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CHAPTER 1
Geometry 1Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Lesson 1.1
Think & Discuss (p. 1)
1. Answers may vary
Sample answer:
A consistent runway naming scheme could prevent acci-dents due to confusion of which runway to use.
2. The missing runway numbers are 50 � 10, or 5, and 230 � 10, or 23.
Skill Review (p. 2)
1. 2.
3. 4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
1.1 Guided Practice (p. 6)
1. A conjecture is an unproven statement that is based onobservations.
2. A conjecture can be proven false by finding a counterex-ample.
3. 4.
5. Each number is 3 times the previous number. The nextnumber is or 162.
6. The numbers are consecutive perfect squares. The nextnumber is or 16.
7. Each number is the previous number. The next numberis or 1.4 � 4
14
42
54 � 3
�9 � 9 � �18 � 4.24
�225 � 100 � �325 � 18.03
�1 � 49 � �50 � 7.07
�36 � 4 � �40 � 6.32
��5�2 � 02 � 25 � 0 � 25
��1�2 � 12 � 1 � 1 � 2
52 � ��2�2 � 25 � 4 � 29
22 � 42 � 4 � 16 � 20
�5 � ��6� � �5 � 6 � 1
�6 � ��5� � �6 � 5 � �1
�7 � ��2� � �7 � 2 � �5
�7 � 2 � �7 � ��2� � �9
3 � ��5� � 3 � 5 � 85 � ��3� � 5 � 3 � 8
9 � 17 � �817 � 9 � 8
8. Every other number is zero. The other numbers alternatebetween 3 and The next number is
9. Each number is 0.5 greater than the previous number. Thenext number is or
10. Each number is 6 less than the previous number. The nextnumber is or
11. The sum of any three consecutive positive integers is 3times the middle integer.
1.1 Practice and Applications (pp. 6–9)
12. 13.
14. 15.
16. Each number is 3 more than the previous number. Thenext number is or 13.
17. Each number is half the previous number. The next num-ber is or
18. Each number is 11 times the previous number. The nextnumber is or 14,641.
19. Each number is 5 less than the previous number. The nextnumber is or
20. Numbers after the first are found by adding consecutiveeven integers. The sixth number is 10 more than the fifthnumber, so it is or 37.
21. Numbers after the first are found by adding consecutivewhole numbers. The sixth number is 6 more than the fifthnumber, so it is or 21.
22. Each number is the square root of the previous number.The next number is
23. Numbers after the first are found by adding a zero afterthe decimal point of the previous number. So the nextnumber is 1.00001.
24. 16 blocks 25. 28 blocks
26.
27. Each distance is 4 times the figure number.
28. The twentieth figure would have a distance of or80 units.
4 � 20
�2.
15 � 6
27 � 10
�15.�10 � 5
1331 � 11
0.625.1.25 � 2
10 � 3
�11.�5 � 6
9.0.8.5 � 0.5
�3.�3.
figure 1 2 3 4 5
distance 4 8 12 16 20
MCRBG-0101-SK.qxd 5-25-2001 11:10 AM Page 1
Chapter 1 continued
29. The sum of any two odd numbers is an even number.
30. The product of any two odd numbers is an odd number.
31. The product of a number and the number is always equal to the difference of the square of thenumber and
32.
The product of 101 and any two digit number is the four-digit number formed by writing the two digits in ordertwice.
33.
The square of the n-digit number consisting of all ones isthe number obtained by writing the digits from 1 to n inincreasing order, then the digits from n � 1 to 1 in decreas-ing order. This pattern does not continue beyond n � 9.
34. The counterexample is 2. The number 2 is prime, but it isnot odd.
35.–39. Sample answers are given.
35.
Three is not larger than 5, which is the larger number.
36.
The product is even, but 3 is not even.
37.
The product is positive, but neither factor is positive.
38. but is not less than
39. Let
is not greater than 1.
40. Answers may vary.
Sample answer:
30 � 13 � 17
28 � 11 � 17
26 � 7 � 19
24 � 7 � 17
22 � 5 � 17
20 � 3 � 17
12
m � 1m
��2 � 1
�2�
�1�2
�12
m � �2.
14.1
2�14 �
12
��2� � ��3� � 6
2 � 3 � 6
�2 � 5 � 3
11,111 � 11,111 � 123,454,321
1111 � 1111 � 1,234,321
111 � 111 � 12,321
11 � 11 � 121
101 � 49 � 4949
101 � 97 � 9797
101 � 25 � 2525
101 � 34 � 3434
1 �n2 � 1�.
�n � 1��n � 1�
41. Answers may vary.
Sample answer:
These are all of the possibilities for the number 17. Noneof these have two addends that are prime.
42. After 8 doubling periods, there will be 3 � 28 � 768billion bacteria.
43.
44. The pattern is that the y-coordinate is half the opposite of thex-coordinate. So the y-coordinate is
45. The y-coordinate is more than the opposite of the x-coordinate. So the y-coordinate is
46. The y-coordinate is one less than half of the x-coordinate.The y-coordinate is
47. E 48. D
49.
50. Conjecture: For n points on the circle, there are regions in the circle. (This conjecture is not true.)
51. There are only 31 sections with 6 points on the circle. Sothe conjecture is false.
1.1 Mixed Review (p. 9)
52.–59.
60. 61.
62. 63.
64.
65.
66. ��2�2 � 22 � ��2���2� � 2 � 2 � 4 � 4 � 8
52 � 122 � 5 � 5 � 12 � 12 � 25 � 144 � 169
32 � 42 � 3 � 3 � 4 � 4 � 9 � 16 � 25
�72� ��7 � 7� � �49��4�2 � ��4���4� � 16
52 � 5 � 5 � 2532 � 3 � 3 � 9
y
x
2
�22�2
(3, �8)
(5, 2)
(1, �10)
(�4, �6)
(�2, 7)
(4, �1)(�2, �6)
(�3, 8)
2n�1
12 �3� � 1 �
32 � 1 �
12.
12 � ��3� � �21
2.
12
12 � ��3� � �11
2.
C6 F14C5 F12
F F F F F F
F
F
F F F F F
C C C C C C F
F F F F F
F
F
F F F F
C C C C C F
17 � 4 � 13
17 � 3 � 14
17 � 2 � 15
17 � 1 � 16
2 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
40 � 17 � 23
38 � 7 � 31
36 � 17 � 19
34 � 11 � 23
32 � 13 � 19
17 � 8 � 9
17 � 7 � 10
17 � 6 � 11
17 � 5 � 12
Number of points on circle 2 3 4 5 6
Maximum number of regions 2 4 8 16
MCRBG-0101-SK.qxd 5-25-2001 11:10 AM Page 2
Geometry 3Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
67.
68. 625 69. 40,000.4 70. 19 71.
Lesson 1.2
Developing Concepts Activity (p. 12)
1. The intersection of and is point G.
The intersection of and is point G.
2. The intersection of and is point G.
3. The intersection of planes M and N is
4. Yes; Sample answer: They lie in plane CEG.
1.2 Guided Practice (p. 13)
1. The symbol means the line segment PQ or the end-points, P and Q, and all the points on line PQ that arebetween P and Q.
The symbol means the ray with initial point P and all
the points on line PQ that lie on the same side of P as Q.
The symbol PQ means the line that passes through P and Q.
The symbol means the ray with initial point Q and all
the points on line PQ that lie on the same side of Q as P.
2.
A. This is true because points R and T are on the sameside of S.
B. This is true because the three points are collinear.
C. This is false because the points R and T cannot bothbe an initial point of the ray unless they are the samepoint.
D. This is true because point R is between point S and T.
E. This is true because they both mean the points S and T
and all the points on between S and T.
F. This is false because the rays go in opposite directionseven though they share the points on
3. False 4. True 5. False 6. True 7. True 8. False
1.2 Practice and Applications (pp. 13–16)
9. False 10. False 11. True 12. True 13. True
14. True 15. False 16. True 17. K 18. N 19. M
20. F 21. L 22. F 23. J 24. M
25. N, P and R; N, Q, and R; R, P, and Q
26. R, S, and T; S, T, and U; T, U, and V; V, T, and S; V, T,and R; U, T, and R
27. A, W, and X; A, W, and Z; A, X, and Y; A, Y, and Z; W, X,and Y; W, X, and Z; W, Y, and Z; X, Y, and Z
ST.
↔ST
S R T
→QP
→PQ
PQ
↔AB.
EFCD
EFAB
CDAB
�3
� 100 � 25 � 125
��10�2 � ��5�2 � ��10���10� � ��5���5� 28. D 29. G 30. H 31. H 32. E 33. E 34. G
35. H 36. P, Q, R, and S 37. K, R, Q, and N
38. K, L, R, and S 39. M, N, P, and Q
40. K, L, M, and N 41. L, M, P, and S
42. L, M, R, and Q 43. M, N, R, and S
44. consists of the endpoints A and B and all the pointson the line AB that lie between A and B.
45. consists of the initial point C and all the points onthe line CD that lie on the same side of C as point D.
46. Two rays or segments are collinear if they are on thesame line.
47. and are opposite rays if A, B, and C are collinearand C is between A and B.
48.–51. Sample figures are given.
48. 49.
50. 51.
52. The railroad tracks illustrate the intersection of two lines.
53. The dart and dartboard illustrate the intersection of a lineand a plane.
54. The two mirrors illustrate the intersection of two planes.
55. and intersect at B.
56. and intersect at A.
57. and intersect at H.
58. Plane ABC and plane DCG intersect at line DC.
59. Plane GHD and plane DHE intersect at line DH.
60. Plane EAD and plane BCD intersect at line AD.
61.–67. Sample figures are given.
61. 62.
63. 64.
65. 66.
↔DH
↔HG
↔AE
↔AD
↔BC
↔AB
A B CX W Y
R S
P
Q T
K L
J
M
→CB
→CA
→CD
AB
MCRBG-0101-SK.qxd 5-25-2001 11:10 AM Page 3
Chapter 1 continued
67.
68. Lines CA and DB intersect at the vanishing point V.
69. Lines CE and DF intersect at the vanishing point W.
70.–72.
The dashed lines are the hidden lines of the house.
73. C 74. B 75. D
76.
5 lines have 10 6 lines have 15intersections intersections
Yes there is a pattern. Each time a line is added to a figure with n lines, n points of intersection are added.
1.2 Mixed Review (p. 16)
77. Each number is 6 times the previous number. So the nextnumber is or 1296.
78. The numbers alternate between 2 and Since the lastnumber is 2, the next number is
79. Numbers after the first are found by adding an 8 immedi-ately before the decimal point of the previous number anda 1 immediately after the decimal point. Since the lastnumber had four eights and four ones, the next number is88,888.11111.
80. Numbers after the first are found by adding consecutivemultiples of 3. So the sixth number is 15 more than thefifth or or 45.
81.
82.
83.
84.
85. 86.
87.
88.
89.
90.
91. 92. �9 � 16 � �25 � 5�25 � 144 � �169 � 13
�40 � 60 � �100 � 10
�21 � 100 � �121 � 11
�7 � ��5� � �7 � 5 � �2
3 � ��8� � 3 � 8 � 11
4 � 7 � 4 � ��7� � �35 � 0 � 5
�5 � ��2� � �5 � 2 � �3
9 � ��4� � 9 � 4 � 13
3 � 9 � 3 � ��9� � �6
0 � 2 � 0 � ��2� � �2
15 � 30
�2.�2.
216 � 6
C
BV W
D F
A E
HG
93.
94.
95.
96.
Lesson 1.3
1.3 Guided Practice (p. 21)
1. A postulate is a geometric rule that is accepted withoutproof.
2. Sample Answer:
3.
Subtract AB from both sides, and we get
4.
5.
6.
7.
8.
� 10
� �100
� �36 � 64
� ���6�2 � ��8�2
ST � ��1 � 7�2��5 � 3�2
� �61
� �25 � 36
� �52 � ��6�2
PQ � ���3 � ��8��2 � ��6 � 0��2
� 2�17
� �4 � �17
� �68
� �4 � 64
� �22 � 82
MN � ��3 � 1�2 � �5 � ��3��2
� 5�5
� �25 � �5
� �125
� �25 � 100
� �52 � 102
GH � ��8 � 3�2 � �10 � 0�2
� �29
� �25 � 4
� �52 � 22
CD � ��5 � 0�2 � �2 � 0�2
BD � AD � AB.
AB � BD � AD
BD � BC � CD
AB � BC � AC.A B C
���5�2 � 102 � �25 � 100 � �125 � 11.18
���3�2 � 32 � �9 � 9 � �18 � 4.24
�32 � ��2�2 � �9 � 4 � �13 � 3.61
�52 � 72 � �25 � 49 � �74 � 8.60
4 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0101-SK.qxd 5-25-2001 11:10 AM Page 4
Geometry 5Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
9.
10.
Yes, because they have the same length.
11.
No, and are not congruent because they do nothave the same length.
12.
Yes, because they have the same length.
1.3 Practice and Applications (pp. 21–24)
13. 30 mm 14. 33 mm 15. 24 mm 16. 27 mm
17. 18 mm 18. 34 mm
JK � KL
� �34
� �9 � 25
� ���3�2 � ��5�2
KL � ��4 � 7�2 � ��8 � ��3��2
� �34
� �9 � 25
� ���3�2 � ��5�2
JK � ��7 � 10�2 � ��3 � 2�2
KLJK
� 2�34
� �4 � �34
� �136
� �36 � 100
� ���6�2 � ��10�2
KL � ���2 � 4�2 � ��7 � 3�2
� �137
� �16 � 121
� �42 � 112
JK � ��4 � 0�2 � �3 � ��8��2
JK � KL
� �65
� �16 � 49
� ���4�2 � ��7�2
� ���5��1��2 � ��5 � 2�2
� �65
� �16 � 49
� ���4�2 � 72
JK � ���1 � 3�2 � �2 � ��5��2
� 5
� �25
� �9 � 16
� �32 � 42
VW � ��1 � ��2��2 � ��2 � ��6��2 19. 20.
21. 22.
23.
25.
26.
27.
28.
29.
30.
31.
32.
33.
� 412� 3 � 11
2� 3 � 1 �32 MN � 3z �
32
z � 1
3z � 3
10z � 7z � 3
10z � 4 � 7z � 7
5z � 2 �72 z �
72
5z � 2 �12 z � 2 � 3z �
32
LN � LM � MN
� 43� 39 � 4� 3 � 13 � 4MN � 3y � 4
LN � 143
13 � y
130 � 10y
143 � 10y � 13
143 � 7y � 9 � 3y � 4
LN � LM � MN
� 3� 8 � 5� 2 � 4 � 5MN � 2x � 5
4 � x
20 � 5x
23 � 5x � 3
23 � 3x � 8 � 2x � 5
LN � LM � MN
� 17� 3 � 3 � 11QT � QR � RS � ST
� 9� 3 � 3 � 3SP � PQ � QR � RS
� 14� 3 � 11RT � RS � ST
� 6� 3 � 3RP � PQ � QR
ST � 11
9 � ST � 20
3 � 3 � 3 � ST � 20
PQ � QR � RS � ST � PT
PQ � 3
PQ � QR
3 � QR
6 � 2�QR�
6 � QR � QR
QS � QR � RS
QR � RS � QSNM � MP � NP
Q R SN M P
GH � HJ � GJDE � EF � DF
G H JD E F
24.
3 � RS
QR � RS
� 20
� 12 � 8
� 3 � 4 � 8
LM � 3x � 8
LN � 23
� 100
� 91 � 9
� 7 � 13 � 9
LM � 7y � 9
� 212
� 12 � 2
� 12 � 1 � 2
LM �12 z � 2
� 7
� 5 � 2
� 5 � 1 � 2
LN � 5z � 2
MCRBG-0101-SK.qxd 5-25-2001 11:10 AM Page 5
Chapter 1 continued
34.
35.
36.
� �37
� �1 � 36
� ���1�2 � ��6�2
HJ � ��4 � 5�2 � ��1 � 5�2
� 5�2
� �25 � �2
� �50
� �49 � 1
� �72 � 12
GH � ��5 � ��2��2 � �5 � 4�2
� 5
� �25
� �9 � 16
� �32 � ��4�2
DF � ��0 � ��3��2 � �2 � 6�2
� 6�2
� �36 � �2
� �72
� �36 � 36
� ���6�2 � ��6�2
EF � ��0 � 6�2 � �2 � 8�2
� �85
� �81 � 4
� �92 � 22
DE � ��6 � ��3��2 � �8 � 6�2
� �130
� �49 � 81
� �72 � ��9�2
AC � ��3 � ��4��2 � ��2 � 7�2
� 5
� �25
� �9 � 16
� ���3�2 � ��4�2
BC � ��3 � 6�2 � ��2 � 2�2
� 5�5
� �25 � �5
� �125
� �100 � 25
� �102 � ��5�2
AB � ��6 � ��4��2 � �2 � 7�2
37.
and have the same length.
38.
and have the same length.EGFG
� �26
� �1 � 25
� �12 � 52
GH � ��5 � 4�2 � �1 � ��4��2
� 5
� �25
� �16 � 9
� ���4�2 � 32
EG � ��1 � 5�2 � �4 � 1�2
� 5
� �25
� �02 � 52
FG � ��5 � 5�2 � �6 � 1�2
BCAC
� 2�10
� �4 � �10
� �40
� �4 � 36
� �22 � ��6�2
CD � ��2 � 0�2 � ��4 � 2�2
� 3�5
� �9 � �5
� �45
� �36 � 9
� ���6�2 � ��3�2
BC � ��0 � 6�2 � �2 � 5�2
� 3�5
� �9 � �5
� �45
� �9 � 36
� �32 � ��6�2
AC � ��0 � ��3��2 � �2 � 8�2
� �61
� �36 � 25
� �62 � ��5�2
GJ � ��4 � ��2��2 � ��1 � 4�2
6 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0102-SK.qxd 5-25-2001 11:12 AM Page 6
Geometry 7Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
39.
No two segments have the same length.
40.
because they have the same length.
41.
because they have the same length.
42.
� 2�41
� �4 � �41
� �164
� �100 � 64
� ���10�2 � ��8�2
PQ � ���5 � 5�2 � ��7 � 1�2
PQ � QR
� �170
� �121 � 49
� �112 � ��7�2
QR � ��3 � ��8��2 � ��2 � 5�2
� �170
� �49 � 121
� ���7�2 � 112
PQ � ���8 � ��1��2 � �5 � ��6��2
PQ � QR
� �13
� �4 � 9
� ���2�2 � 32
QR � ���1 � 1�2 � ��3 � ��6��2
� �13
� �9 � 4
� ���3�2 � ��2�2
PQ � ��1 � 4�2 � ��6 � ��4��2
� 3�10
� �9 � �10
� �90
� �81 � 9
� ���9�2 � 32
PN � ���2 � 7�2 � ��3 � ��6��2
� �109
� �9 � 100
� ���3�2 � ��10�2
MN � ���2 � 1�2 � ��3 � 7�2
� 3�13
� �9 � �13
� �117
� �36 � 81
� �62 � ��9�2
LN � ���2 � ��8��2 � ��3 � 6�2
and are not congruent because they do not havethe same length.
43.
because they have the same length.
44.
45.
feet
The length of the track is about 896 feet.
46.
The distance from Alexandria to Eunice by flying directly is 55 miles.
� 55
� �3025
� �0 � 3025
� �02 � ��55�2
AE � ��26 � 26�2 � �1 � 56�2
� 896
� �802,952
� �538,756 � 264,196
� �7342 � 5142
length of track � ��734 � 0�2 � �514 � 0�2
y
x200 400 600(0, 0)
(734, 514)
200
400
600
PQ � QR
� 2�85
� �4 � �85
� �340
� �196 � 144
� ���14�2 � 122
QR � ���4 � 10�2 � ��2 � ��14��2
� 2�85
� �4 � �85
� �340
� �144 � 196
� �122 � ��14�2
PQ � ��10 � ��2��2 � ��14 � 0�2
QRPQ
� �173
� �4 � 169
� �22 � 132
QR � ���3 � ��5��2 � �6 � ��7��2
MCRBG-0102-SK.qxd 5-25-2001 11:12 AM Page 7
Chapter 1 continued
47.
� 62 � 26 � 88 miles
The approximate shortest driving distance fromAlexandria to Eunice is 63 miles by way of Bunkie andVille Platte.
48. Buffalo and Dallas
49. Chicago and Seattle
50. Miami and Omaha
51. Providence and San Diego
� 8079 units
� �65,274,824
� �24,186,724 � 41,088,100
� ��49182 � 64102
� ��9468 � 4550�2 � �7629 � 1219�2
� 4395 units
� �19,317,520
� �2,768,896 � 16,548,624
� ���1664�2 � 40682
� ��6687 � 8351�2 � �4595 � 527�2
� 5481 units
� �30,043,400
� �122,500 � 29,920,900
� �3502 � 54702
� ��6336 � 5986�2 � �8896 � 3426�2
� 3770 units
� �14,213,585
� �11,296,321 � 2,917,264
� �33612 � 17082
� ��8436 � 5075�2 � �4034 � 2326�2
� 63 miles
� 28 � 20 � 15
� �772 � �416 � �221
� �100 � 121
� �196 � 576 � �16 � 400
� ���10�2 � ��11�2
� �142 � ��24�2 � ���4�2 � ��20�2
��36 � 40�2 � �12 � 32�2 � ��26 � 36�2 � �1 � 12�2
AB � BV � VE � ��40 � 26�2 � �32 � 56�2 �
� �3812 � �677
� �676 � 3136 � �676 � 1
� ���26�2 � ��56�2 � �262 � 12
��26 � 0�2 � �1 � 0�2
AK � KE � ��0 � 26�2 � �0 � 56�2 � 52. Buffalo and Dallas: miles
Chicago and Seattle: miles
Miami and Omaha: miles
Providence and San Diego: miles
53.
54.
55. C
� 52 yards
� �2725
� �2500 � 225
� �502 � ��15�2
CA � ��0 � ��50�2 � �0 � 15�2
� 67 yards
� �4450
� �4225 � 225
� ���65�2 � ��15�2
BC � ���50 � 15�2 � �15 � 30�2
� 34 yards
� �1125
� �225 � 900
� �152 � 302
AB � ��15 � 0�2 � �30 � 0�2
� 65 yards
� 15 � 50
� �225 � �2500
� �0 � 225 � �2500 � 0
� �02 � ��15�2 � �502 � 02
� ��0 � ��50��2 � �0 � 0�2
� ���50 � ��50��2 � �0 � 15�2
CA � CG � GA
� 80 yards
� 65 � 15
� �4225 � �225
� ���65�2 � 02 � �02 � ��15�2
� ���50 � ��50��2 � �15 � 30�2
� ���50 � 15�2 � �30 � 30�2
BC � BF � FC
� 115 yards
� 50 � 30 � 35
� �2500 � 0 ��0 � 900 ��1225 � 0
� �502 � 02 ��02 � 302 ��352 � 02
� ��50 � 15�2 � �30 � 30�2
� ��50 � 0�2 � �0 � 0�2 ���50 � 50�2 � �30 � 0�2
AB � AD � DE � EB
8079 � �0.1 � 2555
4395 � �0.1 � 1390
5481 � �0.1 � 1733
3770 � �0.1 � 1192
8 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0102-SK.qxd 5-25-2001 11:12 AM Page 8
Geometry 9Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
56.
B
57.
58.
59.
1.3 Mixed Review (p. 24)
60. 61.
62. True 63. False 64. False 65. True 66. True
67. True 68. 69. 70. and
71. and
Quiz 1 (p. 25)
1. 8 2. 6 3.–6. Sample answers are given.
3.
4. 5.
6.
→NQ
→NM
→PQ
→PM
→NQ,
→NM
→NM,
→PM
� 3�174
� �1566
� �121 � 961 � 484
� ���11�2 � 312 � ��22�2
FG � ���7 � 4�2 � ��11 � ��42��2 � �38 � 60�2
� 2�131
� �524
� �324 � 196 � 4
� �182 � ��14�2 � ��2�2
AB � ��10 � ��8��2 � �1 � 15�2 � ��6 � ��4��2
� 2�262
� �1048
� �4 � 900 � 144
� �22 � ��30�2 � 122
PQ � ��2 � 0�2 � ��10 � 20�2 � ��20 � ��32��2
MD � 6
3�MD� � 18
3�MD� � CD
2�MD� � MD � CD
CM � MD � CD 7.
Math and History (p. 25)
1.–2.
3. and because they are radii of the
circles. It appears that and intersect at right
angles.
Lesson 1.4
1.4 Guided Practice (p. 29)
1. C 2. D 3. B 4. A
5. Yes, because their measures are equal.
6. Yes, because their measures are equal.
7. Yes, and are adjacent because they share a
common vertex, E, share a common side, and do notshare any interior points.
8. No, and are not adjacent because theyshare the points in the interior of
9. E, about 10. M, about
11. J, about 12. S, about
13. straight 14. right 15. obtuse 16. acute
Practice and Applications (pp. 29–32)
17. X, 18. N, 19. Q,
20. 21.
22. �T, �PTS, �STP
�C, �BCD, �DCB�A, �EAU, �UAE
→QR,
→QS
→NK,
→NE
→XF,
→XT
90�→SR,
→ST;75�
→JH,
→JK;
120�→ML,
→MN;35�
→EF;
→ED,
�DEF.�DEF�GED
→EF,
�FEH�DEF
�DEG � �HEG
�DEF � �FEG
↔CD
↔AB
BC � BDAC � AD
B
C
D
A
� 5 feet
� �25
� �16 � 9
� ���4�2 � 32
BC � ���1 � 3�2 � �7 � 4�2
� 5 feet
� �25
� �9 � 16
� �32 � 42
y
x2
2
C(�1, 7)
T(0, 0)
B(3, 4)
TB � ��3 � 0�2 � �4 � 0�2
MCRBG-0102-SK.qxd 5-25-2001 11:12 AM Page 9
Chapter 1 continued
23. 24.
25.
26.
27.
28.
Figure for 29–34
29.
30.
31.
32.
33.
34.
35. acute; about 36. right; about
37. obtuse; about
38. 39. Sample answer:
40. acute; answers may vary
Sample answer:
is in the interior of
is in the exterior of �ABC.�2, 1�
�ABC.�2, �4�
y
C
AB x
1
�11�1
C
ED
B
A
C
A
BD E
150�
90�40�.
� 130�� 50� � 80�m�BAE � m�BAD � m�DAE
� 130�� 50� � 80�m�FAD � m�EAF � m�DAE
�m�BAD � m�EAF � m�FAC�
� 80�
� 180� � 50� � 50�
m�DAE � m�FAB � m�BAD � m�EFA
� 180�� 130� � 50�m�FAB � m�BAC � m�FAC
� 50�m�BAD � m�FAC
m�FAC � 50�
2m�FAC � 100�
�m�EAF � m�FAC� m�FAC � 100� � m�FAC
m�FAC � m�EAC � m�EAF
C
F
E
A
D
B130°
50°50°80°
50°
� 140�
� 160� � 20�
m�PQR � m�PQS � m�RQS
� 180�
� 60� � 120�
m�DEF � m�DEG � m�GEF
� 105�
� 45� � 60�
m�ABC � m�ABD � m�DBC
m�DEF � 140�
m�XYZ � 25�m�ABC � 55� 41. right; answers may vary.
Sample answer:
is in the interior of
is in the exterior of
42. obtuse; answers may vary.
Sample answer:
is in the interior of
is in the exterior of
43. obtuse; answers may vary.
Sample answer:
is in the interior of
is in the exterior of
44. about 68° 45. about 148° 46. about 38°
47. about 140° 48. about 22° 49. about 132°
50. 14 points 51. 12 points 52. 18 points 53. 40 points
54. a.
and
b.
and
c.
and
d. Answers may vary.
Sample answer: and
55.
56.
57.
58.
59. The difference between the numbers on each end of arunway is 18. So the runway opposite that of runway 3would be
60. The difference between the numbers at the opposite endsof a runway is always 18 because they form a straightline and the measure of the angle formed is Runway numbers are determined by angle measurementsdivided by 10. Since the opposite runways differ by their numbers differ by 180 � 10 � 18.
180�,
180�.
3 � 18 � 21.
� 60�
� 210� � 150�
� 21�10� � 150
m�4 � �3 � 18��10� � 15�10�
� 120�� 150� � 30�m�3 � 15�10� � 3�10�
� 150�� 180� � 30�m�2 � 18�10� � 3(10�
� 30�� 180� � 150�m�1 � 18�10� � 15�10�
�BOE�AOB
�HOC�GOB,�FOA,�AOD, �BOE, �COF, �DOG, �EOH,
�HOB�GOA,�AOC, �BOD, �COE, �DOF, �EOG, �FOH,
�HOA�GOH,�FOG,�AOB, �BOC, �COD, �DOE, �EOF,
�ABC.�1, 1�
�ABC.��3, 3�
yC
AB
x
1
�11�1
�ABC.�0, �3�
�ABC.�3, 0�
y
C AB
x1
1
�ABC.�1, �5�
�ABC.�8, �3�
y
C
AB
x1
�11�1
10 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0102-SK.qxd 5-25-2001 11:12 AM Page 10
Geometry 11Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
1.4 Mixed Review (p. 32)
61.
63.
65.
67.
69.
70. true 71. false 72. false 73. false
74.
75.
76.
77.
� �221
� �100 � 121
� ���10�2 � 112
GH � ��0 � 10�2 � �9 � ��2��2
� 7�2
� �49 � �2
� �98
� �49 � 49
� �72 � ��7�2
EF � ��4 � ��3��2 � �4 � 11�2
� �89
� �64 � 25
� ���8�2 � ��5�2
CD � ���8 � 0�2 � �3 � 8�2
� 13
� �169
� �25 � 144
� ���5�2 � ��12�2
AB � ���2 � 3�2 � ��2 � 10�2
x � �5
x � 3 � �8
x � ��3�
2� �4
x � 15
x � 1 � 14
x � ��1�
2� 7
x � �27
x � 7 � �20
x � 7
2� �10
x � �12
x � 4 � �8
x � 4
2� �4
x � 3
x � 3 � 6
x � 3
2� 3
78.
79.
Lesson 1.5
Drawing Conclusions (p. 33)
1. The segments have the same length.
2. The angles have the same measure.
1.5 Guided Practice (p. 38)
1. An angle bisector is a ray.
2. Congruent segments in a diagram are indicated by match-ing congruence marks.
Congruent angles in a diagram are indicated by matchingcongruence arcs.
3. If and are points in a coordinate plane,
then the midpoint of has coordinates
4.
5.
6.
7.
8.
x � 9
x � 5 � 14
x � 5
2� 7
x � 3
x � 3 � 6
x � 3
2� 3
� �72
, 2�� �72
, 42�M � �6 � 1
2,
�4 � 82 �
� �5, �7�� �102
, �14
2 �M � ��1 � 112
, �9 � ��5�
2 �
� �1, 3�� �22
, 62�M � �5 � ��3�
2,
4 � 22 �
�x2
, y2�.AB
B�x, y�A�0, 0�
� 3�2
� �9 � �2
� �18
� �9 � 9
� ���3�2 � 32
LM � ���3 � 0�2 � �0 � ��3��2
� 2�2
� �4 � �2
� �8
� �4 � 4
� �22 � ��2�2
JK � ��7 � 5�2 � �5 � 7�2
62.
64.
66.
68.
x � �10
8 � x � �2
8 � x
2� �1
x � �5
�9 � x � �14
�9 � x
2� �7
x � 32
�8 � x � 24
�8 � x
2� 12
x � 5
5 � x � 10
5 � x
2� 5
�9, 10�
y � 10
y � 2 � 12
y � 2
2� 6
�3, 8�
y � 8
y � 0
2� 4
MCRBG-0103-SK.qxd 5-25-2001 11:12 AM Page 11
Chapter 1 continued
9.
10.
11.
12.
13.
1.5 Practice and Applications (pp. 38–41)
14. 15.
is the segment is the segmentbisector of bisector of
16. is the segment bisector of
17.
18.
19.
20.
21.
22. � �4, �7�� �82
, �14
2 �M � �4 � 42
, 4 � ��18�
2 �
� ��3, 3�� ��62
, 62�M � �0 � ��6�
2,
�8 � 142 �
� ��5, 12�� ��10
2,
12�M � ��12 � 2
2,
�9 � 102 �
� �4, 132 �� �8
2,
132 �M � �10 � ��2�
2,
8 � 52 �
� �1, 2�� �22
, 42�M � ��1 � 3
2,
7 � ��3�2 �
� ��4, 3�� ��82
, 62�M � �0 � ��8�
2,
0 � 62 �
EF.
↔GH
G H
E
F
CD.AB.
↔EF
↔CD
C
D
E
F
A
C
D
B
� 104�
� 2 � 52�
m�PQR � 2 � m�SQR
m�PQS � m�SQR � 52�
m�PQS � m�SQR �m�PQR
2�
64�
2� 32�
� 80�
� 2�40��
m�PQR � 2m�SQR
m�SQR � m�PQS � 40�
m�JKM � m�LKM �m�JKL
2�
90�
2� 45�
x � �2
x � 4 � �6
x � ��4�
2� �3 23.
24.
25.
27.
29.
31.
33.
34.
is the angle bisector of �ABC.
→BD
A
B C
D
�WXZ � �YXZ
WX � XY
�A � �B
AC � BC,
�14, �21�
y � �21
y � 7 � �14
y � 7
2� �7
x � 14
x � 6 � 20
x � 6
2� 10
�1, 10�
y � 10
y � 12 � �2
y � ��12�
2� �1
x � 1
x � 3 � 4
x � 3
2� 2
��4, �4�
y � �4
y � 6 � 2
y � 6
2� 1
x � �4
x � 2 � �2
x � 2
2� �1
� ��3, 32�
M � ��5.5 � ��0.5�2
, �6.1 � 9.1
2 � � ��62
, 32�
� ��0.625, 3.5�
M � ��1.5 � 0.252
, 8 � ��1�
2 � � ��1.252
, 72�
12 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
26.
28.
30.
32.
�DGE � �EGF
DG � FG
�6.5, 15�
y � 15
y � 6 � 9
y � ��6�
2� 4.5
x � 6.5
x � 3.5 � 3
x � ��3.5�
2� 1.5
��11, �13�
y � �13
y � 9 � �4
y � 9
2� �2
x � �11
x � 5 � �16
x � ��5�
2� �8
�8, 7�
y � 7
y � 1 � 6
y � ��1�
2� 3
x � 8
x � 8 � 0
x � ��8�
2� 0
35.
is the angle bisector of �ABC.
→BD
A
B
D
C
��2, �6�
y � �6
y � 2 � �4
y � 2
2� �2
MCRBG-0103-SK.qxd 5-25-2001 11:12 AM Page 12
Geometry 13Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
36. is the angle bisector of
37.
38.
39.
40.
41.
42.
43. No; yes; the angle bisector of an angle of a triangle pass-es through the midpoint of the opposite side if the twosides of the triangle contained in the angle are congruent.
44.
46.
48.
4 � x
32 � 8x
15x � 32 � 23x
�15x � 18�� � �23x � 14��
m�ABD � m�DBC
10 � x
�40 � �4x
10x � 40 � 6x
�10x � 51�� � �6x � 11��
m�ABD � m�DBC
20 � x
60 � 3x
x � 60 � 4x
�x � 15�� � �4x � 45��
m�ABD � m�DBC
m�PQS � m�SQR �m�PQR
2�
124�
2� 62�
� 90�
� 2 � 45�
m�PQR � 2 � m�PQS
m�SQR � m�PQS � 45�
m�PQS � m�SQR �m�PQR
2�
75�
2� 37.5�
� 160�
� 2 � 80�
m�PQR � 2 � m�PQS
m�SQR � m�PQS � 80�
m�PQS � m�SQR �m�PQR
2�
91�
2� 45.5�
� 44�
� 2 � 22�
m�PQR � 2 � m�SQR
m�PQS � m�SQR � 22�
�ABC.→BD
A
BC
D
49.
51.
52.
53.
54.
55. Sample Answer:
�BAC, �CAN, �NAG,
�GAM, �MAK, and �KAL; �DNE, �ENF, �HMI,and �JMI.
56. Sample answer: To divide a line segment into 4 congru-ent segments using a compass and a straightedge, followthese steps.
1. Place a compass point at A. Use a compass settinggreater than half the length of Draw an arc.
2. Keep the same compass setting. Place the compasspoint at B. Draw an arc. It should intersect the otherarc in two places.
3. Use a straightedge to draw a segment through thepoints of intersection. This segment bisects at M,the midpoint of
4. Place the compass point at A. Use a compass settinggreater than half the length of Draw an arc.
5. Keep the same compass setting as in Step 4. Placethe compass point at M. Draw an arc. It should inter-sect the other arc in two places.
6. Use a straightedge to draw a segment through thepoints of intersection from Step 5. This segmentbisects at N, the midpoint of
7. Place a compass point at M. Use a compass settinggreater than half the length of Draw an arc.
8. Keep the same compass setting from Step 7. Placethe compass point at B. Draw an arc. It should inter-sect the other arc in two places.
—CONTINUED—
MB.
AM.AM
AM.
AB.AB
AB.
BE � LI,BD � LJ,BC � LK,CD � KJ,FG � GH, DE � JI,NE � MI, NF � MH,
DN � MJ, AE � AI,AB � AL, AC � AK, AN � AM,
m�1 � m�2 � 90� � 60� � 30�
m�4 � m�3 � 60�
m�3 � m�4 � 90� � 65� � 25�
m�1 � m�2 �130�
2� 65�
m�3 � m�4 � 90� � 53� � 37�
m�1 � m�2 �106�
2� 53�
� 54�1082
T �45 � 63
2
42 � x
210 � 5x
x � 210 � 6x
x � 40 � 6x � 170
�12
x � 20�� � �3x � 85��
m�ABD � m�DBC
45.
19 � x
57 � 3x
2x � 57 � 5x
�2x � 35�� � �5x � 22��
m�ABD � m�DBC
47.
8 � x
16 � 2x
2x � 16 � 4x
�2x � 7�� � �4x � 9��
m�ABD � m�DBC
50.
� 51
�1022
T �42 � 60
2
MCRBG-0103-SK.qxd 5-25-2001 11:12 AM Page 13
Chapter 1 continued
56. —CONTINUED—
9. Use a straightedge to draw a segment through thepoints of intersection from Step 8. This segmentbisects at P, the midpoint of
10. This should result with
To divide a line segment into 4 congruent segmentsusing the Midpoint Formula, start with and
as the endpoints of the segment. The mid-point, M, of would have coordinates
Now we must find the midpoint N, of
The coordinates of
.
Lastly, the coordinates of
57. (17)
(18)
� �1, 2�
� ��1 � 2, 7 � ��5� � ��1 �
12 �4�, 7 �
12 ��10�
M � ��1 �12 �3 � ��1��, 7 �
12 ���3� � 7
� ��4, 3�
� �0 � ��4�, 0 � 3 � �0 �
12 ��8�, 0 �
12 �6�
M � �0 �12 ���8 � 0�, 0 �
12 �6 � 0�
� �x1 � 3x2
4,
y1 � 3y2
4 �.
� �x1 � x2 � 2x2
22
,
y1 � y2 � 2y2
22 �
P � �x1 � x2
2� x2
2,
y1 � y2
2� y2
2 �
� �3x1 � x2
4,
3y1 � y2
4 �
� �2x1 � x1 � x2
22
,
2y1 � y1 � y2
22 �
N � �x1 �x1 � x2
22
, y1 �
y1 � y2
22 �
AM.
�x1 � x2
2,
y1 � y2
2 �.
ABB�x2, y2�
A�x1, y1�
AN � NM � MP � PB.
MB.MB
(19)
(20)
(21)
(22)
(23)
(24)
Yes, the answers came out the same.
(This is the x-coordinate of the midpoint when theMidpoint formula is used.)
—CONTINUED—
�x1 � x2
2
�2x1 � x2 � x1
2
�2x1
2�
x2 � x1
2
x1 �12
�x2 � x1� � x1 �x2 � x1
2
� ��3, 1.5�
� ��5.5 � 2.5, �6.1 � 7.6�
� ��5.5 �12 �5�, �6.1 �
12 �15.2�
12 �9.1 � ��6.1��M � ��5.5 �
12 ��0.5 � ��5.5��, �6.1 �
� ��0.625, 3.5�
� ��1.5 � 0.875, 8 � ��4.5� � ��1.5 �
12�1.75�, 8 �
12 ��9�
M � ��1.5 �12 �0.25 � ��1.5��, 8 �
12 ��1 � 8�
� �4, �7�
� �4 � 0, 4 � ��11�
� �4 �12 �0�, 4 �
12 ��22�
M � �4 �12 �4 � 4�, 4 �
12 ��18 � 4�
� ��3, 3�
� �0 � ��3�, �8 � 11 � �0 �
12 ��6�, �8 �
12 �22�
M � �0 �12 ��6 � 0�, �8 �
12 �14 � ��8��
� ��5, 12� � ��5, �18
2 �192
� ��12 � 7, �9 �192
� ��12 �12 �14�, �9 �
12 �19�
M � ��12 �12 �2 � ��12��, �9 �
12 �10 � ��9��
� �4, 132 �
� 4, 162
���3�
2 �
� 10 � ��6�, 8 ���3�
2 �
� 10 �12
��12�, 8 �12 ��3��
M � 10 �12
���2� � 10�, 8 �12
�5 � 8��
14 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0103-SK.qxd 5-25-2001 11:12 AM Page 14
Geometry 15Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
57. —CONTINUED—
(This is the y-coordinate of the midpoint when theMidpoint Formula is used.)
58. Sample answer:
a. M is the midpoint of
N is the midpoint of
P is the midpoint of
bisects
bisects
bisects
b. None of the angle bisectors pass through the mid-points of the opposite sides.
c. None of the angle bisectors passed through the mid-points of the opposite sides. This is due to the fact thatall three sides had different length. If an equilateral tri-angle had been used, each angle bisector would havepassed through the midpoint of its opposite side. Iftwo of the sides had been congruent, the angle bisectorof the angle which was made up of the congruentsides would have passed through the midpoint of theopposite side.
59.
It seems to be approaching yards from 0.
60.
is approximately 200 yards.1.5625 � 0.78125
100 � 50 � 25 � 12.5 � 6.25 � 3.125 �
66.6
66.69921875 � 0.48828125 � 66.65039063
66.6015625 � 0.09765625 � 66.69921875
66.796875 � 0.1953125 � 66.6015625
66.40625 � 0.390625 � 66.796875
67.1875 � 0.78125 � 66.40625
65.625 � 1.5625 � 67.1875
68.75 � 3.125 � 65.625
62.5 � 6.25 � 68.75
75 � 12.5 � 62.5
50 � 25 � 75
100 � 50 � 50
�BAC.→AH
�ABC.→BG
�ACB.→CF
AC.
BC.
AB.
CB
M P
F
GA
H N
�y1 � y2
2
�2y1 � y2 � y1
2
�2y1
2�
y2 � y1
2
y1 �12
�y2 � y1� � y1 �y2 � y1
2
1.5 Mixed Review (p. 42)
61. 62.
63.
64.
65.
66.
67.
68.
69. 70. 71. 72.
Quiz 2 (p. 42)
1. If Q is in the interior of then m�QSR � m�PSR.m�PSQ �
�PSR,
35�115�130�20�
� 10
� �100
� �36 � 64
� �62 � 82
LM � ���4 � ��10��2 � �9 � 1�2
� �97
� �81 � 16
� �92 � 42
JK � ��5 � ��4��2 � ��1 � ��5��2
� 3�5
� �9 � �5
� �45
� �9 � 36
� ���3�2 � 62
GH � ��0 � 3�2 � ��2 � ��8��2
� 2�130
� �4 � �130
� �520
� �36 � 484
� ���6�2 � 222
EF � ��2 � 8�2 � �14 � ��8��2
� 4�17
� �16 � �17
� �272
� �16 � 256
� �42 � ��16�2
CD � ���2 � ��6��2 � ��7 � 9�2
� �233
� �64 � 169
� ���8�2 � ��13�2
AB � ���5 � 3�2 � ��1 � 12�2
MCRBG-0103-SK.qxd 5-25-2001 11:12 AM Page 15
Chapter 1 continued
2. acute;
Answers may vary.
Sample answer:
is in the interior of
is in the exterior of
3. obtuse;
Answers may vary.
Sample answer: is in theinterior of
is in the exterior of
4. acute;
Answers may vary.
Sample answer: is in theinterior of
is in the exterior of
5. right;
Answers may vary.
Sample answer: is in theinterior of
is in the exterior of
6.
Lesson 1.6
Technology Activity (p. 43)
1. Nonadjacent angles have the same measure.
2. Answers may vary.
Sample answer:
3. Answers may vary.
Sample answer:
4. The sum of the measures of adjacent angles formed byintersecting lines is 180�.
m�AEC � m�AED � 71� � 109� � 180�
m�AEC � m�AED � 41� � 139� � 180�
� 42�
� 2 � 21�
m�JKL � 2 � m�JKM
m�MKL � m�JKM � 21�
�DEF.�0, �2�
�DEF.�4, 4�
x
F
D
E
2
2
�2�2
y
�DEF.�0, �2�
�DEF.�0, 3�
y
xF
D
E2
2
�2�2
�DEF.�0, �7�
�DEF.�0, 0�
y
x2
2
FED
�2�2
�DEF.�0, �3�
�DEF.�0, 4�
y
x
E
F
D
�2�2
1.6 Guided Practice (p. 47)
1. Two angles are complementary angles if the sum of theirmeasures is
Two angles are supplementary angles if the sum of theirmeasures is
2. Sample answer:
and are acute verticalangles because their measuresare between and and are obtuse verticalangles because their measures are between and
3. and are adjacent congruentcomplementary angles.
and are adjacentcongruent supplementary angles.
4.
6.
Practice and Applications • (pages 47–50)
8. No 9. Yes 10. No 11. Yes 12. No 13. No
14. never 15. always 16. sometimes 17. always
18. always 19. never
20.
22.
24.
26.
28.
x � 58
2x � 116
�2x � 11�� � 105�
m�6 � 10�
m�6 � 170� � 180�
m�6 � m�9 � 180�
m�8 � 38�
m�8 � 142� � 180�
m�8 � m�7 � 180�
m�8 � 70�
m�8 � 110� � 180�
m�8 � m�9 � 180�
m�7 � 108�
m�7 � 72� � 180�
m�7 � m�6 � 180�
m�1 � 55�
m�1 � 35� � 90�
m�1 � 120�
m�1 � 60� � 180�
�CBD�ABC
B
C
DA
�DBC�ABDA
CB
D
180�.90��4�290�,0�
�3�11 2
4 3
180�.
90�.
16 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
5.
7.
x � 40�
x � 50� � 90�
x � m�1 � 90�
m�1 � 20�
m�1 � 160� � 180�
21.
23.
25.
27.
29.
x � 23
7x � 161
7x � 19 � 180
x� � �6x � 19�� � 180�
m�7 � 154�
m�7 � 26� � 180�
m�7 � m�8 � 180�
m�9 � 167�
m�9 � 13� � 180�
m�9 � m�6 � 180�
m�7 � m�9 � 123�
m�6 � 80�
m�6 � m�8
MCRBG-0104-SK.qxd 5-25-2001 11:12 AM Page 16
Geometry 17Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
30.
31.
32.
33.
34.
35.
x � 8
2x � 16
6x� � �4x � 16��
35 � x
70 � 2x
3x � 70 � 5x
�3x � 20�� � �5x � 50��
x � 48
8x � 384
8x � 204 � 180
�7x � 248�� � �x � 44�� � 180�
y � 31
20y � 620
20y � 440 � 180
�9y � 187�� � �11y � 253�� � 180�
x � 25
7x � 175
7x � 5 � 180
�4x � 10�� � �3x � 5�� � 180�
y � 21
6y � 126
6y � 54 � 180
�2y � 28�� � �4y � 26�� � 180�
x � 29
8x � 232
8x � 52 � 180
�6x � 20�� � �2x � 20� � 180�
y � 50
4y � 200
4y � 20 � 180
�y � 12�� � �3y � 8�� � 180�
x � 16
5x � 80
�5x � 2�� � 78� 36.
37. supplementary 38. neither 39. complementary
40. neither
41.
42.
43.
44.
45.
� 17�� 13 � 4m�B � x � 4
� 73�� 65 � 8� 5�13� � 8m�A � 5x � 8
x � 13
6x � 78
6x � 12 � 90
5x � 8 � x � 4 � 90
m�A � m�B � 90�
� 160�� 8�20�m�D � 8�m�C�
m�C � 20�
9�m�C� � 180�
m�C � 8�m�C� � 180�
m�C � m�D � 180�
m�D � 8�m�C�� 67.5�� 3�22.5�� m�B � 3�m�A�
m�A � 22.5�
4�m�A� � 90�
m�A � 3�m�A� � 90�
m�A � m�B � 90�
m�B � 3�m�A�
y � 108
y � 72 � 180
y � 16 � 56 � 180
y � 2�8� � 7�8� � 180
y� � 2x� � 7x� � 180�
x � 8
7x� � 56�
y � 12
11y � 132
11y � 48 � 180
11y � 6�8� � 180
11y� � 6x� � 180�
y � 55
y � 125 � 180
y � �105 � 20� � 180
y � �3�35� � 20� � 180
y� � �3x � 20�� � 180�
50�57�65�80�88�m�2
40�33�25�10�2�m�1
4�14�28�35�41�m�2
86�76�62�55�49�m�1
90�108�132�164�176�m�2
90�72�48�16�4�m�1
2�11�28�60�81�m�2
178�169�152�120�99�m�1
MCRBG-0104-SK.qxd 5-25-2001 11:12 AM Page 17
Chapter 1 continued
46.
47.
� 89°
48.
49.
50.
51.
x � 13
13x � 169
13x � 11 � 180
12x � 1 � x � 10 � 180
m�A � m�B � 180�
� 73�� 90 � 17� 5�18� � 17m�B � 5x � 17
� 107�� 108 � 1� 6�18� � 1m�A � 6x � 1
x � 18
11x � 198
11x � 18 � 180
6x � 1 � 5x � 17 � 180
m�A � m�B � 180�
� 51�� 43 � 8m�B � x � 8
� 129�� 3�43�m�A � 3x
x � 43
4x � 172
4x � 8 � 180
3x � x � 8 � 180
m�A � m�B � 180�
� 79�� 96 � 17� 3�32� � 17m�B � 3x � 17
� 11�� 24 � 13�34 �32� � 13m�A �
34 x � 13
x � 32
15x � 480
15x � 120 � 360
3x � 52 � 12x � 68 � 360
34 x � 13 � 3x � 17 � 90
m�A � m�B � 90�
� 1�� 12 � 11m�B � x � 11
� 96 � 7� 8�12� � 7m�A � 8x � 7
x � 12
9x � 108
9x � 18 � 90
8x � 7 � x � 11 � 90
m�A � m�B � 90�
� 76�� 77 � 1� 11�7� � 1m�B � 11x � 1
� 14�� 21 � 7� 3�7� � 7m�A � 3x � 7
x � 7
14x � 98
14x � 8 � 90
3x � 7 � 11x � 1 � 90
m�A � m�B � 90�
52.
53. Let x� be the supplement of
Let y� be the supplement of
54.
The measure of the angle between the first base foul lineand the path of the baseball is
55.
The acute angle’s measure is and the obtuse angle’smeasure is
56. Answers may vary.
Sample answer:
An angle of measure does not have a complement.An angle that has a complement must have a measurebetween and 90�.0�
112�
135�.45�
m�2 � 135�
m�2 � 3�45��
m�2 � 3�m�1�
m�1 � 45�
4�m�1� � 180�
m�1 � 3�m�1� � 180�
m�1 � m�2 � 180�
m�2 � 3�m�1�56�.
x� � 56�
x� � 34� � 90�
y� � 156�
y� � 24� � 180�
y� � m�2 � 180�
�2.
x� � 122�
x� � 58� � 180�
x� � m�1 � 180�
�1.
� 103�� 72 � 31m�B � x � 31
� 77�� 27 � 50�38 �72� � 50m�A �
38 x � 50
x � 72
11x � 792
11x � 648 � 1440
3x � 400 � 8x � 248 � 1440
38 x � 50 � x � 31 � 180
m�A � m�B � 180�
� 23�� 13 � 10m�B � x � 10
� 157�� 156 � 1� 12�13� � 1m�A � 12x � 1
18 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0104-SK.qxd 5-25-2001 11:12 AM Page 18
Geometry 19Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
57.
E
58.
B
59.
1.6 Mixed Review (p. 50)
60.
62.
64.
66. E or D 67. C 68. B or C 69. A
r � 6.37
6.28 � r � 40
2 � 3.14 � r � 40
b � 5
3 � b � 15
12 � b � 6 � 15
x � 32
3x � 96
y � 70
y � 20 � 90
10 � y � 10 � 90
x � y � 10 � 90
x � 10
�x � y � �80
2x � y � 90
x � y � 80
2x � y � 90
x � y � 10 � 90
2x � y � 90 � 180
m�F � 24�
15�m�F� � 360�
2�m�F� � 13�m�F� � 360�
m�F �132 �m�F� � 180�
m�F � m�G � 180�
m�G � 612 �m�F�
y � 94
3y � 282
3y � 78 � 360
y � 54 � 2y � 24 � 360
�12 y � 27�� � �y � 12�� � 180�
x � 18
16x � 288
16x � 108 � 180
�7x � 20�� � �9x � 88�� � 180�70.
71.
72.
73.
74.
75.
Lesson 1.7
1.7 Guided Practice (p. 55)
1. The perimeter of a circle is called its circumference.
2. To find the perimeter of the rectangle, find the sum oftwice its length and twice its width.
3.
square units
5.
� 3.14(3)2
� 3.14 � 9
� 28.26 square units
7.
� 2 � 3.14 � 4
� 25.12 in.2
8.
feet
You will need 46 feet of fence.
1.7 Practice and Applications (pp. 55–57)
9.
� 60 square units
� 10 � 6
A � lw
� 32 units
� 20 � 12
� 2 � 10 � 2 � 6
P � 2l � 2w
� 46
� 30 � 16
� 2 � 15 � 2 � 815 ft
8 ftP � 2l � 2w
C � 2�r
A � �r2
� 36
� 12 � 9 � 8
A �12 bh
� �2.6, 7�� �5.22
, 142 M � ��2.4 � 7.6
2,
5 � 92
� �1.75, �2.5�
� �3.52
, �52 M � ��1.5 � 5
2,
4 � ��9�2
� ��7, 1�� ��142
, 22M � ��14 � 0
2,
�9 � 112
� �3, �4�� �62
, �82 M � �8 � ��2�
2,
�6 � ��2�2
� ��4, 6�� ��82
, 122 M � �2 � ��10�
2,
5 � 72
� ��3, �2�� ��62
, �42 M � �0 � ��6�
2,
0 � ��4�2
61.
63.
65.
r � ±10
r � ±100
r2 � 100
3.14 � r2 � 314
s � ±102
s � ±100
s � ±200
s2 � 200
h � 8
5 � h � 40
12 � 5 � h � 20
4. A � lw
� 13 � 7
� 91 square units
6. P � 12
4s � 12
s � 3 m
10.
� 81 square units
� 92
A � s2
� 36 units
� 4 � 9
P � 4s
MCRBG-0104-SK.qxd 5-25-2001 11:12 AM Page 19
Chapter 1 continued
11.
13.
15.
17.
18. Use the Pythagorean Theorem to find b.
19.
� 10 � 52 units
� 5 � 5 � 52
P � a � b � c
� 28 units
� 16 � 12
� 2 � 8 � 2 � 6
P � 2l � 2w
b � 8
b2 � 64
36 � b2 � 100
62 � b2 � 102
a2 � b2 � c2
� 225 square units
� �15�2
A � s2
� 60 units
� 4 � 15
P � 4s
� 126 square units
� 12 � 21 � 12
A �12bh
� 54 units
� 31 � 21 � 20
P � a � b � c
� 84 square units
� 12 � 21 � 8
A �12 bh
� 48 units
� 10 � 21 � 17
P � a � b � c
� 12 square units
� 12 � 6 � 4
A �12 bh
� 16 units
� 5 � 6 � 5
P � a � b � c 20. The perimeter is twice the radius plus half of the circumference.
� 2 � 8 � 3.14 � 8
� 16 � 25.12
� 41.12 units
The area is half the area of a circle.
21.
24.
� 3.14(10)2
� 3.14 � 100
� 314 m2
26.
27.
29.
� 3.14(2)2
� 3.14 � 4
� 12.56 square units
A � �r2
r � �1 � ��3� � 2
� 6 square units
� 12 � 4 � 3
A �12 bh
BD � 5 � 2 � 3
AC � 5 � 1 � 4,
r � 50 ft
2r � 100
d � 100 ft
A � �r2
� 15 cm2
� 12 � 5 � 6
A �12 bh
� 100.48 square units� 12 � 3.14 � 82A �
12 ��r2�
� 2r � �r
P � 2r �12 �2�r�
20 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
12.
� 2 � 3.14 � 7
� 43.96 units
� 3.14(7)2
� 3.14 � 49
� 153.86 square units
14.
� 78.75 square units
16.
� 2 � 3.14 � 5.5
� 34.54 units
� 3.14 � (5.5)2
� 3.14 � 30.25
� 94.985 square units
A � �r2
C � 2�r
� 7.5 � 10.5
A � lw
� 36 units
� 15 � 21
� 2 � 75 � 2 � 10.5
P � 2l � 2w
A � �r2
C � 2�r
� 48 square units
� 8 � 6
A � lw
� 12.5 square units
� 255
� 12 � 5 � 5
A �12 bh
22. 23.
� 64 ft2 � 82
A � s2
� 108 yd2
� 12 � 9
A � lw
25.
� 36 m2
� 62
A � s2
s � 6 m
4s � 24
P � 24 m
� 3.14(50)2
� 3.14 � 2500
� 7850 ft2
A � �r2
28.
square units� 25� 52A � s2
HE � 3 � ��2� � 5
HG � 4 � ��1� � 5,
FG � 3 � ��2� � 5,
EF � 4 � ��1� � 5,
30.
x
B
C
A D
2
2
�2
y
� 6 square units
� 12 � 4 � 3
A �12 bh
CD � 7 � 4 � 3
AB � 7 � 3 � 4
MCRBG-0104-SK.qxd 5-25-2001 11:12 AM Page 20
Geometry 21Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
31.
33.
34.
41 square yards of carpet will be needed to cover the room.
35. The entire width of the window with frame isThe entire length of the
window with frame is
The area of the window, including the frame, is 352square inches.
36.
� 2 � 3.14 � 160
� 1004.8 m
The circumference of the covered land is 1004.8 meters.� 3.14(160)2 � 3.14 � 25,600 � 80,384 m2
The area covered is about 80,384 square meters.A � �r2
C � 2�r
r � 160 m
2r � 320 m
d � 320 m
� 352 in.2� 22 � 16A � lw
2 in. � 18 in. � 2 in. � 22 in.2 in. � 12 in. � 2 in. � 16 in.
23
25 ft �253 yd
15 ft � 5 yd
2
2
X
Y W
Z
y
x
� 5�2
� �25 � �2
� �50
� �25 � 25
� ���5�2 � 52
WX � ��0 � 5�2 � �5 � 0�2
x
S
T
R2
2
�2
y
U
� 28 square units
� 12 � 8 � 7
A �12 bh
TU � 4 � ��3� � 7
RS � 6 � ��2� � 8 37. Perimeter of Rectangle
1.
2.
3.
4.
1.
2.
3.
4.
To find the width, divide 100 by the length.
To find the area, multiply the length and the width.
To find the perimeter, use
Notice the pattern for the perimeters. The numbersdecrease to 40 then increase. The rectangle with thesmallest perimeter has dimensions of 10 m 10 m.
38.
� 3.14(5.5)2
� 3.14 � 30.25
� 94.985 m2
About 95 square meters of cranberries could be gathered.
39.
� 2 � 3.14 � 21
� 131.88 in.
Each time a bicycle tire rotates one complete time it travels a distance of 131.88 in. So to find the number ofrotations, divide the total distance of 420 inches by onecomplete rotation, 131.88 inches. The bicycle tire rotatesabout 3.18 times.
40. Area of ring � Area of larger circle � Area of smaller circle
�
� 3.14 � 42.25 � 3.14 � 25
� 132.665 � 78.5
� 54.165 in.2
The area of the ring is about 54.2 square inches.
41.
� 26 in.
� 18 � 8
� 2 � 9 � 2 � 4
P � 2l � 2w
4 � w
36 � 9w
A � lw
��132 �2
� ��102 �2
C � 2�r
A � �r2
�
P � 2l � 2w.
32.
2
4
N
R
M
L
y
x
� 45 square units
� 9 � 5
A � lw
MN � 7 � ��2� � 9
LM � 1 � ��4� � 5
� 50 square units
� 25 � 2
� �5�2�2
A � s2
� 4123 yd2
� 1253
� 5 � 253
A � lw
A B C D E F G
Length 1.00 2.00 3.00 4.00 5.00 6.00
Width 100.00 50.00 33.33 25.00 20.00 16.67
Area 100.00 100.00 100.00 100.00 100.00 100.00
Perimeter 202.00 104.00 72.67 58.00 50.00 45.33
H I J K L M
7.00 8.00 9.00 10.00 11.00 12.00
14.29 12.5 11.11 10.00 9.09 8.33
100.00 100.00 100.00 100.00 100.00 100.00
42.57 41.00 40.22 40.00 40.18 40.67
42.
� 400 m
� 4 � 100
P � 4s
100 � s
10,000 � s2
A � s2
MCRBG-0105-SK.qxd 6-14-2001 11:14 AM Page 21
Chapter 1 continued
43.
45.
47. C � 2�r
100 � 2�r
100 � 2 � 3.14 � r
100 � 6.28r
15.92 � r
A � �r2
� 3.14(15.92)2
� 3.14 � 253.4464
� 795.8 yd2
49. a. C � 2�r
� 2 � 3.14 � 20,908,800
� 131,307,264 ft
The length of the cable would be 131,307,264 feet.
b.
C � 2�r
131,307,270 � 2 � 3.14 � r
131,307,270 � 6.28r
20,908,800.96 ft � r
The radius of the circle would be about 20,908,801 ft.
c. height off ground � 20,908,801 � 20,908,800 � 1. Itwould be about 1 foot above the ground.
d. No, the answer to part (c) would not be different for adifferent planet with a different radius. By adding 6 ftto the circumference, you are only adding 6 � 2� orabout 1 to the radius. This will remain constant.
131,307,264 � 6 � 131,307,270 ft
10�2 cm � r
�100 � �2 � r
�200 � r
200 � r2
200� � �r2
A � �r2
6 ft � h
48 � 8h
48 �12 � 16 � h
A �12 bh
This is the new circumference.
Since then
Therefore, the radius changes by feet without
regard to the actual number represents.
50.
Original rectangle:
Enlarged rectangle:
The enlarged rectangle has double the perimeter of theoriginal rectangle and four times the area of the originalrectangle.
Answers may vary.
Sample answer:
Original rectangle:
Enlarged rectangle:
The perimeter of the enlarged rectangle is 32 units whichis twice the area of the original rectangle. The area of theenlarged rectangle is 60 square units which is four timesthe area of the original rectangle.
� 32 units
� 20 � 12
� 2 � 10 � 2 � 6
P � 2l � 2w
� 16 units
� 10 � 6
� 2 � 5 � 2 � 3
P � 2l � 2w
5
36
10
� 2�2l � 2w� � 4 � l � 4 � w
P � 2 � 2l � 2 � 2w
P � 2l � 2w
w2w
2
r1
62�
C2�
� r1.�C � 2�r1,� r1 �6
2�� r2
C2�
�6
2�� r2
C � 6
2�� r2
C � 6 � 2�r2
C � 2�r
22 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
44.
46. A � �r2
1 � �r2
0.32 � r2
� r
0.56 � r
d � 2r
d � 2 � 0.56
d � 1.12 m
48. 7.5 cm is the largest mea-surement, so it must gowith the longest side,which is the hypotenuseof the right triangle.
� 13.5 cm2
� 12 � 6 � 4.5
A �12 bh
�0.32
8 yd � b
104 � 13b
52 �12 � b � 13
A �12 bh
� 4�lw�
� 4 � l � w
A � 2l � 2w
A � lw
� 15 square units
� 5 � 3
A � lw
� 60 square units
� 10 � 6
A � lw
MCRBG-0105-SK.qxd 6-14-2001 11:14 AM Page 22
Geometry 23Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
Mixed Review (p. 58)
51.
53.
obtuse
Answers may vary.
Sample answer:
is in the interior of
is in the exterior of
55.
right
Answers may vary.
Sample answer:
is in the interior of
is in the exterior of
57.
58.
59. � �–52
, 32�M � ��3 � ��2�
2,
4 � ��1�2 �
� �3, 12�� �6
2,
12�M � �2 � 4
2,
�3 � 42 �
� �52
, 32�
M � �0 � 52
, 0 � 3
2 �
�DEF.��2, 3�
�DEF.�2, 0�
y
1
F
E
Dx
�1
�DEF.�4, �5�
�DEF.�4, 0�
y
x1
1
FE
D
CBA
60.
61.
62.
Quiz 3 (p. 58)
1. Let be the complement of
The complement of has a measure of
2. Let be the supplement of
The supplement of has a measure of
3. Let be the supplement of
The supplement of has a measure of
4. Let be the complement of
The complement of has a measure of
5.
m�A � 75
m�A � 5�15�
m�A � 5�m�B�
m�B � 15
6�m�B� � 90
5�m�B� � m�B � 90
m�A � m�B � 90
m�A � 5�m�B�
55.�D
m�C � 55
m�C � 35 � 90
m�C � m�D � 90
�D.�C
158.�C
m�D � 158
22 � m�D � 180
m�C � m�D � 180
�C.�D
53.�B
m�A � 53
m�A � 127 � 180
m�A � m�B � 180
�B.�A
49.�A
m�B � 49
41 � m�B � 90
m�A � m�B � 90
�A.�B
� ��19, 1�� ��382
, 22�M � ��44 � 6
2,
9 � ��7�2 �
� �7, 3�� �142
, 62�M � �0 � 14
2,
5 � 12 �
� ��92
, �3�
� ��92
, �62 �
M � ��2 � ��7�2
, 0 � ��6�
2 �52. Answers may vary.
Sample answer:
54.
acute
Answers may vary.
Sample answer:
is in the interiorof
is in the exterior of
56.
acute angle
Answers may vary.
Sample answer:
is in the interior of
is in the exterior of �DEF.�4, 0�
�DEF.��2, 0�
y
1
F
E
D
x�1
�1
1
�DEF.��4, �2�
�DEF.�3, �1�
x
1
F
E D
y
1�1
W
X
Y
Z
MCRBG-0105-SK.qxd 6-14-2001 11:14 AM Page 23
Chapter 1 continued
6. A � �r2
� 3.14(18)2
� 3.14 � 324
� 1017.36 m2
C � 2�r2
� 2 � 3.14 � 18
� 113.04 m
8.
10. First, we must find the total area of all 4 walls. There are2 walls that are 8 ft by 12 ft and 2 walls that are 8 ft by24 ft.
To find the number of rolls of wallpaper needed, dividethe total area by the number of square feet perroll So or 21 rolls of wallpaperwill be needed.
Chapter 1 Review • (pages 60–62)
1. Each number is 7 more than the previous number.
2. The numbers after the first number are found by addingconsecutive powers of 2.
3. Each number is the previous number multiplied by 3.
4.
5. If 1 is added to the product of four consecutive positiveintegers, n through the sum is equal to the squareof �n�n � 3� � 1.
n � 3,
576 � 28 � 20.6�28 ft2�.�576 ft2�
Total Area � 576 ft2Total Area � 192 � 384
Total Area � 2 � 8 � 12 � 2 � 8 � 24
� 29.2 cm
� 20 � 9.2
� 2 � 10 � 2 � 4.6
P � 2l � 2w
� 46 cm2
� 10 � 4.6
A � lw
6. Answers may vary.
Sample answer:
The cube of is which is not greater than
7. Answers may vary. 8.
Sample answer:
9. Answers may vary.
Sample answer:
10.
11.
because they have the same length.PQ QR
� 2�2
� �4 � �2
� �8
� �4 � 4
� �22 � ��2�2
QR � ��0 � ��2��2 � ��1 � 1�2
� 2�2
� �4 � �2
� �8
� �4 � 4
� �22 � ��2�2
PQ � ���2 � ��4��2 � �1 � 3�2
RP � 16
RP � 8 � 8
RP � PQ � QR
ST � 6
ST � 30 � 8 � 8 � 8
ST � PT � PQ � QR � RS
PQ � QR � RS � ST � PT
PQ � 8
PQ �12 � 16
PQ �12 � QS
PQ � QR
CB
D
E
A
12.1
812
24 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
7.
� 71.5 in.2 � 1
2 � 13 � 11
A �12 bh
9.
2
2
y
x
R(�1, 12)
P (�3, 4)T (�1, 4)
Q(7, 4)
� 40 square units
� 12 � 10 � 8
A �12 bh
RT � 12 � 4 � 8
PQ � 7 � ��3� � 10
MCRBG-0105-SK.qxd 6-14-2001 11:14 AM Page 24
Geometry 25Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
12.
and are not congruent because they do not havethe same length.
13.
and are not congruent because they do not havethe same length.
14. straight
16. acute
17.
18.
19.
20.
21. � �1, 2�� �22
, 42�M � ��1 � 3
2,
7 � ��3�2 �
� ��4, 3�� � �82
, 62�M � �0 � ��8�
2,
0 � 62 �
m�QNM � 70
m�QNM � 110 � 180
m�QNM � m�QNP � m�MNP
m�HJL � 50
m�HJL � 40 � 90
m�HJL � m�LJK � m�HJK
m�DEF � 105
m�DEF � 60 � 45
m�DEF � m�DEG � m�GEF
Y
45°
LK M
QRPQ
� �10
� �1 � 9
� �12 � 32
QR � ��1 � 0�2 � �4 � 1�2
� �13
� �4 � 9
� �22 � 32
PQ � ��0 � ��2��2 � �1 � ��2��2
QRPQ
� �13
� �9 � 4
� �32 � ��2�2
QR � ��4 � 1�2 � �1 � 3�2
� 2�5
� �4 � �5
� �20
� �16 � 4
� �42 � ��2��2
PQ � ��1 � ��3��2 � �3 � 5�2
22.
23.
24.
25.
26. always 27. sometimes 28. never 29. sometimes
30.
32. To find the perimeter, find thesum of AB, BC, and CA.
The perimeter of ABC is or 18 units.
The area of ABC is 12 square units.
33.
The perimeter of the garden is 56 ft.
� 56 ft
� 4�14�
P � 4s
�
� 12 square units
� 12 � 8 � 3
A �12 bh
CD � 0 � ��3� � 3
8 � 5 � 5�
� 5
� �25
� �16 � 9
� ���4�2 � 32
CA � ���6 � ��2��2 � �0 � ��3��2
� 5
� �25
� �16 � 9
� ���4�2 � ��3�2
BC � ���2 � 2�2 � ��3 � 0�2
AB � 2 � ��6� � 8
� 45 cm2
� 10 � 4.5
A � lw
� 29 cm
� 20 � 9
� 2 � 10 � 2 � 4.5
P � 2l � 2w
m�PQR � 92
m�PQR � 2�46�
m�PQR � 2�m�PQS�
m�SQR � m�PQS � 46
m�RQS � m�SQP �m�PQR
2�
50
2� 25
m�PQR � 100
m�PQR � 2�50�
m�PQR � 2�m�PQS�
m�SQR � m�PQS � 50
� ��5, 12�� ��10
2,
12�M � ��12 � 2
2,
�9 � 102 �
2
1
y
x
C (�2, �3)
A (�6, 0)D (�2, 0)
B(2, 0)15. obtuse
A
150°
31.
� 254.34 in.2 � 3.14 � 81
� 3.14�9�2
A � �r2
� 56.52 in.
� 2 � 3.14 � 9
C � 2�r
MCRBG-0105-SK.qxd 6-14-2001 11:14 AM Page 25
Chapter 1 continued
26 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 Test (p. 63)
1. Sample answer:
Q, T, and N
3. Sample answer:
and
5. 6. 7.
8.
10.
11.
12.
13.
14. Sample answer:
is an obtuse angle. is an acute angle.is right angle. and are complemen-
tary angles.
15.
16.
The coordinates of point S are �0, 7�.
� �0, 7�
� �02
, 142 �
S � ��3 � 32
, 8 � 6
2 � � 15
� 4 � 11
QR � 4 � w
w � 11
3w � 33
3w � 1 � 34
2w � 3 � 4 � w � 34
PQ � QR � PR
�EBD�FBE�FBD�DBC�ABD
m�DBA � 130�
m�DBA � 50� � 180�
m�DBA � m�DBC � 180�
m�ABF � 40�
m�ABF � 140� � 180�
m�ABF � m�FBC � 180�
� 140�
m�FBC � 45� � 45� � 50�
m�FBC � m�FBE � m�EBD � m�DBC
m�DBE � 45�
m�DBE � 45� � 90�
m�DBE � m�EBF � m�DBF
NR � 14
12 � NR � 26
4 � 8 � NR � 26
SM � MN � NR � SR
SM � MP � 4� 4�12 �8�MP �
12 �MN�
↔QL
→TN
→TQ
So
17.
18.
19.
20. The distance is 6 more than twice the figure number. Forthe 20th figure, the distance is 2(20) � 6.
2(20) � 6 � 40 � 6
� 46
The distance around the 20th figure is 46 units.
21.
They would have walked about 3517 feet.
22.
The area of the watered region is about 984,704 squarefeet.
Chapter 1 Standardized Test (pp. 64–65)
1. C 2. E
� 984,704
� 3.14 � 313,600
� 3.14�560�2
A � �r2
� 3517
� 2 � 3.14 � 560
C � 2�r
� 65�
� 12 � 130�
m�PQT �12�m�PQR�
m�5 � m�3 � 68�
m�4 � 112�
m�4 � 68� � 180
m�4 � m�3 � 180�
RS � ST.
� �10
� �9 � 1
� �32 � ��1�2
ST � ��3 � 0�2 � �6 � 7�2
� �10
� �9 � 1
� �32 � ��1�2
RS � ��0 � ��3��2 � �7 � 8�2
2. Sample answer:
Q, N, M, and R
4. Sample answer:
and ↔LQ
↔QN
9.
� 22
� 8 � 14
MR � MN � NR
� 19
� 22 � 3
� 2 � 11 � 3
PQ � 2w � 3
figure 1 2 3 4 5
distance 8 10 12 14 16
MCRBG-0106-SK.qxd 5-25-2001 11:11 AM Page 26
Geometry 27Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
3.
E
4.
D
5. 6. A
B
7.
C
8.
E
m�PQR � 78�
m�PQR � 119 � 41
m�PQR � 7 � 17 � 41
m�PQR � 7x � 41
m�PQR � 5x � 46 � 2x � 5
m�PQR � m�PQS � m�SQR
x � 17
3x � 51
5x � 2x � 51
�5x � 46�� � �2x � 5��
m�PQS � m�SQR
C��19, �40�
x � �19
x � 1 � �20
x � ��1�
2� �10
C�x, y� B��1, 8� M��10, �16�
m�5 � 71�
19� � m�5 � 90�
m�4 � m�5 � 90�
CE � 13
15 � CE � 28
5 � 5 � 5 � CE � 28
AB � BD � DC � CE � AE
AB � BD � DC �12 �BC� �
12 � 10 � 5
AC � AB
� �73
� �64 � 9
� �82 � 32
AB � ��6 � ��2��2 � �7 � 4�2
� �85
� �49 � 36
� ���7�2 � 62
AE � ���9��2��2 � �10 � 4�2
� �73
� �9 � 64
� �32 � ��8�2
AC � ��1 � ��2��2 � ��4 � 4�2 9.
C
10.
C
11. Sample answers:
a. is an acute angle.
b. is an obtuse angle.
c. is a straight angle.
d. is a right angle.
12. a. supplementary angles
b. complementary angles
c. supplementary angles
d. vertical angles
13.
14.
m�EFN � �180 � x��
m�EFN � m�QFA
m�AFE � x�
m�AFE � m�QFN
m�QFA � �180 � x��
m�QFA � x� � 180�
m�QFA � m�QFN � 180�
m�GAH � m�BAH �m�GAB
2�
90�
2� 45�
�GAP
�SBE
�FEB
�GAH
1
1
y
x
C (�3, 2)
D (�6, 2)
A (�6, 7)
B (�6, �1)
� 12 square units
� 12 � 8 � 3
A �12 bh
CD � �3 � ��6� � 3
AB � 7 � ��1� � 8
m�1 � 81�
m�1 � 9�9��
m�1 � 9�m�2�
m�2 � 9�
10�m�2� � 90�
9�m�2� � m�2 � 90�
m�1 � m�2 � 90�
m�1 � 9�m�2�
y � �40
y � 8 � �32
y � 8
2� �16
1
1
y
x
D (0, 4) E (6, 4)
F (6, 0)
� 12 square units
� 12 � 6 � 4
A �12 bh
EF � 4 � 0 � 4
DE � 6 � 0 � 6
MCRBG-0106-SK.qxd 5-25-2001 11:11 AM Page 27
Chapter 1 continued
28 GeometryChapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
15.
16. The 6 in. by 6 in. rectangle had the greatest area.
5 in. � l
10 � 2l
24 � 2l � 14
24 � 2l � 2 � 7
P � 2l � 2w
w � 7
� 35 in.2 � 7 � 5
A � lw
7 in. � l
14 � 2l
24 � 2l � 10
24 � 2l � 2 � 5
P � 2l � 2w
w � 5
� 27 in.2 � 9 � 3
A � lw
9 in. � l
18 � 2l
24 � 2l � 6
24 � 2l � 2 � 3
P � 2l � 2w
w � 3
� 11 in.2 � 11 � 1
A � lw
11 in. � l
22 � 2l
24 � 2l � 2
24 � 2l � 2 � 1
P � 2l � 2w
w � 1
17. The sum of the length and width is 12. The length of arectangle with a width of 3.5 inches would be or 8.5 inches.
18. If the perimeter of a rectangle is known, the rectanglewith the greatest area is a square with the length of a sideequal to of the perimeter. To test the conjecture, onecould try many rectangles of different perimeters andmake a chart as in problem 15. Or one could try to makea generalized chart with 4n as a perimeter.
14
12 � 3.5Width (in.) Perimeter (in.) Length (in.) Area
1 24 11 11
2 24 10 20
3 24 9 27
4 24 8 32
5 24 7 35
6 24 6 36
7 24 5 35
�in.2
� 36 in.2 � 6 � 6
A � lw
6 in. � l
12 � 2l
24 � 2l � 12
24 � 2l � 2 � 6
P � 2l � 2w
w � 6
� 32 in.2 � 8 � 4
A � lw
8 in. � l
16 � 2l
24 � 2l � 8
24 � 2l � 2 � 4
P � 2l � 2w
w � 4
� 20 in.2 � 10 � 2
A � lw
10 in. � l
20 � 2l
24 � 2l � 4
24 � 2l � 2 � 2
P � 2l � 2w
w � 2
� 35 in.2 � 5 � 7
A � lw
MCRBG-0106-SK.qxd 5-25-2001 11:11 AM Page 28
Geometry 29Chapter 1 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 1 continued
Chapter 1 Project (pp. 66–67)
1. No, the distances aren’t always the same. In the exampleon page 66, 3 different distances were shown. The short-est distance from A to B is 10.
2. Yes, there are other paths from A to B with a taxicab dis-tance of 10.
Here are a few.
3. If point A is then point B is
Since then the taxicab distance is greaterthan the Euclidean distance.
4. Answers may vary.Sample answer:
Taxicab distance for AB is 11.
The taxicab distance is greater than the Euclidean distance.
The taxicab distance AB is 19.
The taxicab distance isgreater than the Euclidean distance.
The taxicab distance willalways be greater than orequal to the Euclideandistance.
5. If and are two points, then the taxicabdistance from A to B is x2 � x1 � y2 � y1.
B�x2, y2�A�x1, y1�
A(0, 0)
B(9, 10)19
19
� 13.5 � �181
� �81 � 100
� �92 � 102
AB � ��9 � 0�2 � �10 � 0�2
� 8.5 � �73
� �64 � 9
AB � ��8 � 0�2 � �3 � 0�2
A(0, 0)
B(8, 3)
10 > 5�2,
� 5�2 � 7.07
� �25 � �2
� �50
� �25 � 25
� �52 � 52
AB � ��5 � 0�2 � �5 � 0�2
�5, 5�.�0, 0�,
A
B
6. The points are arranged like points on a diamond or squarecentered around the point
7. Total distance blocks
8. diameter blocks
9. Yes, would have a constant value of or 4.
Present Your Results:
A distance in taxicab geometry is greater than or equal to thecorresponding Euclidean distance. A circle is associated witha location and a distance in both Euclidean and taxicab geom-etry, but the circles have different shapes and circumferencesin the two geometries. Their diameters, however, are thesame: twice the given distance.
Extensions:
The answers below assume that A and B do not lie on ahorizontal or verical line.
The points that lie between two points A and B form arectangle with A and B at two corners. All the grid points onthe perimeter of the rectangle (except A and B) and all thegrid points inside the rectangle lie between A and B in taxicabgeometry.
Yes; the set of points lies on a line that is tilted 45° fromhorizontal. The line divides the rectangular region of pointsbetween A and B into two regions. All the points in oneregion are closer to A than to B. All the points in the otherregion are closer to B than to A.
328�
O(0, 0)44
� 2 � 4 � 8
O(0, 0)
88
8 8
� 4 � 8 � 32
O�0, 0�.
O(0, 0)
MCRBG-0106-SK.qxd 5-25-2001 11:11 AM Page 29
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