mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 223
CHAPTER 11
Think & Discuss (p. 659)
Lesson 11.1
1. 6
11.1 Guided Practice (p. 665)
2. The base angles of an equilateral triangle measure 60
each. 260 120, 120 6 720
1. Interior angles are A, B, D, BCD, and AED.
Exterior angles are AEF, BCG, and DCH.
2. There are 2n exterior angles. No; the Polygon Exterior
Skill Review (p. 660)
Angles Theorem specifies only 1 angle at each vertex.
1
1. A bh
2
3. The sum of the measures of the interior angles of a penta-
gon is 5 2180 540.
1
12 in.8 in.
2
540 105 115 120 105 x
540 445 x
48 in.2
95 x
2. mA mB mC 180
4. Sum of measures of interior angles 6 2180
57 mB 79 180
720 mB 136 180
mB 44
Exterior angle to A is 180 57 123. Exterior angle
to B is 180 44 136. Exterior angle to C is
180 79 101.
XY
12 3
3. a.
DE
8
2
b.
20
29 43.6
21
29 12 2180
1440
1800
9. n 2180
18 2180
2880
21
Number
of sides
3
4
5
6
..
.
Number of
triangles
1
2
3
4
..
.
n
n2
11. n 2180
10. n 2180
29
A
Sum of
measures of
interior angles
1 180 180
2 180 360
3 180 540
4 180 720
..
.
n 2 180
The sum of the measures of the interior angles of any convex
n-gon is n 2180.
20 2180
30 2180
3240
5040
13. n 2180
12. n 2180
Developing Concepts Activity (p. 661)
n-gon
10 2180
2340
B
46.4
Polygon
Triangle
Quadrilateral
Pentagon
Hexagon
..
.
7. n 2180
15 2180
C
20
360
360
45
n
8
11.1 Practice and Applications (p. 665–668)
8. n 2180
4. Sample answer:
mC sin1
5. Measure of an exterior angle 6. n 2180
Perimeter of DEF
8
2
Perimeter of XYZ
12 3
mA sin1
720
120
6
40 2180
100 2180
6840
17,640
14. x 113 80 130 90 5 2180
x 413 540
x 127
15.
x 125 147 106 98 143 6 2180
x 619 720
x 101
16. x 102 146 120 124 170 158
7 2180
x 820 900
x 80
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 11 Worked-out Solution Key
223
mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 224
Chapter 11 continued
17. x n 2 180
n
18. x 5 2 180
5
26.–28.
7 2180
7
128.57
108
19. x n 2 180
n
n 2 180
n
8 2 180
8
29. ext. meas. 360
n
30. ext. meas. 360
n
360
12
360
11
135
30
20. x 80 110 80 4 2180
x 270 360
31. ext. meas. 360
n
32. ext. meas. 360
n
360
21
360
15
x 90
21. x 60 80 120 140 5 2180
x 400 540
x 140
144 22.
n 2180
n
144 n n 2180
144 n 180 n 360
36 n 360
n 10
17.14
360
33. n 60
6
35. n 120 n n 2180
120 n 180 n 360
60 n 360
n6
18
360
72
36. n 5
140 n n 2180
140 n 180 n 360
40 n 360
n9
157.5 25.
n 2180
n
157.5 n n 2180
157.5 n 180 n 360
22.5 n 360
n 16
224
360
10
36
37. x 48 52 55 62 68 360
x 285 360
x 75
360
38. ext. meas. 10
36
39. The measures of the interior angles of the regular hexa-
n 2180
140 n
24.
24
360
34. n 20
n 2180
120 n
23.
32.73
Geometry
Chapter 11 Worked-out Solution Key
gons are 120°. Two of the interior angles of the red triangles form a linear pair with the interior angles of the regular polygons. Therefore, the measure of each interior
angle of the triangles is 60°. The measures of the interior
angles of the yellow hexagons are 120°. Two of the interior angles of the yellow pentagons form linear pairs with
angles in the red triangles, so their measures are 120°.
The measure of a third angle of the yellow pentagons is
540 120 120 90 90 120.
40. Regular pentagons have interior angles measuring 108°.
The interior angles of the red quadrilaterals which form
linear pairs with the interior angles of the pentagons measure 72°. The measure of the angle in the red quadrilateral formed by the joining of two pentagons is 144°. The
measure of the 4th angle of the red quadrilateral is
360 72 72 144 72. For the yellow pentagons, two of the angles form linear pairs with a 72°
angle, so their measures are 108°. The measure of the
fifth angle is 540 108 108 90 90 144.
The measure of each angle of the yellow decagon is 144°.
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 225
Chapter 11 continued
41. m9 and m10 are 70 because 9 and 10 form linear
pairs with 4 and 5, respectively. m3 is 140 because
3 forms a linear pair with 8. m7 is 80 because 7
forms a linear pair with 2. m1 is 80 because the sum
of the measures of the interior angles of a pentagon must
equal 540. m6 is 100 because 6 forms a linear pair
with 1.
42. Any n-gon can be divided into n 2 triangles. The sum of
the measures of the interior angles will be equal to
n 2180. The n-gons do not need to be regular or
similar because the sum of the measures of angles of any
triangle will always be 180. What matters is that n is the
same.
49. 3x 90 90 5 2180
3x 180 540
3x 360
x 120
So, m m
90, mB mC mD 120.
50. x x 2x 160 160 150 150
7 2180
4x 620 900
4x 280
x 70
43. Given pentagon ABCDE, where n 5 draw two diagonals
from one vertex to divide the pentagon into three triangles.
The sum of the measures of the angles of a triangle is
180. Therefore, the sum of the measures of the angles of
the three triangles must be 3180 or (n 2180.
So, mP mV 70° and mS 2x° 140°
51.
or . By the Polygon Interior Angles Thm., the sum of the
measures of the interior s of A is n 2 180. That is,
180, or
x angle with the n-gon. The interior angle and its adjacent
exterior angle at any vertex form a linear pair, so the sum
of their measures is 180. Since the n-gon has n sides,
there will be n linear pairs composed of an interior and
exterior angle, so multiply n by 180. Then subtract the
sum of the measures of the interior angles, which is
n 2 180 to get the total measure of the exterior
angles. So, the sum of the measures of the exterior angles
of a convex n-gon is 180n n 2 180 180n 180n 360 360.
46. The measure of each interior angle of a regular n-gon is
n 2180
360
180 .
n
n
The measure of each exterior angle of a regular n-gon is
180 measure of interior angle. Therefore, the measure
of an exterior angle by substitution is
n 2 180
180n 360
180 180 n
n
360
.
n
47. Sample answer: If a regular hexagon were constructed,
each exterior angle would measure 60 and the sum of the
measures of exterior angles would be 360. Answers
will vary.
53.
n 2180
90
n
180n 360 90n
30n 360
90n 360
n 12
n4
Yes. The polygon would
be a dodecagon.
n 2 180
.
n
45. In a convex n-gon, extend each side to make an exterior
52.
180n 360 150n
44. Let A be a regular n-gon and x the measure of each interi-
n x n 2
n 2180
150
n
n 2180
72
n
Yes. The polygon
would be a square.
54.
n 2180
18
n
180n 360 72n
180n 360 18n
108n 360
162n 360
n
1
33
n 2.22
No. Because n must be
an integer, a regular
polygon can’t have
angles of 18.
No. Because n must be an
integer, a regular polygon
can’t have angles of 72.
55. The function is showing the relationship between the
number of sides, n, and the measure of each interior angle
for a regular n-gon. As n gets larger and larger, f n
approaches 180.
56. f n represents the measure of the exterior angle of a reg-
ular polygon. As n gets larger and larger, f n decreases.
57. The exterior angles measure 55 and 17. Each pair of
consecutive exterior angles measures 55° 17° 72°.
Because 360 72 5, there are 5 pairs of exterior ,
so n 10.
58. B. Column A 1440. Column B 2340.
59. C. Both columns equal 360.
60. A. Column A 74. Column B 73.
61. D. If these were regular polygons, the quantities could be
determined.
48. If the measure of an interior angle increases, the measure
of the corresp. exterior angle decreases. If the measure of
an interior angle decreases, the measure of the corresp.
exterior angle increases. The sum of the measures of the
exterior angles is always 360.
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 11 Worked-out Solution Key
225
mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 226
Chapter 11 continued
62. Each interior angle has a
measure of
8 2180
8 135.
Each exterior angle has a
measure of 45 180 135.
Since the sum of the measures of
the angles of a is 180, mR
must be 180 245 90.
S
T
135
Lesson 11.2
R
45
Z
135
U
1. Six equilateral triangles
Y
V
X
W
1
bh
2
64. A 1
115
2
27.5
1
bh
2
A
1
68
2
1
515
2
37.5 square units
24 square units
?
67.
162
?
81 169 256
132
68.
?
212 722 752
?
441 5184 5625
5625 5625
250 256
Is right .
Not a right .
69. 72 52 217
?
2
?
49 25 68
Not a right .
80
70. m DH
mFH
mFE
mDH
71. m ED
180 35 80
65
mFE
mFH
180 35
145
73.
mE
HG m GD
mDE
mFH
mFE
mDH
mGD
180 180 35 80
245
226
s.
14 3 s 2
3
3 s2
2
11.2 Guided Practice (p. 672)
1. J
2. 5
3. Sample answers: BJA, AJE, BJC, EJD, or DJC
4. KJ
5. Divide 360 by the number of sides.
1
6. A 3 s2
4
1
3 32
4
93
4
7.
360
45
8
3.9 in.2
8. Each side length would be 80 inches 8 or 10 inches.
74 68
72.
6
m2
1
bh
2
2
Area hexagon 6 Area one triangle
height 8
1
bh
2
m EH
3
1
1
3
1
bh s
s 3 s2
2
2
2
4
66. base 6
height 15
92
from one side of the hexagon to the center forming two
30-60-90 right triangles. If the side of the hexagon
A
1
4311
2
236.5
in.2
65. base 5
A
2. To find the area of one triangle, one must draw an altitude
is s, the height of the triangle would be
11.1 Mixed Review (p. 668)
63. A Activity in a Lesson (p. 670)
Geometry
Chapter 11 Worked-out Solution Key
The apothem would be half the height of the sign or 12
inches. The apothem creates a 45-45-90 triangle, so the
radius of the octagon would be equal to the hypotenuse
which is 13 inches.
A
1
1
aP 1280 480 in.2
2
2
11.2 Practice and Applications (pp. 672–675)
1
3 s2
4
1
3 52
4
253
4
10.8 sq. units
9. A 1
3 s2
4
1
2
3 75 4
2543
4
106.1 sq. units
11. A 1
3 s2
4
1
3 112
4
1213
4
52.4 sq. units
10. A 12.
360
40
9
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 227
Chapter 11 continued
13.
360
30
12
360
24
15
14.
16. s 282 42
1
A a
2
2
264 32
82
A
2144 36
123
18. s 220 103
2
2
2400 300
20
530 tan 36
22.
ns
1
63123 2
x
sin 30
7
187.1 sq. units
a 7 cos 30
x 7 sin 30
73
a
units
2
x 3.5 units
1
A a
2
ns
1
103 620
2
s 2x 23.5 7 units
P ns
67
42 units
10
A
5
1
aP
2
1 73
42
2 2
1
5303 2
1473
2
127.31 sq. units
20. P ns
23.
442 162 units
1
aP
2
1
22 162 2
2 2
4
2 2
32 sq. units
360
72
5
The segment whose length is
36
the apothem bisects the
18.54
15
central angle. If a radius
54
is drawn, a 36-54-90
10.9
triangle will be formed.
Use trigonometry ratios to
find the length of the radius r 15
cos 36
and a side. The length of a side of the pentagon
would be: 215 tan 36 30 tan 36
Copyright © McDougal Littell Inc.
All rights reserved.
a
x
1
aP
2
753 129.9 sq. units
21.
30
7
5 3
303 52.0 units
360
60
6
a
cos 30
7
3103 A
1125 tan 36
1083
19. One side 103
1
15150 tan 36
2
A segment whose length is the apothem bisects the central angle. So, a 30-60-90 triangle is formed by this
segment and a radius of the hexagon.
1039.2 sq. units
A
1
aP
2
817.36 sq. units
6003
P ns
109.0 units
1
42 4 82 2
1
a
2
A
150 tan 36
ns
128 sq. units
17. s 2122 62
P ns
360
2
180
15.
360
45
8
A segment whose length is the
apothem bisects the central
x
angle. So,
sin 22.5.
11
x 11sin 22.5
11
22.5
a
x
One side of the octagon 2x 22 sin 22.5.
a
cos 22.5
P ns
11
P 822 sin 22.5
a 11cos 22.5
176 sin 22.5
67.35 units
A
1
aP
2
1
11 cos 22.5176 sin 22.5
2
342.24 sq. units
Geometry
Chapter 11 Worked-out Solution Key
227
mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 228
Chapter 11 continued
24.
28. True. The apothem and radius are the lengths of a leg and
360
30
12
The central angle of a
dodecagon is 30. The central
15
angle is bisected by a segment
whose length is the apothem
4.66
forming a 15-75-90
triangle.
a
x
cos 15
sin 15
9
9
9
a
a 9 cos 15
x 9 sin 15
s 2x 29 sin 15 18 sin 15
P ns
A
1218 sin 15
216 sin 15
55.9 units
1
aP
2
29. False. It depends on the polygon. They are the same for a
regular hexagon.
30. A 163 6
33.
30
15
5 3
a
34.
A segment whose
length is the apothem
bisects the
central angle forming
a
15-75-90 triangle.
1
A aP
2
2
15
75
4
a
tan 75, a 2 tan 75
2
P ns
124
48 in.
60
60
1
is A 3 s2. Since a
4
regular hexagon can be
s , the area of a
divided into six equilateral 1
2
hexagon is A 6 3 s .
4
360
26.
30
12
60
a
sin 60
9
a 9 sin 60
30
93
2
9 in.
a
60
7.8 in.
35.
A
36.–37.
B
A
B
1
2 tan 7548
2
1
48 tan 75
angle, and n the number of sides, r the radius, and P the
perimeter. As n grows bigger, θ will become smaller, so
θ
the apothem, which is given by r(cos ), will get larger.
2
The perimeter of the polygon, which is given by
θ
n(2r sin ), will grow larger, too. Although the factor
2
involving the sine will get smaller, the increase in n more
than makes up for it. Consequently, the area, which is
1
given by aP, will increase.
2
Geometry
Chapter 11 Worked-out Solution Key
1
1
179.14 in.2
27. True; let θ (read “theta”) be the measure of the central
228
360
60 measure of central angle
6
A radius bisects each
interior angle, and with a
segment whose length is the
apothem forms an equilateral
. The area of an equilateral
753
129.9 in.2
77.3 sq. units
6 2180
120 measure of interior angle
6
1
9 cos 15216 sin 15
2
10 3
32 tan 67.5
166.3 sq. units
6.9 sq. units
243 sq. units
1
103 15
2
31. A 4 tan 67.58
963
32. A tan 545
972sin 15cos 15
1
25. A bh
2
the hypotenuse of a right respectively. The hypotenuse
is always longer than the legs so the apothem will always
be less than the radius.
x
38. A 6
43 s2
6
1
1
312
4
39. Do the same construc-
tion as in Exs. 35 and
36, but connect every
other compass mark.
2.6 in.2
40. Sample answer:
41. Sample answer:
C
A
Q
C
B
A
D
Q
B
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1101-sk-.qxd
6-18-2001
3:15 PM
Page 229
Chapter 11 continued
42. Sample answer:
A
C
D
Q
E
49. 6 ft 72 in., 8 ft 96 in.
43. Sample answer:
C
B
A
A 7296 6912 in.2
D
Q
E
6912
74 tiles for the floor total
93.6
number tiles B
74
25 tiles of each color
3
44. Sample answer:
50. B. Column A 45. Column B 51.43.
C
51. A. Column A 0.92. Column B 0.90.
D
A
E
Q
52. A. Column A 6.12. Column B 6.07.
B
53. A Sample answer:
360
72 meas. of central angle
5
A segment whose length is the
apothem bisects the central angle so
with a radius, it forms a 36-54-90
triangle. The apothem can be found
by using trigonometry with r 1 in.
1
aP
2
Area triangle a
cos 36
1
x
sin 36
1
a 0.81 in.
x 0.59 in.
36
a
A
51.18
5.9 in.
Area trapezoid 2.4 in.2
45. The radius side length 0.5 m. A segment whose
length is the apothem bisects a side. So, the length of the
1
base of the right triangle 0.5 0.25 m.
2
The triangle formed by a segment whose length is the
apothem and a radius is a 30-60-90 triangle with the
0.25 m side being opposite the 30 angle. The
apothem 3 0.25 0.43 m.
46. P ns
60.5
3m
A
1
aP
2
1
0.433
2
0.65 m2
Note: See problem 45 for explanation.
47. 3 colors
48. The apothem will be 33 in. based on a 30-60-90
triangle. P ns 66 36 in.
A
1
1
aP 3336 543 93.5 in.2
2
2
Copyright © McDougal Littell Inc.
All rights reserved.
1
b b2 h
2 1
1
5 2 5 sin 545 cos 18
2
1
13.094.76 31.12 sq. units
2
Area total 11.89 31.12 43 sq. units
1
aP
2
1
0.81 in.5.9 in.
2
1
5 sin 545 cos 542
2
11.89 sq. units
x
s 2x 1.18 in.
P ns
2.5
43 sq. units
1
P
a
36 72
1 2.5
55
2 tan 36
Answers should be the same. Differences may arise due
to rounding.
11.2 Mixed Review (p. 675)
54.
x
11
6 12
55.
20x 60
12x 66
x
56.
x3
11
2
13
12
x7
x
20 15
4
x
57.
x6
x
9
11
12x 13x 7
11x 6 9x
12x 13x 91
11x 66 9x
x 91
x 91
66 2x
33 x
58. True. The lengths of corresponding sides of similar trian-
gles are proportional.
59. True. The ratio of the perimeters of similar triangles
equals the ratio of the lengths of any pair of
corresponding sides.
60. True. The corresponding angles of similar triangles are
congruent.
Geometry
Chapter 11 Worked-out Solution Key
229
mcrbg-1103-sk-.qxd
5-25-2001
11:35 AM
Page 230
Chapter 11 continued
61. False. The lengths of corresponding sides of similar trian-
gles are proportional, not necessarily equal. Here, the
scale factor is 4:3 not 1:1 so BC ˇ EF.
62. 14
x 7 12
9x 63
64.
Pred
Pblue
x7
x6
Ablue 4 313
1313
Ared
30
90 9
Ablue
40
40 4
3
81 9x 144
84
x
14
5
30
63. 99 x 88 10
14x 84
82
5. Ared 6
94 23
6. The ratio of the lengths of the pieces of paper is 2:1, so
44 x
the ratio of the areas of the piece of paper is 4:1. The cost
1
of the smaller piece should be as much as the larger
4
sheet or about $.11.
64 16 4x
48 4x
12 x
11.3 Practice and Applications (p. 679–681)
Developing Concepts Activity 11.3 (p. 676)
7. The ratio of the perimeters is 2:1.
Exploring the Concept
The ratio of the areas is 4:1.
For example on p. 676-Sample Answer:
Original
Polygon
Rectangle 1
Rectangle 2
Rectangle 3
..
.
Total
8. The ratio of the perimeter is 5:7.
The ratio of the areas is 25:49.
Area
5 sq units
8 sq units
6 sq units
..
.
Similar
Polygon
Rectangle 1
Rectangle 2
Rectangle 3
..
.
Area
20 sq units
32 sq units
24 sq units
..
.
11 sq units
Total
44 sq units
9. The ratio of the perimeter is 5:6.
The ratio of the areas is 25:36.
10. The ratio of the perimeter is 5:3.
The ratio of the areas is 25:9.
11. sometimes
15.
Note: The areas for similar polygons should be 4 times the
area of the original polygons.
12. sometimes
13. always
14. 4:25
49
7
100
10
16. The scale factor of the hypotenuses is 8:20 or 2:5. The
ratio of the areas is the square of the scale factor, or 4:25.
Drawing Conclusions
1. Areas will vary; the ratio of the area of the similar poly-
gon to the area of the original polygon is the square of
the scale factor and should be 4:1.
2. The ratio of the areas of two similar polygons is the
square of the scale factor.
11.3 Guided Practice (p. 679)
2. True; all corresponding angles are congruent and all side
lengths are proportional.
3. False. The scale factor is 1:2, so the ratio of the areas is
1:4 and the area is quadrupled.
Pblue 96
54
1
Pred
18
Pblue 54
3
The ratios of the areas is equal to the square of the ratio
12 1
of the perimeters so 2 .
3
9
230
x 86.875 in.2
17. Sample answer: AB DC, so AEB DEC because
Area of ABE 1. a: b, a2: b2
18
4x 347.5
vertical angles are . BAE ECD because alternate
interior angles are . CDE ~ABE by the AA
Similarity Postulate.
Lesson 11.3
4. Pred 36
4
13.9
25
x
Geometry
Chapter 11 Worked-out Solution Key
73
2
1
123 18 sq units
2
x
18
49
x
9
18
9x 49 18
x 98
Area of †CDE 98 sq units
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1103-sk-.qxd
5-25-2001
11:35 AM
Page 231
Chapter 11 continued
18. DC AB and LK AB, so DC LK because 2 lines to
the same line are parallel. Then K C and
A J, and all corresponding s are . Ratio of the
lengths of any two corresp. sides is 3:1, so the ratio of the
areas is 32 : 12. The area of ABCD is 9 times the area
of JBKL 915.3 137.7 sq. in.
19. Perimeter of ABCDE 305. Perimeter of
QRSTU 40. Ratio of perimeters 305: 40 or
35:4.
28.
Area outer 466,170 1.04
Area inner 446,400
1
Ratio of perimeters x
21. The ratio of areas is 90:25. The ratio of perimeters is
90:25 310:5.
22. Sample answer: Let ABCD and EFGH be similar rectan-
4.77
1.02
Area of ABC
152 225
29. a.
2 Area of DEF
2
4
b.
225
25x
x5
4
gles with the lengths of corresp. sides in the ratio a:b. Let
al be the length of ABCD and aw the width, and let bl be
the length of EFGH and bw the width.
100x 225x 5
100x 225x 1125
125x 1125
area of ABCD
aLaw
a
.
area of EFGH bLbw b 2
2
23. Area small rug 29 in. 47 in. 1363 sq in.
Area large rug 58 in. 94 in. 5452 sq in.
Area small 1363
or 1:4
Area large
5452
x9
c. 15:2
d.
8y
15
3y 19
2
28 y 153y 19
$79
$.06 per sq in.
1363 sq in.
16 2y 45y 285
301 43y
$299
$.05 per sq in.
Large rug:
5452 sq in.
The large rug is a good buy because it costs slightly less
per square inch than a small rug. Also, the area of the
large rug is 4 times the area of the small rug, but the large
rug costs less than 4 times the cost of the small rug
($79 4 $316).
26. AB 1.3 40 ft
1
4041
2
52 ft
820 ft2
27. Area ABC 1.32820 ft2
1385.8 ft2
Area walkway Area of ABC Area of DEF
1385.8 820
565.8 ft2
7y
s
e. Because AB and DE are corresponding sides of ~ the ratio of their lengths is the scale factor. You can
15
22.5
solve the proportion
to find z; z 1.
13z 10
z
30. Sample answer:
PVQ ~ RVT
RVQ ~ PVU
TQR ~ TUS
All pairs are Í by the
AA Similarity Post.
3 VQ
32.
5
VT
3 VQ
5
15
45 5VQ
9 VQ
—CONTINUED—
Copyright © McDougal Littell Inc.
All rights reserved.
1.02
1
467 ft
Ratio of areas 92:42 81:16.
1
25. A bh
2
1.02x 477
Ratio of perimeters 36:16 9:4.
24. Small rug:
1
1.02 477
1
x
20. Perimeter of small square 16.
Then
1.04
31.
PV
RV
259 53
3 PV
5
10
PV 6
33. Sample answer:
Area PVQ
9
Area RVT
25
Area RVQ
45
25
Area PVU 16.2
9
120
25
Area TQR
Area TUS
19.2
4
Geometry
Chapter 11 Worked-out Solution Key
231
mcrbg-1103-sk-.qxd
5-25-2001
11:35 AM
Page 232
Chapter 11 continued
32. —CONTINUED—
5.
VU 3
VQ 5
Ared
169
Ablue 400
7. The scale factor is 3:7
5VU 27
of the areas is
27
VU 5
Pred
3.25 13
Pblue
5
20
6.
Ared
16
Ablue
9
VU 3
9
5
9
.
49
3
; therefore the ratio
219 feet
feet 7 9
480
49
x
525
4
Pred
8
Pblue 6
3
9x 23,520
UT 15 525 935
x $2613
11.3 Mixed Review (p. 681)
34. 80
38.
35. 145
36. 145
11.3 Math & History (p. 682)
37. 215
10x 8x 180
17y 19y 180
18x 180
36y 180
x 10
y5
1. The length of each side of the
inscribed hexagon will be equal
1
to (diameter). So, the perimeter
2
of the inscribed hexagon is equal
mR 100, mS 85, mT 80, mU 95
1
39. m1 160
2
1
40. m1 110 50
2
80
41. m1 80
1
126 40
2
1. 20 2180 3240
2.
1. Arc measure is the number of degrees of the central angle
360
14.4
25
whose endpoints are the ends of the arc. Arc length is the
measure of a portion of the circumference.
1
4
3. A 3 s2
125.1 in.2
4. The central angle of a nonagon is 40. A segment whose
length is the apothem bisects the central angle.
2. Arc length AB
m1
360
3. F
6. B
.
length CD
1
3 172
4
A
so 3 < < 3.46.
11.4 Guided Practice (p. 686)
Quiz 1 (p. 682)
AB 9.58 cm
3
3 23 3.46,
Lesson 11.4
43
9 cm
cos 20
AB
1
2
to 3 diameter or 3 units. If the
1
apothem is unit, then the length
2
of one side of the hexagon is
1
1 3 3
2 tan 30 2 . The
or
2
2
3
3
perimeter of the circumscribed hexagon is
6
1
86
2
30
60
AC
tan 20
9
B
20
5. C
7. A
m2
360
2r arc
8. E
9. False. Arc length depends on the radius of the circle. So
if the radii are different, the arc lengths will be different.
10. False. The circumference is proportional to the radius. If
the radius doubles, the circumference doubles.
11. False. Arc length is determined by the measure of the arc
AC 3.28
9 cm
1
aP
2
1
99 tan 2029
2
4. D
2r A
C
265.33 cm2
and the radius of the circle. If you solve for the measure
of the arc, then it is dependent upon both the arc length
and the radius of the circle.
12. Length AB
m AB
360
140
360
2r
2 29.5
22.9 cm or 72.1 cm
232
Geometry
Chapter 11 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1103-sk-.qxd
5-25-2001
11:35 AM
Page 233
Chapter 11 continued
13. Length CD
m CD
360
160
360
22. Length AB
2r
2 29
25.8 cm or 81.0 cm
14. Length EF
m EF
360
m EF
67.6 cm 360
2 25
36067.6 cm
m EF
50
155 m EF
m AB
12
3
0.6
3.5
5.1
33
45
30
120
192
90
107
Length of
AB
3
0.5
0.4
3.73
2.55
3.09
m XY
360
2 r
30
360
28
24. Length of XY
11.4 Practice and Applications (pp. 686-689)
16.
44 2 r
10
44
r
2
25. Length of AB
5.5 18. C 2 r
215
30 in.
25.1 m
94.2 in.
C 2 r
32 2r
5.5
m AB
360
55
360
2 r
2 r
2 r
360
55 36 units 2 r
m CD
360
26. Length of CD
20 32
r
2
160
360
2 r
2r
7.16 units r
5.1 yd r
20. Length AB
m AB
360
2 r
45
360
23
2.36 cm
21. Length
4.2 units
7.0 ft r
8
AB
C 2 r
25
31.4 in.
19.
210
20.9 ft
Radius
17. C d
120
360
2 r
23.
2r
15. C 2 r
m AB
360
m AB
360
2 r
60
360
27
7.33 in.
27. Length of AB
m AB
360
118
360
2 r
210.14
20.88 units
28. Length of ST
m ST
360
12.4 84
360
2 r
2 r
53.14 units 2 r
29. Length of LM
42.56 m LM
360
240
360
2 r
2r
10.16 units r
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 11 Worked-out Solution Key
233
mcrbg-1103-sk-.qxd
5-25-2001
11:35 AM
Page 234
Chapter 11 continued
39. Tire A 15 in. 4.6 in. 4.6 in.
30. P 2l 2 r
24.2 in.
212 23.5
Tire B 16 in. 4.43 in. 4.43 in.
45.99 units
24.86 in.
31. P 5 5 5 22
Tire A 17 in. 4.33 in. 4.33 in.
15 15.71
25.66 in.
30.71 units
40. 500 ft 6000 in.
32. P 42 22
8 12.57
20.57 units
33. 2x 15 135
Tire A:
6000 in.
79 rev.
24.2 in.
Tire B:
6000 in.
77 rev.
24.86 in.
Tire C:
6000 in.
74 rev.
25.66 in.
2x 120
x 60
Length of arc m arc
360
135
y 3 360
2 r
41. The sidewall width must be added twice to the rim
diameter to get the tire diameter.
42. L straightaways curves
28
L 30 8 12 12 30 20 1
1
1
6
23 23 2
22.25
4
2
2
y36
y9
34. 15y 30 270
100 28.25
15y 300
163.84 m
y 20
Length of arc m arc
360
270
13x 2 360
1609 m
9.8 laps
43.
164 m
lap
2 r
44. measure of arc 4.2, radius of Earth 4000 mi
210
Length of arc 13x 2 15
x1
45. Length of arc 35. 18x 45
(rear sprocket)
x 2.5
14y 3 m arc
360
45
360
2 4000
164
8
360
3.64 in.
2 r
196
Length of arc 22
360
(front sprocket)
2 7
11.98 in.
14y 3 1.75
Length of chain 3.64 16 11.98 16
14y 4.75
47.62 in.
y 0.34
36. r 3
4.2
360
293.22 miles
13x 13
Length of arc 37. r 27
38. r 2
C 2 r
C 2 r
C 2 r
C 6
C 47
C 4
46. number of teeth 47.62 in. 0.5 in.
tooth
95 teeth
47. P 3
2 r
4
3
28
4
37.7 ft
234
Geometry
Chapter 11 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1103-sk-.qxd
5-25-2001
11:35 AM
Page 235
Chapter 11 continued
48. B. Length of XZ
m XZ
360
2 r
60
360
2 4
2. n 6; P 26 sin
3. 12-gon: P 6.21 units
15-gon: P 6.24 units
18-gon: P 6.25 units
4
units
3
2360
49. B. x 2 r
24-gon: P 6.27 units
360
y
2 r
Lesson 11.5
2360
2 r
x
y
360
2r
11.5 Guided Practice (p. 695)
1. A sector of a circle is bounded by two radii of the circle
and their intercepted arc.
2
1
2. The radius is one half of the diameter or
50. The four semicircles form 2 complete circles so that the
length would be 22 r 4 r.
2r 4 r.
1
51. 8 segments 4 complete circles 4 2
1
r 4 r.
16 segments 8 complete circles 8 2
4
n
n
1
2 n r 4 r.
n segments complete circles 2
2
4
No. The number of segments doesn’t matter.
52. A r 2
92
3.32
34.21
ft2
55. A 411 2
54. A r 2
91.61
56.
552.92 m2
2
57.
2
9 b
3
254.47
45.36 cm2
6. A 5. A r2
12 12
2
m AB
360
r2
70
360
102
62. 31
540 15y
36 y
Technology Activity 11.4 (p. 690)
1. The perimeter increases as n increases, getting closer and
Copyright © McDougal Littell Inc.
All rights reserved.
9. A 60
360
3
45
82
360
25.13 in.2
2
50.27
14. A 60
360
m2
112
63.36 ft2
16. A 293
360
11. A 0.42
0.50 cm2
13. A 102
12. A 42
closer to 2.
34.56 ft2
8. A Area of circle Area of sector
3019.07 ft2
2
y x8
3
61. 258
62
11.5 Practice and Applications (p. 695–698)
x 2.92
60. 176
110
360
r2
61.09 m2
10. A 312
59. 96
m AB
360
23.56 in.2
6x 17.5
8 b
30
y
15 18
3.82
in.2
28.27 4.71
x
5
3.5 6
2 6 b
58.
4. A r2
32 cm2
y mx b
2 in.2
4. The area
2
92
1
2
12 4 .
3. A r2
7. A 53. A r 2
254.47
is r2 113.10 ft2
11.4 Mixed Review (p. 689)
27
5
12 sin 30° 6
180
6 102
255.69 cm2
314.16 in.2
15. A 80
360
2
7
2
8.55 in.2
17. A 120
360
4.62
22.16 m2
Geometry
Chapter 11 Worked-out Solution Key
235
mcrbg-1103-sk-.qxd
5-25-2001
11:36 AM
Page 236
Chapter 11 continued
250
360
18. A 82
139.63
19. A 102
29.
314.16 ft2
in.2
50 r2
20.
15.92 r2
Measure
of arc, x
Area of
corresponding
sector, y
3.99 m r
A
21.
59 AB
m
r2
360
40
360
Measure
of arc, x
Area of
corresponding
sector, y
r2
30.
169.02 r2
30
60
90
2.4
4.7
7.1
120
150
180
9.4
11.8
14.1
31. Yes it appears to be linear.
y
The equation would be
13.00 in. r
A
22.
277 m arc
360
288
360
1
d
2
y
2
4 d 2
1
x.
40
2
x
30
440.86 d 2
21.00 m d
32. With a 5-inch radius, the area of the sectors would be
23. A Area large circle Area small circle
larger. However the relationship would still be linear.
5
y
x.
72
242 62
1696.46 m2
24. A Area large semicircle Area small semicircle
33. A 1
1
382 192
2
2
1701.17 cm2
12.22 ft
34.
ft 2
27. A Area square Area circles
182 44.52
in.2
28. A Area semicircle Area triangle
196 x 2 324
14
x
x 2 128
18
37.57
x 11.31 miles
light house
2
9 4
69.53
142 x 2 182
28
1
4 sin 54254 cos 54
2
26. Area 32 22
15.7
182
692.72 mi2
25. A Area of circle Area of pentagon
42 245
360
1
1
22 3 22
2
4
4.55 cm2
35. A Area of sector Area of triangle
60
360
1
62 22333
3.26 cm2
36. A Area of sector Area of triangle
90
360
1
142 2 27272
55.94 m2
37. A Area of sector Area of triangle
120
360
1
482 2 242243
1415.08 cm2
236
Geometry
Chapter 11 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1103-sk-.qxd
5-25-2001
11:36 AM
Page 237
Chapter 11 continued
38. A Total area Area of smaller sector
72
360
44. D
72
A
3.52 360 32
62
10.8
2.04 ft2
39. To find the area of the sector, you would need to either
45. A Area of triangle 3
know the arc length or the measure of the central angle,
and the radius. To find the area of the kite you would
have to know the lengths of the diagonals or be given
sufficient information to determine the length of the
diagonals, such as a side with an angle or two sides.
40. Area ABCD Area of sector formed by APB
Area of sector formed by DPC
45
360
45
42 360 22
4.71 ft2
41. If you double the radius, the area is quadrupled because it
is proportional to the square of the radius. If you double
the radius, the circumference doubles also because the
circumference is directly proportional to the radius.
42. As n gets larger, the area of the polygon approaches ,
the area of the circle.
Sample spreadsheet:
A
# of sides
n
3
4
5
6
7
8
9
10
11
12
13
14
15
16
B
Area
.5*cos(180/n)*2*n*sin(180/n)
1.299038106
2
2.377641291
2.598076211
2.736410189
2.828427125
2.892544244
2.938926261
2.973524496
3
3.020700618
3.037186174
3.050524823
3.061467459
Continuing the spreadsheet, you can find that
A 3.14151 when n 500,
A 3.14157 when n 1000,
A 3.14159 when n 2500, etc.
Area of sector
1
60
3122 3 4
360
62
62.35 56.55
5.81 cm2
11.5 Mixed Review (p. 698)
46.
2
5
47.
3
16
48.
4
21
49.
4
11
50. 152 21.21 units
51. DB 18
sin 68
19.4 cm
52. DC 18 cm
53. mDBC 68
54. BC n-gon Data
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
108
360
18
tan 68
7.3 cm
55. x 22 y 72 36
56. x 2 y 92 100
57. x 42 y 52 10.24
58. x 82 y 22 11
59. C 2 12.5
78.54 in.
60. Length of AB
53
360
2 13
12.03 ft
61.
31.6 129
360
2 r
31.6 2.25r
14.04 r
43. C
A 62 32
36 9
27
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 11 Worked-out Solution Key
237
mcrbg-1106-sk-.qxd
5-25-2001
11:35 AM
Page 238
Chapter 11 continued
Lesson 11.6
18. P 11.6 Guided Practice (p. 701)
1. A geometric probability involves geometric measures
such as length and area instead of counting events or
outcomes.
2. I would use the area method because a triangular region
is two-dimensional.
19. P ed to a line.
2
1
AB
11%
AF 18 9
5. P BD
9
1
50%
AF
18 2
6. P BF 16 8
89%
AF 18 9
20. P be on BF. So the sum of the probabilities is
1 8
1.
9 9
Area shaded
Area total
Area triangle
Area rectangle
1
2 16
5
160
40
160
25%
7. AF AB BF. Therefore any point not on AB must
8. P 50.27
64 50.27
44%
3. Length method should be used because time can be relat-
4. P Area circle
Area square Area circle
Area of semicircle
Area of triangle
1
2
2 5
1
2010
2
39.27
100
39.27%
21. P 1
1
2 54 2 44
1
2 7 164
18
39%
46
Area of circle
Area of hexagon
1.52
6 3 142
7.07
509.22
1
4
1.39%
11.6 Practice and Applications (pp. 702–704)
22. Since the probability is between l
GH
2
1
14%
9. P GN
14 7
10. P JL
4
2
29%
GN 14 7
11. P JN
8
4
57%
GN 14 7
GJ
6
3
43%
12. P GN 14 7
and 2%, 1 to 2 darts would probably
hit the bull’s-eye.
24. J
M
PS
28
7
58%
PU 48 12
15. P SU
20
5
42%
PU 48 12
16. P PU
1 100%
PU
17. P Area square Area circle
Area square
238
144 113.1
21.5%
144
Geometry
Chapter 11 Worked-out Solution Key
28.28
509.22
5.55%
K
Let L be the midpoint of JM and N be the midpoint of
MK . Then LN is the part of the line closer to M than to J
1
50%.
2
or K, so P 25.
PQ 12 1
25%
13. P PU 48 4
14. P 23. P 2
2
Area of square 4 sq. units
Area of circle 22 6.28 sq. units
Area shaded 6.28 4 2.28 sq. units
P
26. P Area shaded 2.28
36.31%
Area of circle 6.28
3.5
46.7%
7.5
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1106-sk-.qxd
5-25-2001
11:35 AM
Page 239
Chapter 11 continued
27. P 10 1
16.7%
60 6
28. P 1
3
20%
15 5
41. P 29. A 20005000 10,000,000 yd 2
30. Area of circle P
87,266.4 yd
500
3 If the area is doubled, the probability is doubled.
2
2
42. P 43. a. Area of jar 12.5 2 490.87 cm 2
Area of dish 2.52 19.63 cm 2
P
Area of 10 point region 2.4 2 18.1
18.1
1%
1809.58
19.63
4%
490.87
b. 25 coins
32. Area yellow region 4.8 2 72.38
P
900
1.8%
50,000
The probability is four times as much.
87,266.46
0.87%
10,000,000
31. Area Total 24 2 1809.58
P
450
0.9%
50,000
c.
2505
50 prizes
25
d. Yes, the probability will change because now pennies
just have to touch the circle, not be inside it so the
target area is larger and the probability increases.
72.38
4%
1809.58
33. Area white region Total area Area regions 3–10
44.
y
1809.58 19.22
P
651.46
P
4.5
9
28%
16
32
y3
651.46
36%
1809.58
1
1
x
1
34. Area regions 5–10 14.4
2 651.44
Area regions 6–10 122 452.39
Area regions 1–4 1809.58 651.44
1158.14
Area region 5 1809.58 452.39 1158.14
199.05
P
199.05
11%
1809.58
35. Sample answer: It would not hold because an expert
archer is more likely to hit the bull’s eye, so it is no
longer true that every point on the target has an equal
probability of being hit.
36.
↔
45. AB is not tangent to C because ABC is not a right as follows.
10 2 4 2 112
↔
46. AB is tangent to C. ABC is a right because
↔
52 122 132, and CB AB .
↔
47. AB is tangent to C. ABC is a right as follows.
24 2 7 2 25 2
48. secant
Total Area 1
y
yx
2
2
11.6 Mixed Review (p.705)
P
y1
2
1
50%
2
49. tangent
y
y
2
2
2
x
2
y0
2
x
2
y x
x
x 1
50. tangent
1
360° 120° 2 60°
3
1
38. 360° 90° 2 45°
4
1
39. 360° 60° 2 30°
6
152
225
40. P 0.45%
200250 50,000
y
y
2
37.
Copyright © McDougal Littell Inc.
All rights reserved.
51. secant
2
2
x6
2
x
2
yx1
2
Geometry
Chapter 11 Worked-out Solution Key
x
239
mcrbg-1106-sk-.qxd
5-25-2001
11:35 AM
Page 240
Chapter 11 continued
52. The locus is all points on the bisector of the segment
with endpoints (3, 2) and (1, 2) that are on or inside the
circle with center (1, 1) and radius 2, all points 2, y
where 2 < y < 0.
experimental probabilities will be to each other.
Extension
Sample answer: A calculator simulation would be easier
because it would not involve the physical actions of tossing
and retrieving the darts and recording the results. It would be
more accurate because it would not involve such variables as
skill level, effort, or physical factors such as air currents.
y
1
x
1
4. The more trials there are, the closer the theoretical and
Chapter 11 Review (p. 708–710)
9 2180
1. measure of interior 140
9
measure of exterior 180 140 40
11.6 Quiz 2 (p.705)
8.2 1.
68°
360°
2. measure of interior 2 r
measure of exterior 180 152.3 27.7
43.41 m 2 r
2. Length of AB
88°
360°
3. measure of interior 26
138°
360°
24.6 ft 24.6 ft
4. measure of interior 2 r
5.
502
105°
5. A 360°
7854.0 mi 2
10 ft2
1
4
7. Area of triangle 3
44.9 cm 2
7.
n 45
n8
n 2180
150
n
2.7%
10n 360
n 12
n 36
1
9. A 4 3s2
10. S 143122
Conjecture
64.95 m2
centered at the origin with radius 0.5.
probability is approximately 27.5% with the graphing calculator program and theoretical probability is 19.6%.
3. Answers may vary.
11. A 6
Geometry
A 143s2
1
2
4 3 5
1
2
4 35
143432
20.78 in.2
12. Measure of interior angle 10 2180
144
10
a 0.75 tan 72
A 12aP
120.75 tan 72101.5
17.31 ft2
13. sometimes
Chapter 11 Worked-out Solution Key
6
tan 60
43
62.35 cm2
11.6 Technology Activity (p. 706)
2. Answers may vary. Sample answer: Experimental
180n 360 170n
30n 360
6
1. This is the equation for the region inside the circle
n 2180
170
n
8.
180n 360 150n
5 2 10.83
10.83
400
180n 360 135n
45n 360
7
Area of square 20 2 400
240
n 2180
135
n
6.
8n 360
2
253.07 ft 2
P
n 2180
172
n
180n 360 172n
4. A r 2
145°
360°
24 2180
165
24
measure of exterior 180 165 15
1
r
360°
138° 2 10.21 ft r
6. A 2
16 2180
157.5
16
measure of exterior 180 157.5 22.5
19.97 in.
3.
13 2180
152.3
13
14. sometimes
15. always
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1106-sk-.qxd
5-25-2001
11:35 AM
Page 241
Chapter 11 continued
16.
Perimeter CDE 3 1
Perimeter BDF
9 3
Area ADG 5 2 25
2
Area BDF
3
9
25.13 cm
Length of AB
35
360
2 4
2.44 cm
19. C 2r
2 7.6
A r
29. P Length of LM 12
3
30%
Length of JN
40 10
30. P Length of JL
20 1
50%
Length of JN 40 2
31. P Length of KM 24 3
60%
Length of JN
40 5
1
32. Radius of a circle 2 82 62 5
47.75 m
Length of AB
162
360
2 7.6
21.49 m
20. C 2r
2 12
P
Length of AB
118
360
2 12
24.71 ft
C
2
86
52
48
78.54
33. P 12
2
Area shaded
Area of semicircle
2 r1 2
1
r2 2
2
262
1
242
2
1
4
25%
1.91 in.
34. P C
15
15 m
12 102
157.08 in.2
Area shaded
Total area
722
142
98
196
1
2
23. A 12 r 2
135
24. A 360
Area of rectangle
Area of circle
0.611 or 61.1%
75.4 ft
A r 2
28.
3.57 in.r
2 4
22. d 615.75 ft2
40 r
18. C 2r
62
27. A 142
Perimeter ADG 15 5
Perimeter BDF
9
3
21. r 300
360
94.25 cm2
Area CDE 12 1
2
Area BDF
3
9
17.
26. A 50%
5.5
2
35.64 ft2
25. A 152 82
505.8 cm2
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 11 Worked-out Solution Key
241
mcrbg-1107-sk.qxd
5-25-2001
11:35 AM
Page 242
Chapter 11 continued
Chapter 11 Test (p. 711)
1. x 120 135 90 115 120 6 2180
13. Area sector ARB x 580 720
x 140
14. A r 2
1
152
2
30 2180
30
353.43 ft2
168
360
27
1
13 3
52
1
2
n 2180
n
1
4
4. meas. of exterior 105
360
r2
22.91 cm2
2. 40 60 45 90 65 60 360
3. measure of interior mAB
360
5. A 3s2
1
3102
4
43.30
15. Area of sector 162
547.34 in.2
16. Area of sectors ft2
5 2180
108
6. Measure of interior angle 5
245
360
120
360
72
51.31 m2
17. P 1
A aP
2
1
8
852
2
tan 54
Area of circle
Area of square
18. P 102
202
0.785 or 78.5%
7. A 6
210.44 cm2
19.
8. Measure of interior angle 9 2180
140
9
1
A aP
2
1
2 1010
202
50
400
1
or 12.5%
8
232.49 in.2
1
392
4
Area of triangle
Area of square
boat 2 80 502.65 ft
skiier 2 110 691.15 ft
691.15 502.65 188.50 ft
20. P 1
20 min
16.67%
120 min 6
1
cos 7092 sin 70
2
Chapter 11 Standardized Test (p. 712–713)
2.89 m2
9.
8 4
Perimeter ABCD
Perimeter EFGH 6 3
10.
Area of ABCD
42 16
2
Area of EFGH 3
9
11. C 2r
12.
56 16
x
9
2 5
52
31.42 cm
78.54 cm2
20n 360
x 31.5 cm2
n 18
2. A
180 x 7 2180 130 128 133 139 110 136
180 x 900 776
2r
360
105
2 5
360
9.16 cm
n 2180
160
n
180n 360 160n
16x 504
A r2
m AB
Length of AB
1. C
180 x 124
56 x
3. A
Polygon a: a 15 cos 36 12.14
Polygon b: a 14 cos 30 12.12
242
Geometry
Chapter 11 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
mcrbg-1107-sk.qxd
5-25-2001
11:35 AM
Page 243
Chapter 11 continued
4. A
Polygon A: P 2515 sin 36 88.17 units
Polygon B: P 6142614 sin 30 84 units
5. A
Polygon A: A 1212.1488.17
535.19 sq. units
Polygon B: A 1
2 12.1284
509.04 sq. units
17. Sample answer: One method is to find the area of the
circle and from that subtract the area of the octagon. A
second method is to find the area of a sector. From that,
subtract the area of a triangle. Then multiply by 8.
From problems 12–16, Area J 804.25 m 2 and
area octagon 724.08 m2
Shaded area 804.25 724.08 80.17 m2
45°
360°
162 100.53 m2
6. A
Area of smaller octagon 122 144 4
2
Area of larger octagon
18
324 9
Area sector 7. B
Length of AB
1
Area triangle 216 sin 22.5°16 cos 22.5°
2
86
360
2 8.3
12.46 ft
8. E
Area shaded 360 86
8.32
360
164.72
9. C
Area shaded 90 30
82
360
67.02
10. C
P
ft2
cm2
Area P Area Q
Area P
62 32
62
75%
11. A
P
24 2
0.4
60 5
12. C 2 16
13.
8 2180
135
8
90.51 m2
Shaded Area 8100.53 90.512 80.16 m2
Each method yields the same result.
18. r 3 in.
C 6 18.85 in.
A 32 28.27 in. 2
19. P 32
Area of circle
3.27%
Area of dart board 3624
20. P 225
Area of blue region
12.73%
Area of dart board
3624
21. P area green area red area blue
area dart board
32 212 9 67 225
3624
243.27
864
28.16%
100.53 m
22. P 100% 28.16% answer from question 21)
A 162
804.25 m
2
71.84%
14. 180 135 45
15. P 8216 sin 22.5
97.97 m
A 1216 cos 22.58216 sin 22.5
724.08 m2
16. Length of AB
45
360
2 16
12.57 m
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 11 Worked-out Solution Key
243
mcrbg-1107-sk.qxd
5-25-2001
11:35 AM
Page 244
Chapter 11 continued
Chapter 11 Project (p. 714)
1. yes
2. yes
3. No; any two great circles must intersect.
4. Sample answer: No. There are no parallel lines in
geometry on a sphere.
5. Yes; for example, a line of longitude and the equator on
a globe
6. Answers may vary. Sample answer: 90, 90, 90
7. Answers may vary. The sum should be larger than 180.
Sample answer: 270
8. Answers may vary. The angles will have measures
between 60° and 180°. Sample answer: 90, 90, 90
9. no; yes; yes
10. Answers may vary. The sum of the angles is between
180 and 540.
244
Geometry
Chapter 11 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.