chapter 1 · 2013-04-18 · 3. 4 2 6 5 22 5 2 4. 3 2 11 5 28 5 8 5. 24 1 5 5 1 5 1 6. 28 2 10 5 218...

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1Geometry

Worked-Out Solution Key

Prerequisite Skills (p. 1)

1. The distance around a rectangle is called its perimeter, and the distance around a circle is called its circumference.

2. The number of square units covered by a fi gure is called its area.

3. 4 2 6 5 22 5 2

4. 3 2 11 5 28 5 8

5. 24 1 5 5 1 5 1

6. 28 2 10 5 218 5 18

7. 5x 5 5(2) 5 10

8. 20 2 8x 5 20 2 8(2) 5 20 2 16 5 4

9. 218 1 3x 5 218 1 3(2) 5 218 1 6 5 212

10. 25x 2 4 1 2x 5 25(2) 2 4 1 2(2)

5 210 2 4 1 4 5 210

11. 274 5 22z

2137 5 z

12. 8x 1 12 5 60 13. 2y 2 5 1 7y 5 232

8x 5 48 9y 2 5 5 232

x 5 6 9y 5 227

y 5 23

14. 6p 1 11 1 3p 5 27 15. 8m 2 5 5 25 2 2m

9p 1 11 5 27 10m 2 5 5 25

9p 5 218 10m 5 30

p 5 22 m 5 3

16. 22n 1 18 5 5n 2 24

42 5 7n

6 5 n

Lesson 1.1

1.1 Guided Practice (pp. 3–5)

1. Other names for @##$ ST are @##$ TS and line m. Point V is not coplanar with points Q, S, and T.

2. Another name for } EF is } FE .

3. ###$ HJ and ###$ JH are not the same ray. They have different endpoints and continue in different directions. ###$ HJ and ###$ HG are the same ray because they have the same endpoint and continue in the same direction.

4.

P

n m

5. @##$ PQ intersects line k at point M.

6. Plane A intersects plane B at line k.

7. Line k intersects plane A at line k.

1.1 Exercises (pp. 5 – 8)

Skill Practice

1. a. Point Q b. Segment MN

c. Ray ST d. Line FG

2. Collinear points lie on the same line, so they are also coplanar. Coplanar points lie in the same plane, but not necessarily on the same line, so they may not be collinear.

3. Other names for @###$ WQ are @###$ QW and line g.

4. Another name for plane V is plane QST.

5. Points R, Q, and S are collinear. Point T is not collinear with those points.

6. Point W is not coplanar with points R, S, and T.

7. Point W is coplanar with points Q and R because there is only one plane through any 3 points not on the same line.

8. Another name for } ZY is } YZ .

9. ###$ VY , ###$ VX , ###$ VZ , ###$ VW

10. ###$ VY and ###$ VZ ; ###$ VX and ###$ VW

11. Another name for ###$ WV is ###$ WX .

12. ###$ VW and ###$ VZ do have the same endpoints, but points W and Z are not on the same line, so the rays are not opposite.

13. B; C, D, E, and G are coplanar.

14. Pn mA

15.

PAk

16. A; ###$ EC and ###$ ED are opposite rays.

17. @##$ PR intersects @##$ HR at point R.

18. Plane EFG and plane FGS intersect at @##$ FG .

19. Plane PQS and plane HGS intersect at @##$ RS .

20. P, Q, and F are not collinear, but they are coplanar.

21. P and G are neither collinear nor coplanar.

22. Planes HEF, PEF, and PEH, intersect at point E.

23.

J

l

K

24. Sample answer: ###$ CA , ###$ CD , ###$ AC , ###$ AB , ###$ BA , ###$ BC , ###$ BE , ###$ CB , ###$ EB , ###$ DC ; ###$ CA and ###$ CD are opposite rays and ###$ BA and ###$ BE are opposite rays.

Chapter 1

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2GeometryWorked-Out Solution Key

25.

L

JM

K

26.

RP

27. y 5 x 2 4; A(5, 1) 28. y 5 x 1 1; A(1, 0)

1 0 5 2 4 0 0 1 1 1

1 5 1 ✓ 0 Þ 2

A(5, 1) is on the line. A(1, 0) is not on the line.

29. y 5 3x 1 4; A(7, 1) 30. y 5 4x 1 2; A(1, 6) 1 0 3(7) 1 4 6 0 4(1) 1 2

1 Þ 25 6 5 6 ✓

A(7, 1) is not on the line. A(1, 6) is on the line.

31. y 5 3x 2 2; A(21, 25) 32. y 5 22x 1 8; A(24, 0)

25 0 3(21) 2 2 0 0 22(24) 1 8

25 5 25 ✓ 0 Þ 16

A(21, 25) is on the line. A(24, 0) is not on the line.

33. x ≤ 3

0 1 2 321

The graph is a ray.

34. x ≥ 24

0 1212224 23

The graph is a ray.

35. 27 ≤ x ≤ 4

42028 26 24 22

The graph is a segment.

36. x ≥ 5 or x ≤ 22

0 2 4 62224

The graph is a pair of rays.

37. x ≥ 21 or x ≤ 5

0 2 4 62224

The graph is a line.

38. x ≤ 0

0 1 22122

The graph is a point.

39. a. Three planes that do not intersect are possible if the planes are all parallel.

A

B

C

b. It is possible to have three planes that intersect in one line. The planes intersect at @##$ AB .

B

A

C

c. This is not possible because planes intersect at a line, not a point.

d. It is possible for a third plane to intersect two planes that do not intersect, if the two planes are parallel and the other plane intersects those two.

BA

C

e. This is not possible because the third plane cannot be parallel to both of the other intersecting planes, so it must intersect at least one of them.

Problem Solving

40. Intersections of several lines

41. Intersection of a line with a plane

42. Planes intersecting with planes

43. A four-legged table may rock from side to side because four points are not necessarily coplanar. A three-legged table would not rock because three points determine a unique plane.

44. a. When the tripod is on a level surface, the tips of the legs are coplanar.

b. The tips of the legs are still coplanar because three points determine a unique plane.

45. a–c.

B

V W

DF

AC

G

H

E

46. a. If there are 5 streets, there must be 10 traffi c lights. If there are 6 streets, there must be 15 traffi c lights.

b. Each time a street is added, the number of additional traffi c lights that are needed is equal to the previous number of streets.

Mixed Review

47. 215 2 9 5 224 48. 6 2 10 5 24

49. 225 2 (212) 5 213 50. 13 2 20 5 27

Chapter 1, continued

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3Geometry


51. 16 2 (24) 5 20 52. 25 2 15 5 220

53. 5 p 22 1 1 5 5 p 21 5 5 p 1 5 5

54. 28 1 7 2 6 5 21 2 6 5 1 2 6 5 25

55. 27 p 8 2 10 5 27 p 22 5 27 p 2 5 214

56–57.

1

x

y

21

A(2, 4)

B(23, 6)E(6, 7.5)

Lesson 1.2


1. 1 5 }

8 inches 2. 1

3 }

8 inches

3. XY 1 YZ 5 XZ

23 1 50 5 73

The length of } XZ is 73 units.

4. You cannot use the Segment Addition Postulate to fi nd the length of } WZ given WY 5 30 because Y is not between W and Z.

5. VW 1 WX 5 VX

37 1 WX 5 144

WX 5 107

6.

x

y

1

1

A(22, 4) B(3, 4)

C(0, 2)

D(0, 22)

}

AB and }

CD are not congruent, because AB 5 3 2 (22) 5 5 and CD 5 22 2 2 5 4.


Skill Practice

1. } MN means the line segment MN, and MN means the distance between M and N.

2. You can fi nd PN by adding PQ and QN. You can fi nd PN by subtracting MP from MN.

3. 2.1 cm 4. 3.2 cm 5. 3.5 cm

6. MN 1 NP 5 MP 7. RS 1 ST 5 RT

5 1 18 5 MP 22 1 22 5 RT

23 5 MP 44 5 RT

8. UV 1 VW 5 UW 9. XY 1 YZ 5 XZ

39 1 26 5 UW XY 1 7 5 30

65 5 UW XY 5 23

10. AB 1 BC 5 AC 11. DE 1 EF 5 DF

27 1 BC 5 42 DE 1 50 5 63

BC 5 15 DE 5 13

12. The Segment Addition Postulate was used incorrectly.

AB 1 BC 5 AC

9 1 BC 5 14

BC 5 5

13.

x

y

3

1

A(0, 1) B(4, 1)C(1, 2)

D(1, 6)

AB 5 4 2 0 5 4 and CD 5 6 2 2 5 4, so

}

AB > }

CD .

14.

x

y2

2

J(26, 28)

K(26, 2)

L(22, 24)

M(26, 24)

Because JK 5 28 2 2 5 10 and LM 5 22 2 (26) 5 4, } JK is not congruent to } LM .

15.

x

y

100

100

T (300, 2200)

U(300, 100)

R(2200, 300) S(200, 300)

Because RS 5 200 2 (2200) 5 400 and TU 5 2200 2 100 5 300,

} RS is not

congruent to } TU .

16. JK 5 23 2 (26) 5 23 1 6 5 3

17. JL 5 1 2 (26) 5 1 1 6 5 7

18. JM 5 6 2 (26) 5 6 1 6 5 12

19. KM 5 6 2 (23) 5 6 1 3 5 9

20. Yes, it is possible to show that FB > CB using the Segment Addition Postulate. FC 1 CB 5 FB, so FB must be greater than FC and CB individually.It is not possible to show that AC > DB using the Segment Addition Postulate because B is not between A and C.

21. XY 5 YZ 5 WX 22. VW 1 WX 1 XZ 5 VZ

XY 1 YZ 5 XZ VW 1 10 1 20 5 52

WX 1 WX 5 20 VW 5 22

2(WX) 5 20

WX 5 10


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23. WY 5 WX 1 XY 24. VX 1 XZ 5 VZ

WY 5 10 1 10 VX 1 20 5 52

WY 5 20 VX 5 32

25. WZ 5 WX 1 XY 1 YZ 26. VY 5 VW 1 WY

WZ 5 10 1 10 1 10 5 30 VY 5 22 1 20 5 42

27. C; EF 1 FG 5 EG 28. RS 1 ST 5 RT

6 1 x 5 1.6x 2x 1 10 1 x 2 4 5 21

6 5 0.6x 3x 1 6 5 21

10 5 x 3x 5 15

x 5 5

RS 5 2(5) 1 10 5 20

ST 5 5 2 4 5 1

29. RS 1 ST 5 RT 30. RS 1 ST 5 RT

3x 2 16 1 4x 2 8 5 60 2x 2 8 1 3x 2 10 5 17

7x 2 24 5 60 5x 2 18 5 17

7x 5 84 5x 5 35

x 5 12 x 5 7

RS 5 3(12) 2 16 5 20 RS 5 2(7) 2 8 5 6

ST 5 4(12) 2 8 5 40 ST 5 3(7) 2 10 5 11

31. AC 1 CD 5 12

AC 5 CD 5 6

AB 1 BC 5 6

AB 5 BC 5 3

AB 5 3, BC 5 3, AC 5 6, CD 5 6, BD 5 9,AD 5 12

Because 4 of the 6 segments in the fi gure are longer than 3 units, the probability of choosing one of these

is 4 }

6 or

2 }

3 .

Problem Solving

32. Abdomen 5 2 1 }

4 2 0 5 2 1 }

4

thorax 5 4 2 2 1 }

4 5 1

3 }

4

Its abdomen is 2 1 }

4 2 1

3 }

4 5

1 } 2 inch longer than its thorax.

33. a. AB 1 BC 5 AC

1282 1 601 5 1883

The total distance was 1883 miles.

b. d 5 rt

1883 5 r(38)

49.6 ø r The airplane’s average speed was about 50 miles per

hour.

34. a. 2003: 11 2 6 5 5

2004: 12 2 7 5 5

2005: 13 2 8 5 5

The length of the yellow bar represents the number of losses in that year.

b. 2003: 5 }

11 5 0.45 5 45%

The team lost 45% of their games in 2003.

2004: 5 }

12 5 0.42 5 42%


2005: 5 }

13 5 0.38 5 38%


c. You apply the Segment Addition Postulate by subtracting one color of the stacked bar from the entire length of the bar, as you would subtract the length of a short line segment from the length of a longer segment that contains it.

35. a. A

B

C

31 ft

52 ft

b. AC 2 AB 5 BC

52 2 31 5 BC

21 5 BC

The climber must descend 21 feet farther to reach the bottom.

36. City A City B City C City D

City A 12.5 mi 15 mi 25 mi

City B 12.5 mi 2.5 mi 12.5 mi

City C 15 mi 2.5 mi 10 mi

City D 25 mi 12.5 mi 10 mi

AB 5 5(BC)

AD 5 2(AB)

CD 5 10

A B C D

10 mixy

y 5 5x

y 1 x 1 10 5 2y

x 1 10 5 y

x 1 10 5 5x

10 5 4x

2.5 5 x

y 5 5(2.5) 5 12.5

AC 5 12.5 1 2.5 5 15

AD 5 15 1 10 5 25

BD 5 2.5 1 10 5 12.5


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5Geometry


Mixed Review

37. Ï}

45 1 99 5 Ï}

144 5 12

38. Ï}

14 1 36 5 Ï}

50 5 5 Ï}

2

39. Ï}

42 1 (22)2 5 Ï}

42 1 4 5 Ï}

46

40. 4m 1 5 5 7 1 6m 41. 13 2 4h 5 3h 2 8

22 5 2m 13 5 7h 2 8

21 5 m 21 5 7h

3 5 h

42. 17 1 3x 5 18x 2 28

45 5 15x

3 5 x

43. True 44. True 45. False

Lesson 1.3


1. ####$ MN is a segment bisector of }

PQ .

PQ 5 2 1 1 7 }

8 2 5 3

3 }

4

2. Line l is a segment bisector of }

PQ .

5x 2 7 5 11 2 2x

7x 2 7 5 11

7x 5 18

x 5 18

} 7

5 1 18 } 7 2 2 7 5

41 } 7

PQ 5 2 1 41 } 7 2 5

82 } 7

3. M 1 1 1 7 }

2 ,

2 1 8 }

2 2 5 (4, 5)

4. 4 1 x

} 2 5 21

4 1 y }

2 5 22

4 1 x 5 22 4 1 y 5 24

x 5 26 y 5 28

The coordinates of endpoint V are (26, 28).

5. It does not matter which ordered pair you substitute for (x1, y1

) or which you substitute for (x2, y2) because the

distance between the two points is the same no matter which you start with.

6. B; AB 5 Î}}

(x2 2 x1)2 1 ( y2 2 y2

)2

5 Ï}}}

(1 2 (23))2 1 (24 2 2)2

5 Ï}

16 1 36

5 Ï}

52 ø 7.2

The approximate length of }

AB is 7.2 units.


Skill Practice

1. To fi nd the length of }

AB , with endpoints A(27, 5) and B(4, 26), you can use the distance formula.

2. To bisect a segment means to intersect a segment at its midpoint. You cannot bisect a line because it continues forever in both directions and, therefore, has no midpoint.

3. RS 5 ST 5 5 1 }

8 in. 4. UV 5 VW 5

5 } 8 in.

RS 1 ST 5 RT UV 1 VW 5 UW

5 1 }

8 1 5

1 }

8 5 RT

5 }

8 1

5 } 8 5 UW

RT 5 10 1 }

4 in. UW 5 1

1 }

4 in.

5. EF 5 FG 5 13 cm

EF 1 FG 5 EG

13 1 13 5 EG

EG 5 26 cm

6. AB 5 BC 5 1 } 2 (AC) 7. PQ 5 QR 5

1 } 2 (PR)

BC 5 1 } 2 (19) 5 9

1 }

2 cm QR 5

1 } 2 1 9

1 }

2 2 5 4

3 }

4 in.

8. LM 5 MN 5 1 } 2 (LN) 9. RQ 5

1 } 2 (PQ)

LM 5 1 } 2 (137) 5 68

1 }

2 mm RQ 5

1 } 2 1 4

3 }

4 2 5 2

3 }

8 in.

10. UV 5 2(UT ) 11. x 1 5 5 2x

UV 5 2 1 2 7 }

8 2 5 5

3 }

4 m 5 5 x

AM 5 x 1 5

5 5 1 5 5 10

12. 7x 5 8x 2 6 13. 6x 1 7 5 4x 1 5

2x 5 26 2x 1 7 5 5

x 5 6 2x 5 22

EM 5 7x 5 7(6) 5 42 x 5 21

JM 5 6x 1 7

5 6(21) 1 7 5 1

14. 6x 2 11 5 10x 2 51

211 5 4x 2 51

40 5 4x

10 5 x

PR 5 6x 2 11 1 10x 2 51

5 6(10) 2 11 1 10(10) 2 51

5 60 2 11 1 100 2 51 5 98

15. x 1 15 5 4x 2 45

60 5 3x

20 5 x

SU 5 x 1 15 1 4x 2 45

5 20 1 15 1 4(20) 2 45 5 70


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16. 2x 1 35 5 5x 2 22

35 5 3x 2 22

57 5 3x

19 5 x

XZ 5 2x 1 35 1 5x 2 22

5 2(19) 1 35 1 5(19) 2 22 5 146

17. M 1 3 1 7 }

2 ,

5 1 5 }

2 2 5 M (5, 5)

18. M 1 0 1 4 }

2 ,

4 1 3 }

2 2 5 M (2, 3.5)

19. M 1 24 1 6 }

2 ,

4 1 4 }

2 2 5 M (1, 4)

20. M 1 27 1 (23) }

2 ,

25 1 7 }

2 2 5 M (25, 1)

21. M 1 28 1 11 }

2 ,

27 1 5 }

2 2 5 M (1.5, 21)

22. M 1 23 1 (28) }

2 ,

3 1 6 }

2 2 5 M (25.5, 4.5)

23. Substitute the given numbers and variables into the midpoint formula.

M 1 x1 1 x2 }

2 ,

y1 1 y2 }

2 2

M 1 0 1 m }

2 ,

0 1 n }

2 2

Simplify the expression.

M 1 m } 2 ,

n }

2 2

24. When calculating the averages of the x-coordinates and of the y-coordinates, x1 and x2 and y1 and y2 should have been added, not subtracted.

1 8 1 2 }

2 ,

3 1 (21) }

2 2 5 (5, 1)

25. B(3, 0), M(0, 5)

3 1 x

} 2 5 0

0 1 y }

2 5 5

x 5 23 y 5 10

The other endpoint is S (23, 10).

26. B(5, 1), M(1, 4)

5 1 x

} 2 5 1

1 1 y }

2 5 4

5 1 x 5 2 1 1 y 5 8

x 5 23 y 5 7

The other endpoint is S(23, 7).

27. R(6, 22), M(5, 3)

6 1 x

} 2 5 5

22 1 y }

2 5 3

6 1 x 5 10 22 1 y 5 6

x 5 4 y 5 8


28. R(27, 11), M(2, 1)

27 1 x

} 2 5 2

11 1 y }

2 5 1

27 1 x 5 4 11 1 y 5 2

x 5 11 y 5 29


29. R(4, 26), M(27, 8)

4 1 x

} 2 5 27

26 1 y }

2 5 8

4 1 x 5 214 26 1 y 5 16

x 5 218 y 5 22


30. R(24, 26), M(3, 24)

24 1 x

} 2 5 3

26 1 y }

2 5 24

24 1 x 5 6 26 1 y 5 28

x 5 10 y 5 22

The other endpoint is S(10, 22).

31. PQ 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}

(5 2 1)2 1 (4 2 2)2

5 Ï}

16 1 4 5 Ï}

20 ø 4.5 units

32. QR 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}

(2 2 (23))2 1 (3 2 5)2

5 Ï}

25 1 4 5 Ï}

29 ø 5.4 units

33. ST 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}

(3 2 (21))2 1 (22 2 2)2

5 Ï}

16 1 16 5 Ï}

32 ø 5.7 units

34. D; MN 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}}

(4 2 (23))2 1 (8 2 (29))2

5 Ï}

49 1 289 5 Ï}

338 ø 18.4 units.

35. The length of the segment is

3 2 (24) 5 7 5 7 units.

The midpoint has the coordinate x1 1 x2

} 2 5

24 1 3 } 2 5 2

1 } 2 .


2 2 (26) 5 8 5 8 units.


} 2 5

26 1 2 } 2 5 22.


25 2 (215) 5 40 5 40 units.


} 2 5

215 1 25 } 2 5 5.


25 2 (220) 5 15 5 15 units.

The midpoint has the coordinate

x1 1 x2

} 2 5

220 1 (25) } 2 5 212.5.


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7Geometry


39. The length of the segment is 1 2 (28) 5 9 units.


x1 1 x2

} 2 5

28 1 1 } 2 5 23.5.

40. The length of the segment is 22 2 (27) 5 5 units.


x1 1 x2

} 2 5

27 1 (22) } 2 5 24.5.

41. B; LF 5 Ï}}

(3 2 (22))2 1 (1 2 2)2

5 Ï}

25 1 1 5 Ï}

26 ø 5.10

JR 5 Ï}}}

(2 2 1)2 1 (23 2 (21))2

5 Ï}

1 1 4 5 Ï}

5 ø 2.24

The approximate difference in lengths is 5.10 2 2.24 5 2.86.

42. Substitute the coordinates of point P and of the midpoint into the distance formula to fi nd the length of the segment from P to M.

PM 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}

(1 2 (22))2 1 (0 2 4)2

5 Ï}

9 1 16 5 Ï}

25 5 5

If the distance from point P to the midpoint is 5 units, then the length of

} PQ is twice that, or 2(5) 5 10 units.

43. AB 5 Ï}}

(23 2 0)2 1 (8 2 2)2

5 Ï}

9 1 36 5 Ï}

45 ø 6.71

CD 5 Ï}}}

(0 2 (22))2 1 (24 2 2)2

5 Ï}

4 1 36 5 Ï}

40 ø 6.32

The segments are not congruent.

44. EF 5 Ï}}

(5 2 1)2 1 (1 2 4)2

5 Ï}

16 1 9 5 Ï}

25 5 5

GH 5 Ï}}

(1 2 (23))2 1 (6 2 1)2

5 Ï}

16 1 25 5 Ï}

41 ø 6.40


45. JK 5 Ï}}

(4 2 (24))2 1 (8 2 0)2

5 Ï}

64 1 64 5 Ï}

128 ø 11.31

LM 5 Ï}}}

(3 2 (24))2 1 (27 2 2)2

5 Ï}

49 1 81 5 Ï}

130 ø 11.40


46. SP 5 PT, so P is the midpoint of }

ST .

0 1 221

S P T

x

SP 5 PT

x 2 0 5 1 2 x

2x 5 1

x 5 1 } 2

47. 2(JM) 5 JK

2 1 x } 8 2 5

3x } 4 2 6

4 1 x } 4 2 5 1 3x

} 4 2 6 2 4 x 5 3x 2 24

22x 5 224

x 5 12

JM 5 MK 5 x } 8 5

12 } 8 5 1

1 }

2

Problem Solving

48. MR 5 QM 5 18 1 }

2 feet

QR 5 2(QM) 5 2 1 18 1 }

2 2 5 37 feet

49.

5.7 km

House SchoolLibrary

2.85 km

The library is 2.85 kilometers from the house.

50. a. A(1, 1), B(4, 2)

AB 5 Ï}}

(4 2 1)2 1 (2 2 1)2

5 Ï}

9 1 1 5 Ï}

10 ø 3.2 m

b. B(4, 2), C(2, 6)

BC 5 Ï}}

(2 2 4)2 1 (6 2 2)2

5 Ï}

4 1 16 5 Ï}

20 ø 4.5 m

c. C(2, 6), D(5, 4)

CD 5 Ï}}

(5 2 2)2 1 (4 2 6)2

5 Ï}

9 1 4 5 Ï}

13 ø 3.6 m

d. A(1, 1), D(5, 4)

AD 5 Ï}}

(5 2 1)2 1 (4 2 1)2

5 Ï}

16 1 9 5 Ï}

25 5 5 m

e. B(4, 2), D(5, 4)

BD 5 Ï}}

(5 2 4)2 1 (4 2 2)2

5 Ï}

1 1 4 5 Ï}

5 ø 2.2 m

f. A(1, 1), C(2, 6)

AC 5 Ï}}

(2 2 1)2 1 (6 2 1)2

5 Ï}

1 1 25 5 Ï}

26 ø 5.1 m

51. The objects at B and D are closest together. The objects at A and C are farthest apart.


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52. AB 5 Ï}}

(18 2 8)2 1 (7 2 4)2

5 Ï}

100 1 9 5 Ï}

109 ø 10.4

Player A threw the ball about 10.4 meters.

BC 5 Ï}}

(24 2 18)2 1 (14 2 7)2

5 Ï}

36 1 49 5 Ï}

85 ø 9.2

Player B threw the ball about 9.2 meters.

AC 5 Ï}}

(24 2 8)2 1 (14 2 4)2

5 Ï}

256 1 100 5 Ï}

356 ø 18.9

Player A would have thrown the ball about 18.9 meters to player C.

53. a. P(10, 50), Q(10, 10), R(80, 10)

PQ 5 Ï}}}

(10 2 10)2 1 (10 2 50)2 5 Ï}

1600 5 40

QR 5 Ï}}}

(80 2 10)2 1 (10 2 10)2 5 Ï}

4900 5 70

RP 5 Ï}}}

(80 2 10)2 1 (10 2 50)2 5 Ï}

6500 ø 80.62

40 1 70 1 81 5 191

The distance around the park is about 191 yards.

b. M 1 x1 1 x2 } 2 ,

y1 1 y2 }

2 2

M 1 10 1 80 }

2 ,

50 1 10 }

2 2

M (45, 30)

QM 5 Ï}}}

(45 2 10)2 1 (30 2 10)2

5 Ï}

1225 1 400 5 Ï}

1625 ø 40.3

QM is about 40 yards.

c. It takes him about 1.5 minutes.

PQ 5 40, QM 5 40, MR 5 1 } 2 (80) 5 40, RQ 5 70

40 1 40 1 40 1 70 1 40 5 230

Divide the total distance, about 230 yards, by 150 yards per minute.

54.

A B

D

C

M

AB 5 2AM

CM 5 1 } 2 CD

AB 5 4 p CM

2AM 5 4 1 1 } 2 2 CD

2AM 5 2CD

AM 5 CD

AM and CD are equal.

Mixed Review

55. 56 1 16 5 72

72% of the students belong to families with 2 or more children.

56. 25(0.56) 5 14

14 students belong to families with two children.

57. 3x 1 12 1 x 5 20 58. 9x 1 2x 1 6 2 x 5 10

4x 5 8 10x 1 6 5 10

x 5 2 10x 5 4

x 5 2 } 5

59. 5x 2 22 2 7x 1 2 5 40

22x 2 20 5 40

22x 5 60

x 5 230

60. ###$ BA , ###$ BC , ###$ BE

61. ###$ BE , ###$ BC , ###$ CE , ###$ CB , ###$ EB , ###$ CD , ###$ DC , ###$ EC

62. ###$ CB , and ###$ CE

63. @##$ AB and @##$ BC intersect at point B.

64. @##$ BC intersects plane P at @##$ AB .

Quiz 1.1–1.3 (p. 22)

1. Sample answer:

PB

C

A

k l

2. DE 5 AE 2 AD

DE 5 26 2 15 5 11

3. AB 5 AD

} 3 5 15

} 3 5 5

4. AC 5 2AB 5 2(5) 5 10

5. BD 5 AC 5 10

6. CE 5 CD 1 DE 5 5 1 11 5 16

7. BE 5 BD 1 DE 5 10 1 11 5 21

8. M 1 x1 1 x2 }

2 ,

y1 1 y2 }

2 2

M 1 22 1 2 }

2 ,

21 1 3 }

2 2 5 M (0, 1)

The coordinates of the midpoint are (0, 1). The distance between R and S is about 5.7 units.

RS 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}}

(2 2 (22))2 1 (3 2 (21))2

5 Ï}

16 1 16 ø 5.66 units


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9Geometry


Mixed Review of Problem Solving (p. 23)

1. a. B(1, 7), C (7, 7), D (10, 7), E (10, 3)

BD 5 Ï}}

(10 2 1)2 1 (7 2 7)2 5 Ï}

81 5 9

DE 5 Ï}}

(10 2 10)2 1 (3 2 7)2 5 Ï}

16 5 4

9 1 4 5 13

You travel 13 miles using existing roads.

b. BC 5 Ï}}

(7 2 1)2 1 (7 2 7)2 5 Ï}

36 5 6

CE 5 Ï}}

(10 2 7)2 1 (3 2 7)2 5 Ï}

9 1 16 5 5

6 1 5 5 11

You travel 11 miles using the new road.

c. 13 2 11 5 2

The trip is 2 miles shorter.

2. 23x 1 5 5 25x 2 4

5 5 2x 2 4

9 5 2x

4.5 5 x

PQ 5 23(4.5) 1 5 1 25(4.5) 2 4

5 103.5 1 5 1 112.5 2 4 5 217

5 217

3. d 5 rt

r 5 2.4 km/h

t 5 45 min 5 0.75 h

d 5 2.4(0.75) 5 1.8

5.4 2 1.8 5 3.6

You need to hike 3.6 kilometers to reach the end of the trail.

4. FH 5 2.8 m 5 280 cm

280

} 2 5 140

The midpoint of } FH is 140 cm from F. }

EG intersects } FM 143 cm from F, so it is not a bisector.

5. Using the midpoint formula.

M 1 x1 1 x2 }

2 ,

y1 1 y2 }

2 2

M 1 24 1 6 }

2 ,

5 1 (25) }

2 2

M (1, 0)

These are the coordinates of point E (1, 0).

To fi nd point D, substitute the coordinates of point C into the midpoint formula, and set each coordinate equal to the corresponding coordinate from the midpoint E.

2 1 x

} 2 5 1

8 1 y }

2 5 0

2 1 x 5 2 8 1 y 5 0

x 5 0 y 5 28

(0, 28) are the coordinates for point D.

6. Sample answer:

x

y

A(0, 2) B(6, 2)

C(6, 0)D(0, 0)21

1

Perimeter 5 AB 1 BC 1 CD 1 AD

AB 5 Ï}}

(6 2 0)2 1 (2 2 2)2 5 Ï}

36 5 6

BC 5 Ï}}

(6 2 6)2 1 (2 2 0)2 5 Ï}

4 5 2

CD 5 Ï}}

(6 2 0)2 1 (0 2 0)2 5 Ï}

36 5 6

AD 5 Ï}}

(0 2 0)2 1 (2 2 0)2 5 Ï}

4 5 2

Perimeter 5 6 1 2 1 6 1 2 5 16

x

y

E(0, 7) F(1, 7)

G(1, 0)H(0, 0)

222

Perimeter 5 EF 1 FG 1 GM 1 EH

EF 5 Ï}}

(1 2 0)2 1 (7 2 7)2 5 Ï}

1 5 1

FG 5 Ï}}

(1 2 1)2 1 (7 2 0)2 5 Ï}

49 5 7

GH 5 Ï}}

(1 2 0)2 1 (0 2 0)2 5 Ï}

1 5 1

EH 5 Ï}}

(0 2 0)2 1 (7 2 0)2 5 Ï}

49 5 7

Perimeter 5 1 1 7 1 1 1 7 5 16

7. The plane that contains B, F, and C can be called plane BFG, plane FGC, plane GCB, or plane CBF. Plane ABC intersects plane BFE at

} AB .

8. a. AB 5 18.7 km

BC 5 2AB 5 2(18.7) 5 37.4 km

AC 5 AB 1 BC 5 18.7 1 37.4 5 56.1 km

AB 1 BC 1 CA 5 18.7 1 37.4 1 56.1 5 112.2 km

Jill travels 112.2 kilometers.

b. d 5 rt

112.2 5 70t

1.6 5 t

She spends about 1.6 hours driving.

c. No. Sample answer: If she spends 2.5 hours in each town. 2.5(3) 5 7.5 hours spent in towns. The time spent in the three towns plus the total driving time is 7.5 1 1.6 5 9.1 hours.

Lesson 1.4


1. ∠PQR

∠RQS

∠PQS

∠PQS is a right angle.


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2. PR

The rays form a straight angle.

3. m∠KLN 1 m∠NLM 5 1808

10x 2 5 1 4x 1 3 5 180

14x 2 2 5 180

14x 5 182

x 5 13

10(13) 2 5 5 125

m∠KLN 5 1258

14(13) 1 3 5 55

m∠NLM 5 558

4. m∠EFM 1 m∠HFG 5 908

2x 1 2 1 x 1 1 5 90

3x 1 3 5 90

3x 5 87

x 5 29

m∠EFM 5 2(29) 1 2 5 608

m∠MFG 5 29 1 1 5 308

5. ∠QPT > ∠QRS

∠PTS > ∠RST

6. If m∠QRS 5 848 and ∠QRS > ∠QPT, then m∠QPT 5 848. If m∠TSR 5 1218 and ∠TSR > ∠STP, then m∠STP 5 1218.

7.

PNM

m∠MNQ 5 908

m∠PNQ 5 908

1.4 Exercises (pp. 28–32)

Skill Practice

1. Answers may vary. Sample answers:

CB

A

FE

D

KJ

H NML

2. The measure of ∠PQR is equal to the absolute value of the difference between the degree measures of ###$ QP and ###$ QR .

3. ∠ABC, ∠CBA, ∠B

B is the vertex, and ###$ BA and ###$ BC are the sides.

4. ∠NQT, ∠TQN, ∠Q

Q is the vertex, and ###$ QN and ###$ QT are the sides.

5. ∠MTP, ∠PTM, ∠T

T is the vertex, and ###$ TM and ###$ TP are the sides.

6. ∠QRS, ∠TRS, ∠QRT

7. Straight 8. Acute 9. Right 10. Obtuse

11–14. G

F

L

H

K

J

11. m∠JFL 5 908, Right

12. m∠GFM 5 608, Acute

13. m∠GFK 5 1358, Obtuse

14. m∠GFL 5 1808, Straight

15. Another name for ∠ACB is ∠BCA. The angle is a right angle because it is labeled with a red square.

16. Another name for ∠ABC is ∠CBA. The angle is an acute angle because its measure is less than 908.

17. Another name for ∠BFD is ∠DFB. The angle is a straight angle because its measure is 1808

18. Another name for ∠AEC is ∠CEA. The angle is an obtuse angle because its measure is between 908 and

1808.

19. Another name for ∠BDC is ∠CDB. The angle is an acute angle because its measure is less than 908.

20. Another name for ∠BEC is ∠CEB. The angle is an acute angle because its measure is less than 908.

21. B; m∠ACD is between 908 and 1808.

22. m∠QST 5 m∠QSR 1 m∠RST 5 528 1 478 5 998

23. m∠ADC 5 m∠ADB 1 m∠CDB 5 218 1 448 5 658

24. m∠NPM 5 m∠LPM 2 m∠LPN 5 1808 2 798 5 1018

25. x 1 5 1 3x 2 5 5 80

4x 5 80

x 5 20

m∠YXZ 5 3(20) 2 5 5 558

26. 6x 2 15 1 x 1 8 5 168

7x 2 7 5 168

7x 5 175

x 5 25

m∠FJG 5 6(25) 2 15 5 1358

27. A; 2x 1 6 1 80 5 140

2x 5 54

x 5 27


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11Geometry


28. ∠AED > ∠ADE > ∠BDC > ∠BCD > ∠DAB > ∠ABD; ∠EAD > ∠DBC > ∠ADB

m∠BDC 5 348

m∠ADB 5 1128

29. m∠ZWY 5 m∠XWY 5 528

m∠XWY 5 2(52) 5 1048

30. m∠ZWX 5 1 } 2 m∠XWY 5 688

m∠XWY 5 2(68) 5 1368

31. m∠ZWY 5 1 } 2 ∠XWY 5 35.58

m∠XWZ 5 718 2 35.58 5 35.58

32.

L

M

K

J

308

m∠JKL is twice the measure of ∠JKM, not half of it. m∠JKL 5 608

33. a8 5 180 2 142 5 388

34. b8 > a8 5 388

35. c8 5 1428

36. d 8 5 180 2 53 2 90 5 378

37. e8 5 538

38. f 8 > d > 378

39. For a ray to bisect ∠AGC, the endpoint of the ray must be at point G.

CG

A

40. 4x 2 2 5 3x 1 18

x 5 20

m∠ABC 5 4(20) 2 2 1 3(20) 1 18 5 1568

41. 2x 1 20 5 4x

20 5 2x

10 5 x

m∠ABC 5 2(10) 1 20 1 4(10) 5 808

42. x }

2 1 17 5 x 2 33

x 1 34 5 2x 2 66

100 5 x

m∠ABC 5 100

} 2 1 17 1 100 2 33 5 1348

43. ###$ QP lines up with the 758 mark. The new mark for ###$ QR is 58 less than before. The difference between the marks that ###$ QR and ###$ QP line up with on the protractor must remain the same.

44.

1

x

y

A

B C22

The angle is acute.

Sample answer: point (2, 1) lies on the interior of the angle.

45.

3

x

yA B

C

22

∠ABC is acute.

Sample answer: point (22, 2) lies in the interior of the angle.

46.

1

x

yA

BC

21

∠ABC is obtuse.

Sample answer: point (21, 21) lies in the interior of the angle.

47.

1

x

y

A

B

C

1

∠ABC is obtuse.

Sample answer: Point (2, 21) lies in the interior of the angle.

48. 0 < (2x 2 12)8 < 908

0 1 12 < 2x < 90 1 12

12 < 2x < 112

68 < x < 518

49. 688

Sample answer: Since m∠VSP 5 178, m∠RSP 5 348.

Since m∠RSP 5 348, m∠RSQ 5 688, which is equal to m∠TSQ.

50. m∠AEB 5 1 } 2 p m∠CED

1 }

2 p m∠CED 1 90 1 m∠CED 5 180

1 1 }

2 p m∠CED 1 90 5 180

1 1 }

2 p m∠CED 5 90

m∠CED 5 608

m∠AEB 5 1 } 2 (60) 5 308


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Problem Solving

51. m∠LMP 5 m∠LMN 2 m∠PMN

m∠LMP 5 79 2 47

m∠LMP 5 328

52.

Sydney Street Malcolm Way

Park Road

162°

87°

1628 2 878 5 758

Malcom Way intersects Park Road at an angle of 758.

53. a. m∠DEF 5 m∠ABC 5 1128

b. Because ###$ BG bisects ∠ABC, m∠ABG 5 1 } 2 p m∠ABC

5 1 } 2 (1128) 5 568.

c. Because ###$ BG bisects ∠ABC, m∠CBG 5 1 } 2 p m∠ABC

5 1 } 2 (1128) 5 568.

d. Because ###$ BG bisects ∠DEF, m∠DEG 5 1 } 2 p m∠DEF

5 1 } 2 (1128) 5 568.

54. Because ∠DGF is a straight angle, m∠DGF 5 1808.

m∠DGE 5 908

m∠FGE 5 908

55. Sample answer: Acute: ∠ABG, Obtuse: ∠ABC, Right: ∠FGE, Straight: ∠DGF

56. about 1588 57. about 1408 58. about 1678

59. about 628 60. about 398 61. about 1078

62. a. ∠AFB, ∠BFC, ∠CFD, ∠DFE are acute.

∠AFD, ∠BFD, ∠BFE are obtuse.

∠AFC and ∠CFE are right.

b. ∠AFB > DFE, ∠AFC > ∠CFE, ∠BFC > ∠DFC

c. m∠AFB > m∠DFE 5 268

m∠BFC 5 908 2 m∠AFD 5 908 2 268 5 648

m∠BFC > m∠CFD 5 648

m∠AFC 5 908

m∠AFD 5 908 1 m∠CFD 5 908 1 648 5 1548

m∠BFD 5 m∠BFC 1 m∠CFO 5 648 1 648 5 1288

∠DFE > ∠AFB, so they both have an angle measure of 268.

∠BFC and ∠AFB form a 908 angle so their measurements add up to 908, making m∠BFC 5 648. ∠CFD > ∠BFC, so it also has an angle measurement of 648. ∠AFC is a right angle, so m∠AFC 5 908. m∠AFD 5 m∠AFC 1 m∠CFD 5 908 1 648 5 1548, m∠BFD 5 m∠BFC 1 m∠CFD 5 648 1 648 5 1288.

63. Sample answer: In your drawer you have 4 pairs of brown socks, 4 pairs of black socks, 4 pairs of gray socks, 6 pairs of white socks, and 6 pairs of blue socks.

The brown, black and gray socks each represent 1 }

6 , and

the white and blue socks each represent 1 }

4 .

Mixed Review

64. Let x 5 cost of your friend’s meal and x 1 4 5 cost of your meal.

x 1 (x 1 4) 5 25

2x 5 21

x 5 10.5

x 1 4 5 14.5

Your meal costs $14.50. The cost of your friend’s meal is $14.50 2 $4 5 $10.50.

65. x ≤ 28

121232529 27

The graph is a ray.

66. x ≥ 6

022 2 4 6 8

The graph is a ray.

67. 23 ≤ x ≤ 5

0 4 622224


68. x ≥ 27 and x ≤ 21

0 22628 2224


69. x ≥ 22 or x ≤ 4

0 6422224


70. x ≥ 0

x 5 all real numbers

0 321212223


71. M 5 26 1 (21)

} 2 5 23.5

72. M 5 215 1 45

} 2 5 15

73. M 5 220 1 (24)

} 2 5 212

Investigating Geometry 1.4 (p. 34)

1. Sample answer: Draw a segment more than twice as long as the given segment. Set your compass to the length of the given segment. Using your compass, mark off two adjacent line segments on the line segment you drew.


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13Geometry


2. Sample answer:

A

D

Step 1

A

C

B

D

Step 2

A

C

B

D

Step 3

D

F

E

E

E

Step 4

Step 2

D

F

Step 1

E

D

F

E

Step 3

D

F

E

Step 4

D

F

E

P

Lesson 1.5


1. Because 418 1 498 5 908, ∠FGK and ∠GKL are complementary.

Because 1318 1 498 5 1808, ∠HGK and ∠GKL are supplementary.

Because ∠FGK and ∠HGK share a common vertex and side, they are adjacent.

2. No, they do not share a common vertex.

No, they do have common interior points.

3. m∠1 1 m∠2 5 908 4. m∠3 1 m∠4 5 808

m∠1 1 88 5 908 1178 1 m∠4 5 1808

m∠1 5 828 m∠4 5 638

5. m∠LMN 1 m∠PQR 5 908

(4x 2 2)8 1 (9x 1 1)8 5 908

13x 2 1 5 90

13x 5 91

x 5 7

m∠LMN 5 4(7) 2 2 5 268

m∠PQR 5 9(7) 1 1 5 648

6. No, adjacent angles have their noncommon sides as opposite rays.

Angles 1 and 4, 2 and 5, 3 and 6 are vertical angles because each pair of sides form two pairs of opposite rays.

7. Let x8 be the measure of the angle’s complement.

2x8 1 x8 5 908

3x 5 90

x 5 30

One angle measures 308 and the other angle measures 2(308) 5 608.

1.5 Exercises (pp. 38–41)

Skill Practice

1. Sample answer:

D

C

B

A

No, two angles could have angle measures that add up to 908 without sharing a common vertex and side.

2. All linear pairs are supplementary angles because their noncommon sides are opposite rays which form a straight angle. All supplementary angles are not linear pairs. Two angles can have angle measurements that add up to 1808 without their noncommon sides being opposite rays.

3. ∠ABD and ∠DBC are adjacent.

4. ∠WXY and ∠XYZ are not adjacent.

5. ∠LQM and ∠NQM are adjacent.

6. Because 608 1 308 5 908, ∠STR and ∠VWU are complementary.

Because 1508 1 308 5 1808 ∠QTS and ∠VWU, are supplementary.

7. ∠GLH and ∠HLJ are complementary because their measures add up to 908. ∠GLJ and ∠JLK are supplementary because their measures add up to 1808.

8. m∠1 1 m∠2 5 908 9. m∠1 1 m∠2 5 908

438 1 m∠2 5 908 218 1 m∠2 5 908

m∠2 5 478 m∠2 5 698

10. m∠1 1 m∠2 5 908 11. m∠1 1 m∠2 5 908

898 1 m∠2 5 908 58 1 m∠2 5 908

m∠2 5 18 m∠2 5 858


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12. m∠1 1 m∠2 5 1808 13. m∠1 1 m∠2 5 1808

608 1 m∠2 5 1808 1558 1 m∠2 5 1808

m∠2 5 1208 m∠2 5 258

14. m∠1 1 m∠2 5 1808 15. m∠1 1 m∠2 5 1808

1308 1 m∠2 5 1808 278 1 m∠2 5 1808

m∠2 5 508 m∠2 5 1538

16. B; 908 2 378 5 538

17. m∠DEG 1 m∠GEF 5 1808

(18x 2 9)8 1 (4x 1 13)8 5 1808

22x 1 4 5 180

22x 5 176

x 5 8

m∠DEG 5 18(8) 2 9 5 1358

m∠GEF 5 4(8) 1 13 5 458

18. m∠DEG 1 m∠GEF 5 1808

(7x 2 3)8 1 (12x 2 7)8 5 1808

19x 2 10 5 180

19x 5 190

x 5 10

m∠DEF 5 7(10) 2 3 5 678

m∠GEF 5 12(10) 2 7 5 1138

19. m∠DEG 1 m∠GEF 5 908

6x8 1 4x8 5 908

10x 5 90

x 5 9

m∠DEG 5 6(9) 5 548

m∠GEF 5 4(9) 5 368

20. ∠1 and ∠4 are vertical angles.

21. ∠1 and ∠2 are a linear pair.

22. ∠3 and ∠5 are neither.

23. ∠2 and ∠3 are vertical angles.

24. ∠7, ∠8, and ∠9 are neither.

25. ∠5 and ∠6 are a linear pair.



28. Use the fact that angles in a linear pair are supplementary angles.

x8 1 4x8 5 1808

5x 5 180

x 5 36

4(36) 5 144

One angle is 368, and the other angle is 1448.

29. The angles are complementary so the sum of their measures equals 908.

x8 1 3x8 5 908

4x 5 90

x 5 22.5

30. C; x8 1 (x 1 24)8 5 908

2x 5 66

x 5 33

33 1 24 5 57

31. 7x8 1 (9x 1 20)8 5 1808

16x 5 160

x 5 10

7x8 5 2y8

7(10) 5 2y

70 5 2y

35 5 y

32. 3x8 1 (8x 1 26)8 5 1808

11x 5 154

x 5 14

(5y 1 38)8 5 (8x 1 26)8

5y 1 38 5 8(14) 1 26

5y 1 38 5 138

5y 5 100

y 5 20

33. 2y8 5 (x 1 5)8

3y 1 30 5 4x 2 100

2y 2 5 5 x

3y 1 30 5 4(2y 2 5) 2 100

3y 1 30 5 8y 2 20 2 100

3y 1 30 5 8y 2 120

150 5 5y

30 5 y

2(30) 2 5 5 x

55 5 x

34. Never; the measure of an obtuse angle is greater than 908, so its angle measurement cannot be added to the measurement of another angle to equal 908.

35. Never; the measure of a straight angle is 1808, so its measurement cannot be added to the measurement of another angle to equal 908.

36. Sometimes; an angle that measures less than 1808 has a supplement.

37. Always; for the measurements of two angles to add up to 908, they must be be acute.

38. Always; An acute angle measures less than 908 so its supplement must measure between 908 and 1808 for the two to add up to 1808.

39. m∠A 1 m∠B 5 908

(3x 1 2)8 1 (x 2 4)8 5 908

4x 2 2 5 90

4x 5 92

x 5 23

m∠A 5 3(23) 1 2 5 718

m∠B 5 23 2 4 5 198


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15Geometry


40. m∠A 1 m∠B 5 908

(15x 1 3)8 1 (5x 2 13)8 5 908

20x 2 10 5 90

20x 5 100

x 5 5

m∠A 5 15(5) 1 3 5 788

m∠B 5 5(5) 2 13 5 128

41. m∠A 1 m∠B 5 908

(11x 1 24)8 1 (x 1 18)8 5 908

12x 1 42 5 90

12x 5 48

x 5 4

m∠A 5 11(4) 1 24 5 688

m∠B 5 4 1 18 5 228

42. m∠A 1 m∠B 5 1808

(8x 1 100)8 1 (2x 1 50)8 5 1808

10x 1 150 5 180

10x 5 30

x 5 3

m∠A 5 8(3) 1 100 5 1248

m∠B 5 2(3) 1 50 5 568

43. m∠A 1 m∠B 5 1808

(2x 2 20)8 1 (3x 1 5)8 5 1808

5x 2 15 5 180

5x 5 195

x 5 39

m∠A 5 2(39) 2 20 5 588

m∠B 5 3(39) 1 5 5 1228

44. m∠A 1 m∠B 5 1808

(6x 1 72)8 1 (2x 1 28)8 5 1808

8x 1 100 5 180

8x 5 80

x 5 10

m∠A 5 6(10) 1 72 5 1328

m∠B 5 2(10) 1 28 5 488

45. Given ∠GHJ is a complement of ∠RST; m∠GMJ 1 m∠RST 5 908

x8 1 m∠RST 5 908

m∠RST 5 908 2 x8

Given ∠RST is a supplement of ∠ABC

m∠RST 1 m∠ABC 5 1808, so 908 2 x8 1 m∠ABC 5 1808

m∠ABC 5 908 1 x8

Problem Solving

46. The angles are complementary because the sum of their measures is 908.

47. The angles are neither complementary nor supplementary because the sum of their measures is greater than 1808.

48. The angles are supplementary because the sum of their measure is 1808.

49. Sample answer: ∠FGA and ∠AGC are supplementary.

50. Sample answer: ∠AGB and ∠EGD are vertical angles.

51. Sample answer: ∠FGE and EGC are a linear pair.

52. Sample answer: ∠CGD and ∠DGE are adjacent angles.

53. Sample answer: Because ∠FGB and ∠BGC are supplementary angles, m∠BGC 5 1808 2 m∠FGB

m∠BGC 5 1808 2 1208

m∠BGC 5 608.

54. As the sun rises, the shadow becomes shorter and the angle increases.

55. a. y1 5 90 2 x

Domain: 0 < x < 90

y2 5 180 2 x

Domain: 0 < x < 180

The measure of a complement must be less than 908 and the measure of its supplement must be less than 1808.

b.

040 80 120 1600

40

80

120

160

x

y

y1 5 90 2 x

y2 5 180 2 x

Range of y1: 0 < y < 90

Range of y2: 0 < y < 180

56. Let x8 be the measure of one angle and let y8 be the measure of the other angle.

x8 1 y8 5 908

x8 1 y8 5 x8 2 y8 1 868

x 5 90 2 y

90 2 y 1 y 5 90 2 y 2 y 1 86

90 5 176 2 2y

286 5 22y

43 5 y

x 1 43 5 90

x 5 47

One angle measures 438 and the other angle measures 478.


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Mixed Review

57. y 5 5 2 x

y 5 5 2 x

x

y

1

1

x 22 21 0 1 2

y 7 6 5 4 3

58. y 5 3x

x

y

1

2

y 5 3x x 22 21 0 1 2

y 26 23 0 3 6

59. y 5 x2 2 1

x

y

1

1

y 5 x2 2 1

x 22 21 0 1 2

y 3 0 21 0 3

60. y 5 22x2

x 22 21 0 1 2

y 28 22 0 22 28

x

y

2

2

y 5 22x2

61. } LH > } LK

} HJ > } KJ

∠HLK > ∠HJK

∠LHJ > ∠LKJ

62. } FE > }

EG > }

FG

∠EFG > ∠FEG > ∠EGF

63. }

AB > }

BC > }

CD > } DA

∠A > ∠B > ∠C > ∠D

Quiz 1.4–1.5 (p. 41)

1. m∠ABD 5 m∠DBC

(x 1 20)8 5 (3x 2 4)8

24 5 2x

12 5 x

m∠ABD 5 12 1 20 5 328

m∠DBC 5 3(12) 2 4 5 328


(10x 2 42)8 5 (6x 1 10)8

4x 5 52

x 5 13

m∠ABD 5 10(13) 2 42 5 888

m∠DBC 5 6(13) 1 10 5 888


(18x 1 27)8 5 (9x 1 36)8

9x 5 9

x 5 1

m∠ABD 5 18(1) 1 27 5 458

m∠DBC 5 9(1) 1 36 5 458

4. a. 908 2 478 5 438

The measure of the complement of ∠1 is 438.

b. 1808 2 478 5 1338

The measure of the supplement of ∠1 is 1338

5. a. 908 2 198 5 718


b. 1808 2 198 5 1618

The measure of the supplement of ∠1 is 1618.

6. a. 908 2 758 5 158


b. 1808 2 758 5 1058


7. a. 908 2 28 5 888


b. 1808 2 28 5 1788.


Lesson 1.6


1. Sample answers:

Convex heptagon Concave heptagon

2. Quadrilateral; each of the sides is 2 meters long and all the angles are right angles.

3. 8y8 5 (9y 2 15)8

2y 5 215

y 5 15

8(15) 5 120

9(15) 2 15 5 120

Each angle measures 1208.


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17Geometry



Skill Practice

1. The term n-gon is used to name a polygon, where n is the number of sides of the polygon.

2. Yes, the string will lie on the sides of the figure so it will match the distance around the polygon.

No, because the string cannot lie on the concave sides, the length of the string will be less than the distance around the polygon.

3. The figure is a concave polygon.

4. Part of the figure is not a segment, so it is not a polygon.

5. The figure is a convex polygon.

6. Some segments intersect more than two segments, so it is not a polygon.

7. C; the figure is a polygon and is not convex.

8. The polygon has 8 sides. It is equilateral and equiangular, so it is a regular octogon.

9. The polygon has 5 sides. It is equilateral and equiangular, so it is a regular pentagon.

10. The polygon has 3 sides so the figure is a triangle. It is equilateral and equiangular, so it is regular.

11. The polygon has 3 sides, so the figure is a triangle. It is not equilateral or equiangular, so it is not regular.

12. The polygon has 4 sides, so it is a quadrilateral. It is equilateral but not equiangular, so it is not regular.

13. The polygon is a quadrilateral because it has 4 sides. It is equiangular but not equilateral, so it is not regular.

14. Student A: The error is the hexagon must be convex.

Student B: The error is the hexagon does not have congruent sides.

15. 5x 2 27 5 2x 2 6

3x 5 21

x 5 7

2(7) 2 6 5 8

A side of the pentagon is 8 inches.

16. (9x 1 5)8 5 (11x 2 25)8

30 5 2x

15 5 x

9(15) 1 5 5 140

An angle of the nonagon measures 1408.

17. 3x 2 9 5 23 2 5x

8x 5 32

x 5 4

3(4) 2 9 5 3

A side of the triangle is 3 feet.

18. A triangle is always convex, because no line that contains a side of the triangle contains a point in the interior of the triangle.

19. A decagon is sometimes regular, because all of its sides and all of its angles can be congruent, but they don’t have to be.

20. A regular polygon is always equiangular, because all of its angles in the interior of the polygon are congruent.

21. A circle is never a polygon, because a circle does not have sides.

22. A polygon is always a plane figure, because a polygon is a closed plane figure.

23. A concave polygon is never regular, because a regular polygon is not concave.

24. Sample answer: 25. Sample answer: (Equilateral, (Equilateral, not equiangular) not equiangular)

26. Sample answer: 27. Sample answer: (Equilateral, (Equiangular, not equiangular) not equilateral)

28. x2 1 4 5 x2 1 x

4 5 x

29. x2 1 3x 5 x2 1 x 1 2

2x 5 2

x 5 1

30. x2 1 2x 1 40 5 x2 2 x 1 190

3x 5 150

x 5 50

31. m∠BAC 5 368, m∠ABC 5 728, m∠ACB 5 728; Sample answer: Because the pentagonal tiles are regular, all of their interior angles are congruent. By setting the given expressions for angle measures equal to each other and solving for x, we can fi nd the measure of each angle.

(20x 1 48)8 5 (33x 1 9)8

39 5 13x

3 5 x

20(3) 1 48 5 108

Each interior angle of the pentagonal tiles measures 1088. So, m∠CAD 5 1088. Because ∠BAE is a straight angle, m∠BAC 1 m∠CAD 1 m∠DAE 5 1808. Since ∠BAC > ∠DAE,

2 p m∠BAC 1 1088 5 1808

2 p m∠BAC 5 728

m∠BAC 5 368

∠ACB is supplementary to one of the 1088 interior angles of a pentagon. So 1808 2 1088 5 728 5 m∠ACB. Because the angles of a triangle must add up to 1808, m∠ABC 5 1808 2 728 2 368 5 728.


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Problem Solving

32. a. The red polygon is convex because no line that contains a side of the polygon contains a point in the interior of the polygon.

b. The polygon has 8 sides. It appears to be equilateral and equiangular, so it is a regular octagon.

33. The polygon has 3 sides. It appears to be equilateral and equiangular, so it is a regular triangle.

34. The polygon has 4 sides so it is a quadrilateral. It appears to be equiangular but not equilateral, so it is not regular.

35. The polygon has 8 sides. It appears to be equilateral and equiangular, so it is a regular octagon.

36. The polygon has 12 sides, so it is a decagon. It is concave, so it is not regular.

37. C; AB 5 Ï}}

(0 2 0)2 1 (24 2 4)2 5 Ï}

64 5 8

CD 5 Ï}}

(8 2 8)2 1 (24 2 4)2 5 Ï}

64 5 8

AD 5 Ï}}

(8 2 0)2 1 (4 2 4)2 5 Ï}

64 5 8

BC 5 Ï}}}

(8 2 0)2 1 (24 2 (24))2 5 Ï}

64 5 8

C(8, 24) and D(8, 4)

38. a. ; The polygon is a quadrilateral.

b. It appears to be regular, and it is convex.

a. ; The polygon is an octagon.

b. It appears to be regular, and it is convex.

a. ; The polygon is a pentagon.

b. It is not equilateral or equiangular, and it is convex.

a. ; The polygon is a heptagon.

b. It is not equilateral or equiangular, and it is concave.

a. The polygon is a dudecagon.

b. It appears to be equilateral but it is concave, so it is not regular.

39. 105 mm; Because the button is a regular polygon, set the expressions given for the length of the sides equal to each other and solve for x.

3x 1 12 5 20 2 5x

8x 5 8

x 5 1

Substitute the value for x back into one of the expressions to fi nd the length of one side.

3(1) 1 12 5 15 mm

Because there are 7 sides, multiply the length of one side by 7 to fi nd the length of silver wire needed.

15(7) 5 105

40. a. Type of polygon

DiagramNumber of sides

Number of diagonals

Quadrilateral 4 2

Pentagon 5 5

Hexagon 6 9

Heptagon 7 14

Sample answer: Every time you add another side, you increase the number of diagonals by the amount added to the previous polygon plus 1.

b. An octagon has 20 diagonals. A nonagon has 27 diagonals. Sample answer: The pattern described continues.

c. 60(60 2 3)

} 2 5

3420 } 2 5 1710

A 60-gon has 1710 diagonals.

41. a. A regular triangle has 3 lines of symmetry.

b. A regular pentagon has 5 lines of symmetry.

c. A regular hexagon has 6 lines of symmetry.

d. A regular octagon has 8 lines of symmetry.

42.

Mixed Review

43. 1 }

2 (35)b 5 140 44. x2 5 144

35b 5 280 Ï}

x2 5 Ï}

144

b 5 8 x 5 612

45. 3.14r2 5 314

r2 5 100

Ï}

r2 5 Ï}

100

r 5 610

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19Geometry



46. 1 500 m }

1 2 1 100 cm

} 1 m

2 5 50,000 cm

500 m 5 50,000 cm

47. 1 12 mi }

1 2 1 5280 ft

} 1 mi

2 5 63,360 ft

12 mi 5 63,360 ft

48. 1 672 in. }

1 2 1 1 yd

} 36 in.

2 5 18 2 }

3 yd

672 in. 5 18 2 }

3 yd

49. 1 1200 km }

1 2 1 1000 m

} 1 km

2 5 1,200,000 m

1200 km 5 1,200,000 m

50. 1 4 1 }

2 ft }

1 2 1 1 yd

} 3 ft

2 5 1 1 }

2 yd

4 1 }

2 ft 5 1

1 }

2 yd

51. 1 3800 m }

1 2 1 1 km

} 1000 m

2 5 3.8 km

3800 m 5 3.8 km

52. d 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1)2

5 Ï}}}

(0 2 (213))2 1 (212 2 13)2

5 Ï}

169 1 625 5 Ï}

794 ø 28.2

53. d 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}}

(29 2 (29))2 1 (7 2 (28))2

5 Ï}

0 1 225 5 Ï}

225 5 15

54. d 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

5 Ï}}}

(22 2 10)2 1 (22 2 5)2

5 Ï}

144 1 49 5 Ï}

193 ø 13.9

Lesson 1.7Investigating Geometry 1.7 (p. 48)

1. Sample answer:

4

16

8

8

246810

L1321610.6686.4

L2684033.333232.8

L3

2. The shape of the rectangle with the smallest perimeter is a square.


1. A 5 lw P 5 2l 1 2w

5 13(5.7) 5 2(13) 1 2(5.7)

5 74.1 5 37.4

The area is 74.1 m2. The perimeter is 37.4 m.

2. A 5 s2 P 5 4s

5 (1.6)2 5 4(1.6)

5 2.56 5 6.4

The area is about 2.6 cm2. The perimeter is 6.4 cm.

3. A 5 πr2 C 5 2πr

5 3.14(2)2 5 2(3.14)(2)

5 12.56 5 12.56

The area is about 12.6 yd2. The circumference is about 12.6 yd.

4. The height will be a perpendicular line segment from F to

} EG . Find the length by using the x-coordinate for

F and the x-coordinate for }

EG . The x-cordinate for F is 7 and the x-coordinate for

} EG is 1.

7 2 1 5 6

The height from F to }

EG is 6 units.

5. EF 5 Ï}}

(7 2 1)2 1 (3 2 6)2 5 Ï}

36 1 9 ø 6.71

FG 5 Ï}}

(7 2 1)2 1 (3 2 2)2 5 Ï}

36 1 1 ø 6.08

EG 5 Ï}}

(1 2 1)2 1 (2 2 6)2 5 Ï}

16 5 4

P 5 a 1 b 1 c 5 6.71 1 6.08 1 4 5 16.79

b 5 4, h 5 6

A 5 1 } 2 bh 5

1 } 2 (4)(6) 5 12

The perimeter is about 16.8 units, and the area is 12 square units.

6. No; Doubling the length and width will not double the area. The area will be 4 times greater, so it will take longer than twice the time needed to resurface the original rink.

A 5 1 } 2 bh

64 5 1 } 2 b(16)

8 5 b

The base is 8 meters.


Skill Practice

1. Sample answer: The diameter is twice the length of the radius.

2. Sample answer: Find the perimeter of a yard to fence in; fi nd the area of the ceiling to paint it; ft: ft2.

3. The formula for area of a triangle is A 5 1 } 2 bh, not A 5 bh.

A 5 1 } 2 (52)(9) 5 234 ft2

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4. P 5 2l 1 2w 5 2(18) 1 2(8) 5 52 ft

A 5 lw 5 18(8) 5 144 ft2

5. P 5 2l 1 2w 5 2(7) 1 2(4.2) 5 22.4 m

A 5 lw 5 7(4.2) 5 29.4 m2

6. P 5 4s 5 4(15) 5 60 in.

A 5 s2 5 152 5 225 in.2

7. P 5 a 1 b 1 c 5 30 1 72 1 78 5 180 yd

A 5 1 } 2 bh 5

1 } 2 (72)(30) 5 1080 yd2

8. P 5 a 1 b 1 c 5 15 1 15 1 24 5 54 mm

A 5 1 } 2 bh 5

1 } 2 (24)(9) 5 108 mm2

9. P 5 a 1 b 1 c 5 17 1 9 1 10 5 36 cm

A 5 1 } 2 bh 5

1 } 2 (9)(8) 5 36 cm2

10.

32 ft

16 ft12

A 5 1 } 2 bh 5

1 } 2 (32) 1 16

1 }

2 2 5 264 ft2

11. C 5 πd ø 3.14(27) ø 84.8 cm

A 5 πr2 ø 3.14 1 27 }

2 2 2 ø 572.3 cm2

12. C 5 πd ø 3.14(5) ø 15.7 in.

A 5 πr2 ø 3.14 1 5 } 2 2 2 ø 19.6 in.2

13. C 5 2πr ø 2(3.14)(12.1) ø 76.0 cm

A 5 πr2 ø 3.14 (12.1)2 ø 459.7 cm2

14. C 5 2πr ø 2(3.14)(3.9) ø 24.5 cm

A 5 πr2 ø 3.14 (3.9)2 ø 47.8 cm2

15.

18.9 cm

C 5 πd ø 3.14(18.9) ø 59.3 cm

A 5 πr2 ø 3.14 1 18.9 }

2 2 2 ø 280.4 cm2

16. R(2, 4), S (6, 2), T (2, 1)

RS 5 Ï}}

(6 2 2)2 1 (2 2 4)2 5 Ï}

16 1 4 5 Ï}

20 ø 4.47

ST 5 Ï}}

(2 2 6)2 1 (1 2 2)2 5 Ï}

16 1 1 5 Ï}

17 ø 4.12

TR 5 Ï}}

(2 2 2)2 1 (1 2 4)2 5 Ï}

9 5 3

P 5 RS 1 ST 1 TR 5 4.47 1 4.12 1 3 5 11.59

The perimeter is about 11.6 units.

17. E (1, 1), F (4, 1), G (21, 4)

EF 5 Ï}}

(4 2 1)2 1 (1 2 1)2 5 Ï}

9 5 3

FG 5 Ï}}

(21 2 4)2 1 (4 2 1)2

5 Ï}

25 1 9 5 Ï}

34 ø 5.83

GE 5 Ï}}

(1 2 (21))2 1 (1 2 4)2 5 Ï}

4 1 9

5 Ï}

13 ø 3.6

P 5 EF 1 FG 1 GE 5 3 1 5.83 1 3.6 5 12.43


18. M(22, 1), N(3, 2), P (2, 22), Q (22, 22)

MN 5 Ï}}

(3 2 (22))2 1 (2 2 1)2

5 Ï}

25 1 1 5 Ï}

26 ø 5.1

NP 5 Ï}}

(2 2 3)2 1 (22 2 2)2

5 Ï}

1 1 16 5 Ï}

17 ø 4.12

PQ 5 4, QM 5 3

P 5 MN 1 NP 1 PQ 1 QM

5 5.1 1 4.12 1 4 1 3 5 16.22


19. D; A(0, 2), B(4, 4), C (5, 2), D (1, 0)

AB 5 Ï}}

(4 2 0)2 1 (4 2 2)2

5 Ï}

16 1 4 ø 4.47

BC 5 Ï}}

(5 2 4)2 1 (2 2 4)2

5 Ï}

1 1 4 ø 2.24

A 5 lw ø 4.47(2.24) ø 10.01

20. 187 cm2 p 1 m2 }

10,000 cm2 5 0.0187 m2

21. 13 ft2 p 1 yd2

} 9 ft2

ø 1.44 yd2

22. 18 in.2 p 1 ft2 }

144 in.2 5 0.125 ft2

23. 8 km2 p 1,000,000 m2

} 1 km2 5 8,000,000 m2

24. 12 yd2 p 9 ft2 }

1 yd2 5 108 ft2

25. 24 ft2 p 144 in.2 }

1 ft2 5 3456 in.2

26. D; 2.25 ft2 p 144 in.2 }

1 ft2 5 324 in.2

27. A 5 1 } 2 bh 28. A 5

1 } 2 bh

261 5 1 } 2 (36) h 66 5

1 } 2 (b) 1 8

1 }

4 2

14.5 5 h 16 5 b

The height is 14.5 m. The base is 16 inches.

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21Geometry



29. P 5 2l 1 2w 30. P 5 2l 1 2w

25 5 2(8) 1 2w 102 5 2l 1 2(17)

9 5 2w 68 5 2l

4.5 5 w 68 5 2l

The width is 4.5 inches. 34 5 l

The length is 34 inches.

31. A 5 lw

l 5 2w

18 5 2w(w)

18 5 2w2

9 5 w2

3 5 w l 5 2(3) 5 6

The length is 6 inches, and the width is 3 inches.

32. A 5 1 } 2 bh

h 5 3b

27 5 1 } 2 b(3b)

27 5 1 1 }

2 b2

18 5 b2

Ï}

18 5 b

b 5 3 Ï}

2 h 5 3(3 Ï}

2 ) 5 9 Ï}

2

The height is 9 Ï}

2 feet, and the base is 3 Ï}

2 feet.

33. octagon; dodecagon; the square has four sides, so a polygon with the same side length and twice the perimeter would have to have 2(4) 5 8 sides; a polygon with the same side length and three times the perimeter would have to have 4(3) 5 12 sides.

34. A 5 s2

184 5 s2

2 Ï}

46 5 s

The side length is 2 Ï}

46 centimeters.

35. A 5 s2

346 5 s2

Ï}

346 5 s

The side length is Ï}

346 inches.

36. A 5 s2

1008 5 s2

12 Ï}

7 5 s


7 miles.

37. A 5 s2

1050 5 s2

5 Ï}

42 5 s


42 kilometers.

38. Let x 5 diameter of red circle

2x 5 diameter of yellow circle

The area of the red circle is A 5 π 1 x } 2 2 2 and the area of

the yellow circle is A 5 π 1 2x} 2 2 2 or A 5 πx2.

The fraction that expresses the amount of the red circle

that is covered by the yellow one is π 1 x }

2 2

2

} π x2 5

1 } 4 x2

} x2 5

1 } 4 .

The fraction of the red circle that is not covered by the

yellow circle is 1 2 1 } 4 5

3 } 4 .

39. P 5 2l 1 2w

26 5 2l 1 2w

13 5 l 1 w

13 2 l 5 w

A 5 lw

30 5 lw

30 5 l(13 2 l)

30 5 13l 2 l2

l2 2 13l 1 30 5 0

l 5 10, l 5 3

30 5 10w

3 5 w

13 2 l 5 w

l 5 13 2 3 5 10

The length is 10 centimeters and the width is 3 centimeters.

Problem Solving

40. C 5 πd ø 3.14(60) ø 188.4

A 5 πr2 ø 3.14 1 60 }

2 2 2 ø 2826

The circumference is about 188.4 inches, and the area is about 2826 square inches.

41. A 5 lw 5 45(30) 5 1350

P 5 2l 1 2w 5 2(45) 1 2(30) 5 150

You need to cover 1350 square yards with grass seed,

and you need 150 yd p 3 ft }

1 yd 5 450 feet of fencing.

42. a. l 5 0.84 m, w 5 0.54 m

A 5 lw 5 0.84(0.54) 5 0.4536 m2

The area is 0.4536 m2.

b. If its area were 1 square meter it could generate 125 watts of power. If its area were 2 square meters, it could generate 125(2) 5 250 watts of power.

c. Chris’s solar panel can generate 125(0.4536) 5 56.7 watts of power. This is found by multiplying the watts of power per meter by the area of the panel.

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43. a. C 5 2πr

r 5 21 2 x

94 ø 2(3.14)(21 2 x)

94 ø 6.28(21 2 x)

94 ø 131.88 2 6.28x

237.88 ø 26.28x

6.03 ø x

r ø 21 2 6.03 ø 14.97

The radius of the outer edge is about 15 inches.

b. 6 in.; the spoke is 21 inches long from the center to the tip, and it is 15 inches from the center to the outer edge. So, 21 2 15 5 6 inches is the length of the handle.

44. a. Length, x 1 2 5 10 25

Perimeter, y1 4 8 20 40 100

Area, y2 1 4 25 100 625

b. (1, 4), (2, 8), (5, 20), (10, 40), (25, 100), (1, 1), (2, 4), (5, 25), (10, 100), (25, 265)

Length

Per

imet

er

0 5 10 15 20 25 x

y1

0

20

40

60

80

100

Length

Are

a

0 5 10 15 20 25 x

y2

0

100

200

300

400

500

600

c. Sample answer: As the length increases, the difference between the perimeter and area also increases.

45. a. A 5 lw 5 15.2(7) 5 106.4

The area of the grid is 106.4 m2.

b. 380 rows, 175 columns. Sample answer: The panel is 1520 centimeters high and each module is 4 centimeters square so there are 1520 4 4 5 380 rows; the panel is 700 centimeters wide and each module is 4 centimeters square so there are 700 4 4 5 175 columns.

46. a. C 5 2πr ø 2(3.14)(115,800) ø 727,224

C ø 2(3.14)(120,600) ø 757,368

The circumference of the red ring is about 727,200 kilometers and the circumference of the yellow ring is about 757,400 kilometers.

b. 757,400 2 727,200 5 30,200; The yellow ring’s circumference is about 30,200 kilometers greater than the red ring’s circumference.

47. π

} 2 ; the area of the square is 2r2 and the area of the circle

is πr2, so the circle is π�r2

} 2r2 5

π }

2 times greater than the

area of the square.

48. a. P 5 2l 1 2w

30 5 2x 1 2w

15 5 x 1 w

w 5 15 2 x

y 5 lw

y 5 x(15 2 x)

b. For the greatest possible area, use a length and width of 7.5 yards each. A square maximizes the area.

Mixed Review

49. y 5 2x 1 1

x 1 2 3 4 5

y 3 5 7 9 11

50. 94, 88, 82; 100 2 6 5 94, 94 2 6 5 88, 88 2 6 5 82

51.

8x 2 135x 1 40

5x 1 40 5 8x 2 13

53 5 3x

17 2 }

3 5 x

8 1 17 2

} 3 2 2 13 5 128

1 }

3

A side of the triangle is 128 1 }

3 inches.

52. (10x 1 20)8 12x 8

12x8 5 (10x 1 20)8

2x 5 20

x 5 10

12(10) 5 120

An angle of the hexagon measures 1208.

Quiz 1.6–1.7 (p. 56)

1. The fi gure is a concave polygon.

2. The fi gure is not a polygon because part of it is not a segment.

3. The fi gure is a convex polygon.

4. P 5 2l 1 2w 5 2(16) 1 2(5) 5 42 yd

A 5 lw 5 16(5) 5 80 yd2

5. P 5 a 1 b 1 c 5 5 } 8 1

3 } 8 1

1 } 2 5 1

1 }

2 in.

A 5 1 } 2 bh 5

1 } 2 1 1 }

2 2 1 3 }

8 2 5

3 } 32 in.2

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23Geometry



6. P 5 a 1 b 1 c 5 5 1 4.5 1 8.5 5 18 m

A 5 1 } 2 bh 5

1 } 2 (4.5)(4) 5 9 m2

7. A 5 lw 5 3 1 }

2 1 2

1 }

2 2 5 8

3 }

4 yd2

8 3 }

4 yd2 p 9 ft2

} 1 yd2 5 78

3 }

4 ft2

Divide the area of the yard by the number of square feet one bag of wood chips can cover.

78 3 }

4 4 10 5 7

7 }

8 ft2

You need 8 bags of wood chips.

Problem Solving Workshop 1.7 (p. 57)

1. 1 } 8 mi p 5280 ft

} 1 mi

5 660 ft

Time (hour) Area Plowed (ft2)

1 1 p 180,000 5 180,000

2 2 p 180,000 5 360,000

t t p 180,000 5 A

A 5 6602 5 435,600 ft2

180,000t 5 A

180,000t 5 435,600

t 5 2.42

It takes about 2.4 hours.

2. 660 is the length of the fi eld, not the area.

A 5 6602 5 435,600

435,600 5 180,000t

3. l 5 110 yd 5 330 ft

w 5 45 yd 5 135 ft

Area (ft2) Cost ($)

1 1 p $.60 5 $.60

2 2 p $.60 5 $1.20

A A p $.60 5 C

C 5 $.60A

C 5 $.60(330 p 135) 5 $26,730

It will cost $26,730 to pave the parking lot.

4. d 5 (120π) p 3 ø 376.8 p 3 ø 1130.4

The total distance walked is about 1130.4 meters.

(4 3 1000) 4 60 5 4000 4 60 5 66.67

You walk about 66.67 meters per minute.

1130.4 4 66.67 ø 16.96

So it will take you about 17 minutes to walk around the path 3 times.

Mixed Review of Problem Solving (p. 58)

1. a. l 5 4 yd 5 12 ft

w 5 3 yd 5 9 ft

A 5 lw 5 12(9) 5 108

The area of the roof is 108 square feet.

b. asphalt shingles: 0.75(108) 5 $81

Wood shingles: 1.15(108) 5 $124.20

c. $124.20 2 $81 5 $43.20

You will pay $43.20 more to use wood shingles.

2. Sample answer: GAB, GABCDEF; triangle, heptagon

3. a. ∠DGB and ∠BGH are complementary. ∠HGF and ∠CGF are complementary. ∠DGB and ∠BGF, ∠DGH and ∠HGF, ∠CGF and ∠CGD, ∠DBG and ∠GBH, ∠HCG and ∠GCF are supplementary.

b. m∠FGC 5 218 because m∠DGB 5 218 and ∠FGC > ∠DGB. m∠BGH 5 698 because it is complementary to a 218 angle. m∠HGC 5 698 because it is complementary to a 218 angle.

c. m∠HCG 5 558 because m∠HBG is 558 and ∠HCG > ∠HBG.

m∠DBG 5 1258 because it is supplementary to a 558 angle. m∠FCG 5 1258 because it is supplementary to a 558 angle.

4. m∠1 1 m∠2 5 1808

m∠1 1 m∠3 5 908

m∠1 5 m∠2 2 288

m∠2 2 288 1 m∠2 5 1808

2(m∠2) 5 2088

m∠2 5 1048

m∠1 5 1048 2 288 5 768

768 1 m∠3 5 908

m∠3 5 148

5. a. P 5 4s 5 4(22.5) 5 90 ft

C 5 πd ø 3.14(26) ø 81.64 ft

For the square garden you need enough bricks for

90 ft 5 1080 inches of perimeter. 1080

} 10

5 108 bricks.

You need 108 bricks for the square garden.

The circumference of the circular garden in inches is 81.64 ft(12) ø 979.68 inches.

979.68

} 10

ø 97.97 bricks

You need 98 bricks for the circular garden.

b. You need a total of 108 1 98 5 206 bricks, so 3 bundles are needed.

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6. 7x 2 3 5 4x 1 6

3x 5 9

x 5 3

7(3) 2 3 5 18

18(5) 5 90

90 inches 5 7.5 feet

7.5 feet of bamboo are used in the frame.

7. (2x 1 5)8 1 (9x 2 1)8 5 1808

11x 1 4 5 180

11x 5 176

x 5 16

2(16) 1 5 5 37

m∠ZWX 5 378

8. a. Side length 5 Ï}}

(4 2 0)2 1 (0 2 2)2 5 Ï}

20 5 2 Ï}

5

P 5 4 (2 Ï}

5 ) 5 8 Ï}

5

The perimeter is 8 Ï}

5 units.

b. Area of nABC 5 1 } 2 bh 5

1 } 2 (8)(2) 5 8

Area of nADC 5 1 } 2 (8)(2) 5 8

The area of quadrilateral ABCD is the sum of the areas of triangles ABC and ADC.

8 1 8 5 16

The area of quadrilateral ABCD is 16 square units.

Chapter 1 Review (pp. 60–63)

1. Points A and B are the endpoints of }

AB .

2. Sample answer:

1 2

3. If Q is between points P and R on @##$ PR , and PQ 5 QR, then Q is the midpoint of } PR .

4. Sample answer: Another name for line y is @##$ XY .

5. Sample answer: Points P, X, and N are not collinear.

6. Sample answer: Points N, X, Y, and Z are coplanar.

7. ###$ YX and ###$ YZ are opposite rays.

8. The intersection of line h and plane M is point y.

9. AB 5 AC 2 BC 10. NP 5 NM 1 MP

AB 5 3.2 2 2 NP 5 22 1 8

AB 5 1.2 NP 5 30

11. XY 5 XZ 2 YZ

XY 5 16 2 9

XY 5 7

12. DE 5 11 2 (213) 5 24

GH 5 214 2 (29) 5 5

} DE and }

GH are not congruent because they have different lengths.

13. JM 5 MK

6x 2 7 5 2x 1 3

4x 5 10

x 5 2.5

6(2.5) 2 7 5 8

2(8) 5 16

JK is 16 units.

14. AB 5 Ï}}

(4 2 2)2 1 (3 2 5)2 5 Ï}

4 1 4 ø 2.8

The length is about 2.8 units.

M 1 2 1 4 }

2 ,

5 1 3 }

2 2

M(3, 4)

The coordinates of the midpoint are (3, 4).

15. FG 5 Ï}}

(6 2 1)2 1 (0 2 7)2 5 Ï}

25 1 49 ø 8.6


M 1 6 1 1 }

2 ,

0 1 7 }

2 2

M(3.5, 3.5)

The coordinates of the mid point are (3.5, 3.5).

16. HJ 5 Ï}}

(5 2 (23))2 1 (4 2 9)2 5 Ï}

64 1 25 ø 9.4


M 1 23 1 5 }

2 ,

9 1 4 }

2 2

M (1, 6.5)

The coordinates of the midpoint are (1, 6.5).

17. KL 5 Ï}}

(0 2 10)2 1 (27 2 6)2 5 Ï}

100 1 169 ø 16.4


M 1 10 1 0 }

2 ,

6 1 (27) }

2 2

M (5, 20.5)

The coordinates of the midpoint are (5, 20.5).

18. 21 1 x

} 2 5 3

5 1 y }

2 5 8

21 1 x 5 6 5 1 y 5 16

x 5 7 y 5 11

The coordinates of endpopint B are (7, 11).

19. EF 5 Ï}}

(8 2 2)2 1 (11 2 3)2 5 Ï}

36 1 64 5 10

} EF is 10 units, so } EM is 5 units.

20. m∠LMP 1 m∠PMN 5 m∠LMN

(11x 2 9)8 1 (5x 1 5)8 5 1408

16x 2 4 5 140

16x 5 144

x 5 9

m∠PMN 5 5(9) 1 5 5 508

21. m∠UVW 5 2 p m∠UVZ

m∠UVW 5 2(818)

m∠UVW 5 1628

The angle is obtuse.

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25Geometry



22. m∠2 5 908 2 m∠1 5 908 2 128 5 788

23. m∠2 5 908 2 m∠1 5 908 2 838 5 78

24. m∠2 5 908 2 m∠1 5 908 2 468 5 448

25. m∠2 5 908 2 m∠1 5 908 2 28 5 888

26. m∠4 5 180 2 m∠3 5 1808 2 1168 5 648

27. m∠4 5 1808 2 m∠3 5 1808 2 568 5 1248

28. m∠4 5 1808 2 m∠3 5 1808 2 898 5 918

29. m∠4 5 1808 2 m∠3 5 1808 2 128 5 1688

30. m∠1 1 m∠2 5 908

(x 2 10)8 1 (2x 1 40)8 5 908

3x 1 30 5 90

3x 5 60

x 5 20

m∠1 5 20 2 10 5 108

m∠2 5 2(20) 1 40 5 808

31. m∠1 1 m∠2 5 1808

(3x 1 50)8 1 (4x 1 32)8 5 1808

7x 1 82 5 180

7x 5 98

x 5 14

m∠1 5 3(14) 1 50 5 928

m∠2 5 4(14) 1 32 5 888

∠1 is obtuse.

32. The polygon has 3 sides. It is equilateral and equiangular, so it is a regular triangle.

33. The polygon has 4 sides. Its angles are all the same, so it is an equiangular quadrilateral.

34. The polygon has 8 sides, so it is an octagon. It is equilateral but not regular because it is concave.

35. }

BC > } DE , so BC 5 DE

5x 2 4 5 2x 1 11

3x 5 15

x 5 5

5(5) 2 4 5 21

}

BC > }

AB , so BC 5 AB 5 21 units.

36. C 5 πd ø 3.14(15.6) 5 49.0 m

A 5 πr2 ø 3.14 1 15.6 }

2 2 2 5 191.0 m2

37. P 5 2l 1 2w

P 5 2 1 4 1 }

2 2 1 2 1 2

1 }

2 2 5 14 in.

A 5 lw 5 4 1 }

2 1 2

1 }

2 2 5 11.3 in.2

38. UV 5 Ï}}

(28 2 1)2 1 (2 2 2)2 5 Ï}

81 5 9

VW 5 Ï}}}

(28 2 (24))2 1 (2 2 6)2 5 Ï}

16 1 16 ø 5.7

WU 5 Ï}}

(24 2 1)2 1 (6 2 2)2 5 Ï}

25 1 16 ø 6.4

P 5 UV 1 VW 1 WU 5 9 1 5.7 1 6.4 5 21.1

A 5 1 } 2 bh

A 5 1 } 2 (9)(4) 5 18

39. A 5 1 } 2 bh

46.5 5 1 } 2 b(18.6)

5 5 b

The base is 5 meters.

40. A 5 πr 2

320 ø 3.14r 2

101.9 ø r2

10.1 ø r

The radus is about 10.1 meters.

C 5 2πr ø 2(3.14)(10.1) ø 63.4

The circumference is about 63.4 meters.

Chapter 1 Test (p. 64)

1. True

2. False; point D lies on line l.

3. False; Points B, C, and E are coplanar but point Q is not.

4. False; Points C and E are on line l, but point B is not.

5. False; Point Q is not on the plane G.

6. HJ 5 HK 2 JK

HJ 5 52 2 30

HJ 5 22

7. BC 5 AC 2 AB

BC 5 18 2 7

BC 5 11

8. XZ 5 XY 1 YZ

XZ 5 26 1 45

XZ 5 71

9. TW 5 Ï}}

(2 2 3)2 1 (7 2 4)2 5 Ï}

1 1 9 5 Ï}

10 ø 3.2

10. CD 5 Ï}}

(6 2 5)2 1 (21 2 10)2

5 Ï}

1 1 121 5 Ï}

122 ø 11.0

11. MN 5 Ï}}}

(21 2 (28))2 1 (3 2 0)2 5 Ï}

49 1 9

5 Ï}

58 ø 7.6

12. 3 1 x

} 2 5 9

9 1 y }

2 5 7

3 1 x 5 18 9 1 y 5 14

x 5 15 y 5 5

The coordinates of endpoint B are (15, 5).

13. CM 5 MD

3x 5 27

x 5 9

CD 5 3(9) 1 27 5 54

14. L

J

KG

H

m∠GMJ 5 1258; ∠GMJ is obtuse.

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15. m∠KHL 1 m∠CMJ 5 m∠KHJ

(8x 2 1)8 1 (4x 1 7)8 5 908

12x 1 6 5 90

12x 5 84

x 5 7

m∠LMJ 5 4(7) 1 7 5 358

16. m∠QRT 5 m∠SRT 5 1 } 2 m∠QRT

m∠QRT 5 1 } 2 (1548) 5 778

m∠QRS 5 m∠SRT 5 778

17. ∠2 and ∠3, ∠3 and ∠4, ∠1 and ∠4, ∠1 and ∠2 are all linear pairs.

18. ∠2 and ∠4, ∠1 and ∠3 are vertical angles.

19. 908 2 648 5 268

Its complement is 268.

1808 2 648 5 1168

Its supplement is 1168.

20. Sample answer:

concave decagon

convex pentagon

21. 6x 2 12 5 3x 1 6

3x 5 18

x 5 6

6(6) 2 12 5 24

P 5 5(24) 5 120

The perimeter is 120 units.

22. l 5 5.5 yd 5 16.5 ft

w 5 4.5 yd 5 13.5 ft

A 5 lw 5 16.5(13.5) 5 222.75

The total area is 222.75 square feet. The total cost of carpeting is found by multiplying the cost per square yard and the total area.

$1.50 (222.75) 5 $334.13

Since this is more than $300, you cannot afford to buy this carpet.

Chapter 1 Algebra Review (p. 65)

1. 9y 1 1 2 y 5 49 2. 5z 1 7 1 z 5 28

8y 5 48 6z 5 215

y 5 6 z 5 22 1 }

2

3. 24(2 2 t) 5 216 4. 7a 2 2(a 2 1) 5 17

28 1 4t 5 216 7a 2 2a 1 2 5 17

4t 5 28 5a 5 15

t 5 22 a 5 3

5. 4x

} 3 1 2(3 2 x) 5 5 6.

2x 2 5 } 7 5 4

4x

} 3 1 6 2 2x 5 5 2x 2 5 5 28

4x 1 18 2 6x 5 15 2x 5 33

22x 5 23 x 5 16 1 }

2

x 5 1 1 }

2

7. 9c 2 11 5 2c 1 29 8. 2(0.3r 1 1) 5 23 2 0.1r

10c 5 40 0.6r 1 2 5 23 2 0.1r

c 5 4 0.7r 5 21

r 5 30

9. 5(k 1 2) 5 3(k 2 4)

5k 1 10 5 3k 2 12

2x 5 222

x 5 211

10. Let x represent the number of boxes of stationary.

(Cost of each box 3 Number of boxes) 1 Cost of book 5 Gift certifi cate

4.59x 1 8.99 5 50

4.59x 5 41.01

x 5 8.93

You can buy 8 boxes of stationary.

11. Let x represent the number of people.

(Charge per person 3 Number of people) 1 Cost to rent the room 5 Amount to spend

8.75x 1 350 5 500

8.75x 5 150

x 5 17.14

17 people can come to the party.

12. Let x represent the number of smaller beads.

(Length of smaller bead 3 Number of smaller beads) 1 Length of larger bead 5 Length of necklace

3 }

4 x 1 1

1 }

2 5 18

3 }

4 x 5 16

1 }

2

x 5 22

You need 22 smaller beads.

Standardized Test Preparation (p. 67)

1. Full credit. Sample answer: The solution is complete and correct.

2. Partial credit. Sample answer: The reasoning is correct but an incorrect conversion leads to an incorrect solution.

3. No credit. Sample answer: The reasoning and solution are incorrect and do not make sense.

4. Partial credit. Sample answer: The reasoning and computations are correct so far, but a solution is not given.

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27Geometry



Standardized Test Practice (pp. 68–69)

1. Find the area of the fl oor in square feet.

l 5 6 yd 5 18 ft

w 5 4.5 yd 5 13.5 ft

A 5 lw 5 18(13.5) 5 243 ft2

Because this is less than 300 ft2, the cost per square foot is $2. The total cost is 243(2) 5 $486.

2. Find the coordinates of point L and point T.

L (22, 24), T (1, 1)

Use the midpoint formula to fi nd the coordinates of M.

M 1 x1 1 x2 }

2 ,

y1 1 y2 }

2 2

M 1 22 1 1 }

2 ,

24 1 1 }

2 2

M 1 2 1 } 2 , 2

3 } 2 2

Use the distance formula to fi nd the distance between point L and Point M.

d 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1

)2

LM 5 Î}}}

1 2 1 } 2 2 (22) 2 2 1 1 2

3 } 2 2 (24) 2 2

5 Ï}

9 }

4 1

25 } 4

5 Î}

17

} 2 ø 2.92

The distance between the library and your house is about 2.92 km.

3. Find the area of the surface of the water.

l 5 7.5 yd 5 22.5 ft

w 5 3.5 yd 5 10.5 ft

A 5 lw 5 22.5(10.5) 5 236.25 ft2

Multiply the area exposed by 17.6 gallons of water.

236.25(17.6) 5 4158

In one year 4158 gallons of water would evaporate.

4. Find the area of the cover in square yards.

d 5 20 ft 5 6.67 yd

A 5 πr2

A ø 3.14 1 6.67 }

2 2 2 ø 34.9 yd2

Multiply the total area by the cost per square yard.

34.9(4) 5 139.6

The total cost is about $140.

5. To fi nd the length of one side, set the expressions given for the lengths of the sides equal to each other and solve for x.

x 1 5 5 3x 2 19

24 5 2x

12 5 x

12 1 5 5 17

The length of one side is 17 cm.

Since there are 5 sides, the perimeter is the length of one side multiplied by 5.

17(5) 5 85 cm.

30.48 centimeters are in a foot, so 3(30.48) 5 91.44 centimeters are in a yard.

The length of the perimeter in yards is 85 cm p 1 yd }

91.44 cm

5 0.93 yard. The total cost is $1.50(0.93) 5 $1.40.

6. m∠A 1 m∠B 5 908

(2x 2 4)8 1 (4x 2 8)8 5 908

6x 2 12 5 90

6x 5 102

x 5 17

m∠B 5 4(17) 2 8 5 608

The supplement of ∠B is 1808 2 608 5 1208.

7. Use the distance formula to find the distances between each of the towns.

Atkins to Baxton, d1:

d1 5 Ï}}

(5 2 0)2 1 (2 2 2)2 5 Ï}

25 5 5

Baxton to Canton, d 2:

d2 5 Ï}}

(5 2 5)2 1 (5 2 2)2 5 Ï}

9 5 3

Atkins to Canton, d3:

d3 5 Ï}}

(5 2 0)2 1 (5 2 2)2 5 Ï}

25 1 9 ø 5.83

The distance if you go through Baxton is 5 1 3 5 8 km. The distance if you go directly to canton is about 5.83 km. The difference between the two routes is 8 km 2 5.83 km 5 2.17 km.

The trip is about 2.17 km shorter if you do not go through Baxton.

8. Gold wire needed for one earring:

Circumference 5 π p d 5 25π ø 78.54 mm

78.54 mm 1 1 m }

1000 mm 2 5 0.0785 m

Gold wire needed for a pair of earrings:

2(0.0785) 5 0.157 m

Number of pairs of earrings:

2 meters 1 1 pair of earrings }}

0.157 m 2 5 12.74 pairs of earrings.

So, the jeweler can make 12 pairs of earrings.

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9. C;

AB 5 2 p AM

AM 5 Ï}}}

(4 2 (22))2 1 (22 2 6)2 5 Ï}

36 1 64

5 Ï}

100 5 10 units

AB 5 2 p 10 5 20 units

10. C;

Let P be the perimeter, l be the length, and w be the width.

P 5 2l 1 2w

When P 5 85 and l 5 4 1 w:

85 5 2(4 1 w) 1 2w

85 5 8 1 2w 1 2w

85 5 4w 1 8

85 5 2(2w 1 4)

11. m∠XYW 5 m∠WYZ

(6x 2 9)8 5 (5x 1 2)8

x 2 9 5 2

x 5 11

m∠XYZ 5 m∠XYW 1 m∠WYZ

5 6(11) 2 9 1 5(11) 1 2 5 1148

12. m∠A 1 m∠B 5 908

m∠A 5 8 p m∠B

8 p m∠B 1 m∠B 5 908

9 p m∠B 5 908

m∠B 5 108

m∠A 1 108 5 908

m∠A 5 808

1808 2 808 5 1008

The supplement of ∠A is 1008.

13. Perimeter 5 a 1 b 1 c

When a 5 x, b 5 x 1 70, c 5 x 1 90, and the perimeter is 400:

400 5 x 1 (x 1 70) 1 (x 1 90)

400 5 3x 1 160

80 5 x

Area 5 1 } 2 a p b

5 1 } 2 (80) p (80 1 70)

5 6000

The area is 6000 square feet.

14. a. $50,000 1 1 ft2 }

$4 2 5 12,500 ft2

The largest area is 12,500 square feet

b. let A be the area, l be the length, and w be the width.

When A 5 12,500 and l 5 2w:

12,500 5 (2w)w

12,500 5 2w2

6250 5 w2

79 ø w l 5 2(79) 5 158

The field is 158 feet long and 79 feet wide.

15. Amount of black ink for one circle:

Circumference 5 πd 5 π(1) 5 π

Amount of black ink for 30 circles is 30(π) 5 30π ø 94.2

Amount of black ink for one square:

Perimeter 5 4 p s 5 4 p 1 5 4

Amount of black ink for 30 squares is 30(4) 5 120

So, the 30 squares use more black ink.

Amount of red ink for one circle:

Area 5 π(r)2 5 π 1 1 } 2 2 2 5

π }

4 ø 0.8

Amount of red ink for 30 circles is 30(0.8) 5 24

Amount of red ink for one square:

Area 5 s2 5 12 5 1

Amount of red ink for 30 squares is 30(1) 5 30

So, the 30 squares use more red ink.

16. a. AB 5 Ï}}

(7 2 1)2 1 (5 2 7)2 5 Ï}

36 1 4 5 Ï}

40 ø 6.3

CB 5 Ï}}

(7 2 2)2 1 (5 2 2)2 5 Ï}

25 1 9 5 Ï}

34 ø 5.8

The distance from the boat at point A to the buoy is about 6.3 kilometers. The distance from the boat at point C to the buoy is about 5.8 kilometers.

b. Let t be the time (in hours) for a boat to reach the buoy.

Boat at point A:

Distance 5 Rate p Time

6.3 5 5.8 p t

1.26 5 t

Boat at point C:

Distance 5 Rate p Time

5.8 5 5.2 p t

1.12 ø t

Because 1.12 < 1.26, the boat at point C reaches the buoy first.

ngws-01.indd 28 7/11/06 10:58:41 AM

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