chapter 10 - north allegheny school district / district ... 10continued 10.2 guided practice (p....
TRANSCRIPT
CHAPTER 10
Geometry, Concepts and Skills 163Chapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter Opener
Chapter Readiness Quiz (p. 536)
1. B; 202 � 212 � 2922. G; x� � 60� � 90�
400 � 441 � 841 x � 30
841 � 841
3. B; 52 � z2 � 92
25 � z2 � 81
z2 � 56
z � �5�6�
� 7.5
Lesson 10.1
10.1 Checkpoint (p. 538)
1. �2�7� � 5.2;This is reasonable because 27 is between the perfectsquares 25 and 36. So, �2�7� should be between �2�5� and�3�6�, or 5 and 6. The answer 5.2 is between 5 and 6.
2. �4�6� � 6.8;This is reasonable because 46 is between the perfectsquares 36 and 49. So, �4�6� should be between �3�6� and�4�9�, or 6 and 7. The answer 6.8 is between 6 and 7.
3. �8� � 2.8;This is reasonable because 8 is between the perfectsquares 4 and 9. So, �8� should be between �4� and �9�,or 2 and 3. The answer 2.8 is between 2 and 3.
4. �9�7� � 9.8;This is reasonable because 97 is between the perfectsquares 81 and 100. So, �9�7� should be between �8�1� and�1�0�0�, or 9 and 10. The answer 9.8 is between 9 and 10.
5. �3� � �5� � �3� �� 5� 6. �1�1� � �6� � �1�1� �� 6�
� �1�5� � �6�6�
7. �3� � �2�7� � �3� �� 2�7� 8. 5�3� � �3� � 5�3� �� 3�
� �8�1� � 5�9�
� 9 � 5 � 3
� 15
9. �2�0� � �4� �� 5� 10. �8� � �4� �� 2�
� �4� � �5� � �4� � �2�
� 2�5� � 2�2�
11. �7�5� � �2�5� �� 3� 12. �1�1�2� � �1�6� �� 7�
� �2�5� � �3� � �1�6� � �7�
� 5�3� � 4�7�
10.1 Guided Practice (p. 539)
1. The radicand in the expression �2�5� is 25.
2. D; �3�6� � 6 3. A; �3� � �2� � �3� �� 2� � �6�
4. B; �3� � �6� � �3� �� 6� � �1�8� � �9� �� 2� � �9� � �2� � 3�2�
5. C; �3�2� � �1�6� �� 2� � �1�6� � �2� � 4�2�
6. a2 � b2 � c2
22 � 42 � c2
4 � 16 � c2
20 � c2
�2�0� � c
�4� �� 5� � c
�4� � �5� � c
2�5� � c
7. c � 2�5� � 4.5 8. �4�9� � 7
9. �2�8� � �4� �� 7� 10. �7�2� � �3�6� �� 2�
� �4� � �7� � �3�6� � �2�
� 2�7� � 6�2�
11. �5�4� � �9� �� 6�
� �9� � �6�
� 3�6�
10.1 Practice and Applications (pp. 539–541)
12. �1�3� � 3.6;This is reasonable because 13 is between the perfectsquares 9 and 16. So, �1�3� should be between �9� and�1�6�, or 3 and 4. The answer 3.6 is between 3 and 4.
13. �6� � 2.4;This is reasonable because 6 is between the perfectsquares 4 and 9. So, �6� should be between �4� and �9�, or2 and 3. The answer 2.4 is between 2 and 3.
14. �91� � 9.5;This is reasonable because 91 is between the perfectsquares 81 and 100. So, �9�1� should be between �8�1�and �1�0�0�, or 9 and 10. The answer 9.5 is between 9and 10.
15. �3�4� � 5.8;This is reasonable because 34 is between the perfectsquares 25 and 36. So, �3�4� should be between �2�5� and�3�6�, or 5 and 6. The answer 5.8 is between 5 and 6.
16. �10�6� � 10.3;This is reasonable because 106 is between the perfectsquares 100 and 121. So, �1�0�6� should be between�1�0�0� and �1�2�1�, or 10 and 11. The answer 10.3 isbetween 10 and 11.
Chapter 10 continued
17. �1�4�8� � 12.2;This is reasonable because 148 is between the perfectsquares 144 and 169. So, �1�4�8� should be between �1�4�4�and �1�6�9�, or 12 and 13. The answer 12.2 is between 12and 13.
18. �6�2� � 7.9;This is reasonable because 62 is between the perfectsquares 49 and 64. So, �6�2� should be between �4�9� and�6�4�, or 7 and 8. The answer 7.9 is between 7 and 8.
19. �1�8�6� � 13.6;This is reasonable because 186 is between the perfectsquares 169 and 196. So, �1�8�6� should be between �1�6�9�and �1�9�6�, or 13 and 14. The answer 13.6 is between 13and 14.
20. a2 � b2 � c2
��2��2 � ��5��2 � c2
2 � 5 � c2
7 � c2
�7� � c
21. a2 � b2 � c2
��1�3��2 � 62 � c2
13 � 36 � c2
49 � c2
�4�9� � c
7 � c
22. a2 � b2 � c223. 42 � 12 � x2
��1�9��2 � ��1�0��2 � c2 16 � 1 � x2
19 � 10 � c2 17 � x2
29 � c2 �1�7� � x
�2�9� � c 4.1 � x
24. x2 � ��1�1��2 � 6225. x2 � ��5��2 � ��1�3��2
x2 � 11 � 36 x2 � 5 � 13
x2 � 25 x2 � 8
x � �2�5� x � �8�x � 5 x � 2.8
26. �7� � �2� � �7� �� 2�� �1�4�
27. �5� � �5� � �5� �� 5� 28. �3� � �1�1� � �3� �� 1�1�� �2�5� � �3�3�� 5
29. 2�5� � �7� � 2�5� �� 7� 30. �1�0� � 4�3� � 4�3� �� 1�0�� 2�3�5� � 4�3�0�
31. �1�1� � �2�2� � �1�1� �� 2�2� 32. �6�5��2 � 6�5� � 6�5�� �2�4�2� � 6 � 6 � �5� � �5�� �1�2�1� �� 2� � 36 � 5
� �1�2�1� � �2� � 180
� 11�2�
33. �5�3��2 � 5�3� � 5�3� 34. �7�2��2 � 7�2� � 7�2�� 5 � 5 � �3� � �3� � 7 � 7 � �2� � �2�� 25 � 3 � 49 � 2
� 75 � 98
35. �1�8� � �9� �� 2� 36. �5�0� � �2�5� �� 2�� �9� � �2� � �2�5� � �2�� 3�2� � 5�2�
37. �4�8� � �1�6� �� 3� 38. �6�0� � �4� �� 1�5�� �1�6� � �3� � �4� � �1�5�� 4�3� � 2�1�5�
39. �5�6� � �4� �� 1�4� 40. �1�2�5� � �2�5� �� 5�� �4� � �1�4� � �2�5� � �5�� 2�1�4� � 5�5�
41. �2�0�0� � �1�0�0� �� 2� 42. �1�6�2� � �8�1� �� 2�� �1�0�0� � �2� � �8�1� � �2�� 10�2� � 9�2�
43. �4�4� � �4� �� 1�1�� �4� � �1�1�� 2�1�1�
44. Yes, the expression can be simplified further.
�8�0� � �4� �� 2�0�� �4� � �2�0�� 2�2�0�� 2�4� �� 5�� 2�4� � �5�� 2 � 2 � �5�� 4�5�
45. No, the expression cannot be simplified further.
46. A � lw 47. A � lw
� �1�4� � �1�0� � 8�3� � 2�5�� �1�4� �� 1�0� � 8 � 2 � �3� � �5�� �1�4�0� � 16�3� �� 5�� �4� �� 3�5� � 16�1�5�� �4� � �3�5� � 62.0
� 2�3�5�� 11.8
48. A � lw 49. A � lw
� 9�2� � 4�6� � 4�3� � 4�3�� 9 � 4 � �2� � �6� � 4 � 4 � �3� � �3�� 36�2� �� 6� � 16 � 3
� 36�1�2� � 48
� 36�4� �� 3�� 36�4� � �3�� 72�3�� 124.7
164 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
50. A � lw 51. A � lw
� 8�3� � 4�6� � 3�2� � �1�4�
� 8 � 4 � �3�� �6� � 3�2� �� 1�4�
� 32�3� �� 6� � 3�2�8�
� 32�1�8� � 3�4� �� 7�
� 32�9� �� 2� � 3�4� � �7�
� 32�9� � �2� � 3 � 2 � �7�
� 32 � 3 � �2� � 6�7�
� 96�2� � 15.9
� 135.8
52. A � �14
�s2�3�
� �14
�(30)2�3�
� �14
�(900)�3�
� 225�3�
� 389.7 ft2
53. (1 m)2 � (100 cm)2 � 10,000 cm;
A � �14
�s2�3�
10,000 � �14
�s2�3�
�40
�,03�00
� � s2
23,094 � s2
�2�3�,0�9�4� � s
152 cm � s
10.1 Standardized Test Practice (p. 541)
54. C; �1�6�9� � 13 55. H; �2�2�0� � 14.8, 14 � 14.8 � 15
56. D; a2 � b2 � c2
�2�1�0��2 � 82 � c2
40 � 64 � c2
104 � c2
�1�0�4� � c
�4� �� 2�6� � c
�4� � �2�6� � c
2�2�6� � c
10.2 � c
10.1 Mixed Review (p. 541)
57. 45� � ma1 � 90�
ma1 � 45�
58. 180� � 61� � 51� � ma1
180� � 112� � ma1
68� � ma1
59. 180� � 87� � 25� � ma1
180� � 112� � ma1
68� � ma1
60. 9 � x � 4 61. 64� � 4x� 62. 2x � x � 3
13 � x 16 � x x � 3
10.1 Algebra Skills (p. 541)
63. x(x � 5) � x2 � 5x 64. 4(2x � 1) � 8x � 4
65. x(3x � 4) � 3x2 � 4x 66. 5x(x � 2) � 5x2 � 10x
67. �3(1 � x) � �3 � 3x
68. 2x � (x � 6) � 2x � x � 6
� x � 6
Lesson 10.2
10.2 Geo-Activity (p. 542)
2. 45�, 45�, 90� 3. Answers may vary.
4. Answers may vary, but will be �2� � 1.4 times the answerin Step 3.
5. The answer should be the same as in Step 4.
10.2 Checkpoint (pp. 543–544)
1. hypotenuse � leg � �2� 2. hypotenuse � leg � �2�
x � 4�2� x � 5�2�
3. hypotenuse � leg � �2� 4. hypotenuse � leg � �2�
3�2� � x�2� 6�2� � x�2�
�3��2�2�
� � �x��2�2�
� �6��
2�2�
� � �x��2�2�
�
3 � x 6 � x
5. The triangle is an isosceles right triangle. By the BaseAngles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, x� � x� � 90� � 180�. So, 2x� � 90�, and x � 45. So, the triangle is a 45�– 45�– 90� triangle. By the 45�– 45�– 90�Triangle Theorem, hypotenuse � leg � �2�
8 � x�2�
��82�� � x
5.7 � x.
6. The triangle is an isosceles right triangle. By theConverse of the Base Angles Theorem its sides oppositethe congruent angles are congruent. By the TriangleSum Theorem, x� � x� � 90� � 180�. So, 2x� � 90�,and x � 45. So, the triangle is a 45�– 45�– 90� triangle. By the 45�– 45�– 90� Triangle Theorem,hypotenuse � leg � �2�
12 � x�2�
��12
2�� � x
8.5 � x.
Geometry, Concepts and Skills 165Chapter 10 Worked-Out Solution Key
Chapter 10 continued
10.2 Guided Practice (p. 545)
1. An isosceles right triangle has 2 congruent sides.
2. An isosceles right triangle has 2 congruent angles. Themeasures of the angles are 45�, 45�, and 90�.
3. hypotenuse � leg � �2�
x � 6�2�
4. hypotenuse � leg � �2�
2�2� � x�2�
2 � x
5. hypotenuse � leg � �2�
x � �6� � �2�
� �1�2�
� �4� � �3�
� 2�3�
10.2 Practice and Applications (pp. 545–547)
6. hypotenuse � leg � �2� 7. hypotenuse � leg � �2�
� 2�2� � 7�2�
8. hypotenuse � leg � �2�
� 8�2�
9. hypotenuse � leg � �2� 10. hypotenuse � leg � �2�
� �5� � �2� � �3� � �2�
� �1�0� � �6�
11. hypotenuse � leg � �2� 12. hypotenuse � leg � �2�
� �1�0� � �2� 10�2� � x�2�
� �2�0� 10 � x
� �4� � �5�
� 2�5�
13. hypotenuse � leg � �2� 14. hypotenuse � leg � �2�
4�2� � x�2� 5�2� � x�2�
4 � x 5 � x
15. hypotenuse � leg � �2� 16. hypotenuse � leg � �2�
�2� � x�2� 8�2� � x�2�
1 � x 8 � x
17. hypotenuse � leg � �2� 18. hypotenuse � leg � �2�
14�2� � x�2� x � (1.4) � �2�
14 � x x � 2 cm
19. No; you cannot determine the measures of the other twoangles.
20. Yes; the triangle has congruent acute angles, so 2x� � 90and x � 45. So, the triangle is a 45�– 45�– 90� triangle.
21. No; the triangle is isosceles, but there is no informationabout the angle measures.
22. The triangle is an isosceles right triangle. By the BaseAngles Theorem, its acute angles are congruent. By theTriangle Sum Theorem, x� � x� � 90� � 180�. So,2x� � 90�, and x � 45. Since the measure of each acuteangle is 45�, the triangle is a 45�– 45�– 90� triangle.By the 45�– 45�– 90� Triangle Theorem,hypotenuse � leg � �2�
4 � x�2�
��42�� � x
2.8 � x.
23. The right triangle has congruent acute angles. By theTriangle Sum Theorem, x� � x� � 90� � 180�. So,2x� � 90�, and x � 45. Since the measure of each acuteangle is 45�, the triangle is a 45�– 45�– 90� triangle.By the 45�– 45�– 90� Triangle Theorem,hypotenuse � leg � �2�
9 � x�2�
��92�� � x
6.4 � x.
24. By the Triangle Sum Theorem, 45� � 45� � x� � 180�.So, 90� � x� � 180�, and x � 90. Therefore the triangle isa 45�– 45�– 90� triangle. By the 45�– 45�– 90�Triangle Theorem, hypotenuse � leg � �2�
32 � x�2�
��32
2�� � x
22.6 � x.
25. The right triangle has congruent acute angles. By theTriangle Sum Theorem, x� � x� � 90� � 180�. So,2x� � 90�, and x � 45. Since the measure of each acuteangle is 45�, the triangle is a 45�– 45�– 90� triangle.By the 45�– 45�– 90� Triangle Theorem,hypotenuse � leg � �2�
8 � y�2�
��82�� � y
5.7 � y.
26. The triangle is an isosceles right triangle. By the BaseAngles Theorem, its acute angles are congruent. By theTriangle Sum Theorem, x� � x� � 90� � 180�. So,2x� � 90�, and x � 45. Since the measure of each acuteangle is 45�, the triangle is a 45�– 45�– 90� triangle.By the 45�– 45�– 90� Triangle Theorem,hypotenuse � leg � �2�
20 � x�2�
��20
2�� � x
14.1 � x.
166 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
27. By the Triangle Sum Theorem, x� � 45� � 90� � 180�.So, x� � 135� � 180�, and x � 45. Therefore the triangleis a 45�– 45�– 90� triangle. By the 45�– 45�– 90�Triangle Theorem, hypotenuse � leg � �2�
35 � x�2�
��35
2�� � x
24.7 � x.
28. T ADB, T ACD, and T BCD; since D is on the perpen-dicular bisector of AB**** , AD**** � DB**** by the PerpendicularBisector Theorem. maADB � 90�, so T ADB is anisosceles right triangle. By the Base Angles Theorem,aA � aB. By the Triangle Sum Theorem,
x� � x� � 90� � 180�.
2x � 90
x � 45.
Since maA � maB � 45�, T ADB is a 45�– 45�– 90�triangle. maACD � 90�, so by the Triangle Sum Theorem,
maADC � 180� � maACD � maA
� 180� � 90� � 45�
� 45�.
So, T ACD is a 45�– 45�– 90� triangle. maBCD � 90�,so by the Triangle Sum Theorem,
maBDC � 180� � maBCD � maB
� 180� � 90� � 45�
� 45�.
So, T BCD is a 45�– 45�– 90� triangle.
29. Lengths may vary, but AC, CB, and CD are all equal.Since T ACD and T BCD are 45�– 45�– 90� triangles,they are isosceles triangles by the Converse of the BaseAngles Theorem. Therefore, AC � CD and CD � BC.
30. Lengths may vary, but AD � DB and each is equal to �2� � 1.4 times the length in Exercise 29.
31. If the hypotenuse has length 5�2�, the legs should have
length �5��2�2�
�, or 5. If the legs have length �5�, then the
hypotenuse has length �5� � �2�, or �1�0�.
32. r2 � 12 � 12 � 1 � 1 � 2
r � �2�;
s2 � ��2��2 � 12 � 2 � 1 � 3
s � �3�;
t2 � ��3��2 � 12 � 3 � 1 � 4
t � �4� � 2;
u2 � 22 � 12 � 4 � 1 � 5
u � �5�;
v2 � ��5��2 � 12 � 5 � 1 � 6
v � �6�;
w2 � ��6��2 � 12 � 6 � 1 � 7
w � �7�
33. The right triangle with legs of length 1 and hypotenuse oflength r � �2� is a 45�– 45�– 90� triangle.
10.2 Standardized Test Practice (p. 547)
34. a. 3x� � 6x� � 3x� � 180�
12x� � 180�
x � 15;
maA � 3x� � (3 � 15)� � 45�;
maB � 3x� � (3 � 15)� � 45�;
maC � 6x� � (6 � 15)� � 90�
b. b � 12; c � 12�2�
c. Check using the Pythagorean Theorem.
122 � 122 � �12�2��2
144 � 144 � 144 � 2
288 � 288
10.2 Mixed Review (p. 547)
35. a2 � b2 � c236. a2 � b2 � c2
242 � 102 � 262 92 � 142 � 162
676 � 676 277 256
The triangle is right. The triangle is acute.
37. a2 � b2 � c2
92 � 402 � 412
1681 � 1681
The triangle is right.
38. �2�4� � �4� �� 6� 39. �6�3� � �9� �� 7�
� �4� � �6� � �9� � �7�
� 2�6� � 3�7�
40. �5�2� � �4� �� 1�3� 41. �6�4� � 8
� �4� � �1�3�
� 2�1�3�
42. �8�0� � �1�6� �� 5� 43. �1�9�6� � 14
� �1�6� � �5�
� 4�5�
44. �2�5�0� � �2�5� �� 1�0� 45. �1�1�7� � �9� �� 1�3�
� �2�5� � �1�0� � �9� � �1�3�
� 5�1�0� � 3�1�3�
10.2 Algebra Skills (p. 547)
46. �190� � 0.9 47. �
35
� � 0.6
48. �23
� � 0.6� � 0.67 49. �13030
� � 0.33
50. �49
� � 0.4� � 0.44 51. �230� � 0.15
52. �4570� � 0.94 53. �
16
� � 0.16� � 0.17
Geometry, Concepts and Skills 167Chapter 10 Worked-Out Solution Key
Chapter 10 continued
Lesson 10.3
10.3 Activity (p. 548)
1.
2.
3.
4. Lengths will vary, but AC � 2 � AD and CD � �3� � AD � 1.73 � AD.
5. The ratio �AA
DC� is 2:1 and
the ratio �CAD
D� is approximately 1.73:1.
10.3 Checkpoint (pp. 550–551)
1. hypotenuse � 2 � shorter leg
x � 2 � 7
x � 14
2. longer leg � shorter leg � �3�
x � 3�3�
3. longer leg � shorter leg � �3�
x � 10�3�
4. longer leg � shorter leg � �3�
6 � x�3�
��63�� � x
3.5 � x
5. hypotenuse � 2 � shorter leg
42 � 2 � x
21 � x;
longer leg � shorter leg � �3�
y � 21�3�
� 36.4
10.3 Guided Practice (p. 552)
1. Two special right triangles are 45�– 45�– 90� triangle and30�– 60�– 90� triangle.
2. true 3. false 4. false
5. true 6. true 7. true
8. hypotenuse � 2 � shorter leg
x � 2 � 5
x � 10;
longer leg � shorter leg � �3�
y � 5�3�
9. hypotenuse � 2 � shorter leg
8 � 2 � h
4 � h
10. hypotenuse � 2 � shorter leg
4 � 2 � a
2 � a;
longer leg � shorter leg � �3�
b � 2�3�
10.3 Practice and Applications (pp. 552–554)
11. hypotenuse � 2 � shorter leg
x � 2 � 3
x � 6
12. hypotenuse � 2 � shorter leg
x � 2 � 8
x � 16
13. longer leg � shorter leg � �3�
11�3� � shorter leg � �3�
11 � shorter leg
hypotenuse � 2 � shorter leg
x � 2 � 11
x � 22
14. hypotenuse � 2 � shorter leg
x � 2 � 9
x � 18
15. hypotenuse � 2 � shorter leg
x � 2 � 1
x � 2
16. longer leg � shorter leg � �3�
12�3� � shorter leg � �3�
12 � shorter leg
hypotenuse � 2 � shorter leg
x � 2 � 12
x � 24
17. longer leg � shorter leg � �3�
x � 4�3�
18. longer leg � shorter leg � �3�
x � 6�3�
19. longer leg � shorter leg � �3�
x � 13�3�
20. longer leg � shorter leg � �3�
x � 20�3�
168 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
AC AD CD �AA
DC� �
AC
DD�
10 5 8.7 2 1.74
20 10 17.3 2 1.73
50 25 43.3 2 1.732
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
21. longer leg � shorter leg � �3�
x � 16�3�
22. longer leg � shorter leg � �3�
x � 2�3� � �3�
� 2�9�
� 2 � 3
� 6
23. longer leg � shorter leg � �3�
4 � x�3�
��43�� � x
2.3 � x
24. longer leg � shorter leg � �3�
7 � x�3�
��73�� � x
4.0 � x
25. longer leg � shorter leg � �3�
18 � x�3�
��18
3�� � x
10.4 � x
26. longer leg � shorter leg � �3�
10 � x�3�
��10
3�� � x
5.8 � x
27. longer leg � shorter leg � �3�
14�3� � x�3�
14 � x
28. longer leg � shorter leg � �3�
12 � x�3�
��12
3�� � x
6.9 � x
29. hypotenuse � 2 � shorter leg
60 � 2 � h
30 ft � h
30. hypotenuse � 2 � shorter leg
10 � 2 � x
5 � x;
longer leg � shorter leg � �3�
y � 5�3�
31. hypotenuse � 2 � shorter leg
8 � 2x
4 � x;
longer leg � shorter leg � �3�
y � 4�3�
32. hypotenuse � 2 � shorter leg
12 � 2 � x
6 � x;
longer leg � shorter leg � �3�
y � 6�3�
33. hypotenuse � 2 � shorter leg
9 � 2 � x
�92
� � x;
longer leg � shorter leg � �3�
y � �92
��3�
34. hypotenuse � 2 � shorter leg
80 � 2x
40 � x;
longer leg � shorter leg � �3�
y � 40�3�
35. hypotenuse � 2 � shorter leg
15 � 2x
�125� � x;
longer leg � shorter leg � �3�
y � �125��3�
36. hypotenuse � 2 � shorter leg
18 � 2 � h
9 in. � h
Your shoulders should be lifted to a height of 9 inches.
37. Divide the triangle into two 30�– 60�– 90� triangles. Thelength of the shorter leg of each triangle is 15 feet. Thelength of the longer leg of each triangle is 15�3� feet, bythe 30�– 60�– 90� Triangle Theorem.
Area � �12
�bh � �12
� � 30 � 15�3� � 390 ft2
38. Divide the triangle into two 30�– 60�– 90� triangles. Thelength of the shorter leg of each triangle is 9 inches. Thelength of the longer leg of each triangle is 9�3� inches, bythe 30�– 60�– 90� Triangle Theorem.
Area � �12
�bh � �12
� � 18 � 9�3� � 140 in.2
Geometry, Concepts and Skills 169Chapter 10 Worked-Out Solution Key
Chapter 10 continued
39. Divide the triangle into two 30�– 60�– 90� triangles. The length of the shorter leg of each triangle is 3.5 centimeters. The length of the longer leg of each triangleis 3.5�3� centimeters, by the 30�– 60�– 90� TriangleTheorem.
Area � �12
�bh � �12
� � 7 � 3.5�3� � 21 cm2
40. The area of the triangle is given by Area � �12
�bh.
If an equilateral triangle with sides of length s is dividedinto two 30�– 60�– 90� triangles, then the length of the
base is s and the height is �2s
��3�. Then
Area � �12
�bh � �12
� � s � �2s
��3� � �14
� � s2 � �3�.
41. hypotenuse � 2 � shorter leg
� 2 � 15
� 30 cm;
longer leg � shorter leg � �3�
� 15�3� cm
42. Divide one of the six equilateral triangles into two 30�– 60�– 90� triangles. The length of the shorter leg
of each triangle is �12
� cm. The length of the longer leg of
each triangle is ��12
���3�.
x � 2 � longer leg � 2��12
���3� � �3� � 1.73 cm
10.3 Standardized Test Practice (pp. 554–555)
43. C
44. F; hypotenuse � 2 � shorter leg
12 � 2 � shorter leg
6 � shorter leg
longer leg � shorter leg � �3� � 6�3�
perimeter � 12 � 6 � 6�3� � 28.4 cm
10.3 Mixed Review (p. 555)
45. �lwos
isness
� � �160� � �
53
� 46. �lwos
isness
� � �160� � �
35
�
47. �numbe
wr
ionfsgames
�� �10
1�
06
� � �1106� � �
58
�
48. �numbe
lorsosfes
games�� �
106� 6� � �
166� � �
38
�
49. Yes; by the AA Similarity Postulate, T ABC �T FHG.
50. Yes; by the SAS Similarity Theorem, T PQR �T STU.
10.3 Algebra Skills (p. 555)
51. 5x � 4 � 5(�4) � 4 � �20 � 4 � �16
52. 10x � 1 � 10(�4) � 1 � �40 � 1 � �41
53. x2 � 7 � (�4)2 � 7 � 16 � 7 � 9
54. (x � 3)(x � 3) � (�4 � 3)(�4 � 3) � (�1)(�7) � 7
30�
60�
15 cm
55. 2x2 � x � 1 � 2(�4)2 � (�4) � 1 � 32 � 4 � 1 � 37
56. 5x2 � 2x � 3 � 5(�4)2 � 2(�4) � 3 � 80 � 8 � 3 � 69
Quiz 1 (p. 555)
1. �8� � �3� � �8� �� 3� � �24� � �4���6� � �4� � �6� � 2�6�
2. �2� � �15� � �2���15� � �30�
3. �8� � �18� � �8���18� � �14�4� � 12
4. �80� � �5� � �80� �� 5� � �40�0� � 20
5. �27� � �9���3� � �9� � �3� � 3�3�
6. �17�6� � �16� �� 1�1� � �16� � �11� � 4�11�
7. �52� � �4���13� � �4� � �13� � 2�13�
8. �18�0� � �36� �� 5� � �36� � �5� � 6�5�
9. longer leg � shorter leg � �3�
2 � x�3�
��23�� � x
10. hypotenuse � leg�2�
x � �3� � �2�
x � �6�
11. hypotenuse � 2 � shorter leg
28 � 2 � x
14 � x;
longer leg � shorter leg � �3�
y � 14�3�
Lesson 10.4
10.4 Activity (p. 556)
1.
2. Each ratio of the leg lengths is 0.84.
3. Answers will vary, but the ratios in the fourth column willalways be equal; the ratio of leg lengths depends on themeasures of its angles, not the size of the right triangle.
10.4 Checkpoint (pp. 558–560)
1. tan S ��lelgeg
adojpapcoesnittetoaaSS
�� �68
� � �34
� � 0.75;
tan R ��lelgeg
adojpapcoesnittetoaaRR
�� �86
� � �43
� � 1.3333
170 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Triangle longer shorter�slhon
orgteerr
lleegg
�leg leg
A 5 cm 4.2 cm 0.84
B 10 cm 8.4 cm 0.84
C 15 cm 12.6 cm 0.84
D 20 cm 16.8 cm 0.84
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
2. tan S ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaS
S�� �
1204� � �
152� � 0.4167;
tan R ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaR
R�� �
2140� � �
152� � 2.4
3. tan 35� � 0.7002 4. tan 85� � 11.4301
5. tan 10� � 0.1763
6. 90� � 44� � 46�
tan 44� � �8x
� and tan 46� � �8x
�
7. 90� � 37� � 53�
tan 37� � �4x
� and tan 53� � �4x
�
8. 90� � 59� � 31�
tan 59� � �5x
� and tan 31� � �5x
�
9. tan 34� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� 10. tan 55� � �
o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 34� � �7x
� tan 55� � �1x8�
x tan 34� � 7 x tan 55� � 18
x � �tan
734�� x � �
tan1585�
�
x � 10.4 x � 12.6
11. tan 60� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 60� � �2x0�
20 tan 60� � x
34.6 � x
10.4 Guided Practice (p. 560)
1. The acute angles in T DEF are aD and aE.
2. The leg opposite aD is EF**** , and the leg adjacent to aDis DF**** .
3. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
192� � �
43
� � 1.3333
4. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
55
� � 1
5. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
8�8
3�� � �
�13�� � 0.5774
6. tan 25� � 0.4663 7. tan 62� � 1.8807
8. tan 80� � 5.6713 9. tan 43� � 0.9325
10.4 Practice and Applications (pp. 560–562)
10. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
185�
11. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
6�6
3�� � �3�
12. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
23
�
13. tan P � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
274� � 0.2917;
tan R � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
274� � 3.4286
14. tan P � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
1325� � 0.3429;
tan R � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
3152� � 2.9167
15. tan P � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
2105� � �
43
� � 1.3333;
tan R � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
1250� � �
34
� � 0.75
16. tan 28� � 0.5317 17. tan 54� � 1.3764
18. tan 5� � 0.0875 19. tan 89� � 57.2900
20. tan 67� � 2.3559 21. tan 40� � 0.8391
22. tan 12� � 0.2126 23. tan 83� � 8.1443
24. The measure of the other acute angle is 90� � 56� � 34�.
tan 56� � �9x
�
x tan 56� � 9
x � �tan
956��
x � 6.1,
tan 34� � �9x
�
9 tan 34� � x
6.1 � x
25. The measure of the other acute angle is 90� � 39� � 51�.
tan 39� � �3x3�
x tan 39� � 33
x � �tan
3339�
�
x � 40.8,
tan 51� � �3x3�
33 tan 51� � x
40.8 � x
26. The measure of the other acute angle is 90� � 35� � 55�.
tan 35� � �7x0�
x tan 35� � 70
x � �tan
7305�
�
x � 100.0,
tan 55� � �7x0�
70 tan 55� � x
100.0 � x
Geometry, Concepts and Skills 171Chapter 10 Worked-Out Solution Key
Chapter 10 continued
27. tan 41� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� 28. tan 29� � �
o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 41� � �5x2� tan 29� � �
1x2�
x tan 41� � 52 x tan 29� � 12
x � �tan
5421�
� x � �tan
1229�
�
x � 59.8 x � 21.6
29. tan 53� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 53� � �2x0�
x tan 53� � 20
x � �tan
2503�
�
x � 15.1
30. The tangent ratio can only be used for any acute angle ofa right triangle. This triangle is not a right triangle.
31. tan 13� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 13� � �58
h.2�
58.2(tan 13�) � h
13.4 � h
The height h of the slide is about 13.4 meters.
32. tan 70� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 70� � �3x
�
3 tan 70� � x
8.2 � x
33. tan 49� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 49� � �1x4�
x tan 49� � 14
x � �tan
1449�
�
x � 12.2
34. tan 34� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 34� � �1x0�
10(tan 34�) � x
6.7 � x
35. tan 40� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 40� � �3x5�
35(tan 40�) � x
29.4 � x
36. tan 35� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 35� � �2x0�
20(tan 35�) � x
14.0 � x
37. tan 50� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 50� � �oppo
1
si
0
te leg�
10(tan 50�) � opposite leg
11.9 � opposite leg
tan 45� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 45� � �11
x.9�
x(tan 45�) � 11.9
x � �ta
1n14.95�
�
� 11.9
38. tan 42� � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g�
tan 42� � �4d0�
40(tan 42�) � d
36 � d
The distance d is about 36 meters.
10.4 Standardized Test Practice (p. 562)
39. C; tan 38� � �1x0�
x tan 38� � 10
x � �tan
1308�
�
40. F; tan 50� � �6y
�
6 tan 50� � y
7.2 � y
10.4 Mixed Review (p. 562)
41. V � πr2h 42. V � lwh
� π(6)2(9) � 5 � 4 � 8
� π � 36 � 9 � 160 ft3
� 1018 m3
43. V � �13
�πr2h
� �13
� � π � (7)2 � 10
� 513 in.3
172 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
10.4 Algebra Skills (p. 562)
44. 8x � 10 � 3x 45. 4(x � 3) � 32
5x � 10 � 0 4x � 12 � 32
5x � 10 4x � 20
x � 2 x � 5
46. 3x � 7 � x � 11 47. 6x � 5 � 3x � 4
2x � 7 � 11 3x � 5 � �4
2x � 18 3x � �9
x � 9 x � �3
48. 2 � x � 4x � 22 49. 5x � 18 � 2x � 21
2 � 5x � 22 3x � 18 � 21
�5x � 20 3x � 39
x � �4 x � 13
Lesson 10.5
10.5 Checkpoint (pp. 563–565)
1. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
1157�;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
187�
2. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
2245�;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
275�
3. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
180� � �
45
�;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
160� � �
35
�
4. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
4401� � 0.9756;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
491� � 0.2195
5. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
�22�� � 0.7071;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
�22�� � 0.7071
6. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
�83�9�� � 0.7806;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
58
� � 0.625
7. sin 43� � 0.6820 8. cos 43� � 0.7314
9. sin 15� � 0.2588 10. cos 15� � 0.9659
11. cos 72� � 0.3090 12. sin 72� � 0.9511
13. cos 90� � 0 14. sin 90� � 1
15. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�
sin 34� � �a7
�
7(sin 34�) � a
3.9 � a;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�
cos 34� � �b7
�
7(cos 34�) � b
5.8 � b
16. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�
sin 65� � �1a2�
12(sin 65�) � a
10.9 � a;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�
cos 65� � �1b2�
12(cos 65�) � b
5.1 � b
17. sin 43� ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�
sin 43� � �a5
�
5(sin 43�) � a
3.4 � a;
cos 43� ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�
cos 43� � �b5
�
5(cos 43�) � b
3.7 � b
10.5 Guided Practice (p. 566)
1. B; cos D ��leg a
h
d
y
j
p
a
o
ce
te
n
n
t
u
t
s
o
e
aD�� �
DD
EF�
2. C; sin D ��leg o
h
p
y
p
p
o
o
s
te
it
n
e
u
t
s
o
e
aD�� �
DEF
F�
3. A; tan D � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
DEF
E�
4. The value of sin 37� is constant, so sin D � sin A. Thesine of an acute angle of a right triangle depends on themeasure of the angle, not on the size of the triangle.
5. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
45
� � 0.8
6. cos A ��leg a
h
d
y
j
p
a
o
ce
te
n
n
t
u
t
s
o
e
a A�� �
35
� � 0.6
Geometry, Concepts and Skills 173Chapter 10 Worked-Out Solution Key
Chapter 10 continued
7. tan A � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
43
� � 1.3333
8. sin B ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
35
� � 0.6
9. cos B ��leg a
hdyjpaocteenntutsoeaB
�� �45
� � 0.8
10. tan B � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
34
� � 0.75
10.5 Practice and Applications (pp. 566–568)
11. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
1611�;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
6601�
12. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
51�03�
� � ��23��;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
150� � �
12
�
13. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
3369� � �
1123�;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
1359� � �
153�
14. sin P ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
P�� �
1327� � 0.3243;
cos P ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aP�� �
3357� � 0.9459
15. sin P ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
P�� �
61�12�
� � 0.7714;
cos P ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aP�� �
171� � 0.6364
16. sin P ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
P�� �
39
� � �13
� � 0.3333;
cos P ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aP�� �
6�9
2�� � �
2�3
2�� � 0.9428
17. sin 40� � 0.6428 18. cos 23� � 0.9205
19. sin 80� � 0.9848 20. cos 5� � 0.9962
21. sin 59� � 0.8572 22. cos 61� � 0.4848
23. sin 90� � 1 24. cos 77� � 0.2250
25. sin 46� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 46� � �8x
�
8(sin 46�) � x
5.8 � x;
cos 46� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 46� � �8y
�
8(cos 46�) � y
5.6 � y
26. sin 54� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 54� � �1y4�
14(sin 54�) � y
11.3 � y;
cos 54� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 54� � �1x4�
14(cos 54�) � x
8.2 � x
27. sin 29� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 29� � �1x1�
11(sin 29�) � x
5.3 � x;
cos 29� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 29� � �1y1�
11(cos 29�) � y
9.6 � y
28. sin 24� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 24� � �1y6�
16(sin 24�) � y
6.5 � y;
cos 24� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 24� � �1x6�
16(cos 24�) � x
14.6 � x
29. sin 37� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 37� � �1x5�
15(sin 37�) � x
9.0 � x;
cos 37� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 37� � �1y5�
15(cos 37�) � y
12.0 � y
174 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
30. sin 68� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 68� � �2x6�
26(sin 68�) � x
24.1 � x;
cos 68� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 68� � �2y6�
26(cos 68�) � y
9.7 � y
31. sin 70� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 70� � �1h5�
15(sin 70�) � h
14 � h
The ladder reaches about 14 feet up the wall.
32. 8 ft �112
fint.
� � 96 in.
sin 22� � �o
h
p
y
p
p
o
o
s
t
i
e
t
n
e
u
l
s
e
e
g�
sin 22� � �9y6�
96(sin 22�) � y
36 in. � y;
cos 22� � �a
h
d
y
j
p
a
o
ce
te
n
n
t
u
l
s
e
e
g�
cos 22� � �9x6�
96(cos 22�) � x
89 in. � x
33. Answers may vary.
34. (sin A)2 � (cos A)2 � 1
35. When you drag point C, (sin A)2 � (cos A)2 � 1.
36. Sample answer:
In T ABC shown,
sin A � �ac
�, cos A � �bc
�, and tan A � �ab
�.
Then �csoins
AA
� � �ac
� � �bc
� � �ac
� � �bc
� � �abcc� � �
ab
� � tan A.
C
b
a
c
B
A
70�
15 h
37. sin 42� � �3r4�
34(sin 42�) � r
22.8 � r,
cos 48� � �3r4�
34(cos 48�) � r
22.8 � r
Both students get the correct answer.
10.5 Standardized Test Practice (p. 568)
38. C; sin 25� � �8x
�
x(sin 25�) � 8
x � �sin
825��
39. H; the sine of an angle cannot be greater than 1.
10.5 Mixed Review (p. 568)
40. longer leg � shorter leg � �3�
x � 13�3�
41. hypotenuse � 2 � shorter leg
y � 2 � 7
� 14;
longer leg � shorter leg � �3�
x � 7�3�
42. hypotenuse � 2 � shorter leg
16 � 2 � x
8 � x;
longer leg � shorter leg � �3�
y � 8�3�
43. tan 32� � 0.6249 44. tan 88� � 28.6363
45. tan 56� � 1.4826 46. tan 24� � 0.4452
47. tan 17� � 0.3057 48. tan 49� � 1.1504
10.5 Algebra Skills (p. 568)
49. �18, �8, �1.8, �0.8, 0, 0.08, 1.8
50. 2146, 2164, 2416, 2461, 2614, 2641
51. �0.61, �0.6, �0.56, �0.5, �0.47
Lesson 10.6
10.6 Checkpoint (pp. 570–572)
1. maA � tan�1 3.5 � 74.1�
2. maA � tan�1 2 � 63.4�
3. maA � tan�1 0.4402 � 23.8�
Geometry, Concepts and Skills 175Chapter 10 Worked-Out Solution Key
Chapter 10 continued
4. Since tan A � �59
� � 0.5556,
maA � tan�1 0.5556 � 29.1�
5. Since tan A � �1270� � 0.85,
maA � tan�1 0.85 � 40.4�
6. Since tan A � �1160� � 1.6,
maA � tan�1 1.6 � 58.0�
7. maA � sin�1 0.5 � 30�
8. maA � cos�1 0.92 � 23.1�
9. maA � sin�1 0.1149 � 6.6�
10. maA � cos�1 0.5 � 60�
11. maA � sin�1 0.25 � 14.5�
12. maA � cos�1 0.45 � 63.3�
13. 32 � 42 � x2
25 � x2
�2�5� � x
5 � x;
tan A � �34
� � 0.75
maA � tan�1 0.75 � 36.9�;
maB � 90� � maA
� 90� � 36.9�
� 53.1�
14. 42 � y2 � 62
16 � y2 � 36
y2 � 20
y � �2�0�
� 4.5;
cos E � �46
� � 0.6667
maE � cos�1 0.6667 � 48.2�;
maD � 90� � maE
� 90� � 48.2�
� 41.8�
15. z2 � 52 � 72
z2 � 25 � 49
z2 � 24
z � �2�4�
� 4.9;
sin H � �57
� � 0.7143
maH � sin�1 0.7143 � 45.6�;
maG � 90� � maH
� 90� � 45.6�
� 44.4�
10.6 Guided Practice (p. 572)
1. Solving a right triangle means finding the measures ofboth acute angles and the lengths of all three sides.
2. true 3. false
4. x� � 19� � 90�
x � 71
5. 42 � 52 � x26. 92 � x2 � 132
16 � 25 � x2 81 � x2 � 169
41 � x2 x2 � 88
�4�1� � x x � �8�8�
6.4 � x x � 9.4
7. maA � tan�1 5.4472 � 79.6�
8. maA � sin�1 0.8988 � 64.0�
9. maA � cos�1 0.3846 � 67.4�
10. maX � 90� � 60� � 30�;
hypotenuse � 2 � shortest leg
4 � 2 � x
2 � x;
longest leg � shortest leg � �3�
y � 2�3�
� 3.5
11. d2 � 122 � 142
d2 � 144 � 196
d2 � 52
d � �5�2�
� 7.2;
cos D � �1124� � 0.8571
maD � cos�1 0.8571 � 31.0�;
maE � 90� � maD
� 90� � 31�
� 59.0�
10.6 Practice and Applications (pp. 573–575)
12. maA � tan�1 0.5 � 26.6�
13. maA � tan�1 1.0 � 45�
14. maA � tan�1 2.5 � 68.2�
15. maA � tan�1 0.2311 � 13.0�
16. maA � tan�1 1.509 � 56.5�
17. maA � tan�1 4.125 � 76.4�
176 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
18. Use the Pythagorean Theorem;
482 � 552 � (QS)2
5329 � (QS)2
�5�3�2�9� � QS
73 � QS
19. Sample answer: Use the inverse tangent function;
tan Q � �5458� � 1.1458
maQ � tan�1 1.1458 � 48.9�
20. Sample answer: Use the fact that aQ and aS are complements;
maS � 90� � maQ
� 90� � 48.9�
� 41.1�
21. 72 � 72 � x222. 62 � 22 � x2
49 � 49 � x2 36 � 4 � x2
98 � x2 40 � x2
�9�8� � x �4�0� � x
7�2� � x 2�1�0� � x
9.9 � x; 6.3 � x;
tan A � �77
� � 1 tan A � �26
� � �13
�
maA � tan�1 1 � 45� maA � tan�1 �13
� � 18.4�
23. 202 � 212 � x2
400 � 441 � x2
841 � x2
�8�4�1� � x
29 � x;
tan A � �2201� � 0.9524
maA � tan�1 0.9524 � 43.6�
24. maA � sin�1 0.75 � 48.6�
25. maA � cos�1 0.1518 � 81.3�
26. maA � sin�1 0.6 � 36.9�
27. maA � cos�1 0.45 � 63.3�
28. maA � cos�1 0.1123 � 83.6�
29. maA � sin�1 0.6364 � 39.5�
30. (2.5)2 � (horizontal distance)2 � 202
6.25 � (horizontal distance)2 � 400
(horizontal distance)2 � 393.75
horizontal distance � �3�9�3�.7�5�
horizontal distance � 19.8 ft
sin(ramp angle) � �22.05� � 0.125
sin�1 0.125 � 7.2�, so the measure of the ramp angle isabout 7.2�. This ramp meets the standards.
31. Sample answer: 28 ft (Any ramp greater than or equal to27.6 ft will meet the standards)
32. Answers may vary.
33. tan(glide angle) ��distan
acleti
ttoud
reunway
�
tan(glide angle) � �1559.7�
tan(glide angle) � 0.2661
tan�1 0.2661 � 14.9�, so the measure of the glide angleis about 14.9�.
34. sin A � �162� � �
12
� � 0.5
maA � sin�1 (0.5) � 30�
35. cos A � �180� � �
45
� � 0.8
maA � cos�1 0.8 � 36.9�
36. cos A � �68
� � �34
� � 0.75
maA � cos�1 0.75 � 41.4�
37. (LM)2 � 162 � 172
(LM)2 � 256 � 289
(LM)2 � 33
LM � �3�3�
� 5.7;
cosaK � �1167� � 0.9412
maK � cos�1 0.9412 � 19.7�;
maL � 90� � maK
� 90� � 19.7�
� 70.3�
38. (NQ)2 � 42 � 142
(NQ)2 � 16 � 196
(NQ)2 � 180
NQ � �1�8�0�
� 13.4;
sin N � �144� � 0.2857
maN � sin�1 0.2857 � 16.6�;
maP � 90� � maN
� 90� � 16.6�
� 73.4�
Geometry, Concepts and Skills 177Chapter 10 Worked-Out Solution Key
Chapter 10 continued
39. (ST)2 � 62 � 152
(ST)2 � 36 � 225
(ST)2 � 189
ST � �1�8�9�
� 13.7;
sin S � �165� � 0.4
maS � sin�1 0.4 � 23.6�;
maR � 90� � maS
� 90� � 23.6�
� 66.4�
40. tan�1 �ABC
B�; the diagram gives the lengths AB and BC,
but does not give the length AC.
10.6 Standardized Test Practice (p. 574)
41. B 42. G
10.6 Mixed Review (p. 575)
43. C � 2πr � 2π(8) � 16π � 50 cm;
A � πr2 � π(8)2 � 64π � 201 cm2
44. C � 2πr � 2π(15) � 30π � 94 in.;
A � πr2 � π(15)2 � 225π � 707 in.2
45. C � πd � π(34) � 107 yd;
A � πr2 � π(17)2 � 289π � 908 yd2
46. V � �43
�πr347. V � �
43
�πr3
� �43
�π(10)3 � �43
�π(14)3
� 4189 ft3 � 11,494 cm3
48. V �
�
� �2530
�π
� 262 in.3
10.6 Algebra Skills (p. 575)
49. 0.36 � 0.194 � 0.554 50. $8.42 � $2.95 � $5.47
51. 7 4.65 � 32.55 52. 55.40 � 0.04 � 1385
53. 700 � 0.35 � 2000 54. $22.50 0.08 � $1.80
�43
�π(5)3
�2
�43
�πr3
�2
Quiz 2 (p. 575)
1. tan 42� � �6x
�
x tan 42� � 6
x � �tan
642��
x � 6.7
2. tan 63� � �2x
�
2(tan 63�) � x
3.9 � x
3. sin 40� � �1y2� 4. sin 21� � �
1x9�
12(sin 40�) � y 19(sin 21�) � x
7.7 � y; 6.8 � x;
tan 40� � �7x.7� tan 21� � �
6y.8�
x tan 40� � 7.7 y (tan 21)� � 6.8
x � �tan
7.470�
� y � �tan
6.281�
�
x � 9.2 y � 17.7
5. tan 29� � �1x4�
14(tan 29�) � x
7.8 � x
6. sin 35� � �8x
�
8(sin 35�) � x
4.6 � x;
tan 35� � �4y.6�
y tan 35� � 4.6
y � �tan
4.365�
�
y � 6.6
7. tan 72� � 3.0777 8. sin 52� � 0.7880
9. cos 36� � 0.8090
10. maN � 90� � 40� � 50�;
tan 40� � �1m6�
16 tan 40� � m
13.4 � m;
162 � (13.4)2 � q2
435.56 � q2
�4�3�5�.5�6� � q
20.9 � q
178 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
11. q2 � 72 � 82
q2 � 49 � 64
q2 � 15
q � �1�5�
� 3.9;
cos Q � �78
� � 0.875
maQ � cos�1 0.875 � 29.0�;
maP � 90� � maQ
� 90� � 29.0�
� 61.0�
12. k2 � 32 � 12.42
k2 � 9 � 153.76
k2 � 144.76
k � �1�4�4�.7�6�
� 12.0;
sin L � �12
3.4� � 0.2419
maL � sin�1 0.2419 � 14.0�;
maK � 90� � maL
� 90� � 14.0�
� 76.0�
Chapter 10 Summary and Review (pp. 576–579)
1. A radical is an expression written with a radical symbol.
2. The number or expression inside the radical symbol isthe radicand.
3. A trigonometric ratio is a ratio of the lengths of twosides of a right triangle.
4. A right triangle with side lengths 9, 9�3�, and 18 is a30�– 60�– 90� triangle.
5. To solve a right triangle means to determine that lengthsof all three sides of the triangle and the measures of bothacute angles.
6. A right triangle with side lengths 4, 4, and 4�2� is a45�– 45�– 90� triangle.
7. If aF is an acute angle of a right triangle, then
tangent of aF ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaF
F�.
8. If aF is an acute angle of a right triangle, then
cosine of aF ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aF�.
9. If aF is an acute angle of a right triangle, then
sine of aF ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
F�.
10. �1�3� � �1�3� � �1�3� �� 1�3� � �1�6�9� � 13
11. �2� � �7�2� � �2� �� 7�2� � �1�4�4� � 12
12. �7� � �1�0� � �7� �� 1�0� � �7�0�
13. �4�7��2 � 4�7� � 4�7� � 4 � 4 � �7� � �7� � 16 � 7 � 112
14. �3� � �1�9� � �3� �� 1�9� � �5�7�
15. �5� � �5� � �5� �� 5� � �2�5� � 5
16. �6� � �1�8� � �6� �� 1�8� � �1�0�8� � �3�6� �� 3� � �3�6� � �3� � 6�3�
17. �3�1�1��2 � 3�1�1� � 3�1�1� � 3 � 3 � �1�1� � �1�1� � 9 � 11 � 99
18. �2�7� � �9� �� 3� � �9� � �3� � 3�3�
19. �7�2� � �3�6� �� 2� � �3�6� � �2� � 6�2�
20. �1�5�0� � �2�5� �� 6� � �2�5� � �6� � 5�6�
21. �6�8� � �4� �� 1�7� � �4� � �1�7� � 2�1�7�
22. �1�0�8� � �3�6� �� 3� � �3�6� � �3� � 6�3�
23. �8�0� � �1�6� �� 5� � �1�6� � �5� � 4�5�
24. �7�5�0�0� � �2�5�0�0� �� 3� � �2�5�0�0� � �3� � 50�3�
25. �5�0�7� � �1�6�9� �� 3� � �1�6�9� � �3� � 13�3�
26. A � lw � �5�2���2�5�� � 5 � 2 � �2� � �5� � 10�1�0� � 31.6
27. A � lw � �4�6���3�3�� � 4 � 3 � �6� � �3� � 12�1�8� � 50.9
28. A � lw � �2�2����1�0�� � 2�2� � �1�0� � 2�2�0� � 8.9
29. hypotenuse � leg � �2�
x � 15�2�
30. hypotenuse � leg � �2� 31. hypotenuse � leg � �2�
x � 5�2� � �2� x � �7� � �2�
� 5 � 2 x � �7� �� 2�
� 10 x � �1�4�
32. hypotenuse � leg � �2� 33. hypotenuse � leg � �2�
19�2� � x�2� 3�2� � x�2�
19 � x 3 � x
34. hypotenuse � leg � �2�
10 � x�2�
��10
2�� � x
7.1 � x
35. hypotenuse � 2 � shorter leg
x � 2 � 25
x � 50;
longer leg � shorter leg � �3�
y � 25�3�
� 43.3
36. hypotenuse � 2 � shorter leg
x � 2 � 19
x � 38;
longer leg � shorter leg � �3�
y � 19�3�
� 32.9
Geometry, Concepts and Skills 179Chapter 10 Worked-Out Solution Key
Chapter 10 continued
37. longer leg � shorter leg � �3�
94�3� � x�3�
94 � x;
hypotenuse � 2 � shorter leg
y � 2 � 94
y � 188
38. longer leg � shorter leg � �3�
45 � x�3�
��45
3�� � x
26.0 � x;
hypotenuse � 2 � shorter leg
y � 2���45
3���
y � ��90
3��
� 52.0
39. tan A ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaA
A�� �
1360� � �
185� � 0.5333;
tan B ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaB
B�� �
3106� � �
185� � 1.875
40. tan A ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaA
A�� �
202
�0
3�� � �
�13�� � 0.5774;
tan B ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaB
B�� �
202
�0
3�� � �3� � 1.7321
41. tan A ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaA
A�� �
182� � �
32
� � 1.5;
tan B ��le
l
g
eg
ad
o
j
p
a
p
c
o
e
s
n
i
t
te
to
aaB
B�� �
182� � �
23
� � 0.6667
42. tan 17� � 0.3057 43. tan 81� � 6.3138
44. tan 36� � 0.7265 45. tan 24� � 0.4452
46. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
1255� � �
35
� � 0.6;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
2205� � �
45
� � 0.8
47. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
59
� � 0.5556;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
2�91�4�� � 0.8315
48. sin A ��leg
h
o
y
p
p
p
o
o
te
s
n
it
u
e
s
ae
A�� �
306
�0
2�� � �
�22�� � 0.7071;
cos A ��leg a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
t
s
o
e
aA�� �
306
�0
2�� � �
�22�� � 0.7071
49. sin 57� � 0.8387 50. sin 12� � 0.2079
51. cos 31� � 0.8572 52. cos 75� � 0.2588
53. sin 31� � �1x0�
10(sin 31�) � x
5.2 � x;
cos 31� � �1y0�
10(cos 31�) � y
8.6 � y
54. maA � tan�1 3.2145 � 72.7�
55. maA � sin�1 0.0888 � 5.1�
56. maA � cos�1 0.2243 � 77.0�
57. maA � tan�1 1.2067 � 50.4�
58. 32 � 62 � c2
45 � c2
�4�5� � c
6.7 � c;
tan A � �36
� � 0.5
maA � tan�1 0.5 � 26.6�;
maB � 90� � maA
� 90� � 26.6�
� 63.4�
59. 142 � 232 � q2
725 � q2
�7�2�5� � q
26.9 � q;
tan R � �1243� � 0.6087
maR � tan�1 0.6087 � 31.3�;
maP � 90� � maR
� 90� � 31.3�
� 58.7�
60. maZ � 90� � maX
� 90� � 55�
� 35�;
sin 55� � �3x
�
3 sin 55� � x
3(0.8192) � x
2.5 � x;
cos 55 � �3z
�
3 cos 55 � z
3(0.5736) � z
1.7 � z
180 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
Chapter 10 Test (p. 580)
1. �1�5� � �7� � �1�5� �� 7� � �1�0�5�
2. �3�1�1��2 � 3�1�1� � 3�1�1� � 3 � 3 � �1�1� � �1�1� � 9 � 11 � 99
3. �2�3��2 � �2�5��2 � x2
4 � 3 � 4 � 5 � x2
12 � 20 � x2
32 � x2
�3�2� � x
�1�6� �� 2� � x
4�2� � x
4. D 5. A 6. B 7. C
8. hypotenuse � leg � �2�
47 � leg �2�
��47
2�� � leg
33.2 � leg
9. tan 38� � �3x
� 10. tan 70� � �8y
�
3(tan 38�) � x 8(tan 70�) � y
2.3 � x; 22.0 � y;
cos 38� � �3y
� cos 70� � �8x
�
y cos 38� � 3 x cos 70� � 8
y � �cos
338�� x � �
cos870��
y � 3.8 x � 23.4
11. sin 35� � �1y0� 12. sin 50� � �
1x5�
10(sin 35�) � y 15(sin 50�) � x
5.7 � y; 11.5 � x;
cos 35� � �1x0� cos 50� � �
1y5�
10(cos 35�) � x 15 cos 50� � y
8.2 � x 9.6 � y
13. tan 70� � 2.7475 14. cos 14� � 0.9703
15. tan 31� � 0.6009 16. sin 26� � 0.4384
17. cos 30� � 0.8660 18. tan 45� � 1
19. sin 5� � 0.0872 20. tan 10� � 0.1763
21. maA � tan�1 5.2 � 79.1�
22. maA � tan�1 7 � 81.9�
23. maA � sin�1 0.3091 � 18.0�
24. maA � sin�1 0.5318 � 32.1�
25. maA � cos�1 0.6264 � 51.2�
26. maA � cos�1 0.3751 � 68.0�
27. maK � 90� � 30� � 60�;
hypotenuse � 2 � shortest leg
9 � 2 � KL
�92
� � KL
4.5 � KL;
longest leg � shortest leg � �3�
JL � 4 � 5 � �3�
JL � 7.8
28. maF � 90� � 25� � 65�;
tan 25� � �D12
E�
DE(tan 25�) � 12
DE � �tan
1225�
�
DE � 25.7;
sin 25� � �D12
F�
DF(sin 25�) � 12,
DF � �sin
1225�
�
DF � 28.4
29. 42 � QR2 � 62
16 � QR2 � 36
QR2 � 20
QR � �2�0�
QR � 4.5;
cos P � �46
� � �23
� � 0.6667
maP � cos�1 0.6667 � 48.2�;
maR � 90� � maP � 90� � 48.2� � 41.8�
30. tan X � �138200
� � 0.5625
maX � tan�1 0.5625 � 29.4�
Chapter 10 Standardized Test (p. 581)
1. A; �1�2�4� � �4� �� 3�1� � �4� � �3�1� � 2�3�1�
2. J; hypotenuse � leg � �2�
x � �7� � �2�
� �1�4�
3. B; tan J � �o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g� � �
11152
�
4. J; sin�1 0.8170 � 54.8�
5. C; longer leg � shorter leg � �3�
PS � 48�3�
Geometry, Concepts and Skills 181Chapter 10 Worked-Out Solution Key
Chapter 10 continued
6. G; sin A � �161� � 0.5455
maA � sin�1 0.5455 � 33.1�
7. B; sin 32� � �P6R�
PR � sin 32� � 6
PR � �sin
632��
� 11.3
8. J; tan 72� � �3h0�
30(tan72�) � h
92.3 � h
9. D; tan 67� � �8x
�
8(tan 67�) � x
18.8 � x,
cos 67� � �8y
�
y � cos 67� � 8
y � �cos
867��
y � 20.5
Chapter 10 Algebra Review (p. 583)
1. x � y � 1
4x � 5y � 7
Solve for y in Equation 1.
x � y � 1
y � 1 � x
Substitute for y in Equation 2.
4x � 5y � 7
4x � 5(1 � x) � 7
4x � 5 � 5x � 7
5 � x � 7
�x � 2
x � �2
Substitute for x in the revised Equation 1.
y � 1 � x � 1 � (�2) � 1 � 2 � 3
Check that (�2, 3) is a solution.
x � y � 1
(�2) � 3 � 1
1 � 1 ✓
4x � 5y � 7
4(�2) � 5(3) � 7
� 8 � 15 � 7
7 � 7 ✓
The solution is (�2, 3)
2. x � 2y � 9
3x � y � �1
Solve for y in Equation 2.
3x � y � �1
3x � 1 � y
Substitute for y in Equation 1.
x � 2(3x � 1) � 9
x � 6x � 2 � 9
7x � 7
x � 1
Substitute for x in revised Equation 2.
y � 3x � 1 � 3(1) � 1 � 4
Check that (1, 4) is a solution.
1 � 2(4) � 9
1 � 8 � 9
9 � 9 ✓
3(1) � 4 � �1
3 � 4 � �1
�1 � �1 ✓
The solution is (1, 4).
3. 3x � y � 3
7x � 2y � 1
Solve for y in Equation 1.
3x � y � 3
y � 3 � 3x
Substitute for y in Equation 2
7x � 2(3 � 3x) � 1
7x � 6 � 6x � 1
x � 6 � 1
x � �5
Substitute for x in revised Equation 1.
y � 3 � 3x � 3 � 3(�5) � 3 � 15 � 18
Check that (�5, 18) is a solution.
3(�5) � 18 � 3
�15 � 18 � 3
3 � 3 ✓
7(�5) � 2(18) � 1
�35 � 36 � 1
1 � 1 ✓
The solution is (�5, 18).
182 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued
4. x � y � �4
x � y � 16
Solve for y in Equation 2.
x � y � 16
y � 16 � x
Substitute for y in Equation 1.
x � (16 � x) � �4
x � 16 � x � �4
2x � 12
x � 6
Substitute for x in revised Equation 2.
y � 16 � x � 16 � 6 � 10
Check that (6, 10) is a solution.
x � y � �4
6 � 10 � �4
�4 � �4 ✓
x � y � 16
6 � 10 � 16
16 � 16 ✓
The solution is (6, 10).
5. �x � y � 1
2x � y � 4
Solve for y in Equation 1.
�x � y � 1
y � 1 � x
Substitute for y in Equation 2.
2x � y � 4
2x � (1 � x) � 4
2x � 1 � x � 4
3x � 3
x � 1
Substitute for x in revised Equation 1.
y � 1 � x � 1 � 1 � 2
Check that (1, 2) is a solution.
�x � y � 1
�1 � 2 � 1
1 � 1 ✓
2x � y � 4
2(1) � 2 � 4
2 � 2 � 4
4 � 4 ✓
The solution is (1, 2).
6. 6x � y � 2
4x � 3y � �6
Solve for y in Equation 1.
6x � y � 2
6x � 2 � y
Substitute for y in Equation 2.
4x � 3y � �6
4x � 3(6x � 2) � �6
4x � 18x � 6 � �6
22x � 0
x � 0
Substitute for x in revised Equation 1.
y � 6x � 2 � 6(0) � 2 � �2
Check that (0, �2) is a solution.
6x � y � 2
6(0) � (�2) � 2
0 � 2 � 2
2 � 2 ✓
4x � 3y � �6
4(0) � 3(�2) � �6
0 � 6 � �6
�6 � �6 ✓
The solution is (0, �2).
7. 2x � 3y � 5
x � 4y � �3
Solve for x in Equation 2.
x � 4y � �3
x � �3 � 4y
Substitute for x in Equation 1.
2x � 3y � 5
2(�3 � 4y) � 3y � 5
�6 � 8y � 3y � 5
11y � 11
y � 1
Substitute for y in revised Equation 2.
x � �3 � 4y � �3 � 4(1) � 1
Check that (1, 1) is a solution.
2x � 3y � 5
2(1) � 3(1) � 5
2 � 3 � 5
5 � 5 ✓
x � 4y � �3
(1) � 4(1) � �3
1 � 4 � �3
�3 � �3 ✓
The solution is (1, 1).
Geometry, Concepts and Skills 183Chapter 10 Worked-Out Solution Key
Chapter 10 continued
8. �3x � 2y � �5
�x � 3y � �9
Solve for x in Equation 2.
�x � 3y � �9
3y � 9 � x
Substitute for x in Equation 1.
�3(3y � 9) � 2y � �5
�9y � 27 � 2y � �5
�11y � 22
y � �2
Substitute for y in revised Equation 2.
x � 3y � 9 � 3(�2) � 9 � �6 � 9 � 3
Check that (3, �2) is a solution.
�3x � 2y � �5
�3(3) � 2(�2) � �5
�9 � 4 � �5
�5 � �5 ✓
�x � 3y � �9
�(3) � 3(�2) � �9
�3 � 6 � �9
�9 � �9 ✓
The solution is (3, �2).
9. 5x � 2y � 7
2x � 4y � 22
Solve for x in Equation 2.
2x � 4y � 22
2x � 22 � 4y
x � 11 � 2y
Substitute for x in Equation 1.
5x � 2y � 7
5(11 � 2y) � 2y � 7
55 � 10y � 2y � 7
12y � �48
y � �4
Substitute for y in revised Equation 2.
x � 11 � 2y � 11 � 2(�4) � 11 � 8 � 3
Check that (3, �4) is a solution.
5x � 2y � 7
5(3) � 2(�4) � 7
15 � 8 � 7
7 � 7 ✓
2x � 4y � 22
2(3) � 4(�4) � 22
6 � 16 � 22
22 � 22 ✓
The solution is (3, �4).
184 Geometry, Concepts and SkillsChapter 10 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.