x
M(5, 9)
N(4, 4)L(1, 3)
C B
A
EF �12 CA
EF �12 � 2�a2 � b2
EF � �a2 � b2
DF �12 CB
DF �12 �2��a � c�2 � b2�
DF � ��a � c�2 � b2
� 2�a2 � b2
� �4�a2 � b2
� �4�a2 � b2�
� �4a2 � 4b2
� ��2a�2 � �2b�2
CA � ��2a � 0�2 � �2b � 0�2
� 2��a � c�2 � b2
� �4��a � c�2 � b2
� �4�a � c�2 � b2�
� �4�a � c�2 � 4b2
� �22�a � c�2 � �2b�2
CB � ��2a � 2c�2 � �2b � 0�2
� �a2 � b2
EF � ��a � c � c�2 � �b � 0�2
� ��a � c�2 � b2
DF � ��a � c�2 � �b � 0�2
EF � CA.CAEF
�ab
�2a2b
�2a � 02b � 0
�ab
Slope of EF �a � c � c
b � 0
DF � CB.CBDF
Draw a line through N with slope
The lines intersect at and
27.
Draw a line through M with slope
Draw a line through L with slope
Draw a line through N with slope
The lines intersect at and
28.
because
29.
because
� 62 units� 24 � 18 � 20P � PR � QR � PQ
QR � 18
QR � 2 � 9
QR � 2�QU�
QU � UR.QR � QU � QU
QR � QU � UR
24 � PR
12 �12 PR
SU �12 PR
� 40 units� 14 � 16 � 10P � CD � BD � BC
BC � 10
BC � 2 � 5
BC � 2�GC�
BG � GC. BC � GC � GC
BC � BG � GC
16 � BD
8 �12 BD
GF �12 BD
C�7, 9�.B�11, 3�,A�3, �1�,
52
.
Slope of LM �6 � 19 � 7
�52
12
.
�24
�12
Slope of MN �6 � 49 � 5
�32
.
�3
�2� �
32
Slope of LN �4 � 15 � 7
�2
2
y
x
M(9, 6)
L(7, 1)
N(5, 4)
A
C
B
C�2, 8�.A�0, �2�, B�8, 10�,
32
.
Slope of LM �9 � 35 � 1
�64
�32
Chapter 5 continued
92 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
30.
Perimeter of is 4 times the perimeter of
4 times perimeter of
31. The perimeter of the shaded triangle in Stage 1 is because each side of the shaded triangle is of aside in the original triangle.
The total perimeter of the shaded triangle in Stage 2 is
The total perimeter of the shaded triangles in Stage 3 is
32. The bottoms of the legs will be 60 inches apart. Since thecross bar attaches at the midpoints of the legs, the crossbar is a midsegment of the triangle formed by the twolegs and the ground. Since the measure of the midseg-ments is half of the measure of third side, the measure ofthe cross bar, 30 inches, is half of the measure betweenthe bottoms of the legs, 60 inches.
33. is a medsegment of so D is the midpoint ofand By the Midsegment Theorem,
and But F is the midpoint of soThen by transitive
property of equality and the definition of congruent seg-ments, Corresponding angles and
are congruent, so by the SASCongruence Postulate.
�ADE � �DBF�ABC�ADEDE � BF.
BF �12 BC.
BC,DE �12 BC.DE � BC
AD � DB.AB�ABC,DE
� 238.
12 �
14 �
14 �
14 �
18 �
18 �
18 �
18 �
18 �
18 �
18 �
18 �
18
12 �
14 �
14 �
14 � 11
4.
12
12
�GHI � Perimeter of �ABC
Perimeter of �GHI �14 of perimeter of �ABC
� 14 �BC � AC � AB�
� 14 BC �
14 AC �
14 AB
Perimeter of �GHI � GH � HI � GI
GI �14 AB
GI �12 �1
2 AB� GI �
12 FE
HI �14 AC
HI �12 �1
2 AC� HI �
12 DE
GH �14 BC
GH �12 �1
2 BC� GH �
12 FD
�GHI.�ABC
A B
E
C
F H
I G
D
34. In Exercise 33, it was shown that andSo all that is left to show is that
This can be done in the same manner that it was shownthat By using the fact that is a midseg-ment, By using the fact that E is a midpointof we can get Therefore, or
and the triangles would be congruent by SSSCongruence Postulate.
35. Since is closer to than it must be longer thanSince is the midsegment, or
So cannot be 10 or 12 feetlong since it must be longer than which is 12 feetlong. could be 14 feet long but not 24 feet long sincePQ cannot equal RS. So, or
36. a.
b.
c.
d.
The equation of the line containing has slope 3and passes through
The equation of the line is
The equation of the line containing has slope and passes through
The equation of the line is
e.
—CONTINUED—
y �12 x �
12
y � 2 �12 x �
32
y � 2 �12 �x � 3�
y � 3x � 2
y � 4 � 3x � 6
y � 4 � 3�x � 2�
y � 2 �12 �x � 3�.
↔AB
y � 2 �12 �x � 3�
E�3, 2�.
12AB
y � 4 � 3�x � 2�.↔AC
y � 4 � 3�x � 2�
D�2, 4�.AC
�12
slope of FD � m3 �5 � 44 � 2
� 3�31
slope of EF � m2 �5 � 24 � 3
y � �2x � 13
y � 5 � �2x � 8
y � 5 � �2�x � 4�
� �2�2
�1m1 �
4 � 22 � 3
1
1
y
x
F (4, 5)
E(3, 2)A(1, 1)
C (3, 7)
B(5, 3)(2, 4)
D
12 < PQ < 24.MN < PQ < RS,
PQMN
PQMN �12 � 24 � 12 feet.
MN �12 RSMNMN.
MN,RSPQ
AE � DFAE � DFAE �
12 AC.AC,
DF �12 AC.
DEDE � BF.
AE � DF.DE � BF.AD � DB
Geometry 93Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
f. To find A, find the point of intersection of and
has the equation
has the equation
By substitution,
So,
To find B, find the point of intersection of and
has the equation
has the equation
By substitution,
So,
To find C, find the point of intersection of and
has the equation
has the equation
By substitution,
So,
37.
C�3, 7�.
y � 9 � 2 � 7
y � 3 � 3 � 2
y � 3x � 2
x � 3
5x � 15
3x � �2x � 15
3x � 2 � �2x � 13
y � �2x � 13.↔BC
y � 3x � 2.↔AC
↔BC.
↔AC
B�5, 3�.
y � 3
y � �10 � 13
y � �2 � 5 � 13
y � �2x � 13
x � 5
52 x �252
12 x � �2x �255
12 x �
12 � �2x � 13
y � �2x � 13.↔BC
y �12 x �
12.
↔AB
↔BC.
↔AB
A�1, 1�.
y � 3 � 2 � 1
y � 3 � 1 � 2
y � 3x � 2
x � 1
�52 x � �
52
12 x � 3x �52
12 x �
12 � 3x � 2
y � 3x � 2.↔AC
y �12 x �
12.
↔AB
↔AC.
↔AB 38. is the function
that gives the length of themidsection at Stage n. Fromone stage to the next, thelength is multiplied by
5.4 Mixed Review (p. 293)
39.
Addition property of equality
40.
Subtraction property of equality
Division property of equality
41.
Addition property of equality
Subtraction property of equality
Division property of equality
42.
Subtraction property of equality
Subtraction property of equality
Division property of equality
43.
Division property of equality
Addition property of equality
Division property of equality
44.
Division property of equality
Subtraction property of equality
Division property of equality
45.
Subtraction property of equality
Division property of equality
Subtraction property of equality
46.
Distributive property
Simplify
Subtraction property of equality
Division property of equality x � �65
5x � �6
5x � 10 � 4
3x � 2x � 10 � 4
3x � 2�x � 5� � 4
x � �11
x � 1 � �10
�2�x � 1� � 20
�2�x � 1� � 3 � 23
x � �73
3x � �7
3x � 10 � 3
9�3x � 10� � 27
x � 2
4x � 8
4x � 1 � 7
2�4x � 1� � 14
x � 4
�4x � �16
5x � 9x � 16
5x � 12 � 9x � 4
x � 3
6x � 18
8x � 2x � 18
8x � 1 � 2x � 17
x � 11
3x � 33
3x � 13 � 46
x � 14
x � 3 � 11
12.
y � 24 � �12�n
Stage
8
4
0
12
16
20
24
2 4 61 3 5
Mid
seg
men
t
len
gth
m
n
Stage n 0 1 2 3 4 5
Midsegment length 24 12 6 3 1.5 0.75
Chapter 5 continued
47.
48.
49.
50. and because and are angle bisectors and angle bisectors divide anangle into two congruent angles.
51. Point D is the incenter of because it is the inter-section of the angle bisectors for
52. because D is the point of intersection ofthe angle bisectors of and D is equidistant fromthe sides of the triangle.
53.
But because D is equidistant from the sides of
So,
Lesson 5.5
Technology Activity 5.5 (p. 294)
Investigate
1. The longest side is opposite the largest angle.
2. The shortest side is opposite the smallest angle.
3. The answers are the same.
The longest side is opposite the largest angle.
The shortest side is opposite the smallest angle.
4. The longest side will always be opposite the largestangle. The shortest side will always be opposite thesmallest angle. The side with the middle length will beopposite the angle with the middle measure.
DF � DE � 6.
�ABC.DE � DF
DE � 6
�DE�2 � 36
�DE�2 � 64 � 100
�DE�2 � 82 � 102
�DE�2 � �EC�2 � �CD�2
�ABCDE � DG � DF
�ABC.�ABC
→CD
→BD,
→AD,�BCD � �ACD�CAD � �BAD
x � 18
�3x � �54
4x � 7x � 54
4x� � 61� � �7x � 7��
x � 7
17x � 119
17x � 61 � 180
�10x � 22�� � �7x � 1�� � 38� � 180�
x � 23
2x � 46
2x � 134 � 180
�x � 2�� � 132� � x� � 180� Extension
The statement is false because the above example is acounterexample.
5.5 Guided Practice (p. 298)
1. The 1 inch side is opposite the smallest angle of theinches side is opposite the middle angle of and
the inches side is opposite the largest angle of
2. No, it is not possible to draw a triangle with side lengthsof 5 inches, 2 inches, and 8 inches because the sum of thelengths of any two sides must be greater than the lengthof the third side. But is not greater than 8.
3.
The smallest angle is and the largest angle is
4. The shortest side is and the longest side is
5. The distances between Guiuan and Masbate have to bebetween miles and miles (not inclusive).
5.5 Practice and Applications (pp. 298–301)
6.
is the shortest side because it is opposite the smallestangle.
is the longest side because it is opposite the largestangle.
7.
is the shortest side. and are the longest sides
8.
is the shortest side. is the longest side.HJJK
m�J � 55�
m�J � 35� � 90�
m�J � m�H � 90�
�RS � ST�.STRSRT
m�R � 65�
m�R � 115� � 180�
m�R � 50� � 65� � 180�
m�R � m�S � m�T � 180�
AB
AC
m�A � 67�
m�A � 113� � 180�
m�A � 42� � 71� � 180�
m�A � m�B � m�C � 180�
165 � 99 � 264165 � 99 � 66
DE.EF
�F.�D
m�E � 45�
m�E � 135� � 180�
32� � m�E � 103� � 180�
m�D � m�E � m�F � 180�
5 � 2
90�.218
62�,178
28�,
length of shortest sidelength of longest side
�63 mm96 mm
� 0.656
measure of smallest anglemeasure of largest angle
�41�
82�� 0.5
94 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Geometry 95Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
9. is the smallest angle. is the largest angle.
10. is the smallest angle. is the largest angle.
11. is the smallest angle. is the largest angle.
12.
14. and
15.
and
16.
and
17. and 18. and
19. and
20.–23. Answers may vary; sample answers are given.
20.
21. 22.
23. The following combinations of lengths will not producetriangles: 4 inches, 4 inches, and 10 inches; 3 inches, 5inches, and 10 inches; and, 2 inches, 7 inches, and 9inches.
24.
x < 7
�x > �7
2x > 3x � 7
2x � 5 > 3x � 2
x � 2 � x � 3 > 3x � 2
AB � AC > BC
5.5 in.
8.5 in.4 in.
7 in.
6 in.5 in.
2 in.4 in.
8 in. 6 in.
5 in.
8 in.7 in. 7 in. 8 in.
6 in. 6 in.5 in.
�R�T, �S,
�P�N, �Q,�M�L, �K,
HGHJ, JG,
m�G � 25�
m�G � 155� � 180�
m�HGJ � 120� � 35� � 180�
m�G � m�J � m�H � 180�
EFDF, DE,
m�F � 60�
120� � m�F � 180�
90� � 30� � m�F � 180�
m�D � m�E � m�F � 180�
ACAB, BC,
y� � z� � x�
�F�H
�Q�R
�B�C 25.
26. It is shorter to cut across the empty lot because the sumof the lengths of the two sidewalks is greater than thelength of the diagonal across the lot. If the corner ofPleasant Street and Pine Street were labeled point A, thecorner of Pine Street and Union Street were labeled pointB, and the corner of Union Street and Oak Hill Avenuewere labeled point C, could be formed. We know
or walking around the sidewalks islonger than walking through the lot.
27. The sides and angles could not be positioned as they arelabeled; for example, the longest side is not opposite thelargest angle.
28. No, a kitchen triangle cannot have side lengths of 9 feet,3 feet, and 5 feet because 3 feet 5 feet feet is notlarger than 9 feet.
29. The boom is raised when the boom lines are shortened.
30. AB must be less than feet.
31. Yes, when the boom is lowered and length of the boomlines, AB, is greater than 100 feet, then will belarger than
32. The third inequality would be and this isnot helpful because x has to be positive and 14 is alwaysgreater than 10.
33. so is a right triangle. The largest anglein a right triangle is the right angle, so
so (If one angle of atriangle is larger than another then the side oppositethe larger angle is longer than the side opposite the smaller angle.
�,MN > MJ.m�MJN > m�MNJ,
�MJNMJ�JN,
x � 14 > 10
�BAC.�ABC
100 � 50 � 150
� 8�
AB � BC > AC,�ABC
x < 7
�x > �7
2x > 3x � 7
2x � 6 > 3x � 1
x � 2 � x � 4 > 3x � 1
AB � AC > BC
13.
x� > z�
x� > y�
Chapter 5 continued
34.
Statements Reasons
1. 1. Given
2. Extend to D such 2. Ruler Postulatethat
3. 3. Segment Addition Postulate
4. 4. Base Angles Theorem
5. 5. Protractor Postulate
6. 6. Substitution property of equality
7. 7. If the angle of a triangle islarger than another angle, thenthe side opposite the largerangle is longer than the sideopposite the smaller angle.
8. 8. Substitution property of equality
9. 9. Substitution property of equality
35. since the side opposite the angle of is longerthan the side opposite the angle of A
36. is the measure of the exterior angle and B
37. D
38.
Statements Reasons
1. 1. Given
2. Let D be a point on 2. A plane contains at least plane M distinct from C. three noncollinear points
3. 3. Given
4. is a right angle. 4. If two lines are perpendic-ular, then they intersect toform four right angles.
5. is a right triangle. 5. Definition of right triangle.
6. is an acute angle 6.
7. 7. Definition of an acuteangle
8. 8. Definition of a right angle
9. 9. Substitution of equality
10. 10. If one angle of a triangle islarger than another angle,then the side opposite thelarger angle is longer thanthe side opposite thesmaller angle.
PD > PC
m�PDC < m < PCD
m�PCD � 90�
m�PDC < 90�
�PDC
�PCD
�PCD
↔CD�PC
PC�plane M
x� � y� � z�.z�
y��n � 3 > n�.x�x� > y�
m�PDC < m�PCD
DA � AC > BC
DC < BC
m�DBC > m�1
m�DBC > m�2
�1 � �2
AD � AC � DC
AB � AD.AC
�ABC
5.5 Mixed Review (p. 301)
39.–41. Answers may vary. Sample answers are given.
39. The proof for Example 2 on page 230 is a two-columnproof.
40. The proof for Example 1 on page 229 is a paragraphproof.
41. The proof for Example 3 on page 158 is a flow proof.
42. and are corresponding angles. So are and
43. and are vertical angles.
44. and are alternate interior angles. So are and
45. and are alternate exterior angles. So are and
46.
The line containing has slope and passes through
The line containing has slope and passes through
The line containing has slope and passes through
—CONTINUED—
y �12
x �52
y � 1 �12
x �32
y � ��1� �12
�x � 3�
N�3, �1�.
12
,BC
y � �25
x �195
y � 3 � �25
x �45
y � 3 � �25
�x � 2�
M�2, 3�.
�25
AC
y � �4x � 7
y � 1 � �4x � 8
y � 1 � �4�x � 2�
y � 1 � �4�x � ��2��
���2, 1�.�4AB
� �25
�2
�5slope of LN � m3 �
1 � ��1��2 � 3
� �4�4
�1slope of MN � m2 �
3 � ��1�2 � 3
�12
�24
slope of LM � m1 �3 � 1
2 � ��2�
�10.�7�2�7
�11.�6�3�6
�9�12
�9.�5�1�5
96 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Geometry 97Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
46. —CONTINUED—
To find A, find the point of intersection of and
The point A has coordinates
To find B, find the point of intersection of and
The coordinates of B are
To find C, find the point of intersection of and ↔BC.
↔AC
��1, �3�.
y � �3
y � 4 � 7
y � �4��1� � 7
y � �4x � 7
x � �1
�92
x �92
�4x �12
x �92
�4x � 7 �12
x �52
y �12
x �52
y � �4x � 7
↔BC.
↔AB
��3, 5�.
y � 5
y � 12 � 7
y � �4��3� � 7
y � �4x � 7
x � �3
�185
x �545
�4x � �25
x �545
�4x � 7 � �25
x �195
y � �25
x �195
y � �4x � 7
↔AC.
↔AB
The coordinates of C are
47.
The line containing has slope and passes through
The line containing has slope and passes through
—CONTINUED—
y �53
x �163
y � 2 �53
x �103
y � 2 �53
�x � 2�
y � 2 �53
�x � ��2��
M��2, 2�.
53
BC
y �12
x �132
y � 5 �12
x �32
y � 5 �12
�x � 3�
y � 5 �12
�x � ��3��
L��3, 5�.
12
AB
�53
slope of LN � m3 �5 � 0
�3 � ��6��
�12
�24
slope of MN � m2 �2 � 0
�2 � ��6�
� �3�3
�1slope of LM � m1 �
5 � 2�3 � ��2�
�7, 1�.
y � 1
y �72
�52
y �12
� 7 �52
y �12
x �52
x � 7
9x � 63
5x � �4x � 63
5x � 25 � �4x � 38
12
x �52
��25
x �195
y �12
x �52
y � �25
x �195
Chapter 5 continued
47. —CONTINUED—
The line containing has slope and passes through
To find A, find the intersection of and
The coordinates of A are
To find B, find the point of intersection of and
The coordinates of B are �1, 7�.
y � 7
y �142
y �12
�132
y �12
� 1 �132
y �12
x �132
x � 1
�7x � �7
3x � 10x � 7
3x � 39 � 10x � 32
12
x �132
�53
x �163
y �53
x �163
y �12
x �132
↔BC.
↔AB
��7, 3�.
y � 3
y � 21 � 18
y � �3��7� � 18
y � �3x � 18
x � �7
72
x ��49
2
12
x � �3x �492
12
x �132
� �3x � 18
y � �3x � 18
y �12
x �132
↔AC.
↔AB
y � �3x � 18
y � �3�x � 6�
y � 0 � �3�x � ��6��
N��6, 0�.�3AC
To find C, find the intersection of and
The coordinates of C are
48.
The line containing has slope 4 and passes through
The line containing has slope and passes through
The line containing has slope and passes through
—CONTINUED—
y � �16
x �73
y � 1 � �16
x �86
y � 1 � �16
�x � 8�
N�8, 1�.
�16
AC
y � �x � 14
y � 5 � �x � 9
y � 5 � �1�x � 9�
M�9, 5�.�1BC
y � 4x � 6
y � 6 � 4x � 12
y � 6 � 4�x � 3�
L�3, 6�.AB
� �1�5
�5Slope of LN � m3 �
6 � 13 � 8
� 4�41
Slope of MN � m2 �5 � 19 � 8
� �16
�1
�6Slope of LM � m1 �
6 � 53 � 9
��5, �3�.
y � �3
y � 15 � 18
y � �3��5� � 18
y � �3x � 18
x � �5
�14
3x �
703
�3x �53
x �703
�3x � 18 �53
x �163
y �53
x �163
y � �3x � 18
↔BC.
↔AC
98 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Geometry 99Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
48. —CONTINUED—
To find A, find the point of intersection of and
The coordinates of A are
To find B, find the point of intersection of and
The coordinates of B are
To find C, find the point of intersection of and
The coordinates of C are �14, 0�.
y � 0
y � �14 � 14
y � �x � 14
x � 14
56
x �353
�16
x � �x �353
�16
x �73
� �x � 14
y � �x � 14
y � �16
x �73
↔BC.
↔AC
�4, 10�.
y � 10
y � 16 � 6
y � 4 � 4 � 6
y � 4x � 6
x � 4
5x � 20
4x � �x � 20
4x � 6 � �x � 14
y � �x � 14
y � 4x � 6
↔BC.
↔AB
�2, 2�.
y � 2
y � 8 � 6
y � 4 � 2 � 6
y � 4x � 6
x � 2
256
x �253
4x � �16
x �253
4x � 6 � �16
x �73
y � �16
x �73
y � 4x � 6
↔AC.
↔AB
49.
undefined
is vertical.
The line containing has slope and passes through
The line containing is a vertical line passing through
The line containing has slope and passes through
To find A, find the point of intersection of and
The coordinates of A are
—CONTINUED—
�6, �4�.
y � �4
y � �23
� 6
y � �23
x
x � 6
�43
x � �8
�23
x �23
x � 8
y �23
x � 8
y � �23
x
↔AC.
↔AB
y �23
x � 8
y � 6 �23
x � 2
y � ��6� �23
�x � 3�
N�3, �6�.
23
AC
x � 0
M�0, �4�.BC
y � �23
x
y � 2 � �23
x � 2
y � ��2� � �23
�x � 3�
L�3, �2�.
�23
AB
LN
�40
�slope of LN � m3 ��2 � ��6�
3 � 3
� �23
�2
�3slope of MN � m2 �
�4 � ��6�0 � 3
�23
slope of LM � m1 ��2 � ��4�
3 � 0
Chapter 5 continued
49. —CONTINUED—
To find B, find the point of intersection of and
The coordinates of B are
To find C, find the point of intersection of and
The coordinates of C are
Lesson 5.6
5.6 Guided Practice (p. 305)
1. An indirect proof might also be called a proof by contra-diction because in an indirect proof, you prove that astatement is true by first assuming that its opposite istrue. If this assumption leads to a contradiction, then youhave proved that the original statement is true.
2. To use an indirect proof to show that two lines m and nare parallel, you would first make the assumption thatlines m and n are not parallel.
3. 4. 5.
6. In if you wanted to prove that you
would use the two cases and in anindirect proof.
5.6 Practice and Applications (pp. 305–307)
7. 8. 9.
10. 11. 12.
13. 14. 15.
16. The correct answer is C because and so by the Converse of the Hinge Theorem
17. The correct answer is B because and so by the Hinge Theorem AC > BD.m�3 < m�5
AD � AD,AB � DC,
m�4 < m�5.AC > AB
AD � AD,BD � CD,
m�1 > m�2UT > SVAB > CB
m�1 < m�2m�1 � m�2XY > ZY
m�1 > m�2m�1 � m�2RS < TU
BC � ACBC < �C
BC > AC,�ABC,
DC < FEKL < NQm�1 > m > 2
�0, �8�.
y � �8
y � 0 � 8
y �23
� 0 � 8
y �23
x � 8
x � 0
y �23
x � 8
↔BC.
↔AC
�0, 0�.
y � 0
y � �23
� 0
y � �23
x
x � 0
y � �23
x
↔BC.
↔AB
18. because
19. because
20. because
21. Given that and assume that
22. Given with Q the midpoint of assume isnot a median.
23. Given with assume
24. C Assume that there are two points, P and Q, where mand n intersect.
B Then there are two lines (m and n) through points Pand Q.
A But this contradicts Postulate 5, which states that thereis exactly one line through any two points.
D It is false that m and n can intersect in two points, sothey must intersect in exactly one point.
25. Case l: Assume If one side of a triangle islonger than another side, then the angle oppositethe longer side is larger than the angle oppositethe shorter side, so But this con-tradicts the given information that
Case 2: By the Converse of the Base Angles Theorem,But this contradicts the given
information that Since bothcases produce a contradiction, the assumptionthat EF is not greater than DF must be incorrectand
26. Assume Then m and n intersect in a point and a tri-angle is formed. by theTriangle Sum Theorem. Then
by the subtraction property of equality. Butbecause and are supple-
mentary. So by substitution proper-ty of equality. Then by simplifying both sides.But this is not possible; angle measures in a triangle can-not be zero.
So the assumption that must be false. Therefore,m � n.
m � n
m�3 � 0180� � 180� � m�3
�2�1m�1 � m�2 � 180�� m�3
m�1 � m�2 � 180�m�1 � m�2 � m�3 � 180�
m � n
EF > DF.
m�D > m�E.m�E � m�D.
m�D > m�E.
m�D�m�E.
EF < DF.
m�C � 90�.m�A � m�B � 90�,ABC
MQNP,�MNP
RS � 7 in.ST � 5 in.,RS � ST � 12 in.
x < 17.5
4x < 70
2 < 4.�4x � 5�� < 65�
x > 1
2x > 2
3x > x � 2
115� > 45�.3x � 1 > x � 3
70� > 60�.x > 9
100 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Geometry 101Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
27. Case 1: Assume that Then bythe Converse of the Hinge Theorem. But
by the ASA CongruencePostulate, so or This isa contradiction. So
Case 2: Assume Then by theConverse of the Hinge Theorem. But
by the ASA CongruencePostulate, so or Thisis a contradiction. So
Therefore and is an isosceles triangle.
28. The path are described by two triangles in which twosides of one triangle are congruent to two sides of anoth-er triangle, but the included angle in your friend’s triangleis larger than the included angle of your triangle, so theside representing the distance from the airport is longer inyour friend’s triangle.
29. The paths are described by the two triangles in which twosides of one triangle are congruent to two sides of anoth-er triangle, but the included angle in your friend’s triangleis larger than the included angle in your triangle, so theside representing the distance to the airport is longer inyour friend’s triangle.
30. a. As increases, increases because it is theangle opposite
As increases, decreases because increases and and are supplementary.
b. As increases, decreases because as increases increases and decreasesmaking decrease.
c. The cleaning arm illustrates the Hinge Theorembecause the lengths of and remain constantwhile and change. So there are two trian-gles, both being where two sides of the first tri-angle are congruent to two sides of another triangleand the included angle of one triangle is larger thanthe included angle of the other triangle, so the sideopposite the larger angle is longer than the side oppo-site the smaller angle.
EBDEDm�EBD
BDBE
ADm�AEDm�EBD
EDADED
�EBD�DBAm�EBDm�DBAED
ED.m�EBDED
RSTRS � RT
RS < RT.m�S � m�T.�S � �T
RUS � RUT
m�T < m�SRS < RT.
RS > RT.m�S � m�T.�S � LT
RUS � RUT
m�T > m�SRS > RT. 31.
Statements Reasons
1. 1. Given
2. Let P be a point in the 2. Protractor Postulateinterior of suchthat
3. Let P be the point such 3. Ruler Postulate that
4. 4. Definition of congruentangles
5. 5. Definition of congruentsegments
6. 6. SAS Congruence Postulate
7. 7. Corresponding parts of Congruent Triangles arecongruent
8. 8. Definition of congruentsegments
9. 9. Angle Addition Postulate
10. Let be the angle 10. Protractor Postulatebisector of such that H is on
11. 11. Definition of angle bisec-tor
12. 12. Transitive Property of congruence
13. 13. Reflexive Property of congruence
14. 14. SAS Congruence Postulate
15. 15. Corresponding Parts ofCongruent triangles areCongruent.
16. 16. Definition of congruentsegments
17. 17. Segment AdditionPostulate
18. 18. Substitution of equality
19. 19. Triangular Inequality
20. 20. Substitution property ofequality
21. 21. Definition of congruentsegments
22. 22. Substitution property ofequality
AC > DF
PC � DF
AC > PC
PH � HC > PC
AC � PH � HC
AC � AH � HC
AH � PH
AH � PH
ABH � PBH
BH � BH
PB � AB
�PBH � �ABH
AC.�PBA
→BH
m�ABCm�PBC � m�PBA �
m�PBC � m�DEF
BP � ED, PC � DF�PBC � �DEF
PBC � DEF
CP � FD
�BCP � �EFD
CP � FD.
m�BCP � m�EFD�ABC
m�ABC > m�DEFAB � DE, BC � EF,
Chapter 5 continued
5.6 Mixed Review (p. 308)
32. isosceles 33. equilateral, equiangular, and isosceles
34. scalene 35. isosceles
36. equiangular, equilateral, and isosceles
37. isosceles
38.
39.
41.
42. is the median, altitude, anglebisector, and perpendicular bisector.
5.6 Quiz 2 (p. 308)
1. 2. If then
3. If the perimeter of then the perimeter of
4.
5.
6.
MP, NP, MN
m�M � 57�
m�M � 123� � 180�
m�M � 48� � 75� � 180�
m�M � m�N � m�P � 180�
MQ, MP, PQ
m�M � 81�
m�M � 99� � 180�
m�M � 49� � 50� � 180�
m�M � m�P � m�Q � 180�
LQ, LM, MQ
m�M � 31�
m�M � 149� � 180�
75� � m�M � 74� � 180�
m�L � m�M � m�Q � 180�
GHF � 21CDE � 42,
CE � 16.FG � 8,FG � CE
RU
R S
T
U
m�BAC � 84�
m�BAC � 96� � 180
m�BAC � 51� � 45� � 180�
m�BAC � m�B � m�C � 180�
m�B � 51�
m�B � 32 � 19
m�B � �x � 19��
32 � x
2x � 32 � 3x
�x � 13�� � �x � 19�� � 3x�
7. is longer than because two sides of arecongruent to two sides of and the included angle
is larger than included angle so
8. The 2nd Group is farther from the camp because thegroups’ paths form two triangles with 2 pairs of congruent sides and the included angle for the 2nd groupis larger than the included angle for the 1st group.
Review (pp. 310–312)
1. If a point is on the perpendicular bisector of a segment,then it is equidistant from the endpoints of the segment.
2. If then U must be on the perpendicular
bisector of
3. If Q is equidistant from and then Q is on theangle bisector of
4. Let x abe the midpoint of Then oror
But because K is equidistant from R, S,and T.
5.
Since W is equidistant from the sides of
6. The special segments are angle bisectors and the point ofconcurrency is the incenter
7. The special segments are perpendicular bisectors and thepoint of concurrency is the circumcenter.
8. The special segments are medians and the point of con-currency is the centroid.
9. The special segments are the altitudes and the point ofconcurrency is the orthocenter.
WB � WA � 6.XYZ,
WA � 6
�WA�2 � 36
�WA�2 � 64 � 100
�WA�2 � 82 � 102
�WA�2 � �AY�2 � �WY�2
KR � KT � 20
20 � KT
400 � KT 2
144 � 256 � KT 2
122 � 162 � KT 2
�KX�2 � �XT�2 � �KT�2
XT � 16.XT �12 � 32
XT �12 STST.
�RST.
→ST,
→SR
RT.↔SQ
UR � UT,
DE > AB.�BCA�DFEDEF
ABCABDE
102 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
40.
m�C � 45�
m�C � 32 � 13
m�C � �x � 13��
Geometry 103Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
10. midpoint of
midpoint of
midpoint of
The equation is
The equation is
The centroid is the point of intersection of the two equations.
The coordinates of the centroid of are ��43
, 2�.�XYZ
y � 2
y � �32 ��
43�
y � �32
x
�43
� x
3 ��94
x
34
x � 3 ��32
x
y � �32
x
y �34
x � 3
y � �32
x
y � 0 � �32
�x � 0�
m2 �3 � 0
�2 � 0�
3�2
y �34
x � 3
y �34
�x � 4�
y � 0 �34
�x � ��4��
m1 �3 � 0
0 � ��4� �34
� ��2, 3�
� ��42
, 62�
YZ � ��4 � 02
, 0 � 6
2 �
� �0, 3�� �02
, 62�XZ � �0 � 0
2,
6 � 02 �
� ��2, 0�
� ��42
, 02�
XY � ��4 � 02
, 0 � 0
2 � 11. The coordinates of the orthocenter of are since is a right triangle and the two legs are alsoaltitudes of
12. The slope of
The slope of
The slope of
The equation of the line containing has slope andpasses through is
The equation of the line containing has slope 1 passes through is
The equation of the line containing has slope 0 andpasses through is
H is the point of intersection of and
The coordinates of H are
J is the point of intersection of and
The coordinates of J are
—CONTINUED—
�10, 1�.
10 � x
�10 � �x
1 � �x � 11
y � 1
y � �x � 11
↔JK.
↔HJ
�6, 5�
y � 5
y � �6 � 11
y � �x � 11
x � 6
�2x � �12
�x � x � 12
�x � 11 � x � 1
y � x � 1
y � �x � 11
↔HK.
↔HJ
y � 1.
y � 1 � 0
y � 1 � 0�x � 6�
N�6, 1�JK
y � x � 1.
y � 3 � x � 4
y � 3 � 1�x � 4�
��4, 3�HK
y � �x � 11.
y � 3 � �x � 8
y � 3 � �1�x � 8�
M�8, 3��1HJ
LN � m3 �3 � 14 � 6
�2
�2� �1
MN � m2 �3 � 18 � 6
�22
� 1
LM � m1 �3 � 38 � 4
�04
� 0
�XYZ.�XYZ
�0, 0��XYZ
Geometry 103Chapter 5 Worked-out Solution Key
Chapter 5 continued
12. —CONTINUED—
K is the point of intersection of and
The coordinates of K are
13. Let L be the midpoint of M be the midpoint of and N be the midpoint of
The slope of the slope of so
The slope of the slope of so
The slope of the slope of so
14.
16. The angle measurements in order from least to greatestare and The side measurements in orderfrom least to greatest are AB, BC, and AC.
17. The angle measurements in order from least to greatestare and The side measurements in orderfrom least to greatest are EF, DF, and DE.
18.
The angle measurements in order from least to greatestare and The side measurements in orderfrom least to greatest are GJ, GH, and HJ.
19.
The angle measurements in order from least to greatestare and The side measurements in orderfrom least to greatest are KM, LM, and KL.
�M.�L, �K,
m�L � 35�
m�L � 55� � 90�
m�L � m�K � 90�
�G.�H, �J,
m�J � 60�
m�J � 120� � 180�
m�J � 50� � 70� � 180�
m�J � m�H � m�G � 180�
�F.�D, �E,
�B.�C, �A,
P � 64
P � 18 � 22 � 24
P � BC � CD � BD
24 � BD
12 �12 BD
GF �12 BD
18 � BC
9 � 9 � BC
BG � GC � BC
BG � GC � 9
LN � JK.JK,LN � �1 �
MN � HJ.HJ,MN � 1 �
LM � HK.HK,LM � 0 �
HK.JK,HJ,
�2, 1�.
2 � x
1 � x � 1
y � 1
y � x � 1
↔JK.
↔HK
20. The length of the third side must be less than the sum ofthe lengths of the other two sides. So the length of thethird side must be less than 300 feet So themaximum length of fencing needed is 600 feet
of fencing.
21. 22. 23.
24. In a if then is not isosceles.
25. Assume has two right angles at and Thenand, since
This contradicts theTriangle Sum Theorem. Then the assumption that there issuch a must be incorrect and no triangle has tworight angles.
Chapter 5 Test (p. 313)
1. If P is the circumcenter of then PR, PS, and PTare always equal.
2. If bisects then and are sometimescongruent.
3. The incenter of a triangle never lies outside the triangle.
4. The length of a median of a triangle is sometimes equalto the length of a midsegment.
5. If is the altitude to side of then issometimes shorter than
6. a.
b.
c.
d.
BC � 19.8
BC � 9.9 � 9.9
BC � CF � FB
HE � 5
23 HE �103
HE �13 HE �
103
HE �13 �HE � 10�
HE �13 �HE � HB�
HE �13 EB
10 � HB
100 � �HB�2
36 � 64 � �HB�2
62 � 82 � �HB�2
�HG�2 � �GB�2 � �HB�2
HC � 16
13 HC �163
HC �23 HC �
163
HC �23 �HC � 8�
HC �23 �HC � HG�
HC �23 CG
AB.AM�ABC,BCAM
CDAD�ABC,→BD
�RST,
�ABC
m�A � m�B � m�C > 180�.m�C > 0�,m�A � m�B � 180�
�B.�A�ABC
�MPQ�M � �Q,�MPQ,
TU � VSm�1 < m�2AB < CB
�100 � 200 � 300�
�100 � 200�.
104 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
15.
P � 31
P � 9 � 12 � 10
P � ST � TU � SU
TU � 12
TU �12 � 24
TU �12 PQ
ST � 9
ST �12 � 18
ST �12 RQ
18 � RQ
9 � 9 � RQ
RU � UQ � RQ
RU � UQ � 9
Geometry 105Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
7. Point H is the centroid of the triangle.
8. is a(n) median, perpendicular bisector, altitude, andangle bisector of
9. and by the Midsegment Theorem.
10. because the side opposite islonger than the side opposite
11. To locate the pool so that its center is equidistant fromthe sidewalks, find the incenter of the triangle by con-structing angle bisectors of each angle of the triangle andlocating the point of intersection of the bisectors. Thispoint will be equidistant from each sidewalk.
12. The converse of the Hinge Theorem guarantees that theangles between the legs get larger as the legs are spreadapart.
13. The maximum distance between the end of two legs is 10feet because the length of the third side of the trianglemust be less than the sum of the lengths of the other twosides.
14.
If then is longer than because two sides of one triangle are congruent to twosides in another triangle and the measure of the includedangle of one triangle is larger than the measure of theincluded angle of the other triangle.
15.
Statements Reasons
1. 1. Given
2. 2. Definition of congruent segments
3. is an isosceles 3. Definition of isosceles triangle triangle
4. 4. Base Angles Theorem
5. 5. Angle Addition Postulate
6. 6. Substitution property of equality
7. 7. A part is smaller than the whole.
8. 8. Hinge Theorem
16. Assume Then because if twoangles of a triangle are congruent, then the sides oppositethem are congruent. So by the definition ofcongruent segments. But this contradicts the given state-ment that Therefore, the assumption must befalse. So m�D � m�ABC.
AD � AB.
AD � AB
AD � ABm�D � �ABC.
BE < AE
m�CAB < m�ABE
� m�ABEm�CAB � m�CBE
� m�ABE
m�ABC � m�CBE
�CAB � �ABC
�ABC
AC � BC
AC � BC
BCACm�AOC > m�BOC,
A BC
O
�ACB.�BACm�BAC > m�ACB
EF � ABEF �12 AB
�ABC.CG
Chapter 5 Standarized Test (pp. 314–315)
1. 2. D 3. B
B
4. The midpoint of is
The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side.
But
So
The coordinates of C are or
E
5. so
so
So
C
6. A
7. The length of the side has to be between or 12inches and inches. But it cannot be 12inches or 44 inches. A
8. A
9. because the sum of the measures of theacute angles of a right triangle is 90�.x� � y� � 90�
28 � 16 � 4428 � 16
P � 48
P � 12 � 16 � 20
P � NP � PL � NL
20 � NL
400 � �NL�2
144 � 256 � �NL�2
122 � 162 � �NL�2
�NP�2 � �PL�2 � �NL�2
NP � 12MK � 12,
PL � 16KP � 16,
��7, �3�.��7, �11 � 8�
CH � 8
CH �23 � 12
CH �23 MH
MH � 1 � ��11� � 12.
CH �23 MH
� ��7, 1�.
� ��142
, 22�
M � ��12 � ��2�2
, 1 � 1
2 �FG
y � 4
�12 y � �2
92 y � 5y � 2
92 y � 4 � 5y � 6
x � 9
103 x � 30
4x �23 x � 30
4x � 9 �23 x � 21
Chapter 5 continued
10. because the side opposite is longer than theside opposite
11. If then But so
12. The location of the point of intersection of the perpendic-ular bisectors is on because is a right triangle.
13. Let M be the midpoint of
Let N be the midpoint of
Let P be the midpoint of
The slope of
The equation of is
The slope of
The equation of is
The slope of
The equation of is
The centroid is the point of intersection of and
The coordinates of the centroid are �10, 2�.
y � 2
y �15
� 10
y �15
x
x � 10
�95
x � �18
15
x � 2x � 18
y � 2x � 18
y �15
x
↔CM.
↔AN,
↔PB,
y � �14
x �92
y � 0 � �14
�x � 18�↔CM
� �14
�3
�12↔CM � m3 �
3 � 06 � 18
y � 2x � 18
y � 0 � 2�x � 9�↔BP
↔BP � m2 �
6 � 012 � 9
�63
� 2
y �15
x
y � 0 �15
�x � 0�↔AN
�15
�3
15↔AN � m1 �
3 � 015 � 0
� �9, 0�� �182
, 02�P � �0 � 18
2,
0 � 02 �
AC.
� �15, 3�� �302
, 62�N � �12 � 18
2,
6 � 02 �
BC.
� �6, 3�� �122
, 62�M � �0 � 12
2,
0 � 62 �
AB.
�GHJGH
x� > 45�.x� > y�x � 45�.x � y,
�H.�Gx� > y�
14. The slope of
The slope of the line perpendicular is undefined. So the
line perpendicular to that passes through B is the line
The slope of
The slope of the line perpendicular to has slope
because
The equation of the line perpendicular to and passingthrough is
The slope of
The slope of the line perpendicular to has slope 1because
The equation of the line parallel to and passingthrough is
The orthocenter is the point of intersection of
and
The coordinates of the orthocenter are
15. a. The coordinates of the centroid are The coor-dinates of the orthocenter are Find the equa-tion of the line passing through the centroid and the orthocenter then show that the cir-cumcenter is also on the line.
is the equation of the line
Substitute the coordinates of the circumcenter into theequation of the line passing through the centroid andthe orthocenter.
—CONTINUED—
y � 5x � 48
y � 2 � 5x � 50
y � 2 � 5�x � 10�
slope � m �12 � 2
12 � 10�
102
� 5
�9, �3��12, 12�,
�10, 2��12, 12�.
�10, 2�.
�12, 12�.
y � 12
y � �24 � 36
y � �2 � 12 � 36
y � �2x � 36
x � 12
↔CP.
↔BM,
↔AN,
y � x.
y � 0 � 1�x � 0�A�0, 0�
↔BC
1 � ��1� � �1.
↔BC
�6
�6� �1.
↔BC � m3 �
6 � 012 � 18
y � �2x � 36.
y � 0 � �2�x � 18�C�18, 0�
↔AB
12
� ��2� � 1.
�2↔AB
�6
12�
12
↔AB � m2 �
6 � 012 � 0
x � 12.
↔AC
↔AC � m1 �
0 � 018 � 0
�0
18� 0
106 GeometryChapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Geometry 107Chapter 5 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 5 continued
15. —CONTINUED—
Since it is equal, the circumference is on the same lineas the centroid and the orthocenter. Therefore, theyare all collinear.
b. The distance from the circumference C to the centroidD is CD.
The distance from the circumcenter C to the orthocenter P is CP.
So the distance from the circumcenter to the centroidis one third the distance from the circumcenter to theorthocenter.
Investigation
1. The lines are medians because they are the lines that con-tain the line segments whose endpoints are a vertex of thetriangle and the midpoint of the opposite side.
2. The balancing point of the triangle is the centroidbecause it is the point of intersection of the medians.
3. Answers will vary.
4. Conjecture: The balancing point of squares, rectangles,parallelograms, and rhombuses is the point of intersectionof its diagonals.
5. Answer will vary.
Sample answer:
1 tested the conjecture by making more example shapesof each kind. The results were the same each time. Thebalance points were the points of intersection of the diagonals.
�26 � �26
�26 �13
�3�26�
CD �13
CP
� 3�26
� �9�26
� �234
� �9 � 225
� �32 � 152
CP � ��12 � 9�2 � �12 � ��3��2
� �26
� �1 � 25
� �12 � 52
CD � ��10 � 9�2 � �2 � ��3��2
�3 � �3
�3 � 45 � 48
�3 � 5 � 9 � 48
Present Your Results
Projects may vary.
Extension
The conjecture does not work for all four–sided shapes. Thefollowing is an example for which it was not true.