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CHAPTER 8
158 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Think & Discuss (p. 455)
1. Sample answer: pantographs, dilations, and projections
2. About 2 times larger. Measuring each fish from nose totail, the smaller fish is in., the larger in., so
Skill Review (p. 456)
1. 2.
3. 4.
5. 6.
Lesson 8.1
Activity (p. 457)
1. Sample answer:
2.
3. The ratio is about the same as the ratio.
4. Answers may vary.
5. No; if all the units of measure are the same, they do notaffect the ratio.
8.1 Guided Practice (p. 461)
1. means; extremes 2.
3. 4.
5. 6.
x � 6
x � 2�93�
x2
�93
2x
�39
8:104:5
x � 4
6x � 24
10x � 4x � 24
10x � 4�x � 6�
8:24:1
; 2:81:4
10
x � 6�
4x
2:1 36 in.:18 in.
3 �12 in.
1 :18 in.
wrist:thumbneck:wrist
5.75:2.523:10
; 12.5:5.75
50:23
�8 � 3�4 � 0
��11�4
�114
5 � 26 � ��1� �
37
2 � 04 � 0
�24
�12
22 � 18 � 20 � 31 � 91
13 � 11 � 7 � 312 � 10 � 2 � 6 � 32
2216 �
1216 �
116 � 1.83.
1716
1216
7. 8.
9.
8.1 Practice and Applications (pp. 461–464)
10. 11. 12.
13. 14. 15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25. 26.BDCF
�77
� 1ABCD
�23
20 oz4 lb
�20 oz
4 lb�16 oz1 lb �
�20 oz64 oz
�5
16
400 m0.5 km
�400 m
0.5 km�1000 m1 km �
�400 m500 m
�45
2 lb20 oz
�
2 lb�16 oz1 lb �
20 oz�
32 oz20 oz
�85
6 yd10 ft
�
6 yd� 3 ft1 yd�
10 ft�
18 ft10 ft
�95
2 mi3000 ft
�
2 mi�5280 ft1 mi �
3000 ft�
10,560 ft3000 ft
�8825
350 g1 kg
�350 g
1 kg�1000 g1 kg �
�350 g
1000 g�
720
60 cm1 m
�60 cm
1m�100 cm1 m �
�60 cm
100 cm�
35
3 ft12 in.
�
3 ft�12 in.1 ft �
12 in.�
36 in.12 in.
�31
12 in.2 ft
�12 in.
2ft �12 in.1 ft
�12 in.24 in.
�12
7.5 cm10 cm
�34
16 mm20 mm
�45
69
�23
2252
�1126
488
�61
1624
�23
x � 6
9x � 54
x
27�
29
b � �6 �2b � 12
2b � 4b � 12 z � 9.6 or 485
2b � 4�b � 3� 5z � 48
2
b � 3�
4b
58
�6z
Thumb 2.5 in.
Wrist 5.75 in.
Neck 12.5 in.
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 158
Geometry 159Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
27. 28.
29.
31.
32.
33.
35.
37.
39.
x � 12
5x � 4x � 12
5x � 4�x � 3�
5
x � 3�
4x
c �7030
�73
30c � 70
305
�14c
z �17510
�352
z � 7�2510�
z7
�2510
7z
�1025
x �207
7x � 20
x4
�57
15�5� � � 75�; 19�5�� � 95�2�5� � � 10�;
x � 5
36x � 180
2x � 15x � 19x � 180
7�15�� � 105�4�15�� � 60�;1�15�� � 15�;
x � 15
12x � 180
1x � 4x � 7x � 180
2�6� � 12 ft5�6� � 30 ft;
x � 6
14x � 84
2�7x� � 84
2�5x� � 2�2x� � 84
CFAB
�72
BFAD
�119
41.
43.
45.
47.
49. Venus: ; Mars:
Jupiter: ; Pluto:
50. 51.
x � 121 lb
x � 46�10038 �
x
46�
10038
120 ft�12 in.1 ft � � 1440 in.
46x
�38
100
x � 10 lb x � 330 lb
100x � 980 100x � 33,040
x
140�
7100
x
140�
236100
x � 53 lb x � 126 lb
100x � 5320 10x � 1260
x
140�
38100
; x
140�
910
z � 21
2z � 42
2z � 6 � 36
2�z � 3� � 36
z � 3
12�
32
12
z � 3�
23
x �483
� 16
x � 6�83�
x6
�83
6x
�38
y � �485
�5y � 48
30y � 48 � 35y
6�5y � 8� � 35y
5y � 8
7�
5y6
z � 15
7z � 6z � 15
7z � 3�2z � 5�
7
2z � 5�
3z
30.
Since length must be positive
34.
36.
38.
40.
y � 6
�4y � �24
4y � 8y � 24
4y � 8�y � 3�
4
y � 3�
8y
d � 32
3d � 96
3
16�
6d
163
�d6
b �1210
�65
b � 4� 310�
b4
�3
10
4b
�103
y �7210
�365
10y � 72
y8
�9
10
3�3� � 9 cm4�3� � 12 cm;
x � 3.
x � ±3
x2 � 9
12x2 � 108
4x�3x� � 108
42.
44.
46.
48. Venus because is closest to 1.910
y � 25
5y � 125
5y � 35 � 160
5�y � 7� � 160
y � 7
40�
45
z �688
�172
8z � 68
28z � 8 � 20z � 60
4�7z � 2� � 10�2z � 6�
4
2z � 6�
107z � 2
x �8018
�409
18x � 80
30x � 80 � 12x
10�3x � 8� � 6�2x�
3x � 8
6�
2x10
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 159
Chapter 8 continued
160 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
52.
53.
54.
56.
57.
58.
UT � �819 � 3�91 RQ � �91
�UT� 2 � 819 �RQ� 2 � 91
�UT� 2 � 30 2 � 9 2 �RQ� 2 � 10 2 � 32
ST � 30
ST � 10�31� PR � 3
ST10
�31
3�PR� � 9
10ST
�13
PR9
�13
PQ � 13 ST � 39
PQ � �169 ST � �1521
PQ � �52 � 122 ST � �362 � 152
SU � 15
SU � 5�31� RQ � 12
SU5
�31
3�RQ� � 36
5
SU�
13
RQ36
�13
k �143
or 423
3k � 14
7k � 14 � 10k
7�k � 2� � 5�2k�
k � 2
2k�
57
j � 18
4j � 72
j
24�
34
x � 1.0 in.
1440x � 1470
144035
�42x
35
1440�
x42
35 in.
1440 in.�
x
3.5 ft�12 in.1 ft �
x � 2.6 in.
3780 � 1440x
35
1440�
x108
35 in.
1440 in.�
x
9 ft�12 in.1 ft �
59.
61.
62.
63. EF � y � 20;
64.
65. Let x � number of pounds of compost.
D
66.
67.
and
AC � x � 16 � 20 � 16 � 36
x � 20
�10x � �200
216 � 9x � x � 16
9�24 � x� � x � 16
24 � xx � 16
�19
AC � x � 16.CD � 24 � xLet BC � x, then
AB � 16
3�AB� � 48
AB24
�23
2�15�� � 30�, 3�15�� � 45�, 7�15�� � 105� D
x � 15
12x � 180
2x � 3x � 7x � 180
x � 18 lb
5x � 90
53
�30 lb
x
b � 15
6b � 5b � 15
6b � 5�b � 3� GR � b � 3 � 15 � 3 � 18
b
b � 3�
56
GH � HR � b � 15;
y � 20
DF � y � 4 � 20 � 4 � 24 4y � 80
45
�16y
x � 8
AB � x � 2 � 8 � 2 � 10 3x � 24
AC � x � 8; 34
�6x
72 ft
6 in.� 1 ft12 in.�
�72 ft0.5 ft
�1441
12:1
72:6
6 ft�12 in.1 ft �: 6 in.
55.
m � 6
�m � �6
3m � 6 � 4m
3m � 6
m�
41
60.
x � 72 ft
0.5x � 36
x6
�6
0.5
x
6 ft�
6 ft
6 in.� 1 ft12 in.�
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 160
Geometry 161Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
8.1 Mixed Review (p. 464)
68.
69.
70. 71.
72.
73. and
74. and
75. and
76.
Lesson 8.2
8.2 Guided Practice (p. 468)
1. 2. 3. ; or 3
4. and
(cross product property)
The equations are equal, therefore the statement is true.
5. 6.
7.
8. use stripes as a height unit measure � 7 stripes13 stripes�.
713
;
x � 11.4 ft
1
1.9�
6x
x � 6
DE � 9 x � �36
2�DE� � 18 x2 � 36
23
�6
DE 3x
�x
12
15r � 6s 15r � 6s
15s
�6r
rs
�6
15
62
y � 5ax
�xb
y
x
1
�11�1
A�(3, �3)
B(6, �7)B�(�2, �7)
A(1, �3)
�112
, 1���12
, 312�
�1 � 22
, 4 � ��2�
2 ���2 � 12
, 3 � 4
2 ��3
12
, �112��1, �2�
�1 � 62
, 1 � ��4�
2 ��1 � 12
, 1 � ��5�
2 �
�1, 3���112
, 3���1 � 3
2,
5 � 12 ���2 � ��1�
2,
1 � 52 �
WV
m�U � m�V � 95�m�T � m�W � 65�
m�V � 95�
m�V � 85� � 180�
m�V � 20� � 65� � 180�
m�X � m�S � 20�
8.2 Practice and Applications (pp. 468–471)
9. 10. 11.
12.
13. By Property of proportions 4, the statement is true.
14. If then by the reciprocal property.
The statement is false.
15. By Property of proportions 3, the statement is true.
16. Similarily,
Then and the statement
is true by the subtraction property of equality.
17. 18.
19. 20.
21. 22.
23. 24.
25. 26.
x �352
� 1712
�2x � �35
7x � 35 � 9x
7�x � 5� � 9x x �406
�203
� 623
79
�x
x � 5 6x � 40
7
7 � 2�
xx � 5
4x
�6
10
x � 9
6x � 54 x �18016
�454
� 1114
186
�x3
16x � 180
24 � 66
�x3
20x
�169
x � 5�3 x � 4�10
x � �75 x � �160
x2 � 75 x2 � 160
5x
�x
15 8x
�x
20
x � 4�5 x � 14
x � �80 x � �196
x2 � 80 x2 � 196
2x
�x
40 7x
�x
28
x � 8 x � 9
x � �64 x � �81
x2 � 64 x2 � 81
4x
�x
16 3x
�x
27
m12
�3n
1 �m12
� 1 �3n
1 �3n
.
3 � nn
�12 � m
12�
1212
�m12
� 1 �m12
.
43
�rp
34
�pr ,
x � yy
y � 1212
634
�3
17xy
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 161
Chapter 8 continued
162 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
27.
29. Using in. for the 30. Using in. for the
width length
31. 32.
33. 34.
35. Each side of the equation represents the slope of the linethrough two of the points. If the points are collinear, theslopes are the same.
36. Given
Cross product property
Division property of equality
Simplify.
37. Given
Addition property of equality
Inverse property of multiplication
Addition of fractions a � b
b�
c � dd
ab
�bb
�cd
�dd
ab
� 1 �cd
� 1
ab
�cd
ac
�bd
adcd
�bccd
ad � bc
ab
�cd
y � 23 x � 11
92 � 4y 2x � 22
20 � 4y � 72 2x � 2 � 20
20 � 4�y � 18� 2�x � 1� � 20
104
�y � 18
2 25
�4
x � 1
18 � 86 � 2
�y � 188 � 6
1 � ��1�1 � ��4� �
5 � 1x � 1
x � 11,380 rubles x � 198 hits
22.761
�x
500
x643
�0.3081.000
x � 38 � 40 feet x � 26 � 25 feet
x �161 �38
16� x �161 �26
16�
116
x �3816
116
x �2616
x ft
3816
in.�
1 ft116
in.
x ft2616
in.�
1 ft116
in.
3816
2616
x �14421
�487
� 667
�21x � �144
144 � 9x � 12x
9�16 � x� � 12x
16 � x
x�
129
38. 39.
40.
feet
Sample answer: Construct a ramp consisting of tworamps in opposite directions, each 18 ft long. The first
should be 3 ft high at its beginning and ft high at
its end, for a rise:run ratio of . The second would be
ft high at its beginning and at ground level at its end.
The second ramp would also have a rise:run ratio of .
41. The map distance is 42. The map distance is
about or inch. about or inches.
43. If the two sizes share a dimension, the shorter dimensionof A5 paper must be the longer dimension of A6 paper.That is, the length of A6 paper must be 148 mm. Let x bethe width of A6 paper; 148 is the geometric mean of xand 210.
44. If of the fish are tetras, then are not tetras, so
D
45. Sample answer:
C
W � 3�3� � 9 games
L � 3
�2L � �6
3L � 6 � 5L
W � 6 � 5L
W � 6
L�
51
W � 3L
WL
�31
24 78
23
� 14.
78
18
x � 104 mm
210x � 21,904
x
148�
148210
x � 838
miles x � 212
miles
x �54
�6.7� x �38
�6.7�
x mi
114
in.�
6.7 mi1 in.
x mi38
in.�
6.7 mi1 in.
114
2016
38
616
112
112
112
112
x � 36
1
12�
3x
x �5
12�
54
� 114
ft
x � 24 ft 12x � 15
1
12�
2x
112
�x
1528.
x � 9
4x � 36
4x � 24 � 60
4�x � 6� � 60
x � 6
6�
104
x � 6
6�
4 � 64
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 162
Geometry 163Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
46. a.
b.
So 2 is the geometric meanof and .
c.
8.2 Mixed Review (p. 471)
47. 48.
49. 50.
51.
52.
53.
54.
55.
56.
57. A regular pentagon has 5 lines of symmetry (one fromeach vertex to the midpoint of the opposite side) androtational symmetries of and clockwise andcounterclockwise, about the center of the pentagon.
Developing Concepts Activity (p. 472)
Exploring the Concept
1. Photo 1: 4.2 cm; photo 2: 3.0 cm
2. 3. Both have measure .
4.30�
30�� 1
30�4.23.0
� 1.4
144�,72�
m�B � 180� � 120� � 60�
m�A � m�D � 90�
m�C � m�D � 180� � 41� � 139�
m�B � m�A � 41�
m�A � m�B � 180� � 67� � 113�
m�C � m�D � 67�
m�A � m�B � 180� � 80� � 100�
m�C � m�D � 80�
m�C � 180� � 128� � 52�
m�A � 180� � 71� � 109�
m�A � m�D � 180� � 115� � 65�
m�C � m�B � 115�
� 95 ft 2
��3.14��5.5� 2 � 26 cm2
Area � r 2 � 12�13��4�
r � 5.5 ft Area �12bh
� 9 cm2 � 12 m2
� 32 � 4 � 3
Area � s 2 Area � lw
� 1.618034
1 � �5
2�
1 � 2.2360682
�1 � �51 � �5 4 � 4
�1 � 5 � 4
1 � �5
2�
2�1 � �5
1 � �5
2�
2x
�1 � �5 � x
1 � �5 � 2 � x 5. Measurements will vary.
Drawing Conclusions:
1. cm 2. 35
3. Any two corresponding lengths or corresponding perime-ters have the same ratio; corresponding
4. The ratio of two corresponding areas is the square of theratio of corresponding lengths.
Lesson 8.3
8.3 Guided Practice (p. 475)
1. No; the figures are only if the scale factor is 1.
2. No; the lengths of corresponding sides are not propor-
tional;
3. Yes; corresponding can be shown to be by theInterior Angles of a Quad. Thm., and lengths of corre-sponding sides are proportional
4.
5.
6.
7.
8.3 Practice and Applications (pp. 476–478)
8.
9.
JKWX
�KLXY
�LMYZ
�JMWZ
�M � �Z;�J � �W, �K � �X, �L ��Y,
DEPQ
�EFQR
�DFPR
�D � �P, �E ��Q, �F � �R;
m�TUV � m�ABC � 180� � 70� � 110�
x � 10
3x � 30
53
�x6
159
�53
ABTU
�BCUV
�CDVW
�ADTW
�D � �W;�C � �V,�B � �U,�A � �T,
105
�147
�126
�84
�21
��
1510
�83
.
�
� are �.
�5�1.4� � 7
Measurement Photo 1 Photo 2 Ratio
AB 4.2 cm 3.0 cm
AF 7.8 cm 5.4 cm
CD 4.0 cm 2.8 cm
30 30 1
106 106 1
Perimeter of 24.8 cm 17.2 cmPhoto
24.817.2 � 1.4
��m�2
��m�1
42.8 � 1.4
7.85.4 � 1.4
4.23 � 1.4
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 163
Chapter 8 continued
164 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
10.
11. Yes; both figures are rectangles, so all 4 and
12. No; lengths of corresponding sides are not proportional;
13. No; and so corresponding arenot
14. No; and so corresponding are
not and so lengths of corresponding sides are not
proportional.
15. Yes; Sample answers:
16. No;
17. Yes; and
18. No; and so cor-
responding are not and so corresponding
sides are not proportional.
19. 20.
21.
22.
23. 24.
25.
26.
27.
28. 29. No; 1814
�188
�3018
1020
�12
3 � 3 � 2 � 2 � 10
m�G � m�C � 180� � 30� � 150�
x � 2
2x � 4
x4
�12
63
�21
5872.5
�45
� 58 PQRS: 16 � 10 � 12 � 20
� 72.5 JKLM: 25 � 15 � 12.5 � 20
y � 20 x � 12.5 w � 20
4y � 80 4x � 50 5w � 100
y
16�
54
x
10�
54
w25
�45
1512
�54
1215
�45
3014
�2010
,��
m�L � 180� � 118� � 62�,m�R � 65�
�XYZ~�CAB128
�96
�6.754.5
�32
;
�B � �Z�C � �X,�A � �Y,
63
�44
ABCD~FEHG
ABCD~EFGH,
65
�33
,�
�m�Q � 88�,m�K � 90�
�.�m�Q � 88�,m�B � 90�
75
�3.55
.
ABFG
�BCGH
�CDHE
�ADFE
�74
.
� are �
QRAB
�RSBC
�STCD
�TUDE
�QUAE
�T � �D, �U � �E;�Q � �A, �R � �B, �S � �C, 30. Yes; because A and B are parallelograms, the markedangles are to their opposite angles and the angles consecutive to the marked angles are supplements of themarked angles. So corresponding are . Also, becauseall 4 sides of A and all 4 sides of B are , the lengths ofcorresponding sides are proportional.
31. sometimes 32. sometimes 33. sometimes
34. sometimes 35. always 36. never
37. always 38. sometimes
39.
40.
41.
42.
43. Analog: Digital:
9x � 9�1.47� � 13.2 in.3x � 3�5.4� � 16.2 in.
16x � 16�1.47� � 23.5 in.4x � 4�5.4� � 21.6 in.
x � 1.47 x � 5.4
x2 � 2.1632 x2 � 29.16
337x2 � 729 25x2 � 729
256x2 � 81x2 � 729 16x2 � 9x2 � 729
�16x�2 � �9x�2 � 27 2 �4x�2 � �3x�2 � 27 2
y � 166 �152
� 712
y � 73 � 93 x �304
� 93� 4x � 30
�y � 73�� � 360� � 116� � 61� � 90� x5
�64
x �82821
�2767
� 3937
21x � 828
y �48621
�1627
� 2317
21x � 126 � 702
21y � 486 21�x � 6� � 702
y
27�
1821
x � 6
39�
1821
x �19218
�323
� 1023
18x � 192
18x � 72 � 264
18�x � 4� � 264
Because y° is the supplementof 60°; y � 180 � 60 � 120.
x � 422
�1218
y � 9 x � 11
10y � 90 10x � 110
10y � 30 � 120 10x � 20 � 90
10�y � 3� � 120 10�x � 2� � 90
y � 3
8�
1510
x � 2
6�
1510
64
�32
���
�
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 164
Geometry 165Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
44. No; the ratio of length to width for a standard screen
and for a digital screen. Then and
and
45. with a scale factor of The
perimeter of is The
ratio So,
46.
47.
48.
49. a. Figure 1:
b.
y �107
x
0 � b
6 � 6 � b
6 � �107 ��4.2� � b
y
x
1
1
y � mx � b
EA � 2.8
10�EA� � 28
EA4
�7
10
DE � 1.4 CD � 3.5
10�DE� � 14 10�CD� � 35
DE2
�710
CD5
�710
BC � 2.1 AB � 4.2
10�BC� � 21 10�AB� � 42
BC3
�7
10 AB6
�7
10
y � 6.4 cm x � 3 cm
5y � 32 5x � 15
51
�32y
51
�15x
9418.8
�51
so
x � 11.2 in.
5x � 56
25
�x
28
4:15
2:7.5
perimeter of ABCDperimeter of EFGH
�ABEF
.
perimeter of ABCDperimeter of EFGH
�2l � 2w
k�2l � 2w� �1k.
2�kl� � 2�kw� � k�2l � 2w�.EFGH
ABEF
�lkl
�1k.ABCD~EFGH
416
�39
.widthwidth
�39
,
lengthlength
�4
16169
�43
c. it is the reciprocal of the scale factor.
50. Sample sketch:
51. Since if you are standing at a point onEarth at the vertex of the triangles (E), it would appearthat the moon is completely blocking the sun’s light.
52.
8.3 Mixed Review (p. 479)
53. 54.
55. 56.
57. 58.
59.
60.
61.
62.
64.
z � 120
5z � 600
5
24�
25z
x � 2
27x � 54
x9
�6
27
x � 9
9x � 81
9x � 99 � 180
3x� � 105� � �6x � 6�� � 180�
x � 7
14x � 98
14x � 82 � 180
83� � 5x� � �9x � 1�� � 180�
x � 49
x � 131 � 180
x� � 90� � 41� � 180�
�5 � 65 � ��1� � �
116
�2 � 52 � ��4� � �
76
10 � ��3�1 � ��2� �
133
5 � 42 � 9
� �17
�3 � ��7��6 � 0
�4
�6� �
23
8 � 43 � ��1� �
44
� 1
CD � 1120 miles
93,000,000�CD� � 103,800,000,000
93,000,000
240,000�
432,500CD
AECE
�ABCD
CDE~�ABE,�
moon
E D
B
AC
Earthsun
432,500 mi
240,000 mi
93,000,000 mi
m �107
;
63.
65.
b �3213
13b � 32
4
13�
b8
y � 38
2y � 76
4y
�2
19
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 165
Chapter 8 continued
166 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
66. 67.
Quiz 1 (p. 479)
1. 2.
3. 4.
5. 6.
7. 8.
Scale factor � 45:30 � 3:2
Ratio of perimeters �
Scale factor � 3:6 � 1:2
Ratio of perimeters �
9.
None are exactly similiar, but the and wallet sizes
are nearly similar .
10.
x � 318
in.
8x � 25
8
10�
2.5x
� 52.25
� 2.22, 7
3.25� 2.15�
5 7
85
�107
, 8
2.25�
103.25
, 5
2.25�
73.25
12
32
x � 3
18 � 6x
x � 30 24 � 6x � 6
x2 � 900 24 � 6�x � 1�
45x
�x
20 36
�x � 1
8
x � �70 � 8.37 x � �55 � 7.42
x2 � 70 x2 � 55
10x
�x7
5x
�x
11
x �9628
�247
�28x � �96
x � 21 4x � 32x � 96
x2 � 441 4x � 16�2x � 6�
7x
�x
63
42x � 6
�16x
d � 28 p � 10
5d � 140 3p � 30
57
�20d
p
15�
23
x � 21 x � 9
42 � 2x 2x � 18
18x � 42 � 20x 11x � 9x � 18
6�3x � 7� � 20x 11x � 9�x � 2�
3x � 7
5�
4x6
11
x � 2�
9x
Lesson 8.4
Activity (p. 480)
Sample answer:
The angles are both by the Triangle Sum Thm.;yes, the triangles are similar.
8.4 Guided Practice (p. 483)
1.
2. Sample answer: any two points;
or
3. No; corresponding sides of s are not unless thescale factor of the s is Corresponding
4. No; corresponding .
5. Yes; all are so corresponding are
6.
7.
8. by the Reflexive Prop. of Congruence,
so by the AA Similarity Postulate.
8.4 Practice and Applications (pp. 483–487)
9.
10.
11.
12. 13.
14. 15. 15; x15; y
LM; MN; NL�LMN
LMQP
�MNPN
�LNQN
�L and �Q, �M and �P, �LNM and �QNP;
VTST
�TWTU
�VWSU
�V and �S, �T and �T, �W and �U;
JKFG
�KLGH
�JL
FH
�J and �F, �K and �G, �L and �H;
�ABC~�BDC
�C � �C
MP � 10
40 � 4�MP�
84
�MP5
m�P � m�L � 53�
m�N � m�K � 90�;
m�J � m�M � 37�;
�.�60°�
� are not �
� are �.1:1.�
��~
5 � 10 � 4
�4
�4� �1
6 � 3�1 � 2
�3
�3� �1
ABXY
�64
�32
32 �
64 �
4.53 ;
80�
2
3
460�
40�
60�
40�
3 6
4.5
PN � 6
24 � 4�PN�
84
�PN3
mcrbg-0801-sk.qxd 6-15-2001 3:31 PM Page 166
Geometry 167Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
16. 17.
18. No; and so corresponding are not
19. Yes; and so the s are
by AA Similarity Postulate;
20. No; so the lengths of corresponding sides are
not proportional.
21. Yes;
and so the s are by AASimilarity Postulate;
22. No; m ∠E � 94°, m ∠B � 94°, m ∠A � 54°, so corresponding are not
23. Yes; by Vertical Thm.,so the s are by AA
Similarity Postulate;
24. Yes; and so the s are by AA Similarity Postulate;
25. Yes; and so the s
are by AA Similarity Postulate;
26. Yes; and by the ReflexiveProp. so the s are by AA Similarity Postulate;
27. Sample answer:
28. Sample answer:
29. Slope of
so
�10, 0�
x � 10
�1��x� � 10
�12
�5 � 00 � x
↔BC �
0 � 36 � 0
� �12
�1 � ��2�2 � ��1� �
13
0 � ��1�
5 � 2�
13
�3 � ��1�
7 � 2� �
25
1 � 3
�3 � ��8� � �25
�PQR ~ �PST~�
�P � �P�PST � �TQR
�VWX ~ �VYZ.~�m�X � m�VZYm�W � m�VYZ
�ABC ~ �EDC.~�
m�B � m�D � 40�m�A � m�E � 50�
�JMN~�JLK.~�m�N � m�K � 50�
��NJM � �LJK
�.�
�XYZ ~ �GFH~��Z � �H �X � �G
m�Z � 55°
m�Z � 48° � 77° � 180°
2026
�1620
,
�PQR ~ �WPV
~��W � �RPQ �Q � �VPW
�.�
m�D � 47�m�C � 31�
x � 24 y � 16
15x � 360 15y � 240
18x
�1520
y
12�
2015
30. Slope of
so
31. Slope of
so
32. Slope of
so
33.
35.
37.
39.
41.
43.
x � 100
35° � 45° � x° � 180°
y � 27
24y � 648
24y � 72 � 576
24�y � 3� � 576
y � 3
32�
1824
p � 14
8p � 112
168
�p7
x � 20
6x � 120
6
15�
8x
615
�8x
CDE
�4 12, 0�
x � 412
2x � 9
�2��x� � 9
�2 �9 � 00 � x
↔BC �
0 � 84 � 0
� �2
�30, 0�
x � 30
�1��x� � 30
�15
�6 � 00 � x
↔BC �
0 � 15 � 0
� �15
�514, 0�
x � 514
4x � 21
�4��x� � 21
�43
�7 � 00 � x
↔BC �
0 � 43 � 0
��43
34.
36.
38.
40.
42.
z � 36
180 � 5z
45
�z
45
r � 26
11
11r � 28
114
�7r
y � 4
15y � 60
156
�10y
156
�10y
CD; CE; DE
mcrbg-0804-sk.qxd 5-25-2001 11:17 AM Page 167
Chapter 8 continued
168 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
44.
45.
46. 47.
48. Since and areright Since all right By theReflexive Prop. of Cong. so by the AA Similarity Post.
49. Since are both right they are by the Reflexive Prop. of Cong, soby AA Similarity Postulate
50. True; all right so the s are by AA SimilarityPostulate
51. False; all of any 2 equilateral s are so the s areby AA Similarity Postulate
52. True; since the vertex of each isosceles has measurethe measure of each base is
The s are by AA Similarity Postulate
53.
54. Answers will vary, but the slopes are always equal.
55. and so and are right
angles and so by Corresp.
Thm. So by AA Similarity Post.;
therefore so and
56. Sample answer: and thebuilding shadow
x � 31.25 ft
4x � 125
x5
�254
� 25 ftrod shadow � 4 ft,Rod � 5 ft,
PQ � 480 feet.PQ4
�7806.5
PQSR
�QRRT
�PQR ~ �SRT�
�PRQ � �TPR � ST�.
�SRT�QSR � QTPQ � QT
d � 1.5 m
2.4d � 3.6
2.43.6
�1d
2.46 � 2.4
�1d
~�
12�180� � 40�� � 70�.�40�,
��
~��,��
~�� are �,
�ABE ~ �CDE�CED � �AEB
�.�,�ECD and �EAB
�JKL ~ �JMK�J � �J,�JMK � �JKL.� are �,�.
�JMK and �JKLJK � KL,KM � JL
z � 25 y � 80
450 � 18z 12y � 960
378 � 18z � 72 48y � 960 � 36y
378 � 18�z � 4� 48�y � 20� � 36y
14
z � 4�
1827
48y
�36
y � 20
x � 12
12x � 144
1218
�8x
s � 1312
4s � 54
49
�6s
57. a. and are right and andare by the Vertical Angles Thm., so
by AA Similarity Postulate
b.
d. Sample Answer: If point is added to the diagram so that and
then BCDE is a rectangle,and
58. Sample answer:
59. 60.
8.4 Mixed Review (p. 487)
61.
62. 63.
64. is the midpoint of so
65.
66.
68.
70.
x � 69
1656 � 24x
2324
�x
72
x � 51
561 � 11x
17x
�1133
x �92
8x � 36
x
12�
38
MN �12�JL� �
12�24� � 12
MP �12�KL� �
12�32� � 16
KL � 32,KLN
KJ � 2�NP� � 2�23� � 46KJ
�1089 � 961 � �2050 � 5�82
���21 � 12�2 � �14 � ��17��2
h � 0.4 m
d � 25 m 10 � 25h
d
25 mm�
1 m1 mm
10 m
25 mm�
h1 mm
d25
�hr
AD � ��AE� 2 � �ED� 2 � �42 � 32 � �25 � 5.AE � AB � BE � 4,BE � CD � 3,
BE � CD,ED � BCm�E � 90�,
E
x �34
mi
4x � 3
3x � 3 � x
13
�x
3 � x
�ABX ~�DCX��DXC
�AXB��DCX�ABX
c.
AD � 5 mi
AX � DX � AD
DX � 3.75 mi
14.0625 � �DX� 2
32 � �2.25�2 � �DX� 2
AX � 1.25 mi
1.5625 � �AX� 2
12 � �0.75�2 � �AX� 2
67.
69.
71.
x � ±16
x2 � 256
8x
�x
32
x �3611
36 � 11x
102 � 11x � 66
102 � 11�x � 6�
3411
�x � 6
3
y � 8
12y � 96
3y
�1232
mcrbg-0804-sk.qxd 5-25-2001 11:17 AM Page 168
Geometry 169Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
18. The ratios of the lengths of the given sides are the sameand the included are so the s are by SAS
Similarity Thm.
19. by Alt. Int. Thm. 20. by Alt. Int. Thm.
21. by Triangle Sum Thm.
22. 23.
24. 25.
26.
27. and included are So by
SAS Similarity Thm.
28. and so by
SAS Similarity Thm.
29.
30. Since and and
Then and, since
by SAS Similarity Thm.
B
Y
C ZA X
�ABC ~ �XYZ�B � �Y,BAYX
�BCYZ
YX � YZ.BA � BCYX � YZ,BA � BC
h � 140 ft
560 � 4h
5.6h
�4
100
r � 10
6r � 60
6
12�
5r
�EFG ~ �EJH�E � �E6
12�
48
p � 18
10p � 180
1015
�12p
�ABC ~ �BDC�.�1015
�8
12
�ABC and �EDC�BGC and �DFC,�AGC and �EFC,
x � 4�2 x � 21
48�2 � 12x 4x � 84
412
�x
12�2 4
12�
7x
x � 15 x � 12
3x � 45 3x � 36
39
�5x
39
�4x
82�
�45��53�
~����6:1�
Chapter 8 continued
Lesson 8.5
8.5 Guided Practice (p. 492)
1.
2. AA Similarity Post.;
3. SAS Sim. Thm.;
4. are both
and
the s are by the SSS Similarity Thm.
5. all ratios are equal to so the s are
by SSS Similarity Thm.
6. and are � by the SSS Sim. Thm.
so the scale factor is
7. and are � by the SSS Sim. Thm.
so the scale factor is
8. and are by the SSS Sim. Thm.
so the scale factor is
9. Yes; so (or )
by SSS Similarity Thm.
10. Yes; SAS Similarity Thm.
11. No;
12. Yes; SAS Similarity Thm.
13. Yes; SSS or SAS
Similarity Thm.
14. Yes; by reflexive prop, and by Corresp. Post.; AA SimilarityPost.
15. SSS Similarity Thm.
16. If and then byTriangle Sum. Thm. Also, and so So s are by the AA Similarity Post.
17. The ratios of the lengths of corresponding sides are the
same, and the included are so the
s are by SAS Similarity Thm.~�
���186
�248
�31�,
~�m�E � 62�.m�F � 90�m�D � 28�
m�C � 90�m�B � 62�m�A � 28�
84
�168
�189
�21
;
�WXP ~ �WYZ;�
�XPW � �YZW�W � �W
�PQR ~ �DEF;1815
�3025
�2420
�65
;
68
�8
1023
�34
; �ABC ~ �LKJ;
3224
�2015
�1814
�PMS ~ �RMN;
�XZY�JKL ~ �XYZ3510
�3510
�288
�72
,
34
.1824
�2736
�3344
�34
,
~�TSU�XYZ
25
.820
�1025
�4
10�
25
,
�GHJ�DEF
32
.64
�4.53
�7.55
�32
,
�QRS�MNP
~
�16
212
�5
30�
636
�16
;
~�
ABMN
�BCNP
�ACMP
� 2;ABJK
�BCKL
�ACJL
�85
~ to �ABC,�JKL and �MNP
�ABC ~ �DFE
�ABC ~ �DEF
FHRX
�HGXS
�FGRS
mcrbg-0804-sk.qxd 5-25-2001 11:17 AM Page 169
170 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
31. Locate on so that and draw throughCorresp. and are as are corresp.
and so Then
But and so and
GH � DF. By the SAS Cong. Post.,Corresp. and are so by theTransitive Prop. of Cong. by AASimilarity Postulate.
32.
33.
34.
35. Julia and the flagpole are both perpendicular to theground and the two s formed have a shared angle. Then the s are by the AA Similarity Postulate
36.
so perimeter of
The length of is
B
37.
A
XZ � YZ � 70 � 50 � 120
YZ � 50
2�YZ� � 100
25
�20YZ
XY � BC � 105 � 20 � 125;
XY � 105
2�XY� � 210
42XY
�25
XY
�ABC � 28 � 20 � 42 � 80;
AC � 28;
5�AC� � 140
AC70
�25
~�
�
x � 25 ft
7x � 175
7
35�
5x
x � 18 ft
80x � 1440
x
48�
3080
x � 80 m
2000 � 25x
2500 � 25x � 500
2500 � 25�x � 20�
20
x � 20�
25125
�ABC ~ �DEF�F � �C�,BHG� F
�BGH � �EDF.
ACGH
�ACDF
GB � DE,ABDE
�ACDF
ABGB
�ACGH
.�ABC ~ �GBH.BHG,� C
�,BGH� AG � to AC.GHGB � DEABG 38. Assume that the posts and are perpendicular to
the ground, . Then and are all toso they are all to each other. Then ∠ADC, ∠BCD,
∠EFD, and ∠EFC are all right and so are all �. ∠DAC � ∠FEC and ∠DBC � ∠DEF by the Corr.
Post. So ADC EFC and BCD EFD by the
AA Similarity Post. Then and by
the definition of similar triangles. Then add equals to
equals and solve for EF as follows:
Addition Prop. of Equality
Combine fractions.
Simplify.
Simplify.
Combine fractions.
Simplify.
Multiply.
Divide and simplify.
The length of is ft.1717
EF
EF �1207
� 1717
7 � EF � 120
7 � EF120
� 1
3 � EF � 4 � EF120
� 1
EF40
�EF30
� 1
EF40
�EF30
�DCDC
EF40
�EF30
�FC � DF
DC
EF40
�EF30
�FCDC
�DFDC
EF30
�DFDC
EF40
�FCDC
�~��~��
�
�DC�EFAD, BC,DC
BCAD
mcrbg-0804-sk.qxd 5-25-2001 11:17 AM Page 170
Geometry 171Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
8.5 Mixed Review (p. 495)
39.
40.
41.
42. 43. 44. 45.
46. A reflection in the x-axis changes the sign of they-coordinate;
47. A reflection in the y-axis changes the sign of the x-coordinate;
48. A reflection in the y-axis changes the sign of the x-coordinate;
49. A reflection in the x-axis changes the sign of the y-coordinate;
Quiz 2 (p. 496)
1. Yes;
by Alt. Int. Thm.;
by Vert. Thm.;
by Alt. Int Thm.
2. Yes;
by Corresp. Post.
by Corresp. Post.
3. No;
so and socorresp. are not
4. No; 5. Yes;
6. Yes;
7.
Math & History (p. 496)
Sample answer:
M
1 in.
1 in.58
x � 10 mi
7x � 70
5x
�7
14
31
�62
�7
�73�
36
�6
12�
714
43
�76
�87
�.�
m�P � 94�m�A � 43�m�J � m�H � 42�;
180� � 96� � m�J � m�H � 84�;
m�V � 32°
m�V � 47° � 101° � 180°
�m�VSU � 47�
�m�U � 101�
m�P � 47�;
m�P � 101� � 32� � 180�
�m�A � 53�
�m�ANB � 46�
�m�B � 81�
m�E � 81�;
m�E � 53� � 46� � 180°
��5, 1�
�3, �10�
�2, 7�
�0, �5�
�5�7�10�12
m�ABC � 128�m�ABD � m�DBC � 64�;
m�ABD � m�DBC �12�m�ABC� �
12�36°� � 18�
m�ABD � m�DBC �12�m�ABC� �
12�77°� � 38.5�
Technology Activity
Sample answers:
Investiagte
1. Measurements will vary, but
2. The ratios will remain the same.
3. The ratios will remain the same.
Make a Conjecture
4. When a line to one side of a intersects the other twosides, it divides them proportionally.
Investiagte
5. Measurements will vary, but
Make a Conjecture
6. The ratios will remain the same.
7. The bisector of an of a divides the opp. side into 2 segments whose lengths are in the same ratio as thelengths of the other 2 sides of the .
Extension
The s are if and only if the original is isosceles. (If so, the resulting s are not only , they are �.)
Lesson 8.6
Activity (p. 500)
Sample answer:
1–3.
4. The lines divide the transversals proportionally.
Since then
and
8.6 Guided Practice (p. 502)
1. parallel; Proportionality Converse
2.
3. True by Proportionality Thm.
4. True by Corresp. Post., AA Similarity Post.,and def. of s .
5. True by Corresp. Post., AA Similarity Post.,and def. of s .�~
�
�~�
�
BRRC
�ABAC
�
AJ � JK � KL � LB.
AJJK
�JKKL
�KLLB
� 1ADDE
�DEEF
�EFFG
� 1,
�
A B
C
DE
FG
~�
�~�
�
��
BRBQ
�RPQP
.
��
BDDA
�BEEC
.
or about
1:1.618
1:1.625
1:158
mcrbg-0804-sk.qxd 5-25-2001 11:17 AM Page 171
172 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
6. False; statement does not follow from similarity.
7. 8. 9. 10.
8.6 Practice and Applications (pp. 502–505)
11. Yes; 12. Yes;
13. No; 14. Yes;
15. Yes; Proportionality Converse
16. No; only one angle is given (∠N is � to itself), which isnot enough to prove that
17. Yes; Corresp. Converse
18. Yes; Corresp. Converse
19. No; and are sides of and is aside of , but is not a side of a , so this isnot a proportion of lengths of corresponding sides. Theproportion does not give enough information to prove
s or conclude that
20. Yes; def. of s and Corresp. Converse.
21.
23.
25.
27.
f � 29.4
15f � 441
f
21�
2115
p � 14
12p � 168
7p
�1224
x � 6
20x � 120
8
20�
x15
a � 3
15a � 45
9a
�155
��~
LP � MQ.�~
�LM�LNPLP�MNQMQMN
�
�
LP � MO.
�
129
�2015
3415
�3516
12.55
�104
164
�82
EAGEADCE29.
m;
30. Lot since it has the longest distance of ocean frontage.
31. Statements Reasons
32. Draw intersecting at (Through any 2 pointsthere is exactly 1 line.) By the Proportionality Thm.,
since and and
Then by the Transitive property of equality.
33. Draw a line to through and extend to intersect the line at ( is not to because it
would also have to be to .) Then Also,
corresp. and are as are alt. int. and Since by theTransitive prop. of cong. By the Converse of the BaseAngles Thm., or Then by the
Substit. prop. of equality,YWWZ
�XYXZ
.
XA � XZ.XA � XZ
�A � �AZX�YXW � �WXZ,AZX.� WXZ�,A� YXW
YWWZ
�XYXA
.XW�
AZ�XYA.�XY� � Post.�ZXW �
CBBA
�DEEF
DXXA
�DEEF
.CBBA
�DXXA
k2 � k3,k1 � k2
�
X.BEAD
A
Lot C 34.0 m
97�Lot C� � 3294
2797
�Lot C122
Lot B 40.2
3904 � 97�Lot B�
3297
�Lot B122
Lot A 47.8 m;
97�Lot A� � 4636
3897
�Lot A122
22.
24.
26.
28.
q � 12
210 � 17.5q
17.5
1.25q�
1412
q � 27.5
21q � 577.5
21
17.5�
33q
z � 15
300 � 20z
300 � 12z � 8z
12�25 � z� � 8z
25 � z
z�
812
c � 9
288 � 32c
288 � 12c � 20c
12�24 � c� � 20c
24 � c
c�
2012
1.
2.
3.
4.
5.
6.
7.
8.DABD
�ECBE
1 �DABD
� 1 �ECBE
BDBD
�DABD
�BEBE
�ECBE
BD � DABD
�BE � EC
BE
BABD
�BCBE
�DBE ~ �ABC
�DEB � �ACB�EDB � �CAB;
DE � AC 1. Given
2. Corresp Post.
3. AA Similarity Post.
4. Def. of s
5. Segment Addition Post.
6. Addition of fractions
7. Substit. Prop. of Equality
8. Subtraction Prop. of Equality
�~
�
mcrbg-0804-sk.qxd 5-25-2001 11:17 AM Page 172
Geometry 173Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
34.
35.
36.
37.
38. In the diagram for Ex. 31 on page 504, suppose that
Then
So, by
the Segment Addition Post. and the substitution prop.of equality. Since by the Reflexive Prop of Cong., by the SAS Similarity Thm.Then corresp. and A are and bythe Corresponding Angles Converse.
DE � AC �� BDE�ABC ~ �DBE
�B � �B
BABD
�BCBE�If
ab
�cd
, then a � b
b�
c � dd �.
DA � BDBD
�EC � BE
BE.
DABD
�ECBE
.
x 1040 ft
2800x 2,912,000
1120
x
28002600
x 1400 ft
2600x 3,640,000
x
1300
28002600
UR � 21
44�UR� � 924
UR14
�6644
PR � 27 SN � 8
18�PR� � 486 39�SN� � 312
PR18
�2718
SN12
�2639
LN � 8 MT � 8.4
27�LN� � 216 18�MT� � 151.2
LN18
�1227
MT5.6
�2718
HJ � 15.8 FG � 13.8
21.6�HJ� � 341.28 21.6�FG� � 298.08
HJ
31.6�
10.821.6
FG27.6
�10.821.6
BC � 4 GJ � 4
21.6�BC� � 86.4 11.9�GJ� � 47.6
BC10.8
�8
21.6 GJ
13.6�
3.511.9
DF � 5.25 AB � 7
13.6�AB� � 71.4 13.6�AB� � 95.2
DF11.9
�6
13.6
AB11.9
�8
13.6 39. a.
c.
d. Sample answer: If lines are cut by a transversal,corresp. are so and then by AA Similarity Post.
40. Sample construction:
so
8.6 Mixed Review (p. 505)
41.
42.
43.
44.
45.
46.
47. Sample answer:
48. Sample answer:
D � 20
400 � D 2
162 � 122 � D 2
2
2
y
x
D
h � 15
h � �225
225 � h 2
122 � 92 � h 2
2
h
2
y
x
� �92 � ��8�2 � �145��4 � ��5��2 � �0 � 8�2
� �17 2 � 4 2 � �305��7 � ��10��2 � �4 � 0�2
� ���9�2 � ��5�2 � �106��2 � 11�2 � ��5 � 0�2
� �72 � 72 � �98 � 7�2���2 � ��9��2 � �6 � ��1��2
� �72 � ��16�2 � �305��4 � ��3��2 � ��9 � 7�2
� �92 � 162 � �337��5 � ��4��2 � �10 � ��6��2
xy
�zp
B
E
C
Dx
z
p
y A
�ABC ~ �DBE�BED � �C,�BDE � �A�,�
2 �
CE � 12
10�CE� � 120
610
�CE20
BE � 15
8�BE� � 120
68
�BE20
b. Sample answers:
BABD
�ACDE
BDBA
�BEBC
, BDDA
�BEEC
,
mcrbg-0804-sk.qxd 5-25-2001 11:18 AM Page 173
174 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
49. Sample answer:
50. Sample answer:
51. reflection 52. glide reflection 53. rotation
Lesson 8.7
8.7 Guided Practice (p. 509)
1. similar
2. She found , rather than
3. Enlargement; the scale factor is which is
4. similar
5. larger; enlargement
6.
7. yes; Sample answer: A preimage and its image after adilation are
8.7 Practice and Applications (pp. 509–512)
8. Reduction; the dilation has center and a scale factor of
9. Enlargement; the dilation has center and scale factor of
10. Enlargement; the dilation has center and scale factor of
y � 16 x � 8
2y � 32 2x � 16
21
�32y
21
�16x
2814 �
21.
C
249 �
83.
C
614 �
37.
C
~.
y
x
1
1A
A� D�
D
B C
C�B�
�since k > 1�
> 1.42
�21
� 2
CP�
CP.
CPCP�
� � 10
100 � �2
82 � 62 � �2
2
2
y
x
ll
h � 6�2
72 � h2
62 � 62 � h2
1
1
y
x
h
11. Reduction; the dilation has center and a scale factor of
12.
13.
14.
15.
16. same as the scale factor
17.
18.
19.
20. 65:39
5:3
21. 30:22.5
4:3
y � 6.3 x � 7.2
12y � 75.6 12x � 86.4
y
8.4�
912
x
9.6�
912
;
t � 1820 � r
15t � 270180 � 9r
159
�30t
; 159
�r
12;
D
F E
D �
F �E �
HD
F E
D �
F � E �
F E
D
D �
F � E �
G
2; G; 2:1;
V��3, �1� → V� � 4��3, �1� � ��12, �4�
U��1, 1� → U� � 4��1, 1� � ��4, 4�
T��3, 4� → T� � 4��3, 4� � ��12, 16�
S��5, 2� → S� � 4��5, 2� � ��20, 8�
G��3, �4� → G� �13��3, �4� � ��1, �4
3�F�5, �3� → F� �
13�5, �3� � �5
3, �1�E�3, 2� → E� �
13�3, 2� � �1, 23�
D��5, 4� → D� �13��5, 4� � ��5
3, 43�R�1, 1� → R� � 2�1, 1� � �2, 2�
Q�4, 0� → Q� � 2�4, 0� � �8, 0�
P�3, 5� → P� � 2�3, 5� � �6, 10�
M��5, �3� → M� �12��5, �3� � ��21
2, �112�
L�2, �3� → L� �12�2, �3� � �1, �11
2�K�2, 3� → K� �
12�2, 3� � �1, 11
2�J��5, 3� → J� �
12��5, 3� � ��21
2, 112�
z � 25y � 20; x � 20;
2x � 40
25
�8x
1025 �
25.
C
mcrbg-0804-sk.qxd 5-25-2001 11:18 AM Page 174
Geometry 175Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
22. Sample answer:
reduction;
23.
enlargement;
24.
25.
27.
29.
31.
x � 1.7 in.
10x � 17
210
�x
8.5
x � 4.8 in.
1.2x � 5.76
1.27.2
�0.8x
7:1
D 9.2 cm
2.7D � 24.9
2.78.3
�3D
bulb preimage imagecircle
oflight
AC � 9
B�C� � 28 4�AC� � 36
41
�B�C�
7 41
�36AC
;
k �205
�41
;
B�
C �
A�
A
C
B
9 7
5
20
2836
A�C� � 4 B�C� � 3
13
�A�C�
12 13
�B�C�
9;k �
26
�13
;
B�
C �A�
A C
B
6
23
4
12
9
32. The s are All 3 s are to so all the corresp. The scale factor of each dilation is 2,so each side of every image is twice as long as the corresp. side of Then the corresp. sides of the 3 s are and the 3 s are by the def. of s .
33. Sample answer:
I would show the student this figure and explain that Lrepresents the light source, CP the height of the puppet,and SH the height of the image. I would tell the studentthat a dilation transforms a figure in such a way that theoriginal figure (or preimage) and the figure after the dilation (the image) are similar. In the figure, the centerof the dilation is L and the scale factor is the ratio of LSto LC.
H
S
C
P
L
������
�PQR.�
� are �.�PQR,~��.�
26.
28.
30.
x 2.33 cm
12x � 28
x14
�2
12
x � 8.75 in.
1.25x
�17
I 6.1 cm
2.7I � 16.6
2.78.3
�2I
mcrbg-0804-sk.qxd 5-25-2001 11:18 AM Page 175
Chapter 8 continued
176 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
34. Sample answer:
35. since this is a reduction. C
36. or D
37. Sample answer: Let have sides 4, 6, 8 and letThen would have sides of 2, 3, 4. The
second dilation would make twice as large so
would have sides 4, 6, and 8. Therefore thenew image is the original figure.
8.7 Mixed Review (p. 513)
38. 39.
40. 41.
42.
44. Yes; Sample answer: and by theAlt. Int Thm; so the s are by AA Similarity Post.
Quiz 3 (p. 513)
1. 2. 3. 4.
5. Enlargement; the dilation has center and scale factor
6. Reduction; the dilation has center and scale factor
7. reduction; larger
8. k �188
�94
k �21
21 � 42�
2163
�13
.
C
k �5 � 5
5�
105
� 2.
C
FAAFCEBD
~��
�R � �S�P � �T
x � 22
x2 � 484
11x
�x
44
a � 7 a � 11
a2 � 49 a2 � 121
a2 � 1 � 50 a2 � 4 � 125
a2 � 12 � ��50�2 a2 � 22 � �5�5�2
b � 14 c � 13
b2 � 196 169 � c 2
64 � b2 � 260 25 � 144 � c 2
82 � b2 � �2�65�2 52 � 122 � c 2
�A�B�C�
�A�B�C�
�A�B�C�k � 2.�ABC
3313%2
6 � 0.3313
k < 1k �26 �
13;
Technology Activity (p. 514)
Investigate
1. the ratio is the reciprocal of the scale factor of the
dilation.
2. the ratio is the reciprocal of the scale factor of the
dilation.
3. no 4. . 5.
Conjecture
6. If the scale factor of a dilation is a:b, then the ratio of thearea of a polygon to the area of its image after a dilationis
Extension
7. Sample answer: Let be a side of a polygon,
its image after the first dilation, and its imageafter the second. Since the first image is to the pre-image and the second is to the first, the transitive properties of and equality can be used to show the
second image is to the pre-image. Then so
Also Therefore, the scale factor of the dilation that maps theoriginal polygon to the final one is yx.
Chapter 8 Review (pp. 516–518)
1.
3.
5.
7.
8.
x � 45
3x � 135
53
�x
27
m�F � 67� � 180�
m�F � 113�
3018
�53
x � 39 in.
13
�13x
x � 4
�2x � �8
2x � 12 � 4x � 4
2�x � 6� � 4�x � 1�
2x � 1
�4
x � 6
x �212
21 � 2x
3x
�27
A�B� � y � A�B� � y � x � AB.A�B� � x � AB.
ABA�B�
�1x
~
�~
~A�B�A�B�
ABxy or yx.
a 2:b 2.
41
A�B�C�D�E� is inside ABCDE
21
;
21
;
43. Yes; Sample answer:
and
so the s are by the SAS
Similarity Thm.
~�
CALJ
�CBLK
�C � �L
2.
4.
6.
4000 oz� 1 lb16oz� � 250 lb
x � 4000 oz
3x � 12,000
31000
�12x
d � 7
4d � 28
7d � 28 � 3d
7�d � 4� � 3d
d � 4d
�37
a � 9
9a � 9 � 10a
9�a � 1� � 10a
a � 15
�2a9
mcrbg-080R-sk.qxd 5-25-2001 11:17 AM Page 176
Geometry 177Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
9.
10. Yes; so and the s by the AA Similarity
Post;
11. No; andtherefore corresp.
are not
12. Yes; vertical are and so are andThe s by the AA Similarity Post.;
13. No;
14. Yes; and vertical are , so the s are by theSAS Similarity Thm.
15. 16.
17.
Chapter 8 Test (p. 519)
1. 2.
3. 4. 2
5. 36.
7. 8. No;
9. since all right are Since by the Reflexive prop. of cong.,
by the AA Similarity Post.
10.1525
�35
�RSQ ~�RQT �R � �R
�.��RSQ � �RQT
FG � 4.8
2.8�FG� � 13.44
1.52.25
1.44.2
2.83.2
�4.2FG
EF � 1.6
2.8�EF� � 4.48
2.83.2
�1.4EF
3 � y
z � 1
110z � 110
11110
�z
10
y � 24 x � 4
15y � 360 9x � 36
18y
�1520
x3
�129
h � 161659
59h � 960
35h � 960 � 24h
35h � 24�40 � h�
3524
�40 � h
h
y � 10.5 x � 22
10y � 105 12x � 264
15y
�107
1224
�11x
~���2520 �
37.530
2114 �
3926
4227
�PQN ~ �SRQ.are ~�R.
� N�� PQN and SRQ
�.�
m �J � 180� � �64� � 75�� � 41�;m �G � 180� � �38� � 75�� � 67�
�STU ~ �VWXare ~��S � �V
m�S � 180� � �104� � 48�� � 28�,
HJKLDEFG
�2�18� � 2�27�2�45� � 2�30� �
90150
�35
11. Yes; Sample answer: Since is a right isalso a right Also, since corresp.
are Then by the AASimilarity Post.
12.
13. Yes; Proportionality Converse
14. Yes; Corresponding Converse
15. Yes; Sample answer: by SAS SimilarityThm., def. of s , and Corres. Converse.
16.
17. No;
18.
Chapter 8 Standardized Test (pp. 520–521)
1. B 2.
3.
C
5. ;
E
6. B; and are not corresponding sides.
7. 12 to 4
3 to 1
A
TUPR
y �152
12y � 90
12y � 36 � 126
12�y � 3� � 126 x � 106
1218
�7
y � 3 x� � 74� � 180�
7x � 7�3� � 21
2x � 2�3� � 6
x � 3
18x � 54
2�9x� � 54
2�2x � 7x� � 54
20 ft5 yd
�20 ft15 ft
�43
D
k �9 in.3 ft
�9 in.36 in.
�14
78
1418
R�T� � 5 S�T� � 3
4�R�T�� � 20 4�S�T�� � 12
14
�R�T�
2014
�S�T�
12k �
624
�14
��~�LHM ~ �JHK
�
�
x � 12
5x � 60
35
�x
20
�RSQ ~�QST�.� RQS and T�RSQ ~ �RQT,�.
�QST�,�RSQ
18. Reduction;
b � 17.5
4b � 70
47
�10b
k �2442
�47
;
4.
B
x � 64
x2 � 4096
16x
�x
256
mcrbg-080R-sk.qxd 5-25-2001 11:17 AM Page 177
Chapter 8 continued
178 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
8.
;
E
9.
B
11. the dilation is a reduction, so theperimeter of the pre-image is greater.
A
12.
C
13. 14. 15.
16.
17. The colors are given in percents, which stay the sameregardless of the number of cars manufactured.
18. a. 7.5
b.
c.
d.
19.187.5
�3615
�125
m �DGF � m �DCB � 84�
EF � 12.5
18�EF� � 225
EF30
�7.518
BC � 33.6
7.5�BC� � 252
7.518
�14BC
800 0.13 � 104 cars
12,560 0.05 � 628 cars39
�13
820
�25
k �A�B�
AB�
13
AB � 3; A��� � 1;
x �5416
�278
16x � 54
163
�18x
so AB � 35, BC � 30and perimeter � 105
AC � 40
7�AC� � 280
78
�35AC
20.
21.98.441
�125
� 41
perimeter of EFGD � 14 � 7 � 7.5 � 12.5
� 98.4
perimeter of ABCD � 30 � 33.6 � 16.8 � 18
DC � 16.8
7.5�DC� � 126
7DC
�7.518
10.
C
CD � 20
15�CD� � 300
2515
�CD12
mcrbg-080R-sk.qxd 5-25-2001 11:17 AM Page 178
Geometry 179Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
Algebra Review (pp. 522–523)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
� 5�7
� 2�7 � 3�7
�28 � �63 � �4 � �7 � �9 � �7
� �16�3
� �7�3 � 9�3
��147 � �243 � ��49 � �3 � �81 � �3
� 21�2
� 11�2 � 10�2
�242 � �200 � �121 � �2 � �100 � �2
� �5
� 5�5 � 4�5
�125 � �80 � �25 � �5 � �16 � �5
� 4�11
� 2�11 � 2�11
�44 � 2�11 � �4 � �11 � 2�11
�64 � �28 � 8 � �7 � �4 � 8 � 2�7
� 2�2
� 5�2 � 3�2
�50 � �18 � �25 � �2 � �9 � �2
�75 � �3 � �25 � �3 � �3 � 5�3 � �3 � 6�3
�225 � 15
�320 � �64 � �5 � 8�5
�288 � �144 � �2 � 12�2
�243 � �81 � �3 � 9�3
�50 � �25 � �2 � 5�2
�80 � �16 � �5 � 4�5
�27 � �9 � �3 � 3�3
�40 � �4 � �10 � 2�10
�72 � �36 � �2 � 6�2
�45 � �9 � �5 � 3�5
�52 � �4 � �13 � 2�13
�121 � 11
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.2�3�6
�2�3 � �6�6 � �6
�2�18
6�
�9 � �23
�3�2
3� �2
5�7
�5�7
�7 � �7�
5�77
4�3
�4�3
�3 � �3�
4�33
�10�11�2� 102�112 � 100 � 11 � 1100
�5�5�2� 52�52 � 25 � 5 � 125
�2�3�2� 22�32 � 4 � 3 � 12
�8�3�2 � 82�32 � 64 � 3 � 192
�4�2�2� 42�22 � 16 � 2 � 32
�6�5�2� 62�52 � 36 � 5 � 180
�5�4��2�4� � 10 � 4 � 40
� 112
� 56 � 2
� �7�2��8�2� ��98���128� � ��49 � �2���64 � �2���32���2� � �64 � 8
��21���24� � �504 � �36 � �14 � 6�14
� 18 � 2 � 36
� �3�2��6�2� ��18���72� � ��9 � �2���36 � �2��6�2��2�2� � 12 � 2 � 24
� 330
� 110 � 3
� �11�3��10�3�� �363���300� � ��121 � �3���100 � �3�
� 21�10
� 3 � 7 � �10
� 3 � �49 � �10
�3�14���35� � 3�490
��13���26� � �338 � �169 � �2 � 13�2
� 4�5
� 2�5 � 3�5 � �5
�20 � �45 � �5 � �4 � �5 � �9 � �5 � �5
mcrbg-080R-sk.qxd 5-25-2001 11:17 AM Page 179
Chapter 8 continued
180 GeometryChapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
55.
57.
59.
61.
x � ±7
x2 � 49
6x2 � 294
x � ±6
x2 � 36
7x2 � 252
x � ±4
x2 � 16
x2 � 4 � 12
x � ±17
x2 � 289
x � ±3
x2 � 9
�50�75
��50 � �75�75 � �75
��3750
75�
�625 � �675
�25�6
75�
�63
�27�45
��27 � �45�45 � �45
��1215
45�
�81 � �1545
�9�15
45�
�155
�32�5
��32 � �5�5 � �5
��160
5�
�16 � �105
�4�10
5
�18�10
��18 � �10�10 � �10
��180
10�
�36 � �510
�6�510
�3�5
5
�12�24
��12 � �24�24 � �24
��12 � �12 � �2
24�
12�224
��22
9�52
�9�52
�52 � �52�
9�4 � �1352
�18�13
52�
9�1326
3�5�20
�3�5 � �20�20 � �20
�3�100
20�
3020
�32
4�12
�4�12
�12 � �12�
4�1212
��4 � �3
3�
2�33
�5�10
��5 � �10�10 � �10
��5010
��25 � �2
10�
5�210
��22
16�24
�16�24
�24 � �24�
16�2424
�2�4 � �6
3�
2 � 2 � �63
�4�6
3
4�8
�4�8
�8 � �8�
4�88
��82
��4 � �2
2�
2�22
� �2
�183�2
��18 � �23�2 � �2
��363 � 2
�66
� 1
2�3�5
�2�3 � �5�5 � �5
�2�15
5
54.
56.
58.
60.
62.
x � ±�10
x2 � 10
4x2 � 40
4x2 � 5 � 45
x � ±8
x2 � 64
3x2 � 192
x � ±�13
x2 � 13
x2 � 7 � 6
x � ±�10
x2 � 10
x2 � 3 � 13
x � ±25
x2 � 625
mcrbg-080R-sk.qxd 5-25-2001 11:17 AM Page 180
Geometry 181Chapter 8 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 8 continued
63.
65.
67.
69.
71.
73.
x � ±13
169 � x2
25 � 144 � x2
52 � 122 � x2
x � ±4
x2 � 16
x2 � 9 � 25
x2 � 32 � 5 2
x � ±�6
x2 � 6
8x2 � 48
8x2 � 12 � 36
x � ±1
x2 � 1
10x2 � 10
10x2 � 16 � �6
x � ±2
x2 � 4
11x2 � 44
11x2 � 4 � 48
x � ±3
x2 � 9
2x2 � 18
2x2 � 5 � 23 64.
66.
68.
70.
72.
x � ±24
x2 � 576
49 � x2 � 625
7 2 � x2 � 25 2
x � ±5
x2 � 25
5x2 � 125
5x2 � 61 � 64
x � ±�7
x2 � 7
5x2 � 35
5x2 � 6 � 29
x � ±�2
x2 � 2
6x2 � 12
6x2 � 3 � 9
x � ±�5
x2 � 5
9x2 � 45
9x2 � 7 � 52
mcrbg-080R-sk.qxd 5-25-2001 11:17 AM Page 181
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