chapter 1mrjaffe.net/geomref/unit1solutions.pdfchapter 1continued 29. the sum of any two odd numbers...

29
CHAPTER 1 Geometry 1 Chapter 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 on observations. 2. A conjecture can be proven false by finding a counterex- ample. 3. 4. 5. Each number is 3 times the previous number. The next number is or 162. 6. The numbers are consecutive perfect squares. The next number is or 16. 7. Each number is the previous number. The next number is or 1. 4 4 1 4 4 2 54 3 9 9 18 4.24 225 100 325 18.03 1 49 50 7.07 36 4 40 6.32 5 2 0 2 25 0 25 1 2 1 2 1 1 2 5 2 2 2 25 4 29 2 2 4 2 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 8 5 3 5 3 8 9 17 8 17 9 8 8. Every other number is zero. The other numbers alternate between 3 and The next number is 9. Each number is 0.5 greater than the previous number. The next number is or 10. Each number is 6 less than the previous number. The next number is or 11. The sum of any three consecutive positive integers is 3 times 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. The next 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 next number is or 14,641. 19. Each number is 5 less than the previous number. The next number is or 20. Numbers after the first are found by adding consecutive even integers. The sixth number is 10 more than the fifth number, so it is or 37. 21. Numbers after the first are found by adding consecutive whole numbers. The sixth number is 6 more than the fifth number, 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 after the decimal point of the previous number. So the next number 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 or 80 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

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Page 1: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 2: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 3: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 4: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 5: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 6: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

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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

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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

Page 9: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 10: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

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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

Page 12: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 13: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 14: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 15: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 16: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 17: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 18: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 19: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 20: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 21: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 22: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 23: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 24: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 25: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 26: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

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Page 27: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 28: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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

Page 29: CHAPTER 1mrjaffe.net/GeomRef/unit1solutions.pdfChapter 1continued 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

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