# C Chapter 11

by user

on
Category: Documents
149

views

Report

#### Transcript

C Chapter 11
```Chapter 11
Prerequisite Skills (p. 718)
4.
2. diameter 5 6
3. mC
ADB 5 3608 2 708 5 2908
95b
The base is 9 inches.
4. P 5 2* 1 2w
5. The area of the window is 5 + 10 5 50 square feet.
24 5 2(9) 1 2w
Because you don’t paint over the window, you need to
paint about 637 2 50 5 587 square feet.
24 5 18 1 2w
6 5 2w
11.1 Exercises (pp. 723–726)
35w
Skill Practice
The width is 3 centimeters.
5. AC 1 CB 5 AB
2
2
1. Either pair of parallel sides of a parallelogram can be
2
called its bases, and the perpendicular distance between
these sides is called the height.
AC 1 6 5 14
2
2
2
AC 2 1 36 5 196
2. Two formulas for the area of rectangle are A 5 b + h or
AC 2 5 160
A 5 l + w. Because the base is the length and the height
is the width, the two formulas give the same results.
}
AC 5 4Ï 10
3. A 5 bh 5 7(4) 5 28 square units
AC
cos A 5 }
AB
6.
A 5 bh
153 5 b(17)
4. A 5 bh 5 14(2) 5 168 square units
AC
cos 358 5 }
25
5. A 5 s2 5 152 5 225 square inches
1
1
6. A 5 }bh 5 }(10)(13) 5 65 square units
2
2
25(cos 358) 5 AC
20.5 ø AC
7. Pythagorean Theorem: 182 1 b2 5 302
}
b2 5 576
7. BC 5 x and AC 5 xÏ 3 in a 308-608-908 triangle.
}
Because BC 5 5, AC 5 5Ï 3 .
b 5 24
8. Parallelograms, rhombuses, rectangles, and squares have
1
VW
UV
9. } 5 }
YZ
XY
1
1
8. A 5 } bh 5 }(15)(9) 5 67.5 square units
2
2
}
8
12
5
XY
9. Method 1 Use DC as the base. The base is extended to
}5}
}
measure the height AE. So, b 5 10 and
h 5 16. A 5 bh 5 10(16) 5 160 square units
}
}
Method 2 Use AD as the base. Then the height is DF.
So, b 5 20 and h 5 8.
A 5 bh 5 20(8) 5 160 square units.
60 5 8XY
15
2
} 5 XY or 7.5 5 XY
Lesson 11.1
10. The height of the parallelogram is 4 units, not 5.
A 5 bh 5 6(4) 5 24 square units
11.1 Guided Practice (pp. 721–722)
1. P 5 17 1 10 1 21 5 48 units
1
11. The base of the parallelogram is 3 units, not 7.
A 5 bh 5 3(4) 5 12 square units
1
A 5 }2 bh 5 }2 + 21 + 8 5 84 square units
2. p 5 2* 1 2w 5 2(30) 1 2(20) 5 100 units
12.
a2 1 b2 5 c2
122 1 b2 5 152
A 5 bh 5 30(17) 5 510 square units
b2 5 81
3. Pythagorean Theorem: 5 1 b 5 13
2
13
5
1
A 5 }2 bh 5 }2(24)(18) 5 216 square units
diagrams that bisect each other.
2
2
b2 5 144
b 5 12
12
b59
P 5 15 1 12 1 9 5 36 in.
1
1
A 5 }2 bh 5 }2 (9)(12) 5 54 in.2
P 5 5 1 13 1 12 5 30 units
1
1
A 5 }2 bh 5 }2(12)(5) 5 30 square units
Geometry
Worked-Out Solution Key
333
Chapter 11,
continued
13. a2 1 b2 5 c2
162 1 b2 5 342
10 m
b2 5 900
b 5 30
1
16 m
10 m
P 5 34 1 16 1 30 5 80 ft
1
A 5 }2 bh 5 }2 (30)(16) 5 240 ft2
14.
10 m
16 m
16 m
a2 1 b2 5 c2
842 1 b2 5 852
10 m
10 m
10 m
b2 5 169
b 5 13
8m
1
2
1
5 17(8) 1 }2(17)(5) 5 178.5 ft2
}bh 5 }(13)(84) 5 546 m2
23. Area 5 Area of ﬁrst triangle 1 Area of parallelogram
a2 1 b2 5 c2
1 Area of second triangle
20 1 b 5 29
2
2
2
b 5 441
2
1
b 5 21
5 364 cm2
P 5 29 1 20 1 21 5 70 cm
1
24. Base of triangle 5 16 2 base of parallelogram
1
A 5 }2bh 5 }2 (21)(20) 5 210 cm2
5 16 2 11 5 5 m
Area 5 Area of triangle 1 Area of parallelogram
1
A 5 }2 bh
16.
1
5 }2(5)(10) 1 (11)(10)
1
36 5 }2 (12)(x)
5 135 m2
6 in. 5 x
25. Height of triangle: 152 1 h2 5 252
A 5 bh
17.
1
5 }2(9)(13) 1 18(13) 1 }2 (11)(13)
18.
276 5 x(12)
A 5 bh
476 5 x(17)
23 ft 5 x
28 cm 5 x
19. A 5 4 ft
2
h2 5 400
1
h 5 }2 b
h 5 20 in.
Area 5 Area of triangle 1 Area of rectangle
1
5 }2(15)(20) 1 (25)(19)
5 625 in.2
1
A 5 }2 bh
26. Height of triangles and parallelogram: 102 1 h2 5 262
1 2
1 1
4 5 }2 b }2 b
h2 5 576
h 5 24 m
1
4 5 }4 b2
Area 5 Area of ﬁrst triangle 1 Area of parallelogram
16 5 b2
1 Area of second triangle
45b
The base of the triangles is 4 inches, and the height is
1
2
}(4) 5 2 inches.
20.
A 5 507 cm2
h 5 3b
A 5 bh
507 5 b(3b)
507 5 3b2
169 5 b 2
13 5 b
The base of the parallelogram is 13 centimeters, and the
height is 3(13) 5 39 centimeters.
334
Geometry
Worked-Out Solution Key
1
1
5 }2(10)(24) 1 (40)(24) 1 }2 (20)(24) 5 1320 m2
27. Height of triangle 5 8 2 5 5 3 in.
Area 5 Area of square 2 Area of triangle
1
5 82 2 }2 (8)(3) 5 52 in.2
15.
8m
22. Area 5 Area of rectangle 1 Area of triangle
85 1 84 1 13 5 182 m
1
2
8m
Chapter 11,
28.
continued
34. h2 1 42 5 52
y
5
h2 1 16 5 25
B
A
4
4
h53
x
1
1
5
u
h2 5 9
1
h
8
1
A 5 }2bh 5 }2(8)(3) 5 12 ft2
C
D
The area of the triangle is 12 square feet.
3
sin u 5 }5
A 5 bh 5 7(6) 5 42 square units
29.
y
u ø 36.98
F
1
21
1
30. B; 2 ft 3 in. 5 24 in. 1 3 in. 5 27 in.
4 ft 2 in. 5 48 in. 1 2 in. 5 50 in.
A 5 bh 5 50(27) 5 1350 in.2
31. Check students’ work; The base, height, and area of
n ABC remains the same no matter how you move
point C.
Because the original triangle is isosceles, the altitude
from the vertex where u is at is also 4.8 ft. The height of
the original triangle is an altitude. So, the lengths of the
altitudes are 3 ft, 4.8 ft, and 4.8 ft.
35. Construct a rectangle so that A, B, C, and D are all on the
sides of the rectangle. The area of ABCD is equal to the
area of the retangle minus the areas of the right triangles
formed at each corner of the rectangle that are not part
of ABCD.
2
3
When mDAB 5 208:
y
C
h
32. sin DAB 5 }
AB
When mDAB 5 508:
h
sin 208 5 }8
h
sin 508 5 }8
8 sin 208 5 h
8 sin 508 5 h
2.74 ø h
6.13 ø h
A 5 bh ø 15(2.74) ø 41
A 5 bh ø 15(6.13) ø 92
The height is about 2.74 units and the area is about 41
square units when mDAB 5 208. The height is about
6.13 units and the area is about 92 square units when
mDAB 5 508.
1
1
33. A 5 }bh 5 }(12)(35) 5 210 cm2
2
2
D
B
1
4
1
1
x
A
A 5 Arect. 2 An 2 An 2 An 2 An
1
2
3
4
1
1
1
1
5 (11)(7) 2 }(4)(6) 2 }(1)(7) 2 }2(4)(3) 2 }(7)(4)
2
2
2
5 41.5
The area of ABCD is 41.5 square units.
Problem Solving
1
1
36. Sail A: A 5 } bh 5 } (65)(35) 5 1137.5 ft2
2
2
u
a
8
4.8 ft ø a
A 5 }2 bh 5 }2(5)(3) 5 7.5 square units
12 cm
u
a
8 sin 36.98 5 a
G
1
5
5
sin 36.98 5 }9
x
E
a
37 cm
1
1
Sail B: A 5 }2 bh 5 }2(29.5)(10.5) 5 154.875 ft2
35 cm
opp
tan u 5 }
35
tan u 5 }
12
u ø 718
Sail A
Sail B
1138
155
} 5 } ø 7.3
The area of Sail A is about 1138 ft2 and the area of Sail B
is about 155 ft2. The area of Sail A is about 7.3 times as
great as the area of Sail B.
a
sin 718 5 }
12
12 sin 718 5 a
11.3 ø a
The area of the right triangle is 210 cm2. The length of
the altitude drawn to the hypotenuse is about 11.3 cm.
Geometry
Worked-Out Solution Key
335
Chapter 11,
continued
1
1
37. Area of triangle plot: A 5 }bh 5 }(24)(25) 5 300 yd2
2
2
300 yd2 1 min
2
If you mow 10 yd /min: }
+ }2 5 30 min
1
10 yd
Area of rectangular plot: A 5 bh 5 (24)(36) 5 864 yd2
864 yd2 1 min
If you mow 10 yd /min: }
+ }2 5 86.4 min
1
10 yd
2
It takes you 30 minutes to mow the triangular plot and
86.4 minutes to mow the rectangular plot.
and RVS is a right angle. Because PQRS is a
} }
parallelogram, PQ > SR. By the Hypotenuse-Leg
Congruence Theorem, n PQT > n SRV.
b. Because QRVT is a rectangle, its area is bh. Rectangle
QRVT is composed of nSRV and trapezoid QRST.
Because nPQT > nSRV, their areas are equal. So, the
area of nPQT plus the area of trapezoid QRST is equal
to the area of QRVT, which is bh.
parallelogram. The area of a parallelogram is bh. Because
the parallelogram is made up of two congruent triangles,
A 5 bh
360 5 (20)h
1
the area of one triangle, nXYW, is }2 bh.
18 5 h
1 2
1 2
44. a. h2 1 }s 5 s2
2
The height of the new table should be 18 inches.
s2
39. A 4 inch square does not have an area of 4 square inches. A
h2 5 s2 2 }
4
2 inch square has an area of 4 square inches. The formula
for the area of a square is s2, and 22 5 4, so the side length
should be 2 inches to produce an area of 4 square inches.
352
h2 5 }
4
}
40. The dimensions of a 308-608-908 triangle are:
2x
x
sÏ 3
h5}
2
608
x
308
x 3
308
6 ft
12 ft
}
6 5 xÏ3
}
1 sÏ3 2
}
s2Ï3
}
s2Ï3
b. } 5 4
4
}
6Ï3
3
}
16Ï 3
}5x
s2 5 }
3
}
2Ï 3 5 x
Area 5 Area of triangle 1 Area of rectangle
}
5 }2 (12)(2Ï 3 ) 1 12(6.5) ø 98.8 ft2
197.6 ft2
1 hour
200 ft
If you paint 200 ft2 per hour: }
+ }2 5 0.988
1
If you charge \$20 per hour: \$20(0.988) 5 \$19.76
You will get paid about \$20 for painting those two sides
of the shed.
41. height 5 6 1 14 1 14 5 34 cm
base 5 3 1 17 1 3 5 23 cm
Area of paper used 5 Area of small rectangle
1 Area of large rectangle
1 Area of triangle
The side lengths of the triangle is about 3 cm.
and height are not necessarily side lengths. The least
possible perimeter for the parallelogram is
7 1 3 1 7 1 3 5 20 feet. The greatest possible
perimeter cannot be determined.
46. a. RNSV is a square because its base and height are both
h. UVTQ is a square because its base and height are
both b. MNPQ is a square because its base and height
are both b 1 h. Area of RNSV 5 h2, Area of UVTQ 5
b2, and Area of MNPQ 5 (b 1 h)2 5 b2 1 2bh 1 h2.
b. Area of MNPQ 5 Area of MRVU 1 Area of UVTQ
1 Area of RNSV 1 Area of VSPT
b2 1 2bh 1 h2 5 A 1 b2 1 h2 1 A
2bh 5 2A
1
5 17(14) 1 23(14) 1 }2(17)(6)
5 611 cm2
Area of paper that is cut off 5 Area of original rectangle
2 Area of paper used
5 34(23) 2 611 5 171 cm2
34 cm by 23 cm, the area of the paper that is actually
used in the envelope is 611 cm2, and the area of the paper
that is cut off is 171 cm2.
Geometry
Worked-Out Solution Key
s ø 3.04
45. The perimeter cannot be determined because the base
Two sides 5 2(98.8) 5 197.6 ft2
336
1
s2Ï 3 5 16
}
1
1
5}
A 5 }2bh 5 }2(s) }
4
2
bh 5 A
608
}
43. Opposite sides are congruent, so XYZW is a
38. Area of old table: A 5 bh 5 (24)(15) 5 360 in.2
Area of new table:
}
42. a. By the way the segments were drawn, RV > QT
Chapter 11,
}
continued
}
47. Let AB > BC. To get from A to B move 2 units up and
2 units to the right, which is (4, 4). To get from B to C,
move 2 units down and 2 units to the right, which is
(6, 2). The height of nABC is 2 and the base is 4, so the
area of nABC is 4 square units. Find the equation of the
line through B(4, 4) and C(6, 2).
422
2
m5}
5}
5 21
426
22
y 5 2x 1 b
4 5 24x 1 b
85b
y 5 2x 1 8
} }
}
Now, let BC > AC and choose BC to be a vertical line.
So, the base and height of nABC is the same. Since the
area has to be 4 square units, ﬁnd the measure of the
base.
Pencil (14.8 cm)
Textbook (21 cm)
0.1 cm
1 cm
Greatest
possible error
}(0.1) 5 0.05 cm
} (1 cm) 5 0.5 cm
Relative error
}
0.05 cm ø 0.0037
14.8 cm ø 0.37%
0.5 cm
} ø 0.0240
21 cm ø 2.4%
Unit of measure
1
2
The measurement of the pencil is more precise because
the unit of measure is smaller and the greatest possible
error is smaller. The pencil also has the smaller relative
error, so it is more accurate.
1
2. Unit of measure: } inch; greatest possible error:
10
1
1 1
} } 5 } 5 0.05 inch
20
2 10
1 2
3. Unit of measure: 1 meter; greatest possible error:
1
}(1) 5 0.5 meter
2
1
4. Unit of measure: } kilometer; greatest possible error:
1000
1
1 1
} } 5 } 5 0.0005 kilometer
2000
2 1000
1
A 5 }2 bh
1
A 5 }2 b(b)
1
1
4 5 }2 b2
8 5 b2
}
2Ï 2 5 b
}
So, the base and height of nABC
are 2Ï2 units. So,
}
(2 1 2Ï2 , 2) and the equation of
point C is located at }
the line x 5 2 1 2Ï 2 . The other line that could ﬁt this
} }
requirement is when AB > AC.
2
1
5. Unit of measure: } yard; greatest possible error:
16
1
1 1
} } 5 } 5 0.03125 yard
32
2 16
1 1
6. Greatest possible error: } } cm 5 0.05 cm
2 10
1 2
1
2
greatest possible error
measured length
0.05 cm
5 0.0125
Relative error: }} 5 }
4 cm
5 1.25%
Mixed Review for TAKS
1
2
1
7. Greatest possible error: }(1 in.) 5 0.5 in.
2
48. A;
Number
Number
Cost of
Total
of tooth- + Cost of 1 of tubes + 1 tube
5
1 toothcost
of toothbrushes
of toothbrush
paste
bought
paste
3
+
b
1
1
+
p
5 12.7
2
+
b
1
2
+
p
5 13.6
49. J;
4
4
greatest possible error
measured length
ø 1.8%
1
8. Greatest possible error: }(0.1 m) 5 0.05 m
2
greatest possible error
measured length
113 cubic feet.
11.1 Extension (p. 728)
1. The precision of a measurement depends only on the
unit of measure. The accuracy of measurement depends
on both the unit of measure and on the size of the object
being measured.
0.05 m
ø 0.0109
Relative error: }} 5 }
4.6 m
ø 1.1%
4
V 5 }3 :r3 5 }3:(3)3 5 }3 :(27) 5 36: ø 113
0.5 in.
ø 0.0179
Relative error: }} 5 }
28 in.
1
9. Greatest possible error: } (0.01 mm) 5 0.005 mm
2
greatest possible error
measured length
0.005 mm
Relative error: }} 5 }
12.16 mm
ø 0.0004 ø 0.04%
10. 1 inch; You are estimating the amount of paper, so the
1
greatest possible error of }2 inch is precise enough.
11. Such a measurement could be more precise when
measuring larger objects (like a running track) than
smaller objects (like a pencil). Because the actual
measured length goes in the denominator when
determining the relative error, a larger measured length
when the greatest possible error is constant yields a
smaller relative error. Because the larger length has the
smaller relative error, it is more accurate.
Geometry
Worked-Out Solution Key
337
Chapter 11,
continued
Greatest possible side lengths: 1.4 1 0.05 5 1.45 cm
12.
17 cm
12 cm
1 cm
1 cm
Greatest
possible error
1
}(1 cm) 5 0.5 cm
2
1
} (1 cm) 5 0.5 cm
2
Relative error
} ø 0.029
Unit of measure
0.5 cm
17 cm
ø 2.9%
0.5 cm
12 cm
} ø 0.0417
ø 4.2%
5.1 1 0.05 5 5.15 cm
Greatest possible perimeter:
P 5 2(1.45) 1 2(5.15) 5 13.2 cm
Least possible side lengths: 1.4 2 0.05 5 1.35 cm
5.1 2 0.05 5 5.05 cm
Least possible perimeter:
P 5 2(1.35) 1 2(5.05) 5 12.8 cm
The precision is the same. 17 centimeters is more accurate.
Lesson 11.2
13.
18.65 ft
25.6 ft
0.01 ft
0.1 ft
Greatest
possible error
1
}(0.01 ft) 5 0.005 ft
2
1
} (0.1 ft) 5 0.05 ft
2
Relative error
0.005 ft
} ø 0.0003
18.05 ft
} ø 0.002
Unit of
measure
Investigating Geometry Activity 11.2 (p. 729)
1
1. The area of one trapezoid is } the area of the
2
parallelogram formed from two trapezoids.
0.05 ft
25.6 ft
ø 0.03%
ø 0.2%
b 5 b1 1 b2
1
1
A 5 }2 bh 5 }2 (b1 1 b2)h
2. The base of the rectangle is d1 and the height of the
rectangle is }2 d2. A 5 (d1)1 }2 d2 2 5 }2 d1d2
1
18.65 feet is more precise and more accurate.
1
1
14.
6.8 in.
13.4 ft
11.2 Guided Practice (pp. 731–732)
1
1
1. A 5 } h(b1 1 b2) 5 }(4)(6 1 8) 5 28 ft2
2
2
0.1 in.
0.1 ft
Greatest
possible error
}(0.1 in.) 5 0.05 in.
} (0.1 ft) 5 0.05 ft
Relative error
} ø 0.007
1
2
0.05 in.
6.8 in.
1
2
0.05 ft
13.4 ft
} ø 0.004
ø 0.7%
ø 0.4%
6.8 inches is more precise. 13.4 feet is more accurate.
15.
1
1
2. A 5 } d1d2 5 } (6)(14) 5 42 in.2
2
2
3. d1 5 30 m 1 30 m 5 60 m
d2 5 40 m 1 40 m 5 80 m
1
1
4. A 5 } d1d2; when A 5 80 ft2, d1 5 x, and d2 5 4x.
2
1
3.5 ft
35 in.
80 5 }2 (x)(4x)
0.1 ft
1 in.
80 5 }2 4x2
Greatest
possible error
}(0.1 ft) 5 0.05 ft
} (0.1 in.) 5 0.5 in.
Relative error
} ø 0.014
Unit of measure
1
2
0.05 ft
3.5 ft
1
2
ø 1.4%
0.5 in.
35 in.
16. Greatest possible error for 5.1 cm side:
1
40 5 x 2
}
2Ï10 5 x
} ø 0.014
ø 1.4%
35 inches is more precise. The accuracy is about the same.
}
1
1
5. A 5 } d1d2 5 }(4)(8) 5 16 units2
2
2
The area of the rhombus is 16 square units.
y
Greatest possible error for 1.4 cm side:
P 5 2(5.1 cm) 1 2(1.4 cm) 5 13 cm
Greatest possible error for perimeter:
1
2
} (1 cm) 5 0.5 cm
Geometry
Worked-Out Solution Key
}
4(2Ï10 ) 5 8Ï 10 ft.
1
2
1
} (0.1 cm) 5 0.05 cm
2
}
One diagonal is 2Ï10 ft and the other diagonal is
} (0.1 cm) 5 0.05 cm
338
1
A 5 }2 d1d2 5 }2 (60)(80) 5 2400 m2
N
M
P
1
Q
1
x
Unit of
measure
Chapter 11,
continued
11.2 Exercises (pp. 733–736)
Skill Practice
1
15. B; A 5 } d1d2 when d1 5 x, d2 5 3x, and A 5 24 ft2
2
1
1. The perpendicular distance between the bases of a
trapezoid is called the height of the trapezoid.
2. The vertical diagonal is bisected by the horizontal
diagonal. You also know the angles formed by the
intersecting diagonals are right angles.
24 5 }2(x)(3x)
1
24 5 }2 (3x2)
16 5 x2
45x
d1 5 4
d2 5 3(4) 5 12
The diagonals are 4 ft and 12 ft.
1
1
3. A 5 } h(b1 1 b2) 5 } (10)(8 1 11) 5 95 units2
2
2
1
1
4. A 5 } h(b1 1 b2) 5 }(6)(6 1 10) 5 48 units2
2
2
1
1
5. A 5 } h(b1 1 b2) 5 }(5)(4.8 1 7.6) 5 31 units2
2
2
1
108 5 }2(x)(14 1 22)
108 5 18x
6 ft 5 x
1
17. A 5 } h(b1 1 b2)
2
1
300 5 }2(20)(10 1 x)
5.4 cm
6.
1
16. A 5 } h(b1 1 b2)
2
8 cm
300 5 10(10 1 x)
30 5 10 1 x
10.2 cm
1
1
A 5 }2 h(b1 1 b2) 5 }2 (8)(5.4 1 10.2) 5 62.4 cm2
1
1
7. A 5 } d1d2 5 }(50)(60) 5 1500 units2
2
2
1
8. A 5 } d1d2 5
2
1
}(16)(48) 5 384 units2
2
1
9. A 5 } d1d2 5
2
1
}(21)(18) 5 189 units2
2
1
10. A 5 } d1d2 5
2
}(10)(19) 5 95 units2
1
2
11. d1 5 12 1 12 5 24
d2 5 15 1 15 5 30
1
A 5 }2 d1d2 5
1
}(24)(30) 5 360 units2
2
12. d1 5 2 1 2 5 4
d2 5 4 1 5 5 9
1
2
1
2
A 5 } d1d2 5 }(4)(9) 5 18 units2
13. The height is 12 cm, not 13 cm.
1
A 5 }2 (12)(14 1 19) 5 198 cm2
14. The length of the one diagonal is 12 1 12 5 24, not just
12 itself.
1
20 m 5 x
1
18. A 5 } d1d2
2
1
100 5 }2(10)(x)
100 5 5x
20 yd 5 x
19. The ﬁgure is a trapezoid.
b1 5 {4 2 2{ 5 2, b2 5 {5 2 0{ 5 5
h 5 {4 2 1{ 5 3
1
1
A 5 }2h(b1 1 b2) 5 }2 (3)(2 1 5) 5 10.5 units2
20. d1 5 {3 2 (21){ 5 4
d2 5 {23 2 1{ 5 4
The ﬁgure is a rhombus.
1
1
A 5 }2 d1d2 5 }2(4)(4) 5 8 units2
21. The ﬁgure is a kite.
d1 5 {4 2 0{ 5 4
d2 5 {2 2 (23){ 5 5
1
1
A 5 }2 d1d2 5 }2(4)(5) 5 10 cm2
A 5 }2 (24)(21) 5 252 cm2
Geometry
Worked-Out Solution Key
339
Chapter 11,
continued
1
22. A 5 } h(b1 1 b2); b1 5 x, b2 5 2x
2
1
13.5 5 }2 (3)(x 1 2x)
26. Use the Pythagorean Theorem to ﬁnd part of the
second base.
a2 1 202 5 292
a2 5 441
3
13.5 5 }2 (3x)
9
2
20
a 5 21
b2 5 21 1 21 5 42
9
13.5 5 }3 x
35x
1
2
a
1
2
A 5 } h (b1 1 b2) 5 }(20)(21 1 42) 5 630 units2
One base is 3 feet and the other base is 2(3) 5 6 feet.
1
23. A 5 } h(b1 1 b2); b1 5 x, b2 5 x 1 8
2
1
54 5 }2 (6)(x 1 x 1 8)
27. Height of trapezoid 5 7 2 5 5 2
Area 5 Area of trapezoid 1 Area of rectangle
1
hT 5 height of trapezoid,
bR 5 base of rectangle, and
1
h
}
5 2(2)(7 1 10) 1 (10)(5) R 5 height of rectangle
A 5 }2hT (b1 1 b2) 1 bRhR
54 5 3(2x 1 8)
18 5 2x 1 8
5 67 units2
10 5 2x
28. Use the Pythagorean Theorem to ﬁnd the height of
55x
One base is 5 cm and the other base is 5 1 8 5 13 cm.
the trapezoid and half of the shorter diagonal of the
rhombus:
24. Use the Pythagorean Theorem to ﬁnd the length of half
of the shorter diagonal.
20
5
h
a
4
4 1 h 5 52
2
h2 5 9
a 1 16 5 20
2
2
a2 5 144
a 5 12
d1 5 12 1 12 5 24
d2 5 16 1 30 5 46
1
1
A 5 }2 d1 1 d2 5 }2 (24)(46) 5 552 units2
25. Use the Pythagorean Theorem to ﬁnd the height of
the trapezoid.
h53
b1 5 4, b2 5 4 1 4 1 4 5 12, d1 5 3 1 3 5 6,
and d2 5 4 1 4 5 8
1
Area 5 Area of trapezoid 1 }2 Area of rhombus
A 5 }2h(b1 1 b2) 1 }2 1 }2d1d1 2
1
1 1
F
G
1
1 1
5 }2(3)(4 1 12) 1 }2 1 }2 2(6)(8) 5 36 units2
29. Area 5 Area of parallelogram 2 Area of kite
9
1
1
A 5 bh 2 }2 d1d2 5 (12)(7) 2 }2(12)(7) 5 42 units2
h
15
4
3
92 1 h2 5 152
h 5 144
h 5 12
b2 5 9 1 11 5 20
1
1
a 5 }2h(b1 1 b2) 5 }2(12)(8 1 20) 5 168 units2
3
8
2
2
1
A 5 8(3) 5 24 units2
1
A 5 }2 (3)(4 1 12) 5 24 units2
5
2
3
3
1
1
A 5 }2(3)(5 1 11) 5 24 units2
4
1
1
A 5 }2(3)(2 1 14)
5 24 units2
340
Geometry
Worked-Out Solution Key
16
2
2
Chapter 11,
continued
31. Use the Pythagorean Theorem to ﬁnd the length of the
B
B
side of the trapezoid.
7
7
A
C
D
15 2 7
15
E
62 1 82 5 s2
F
H
170 5 BC2
10 5 s
}
Ï170 5 BC
P 5 6 1 7 1 15 1 10 5 38 units
1
To ﬁnd mABC, use trigonometric ratios.
1
A 5 }2 h(b1 1 b2) 5 }2(6)(7 1 15) 5 66 units2
B
u1 u2
6 7
10
32. Use the Pythagorean Theorem to ﬁnd the length of part
of the other diagonal.
A
13
13
a
12
170
8
C
11
8
11
tan u1 5 }6
12
13
tan u2 5 }
7
u 5 tan21 1 }3 2
u2 5 tan211 }
72
u1 ø 53.18
u2 ø 57.58
4
24
a2 1 122 5 132
a2 5 25
11
}
So, mABC ø 53.18 1 57.78 ø 110.68. Extend BC
and drop an altitude down from A to ﬁnd the height
of the parallelogram. By the Linear Pairs Postulate,
} }
the angle made by joining extended BC to AB is
1808 2 110.68 5 69.48. Use a trigonometric function
to ﬁnd the height.
a55
d1 5 5 1 5 5 10
P 5 4(13) 5 52 units
1
A 5 }2 d1d2 5 }2(10)(24) 5 120 units2
}
33. First, make a horizontal line from A to the segment BF.
G
112 1 72 5 BC2
100 5 s2
1
C
11
s
6
6
The length of the new segment is 8 because it is parallel
}
to EF, and EF 5 8. Because BF 5 16 and AE 5 10, the
part of the segment from B to this new segment is
16 2 10 5 6. Using the Pythagorean Theorem, you can
}
ﬁnd out the length of AB.
82 1 62 5 AB2
h
sin 69.48 5 }
10
10 + sin 69.48 5 h
9.36 ø h
B
69.48
h
10
A
So, the area }
of the parallelogram is
A 5 bh ø Ï 170 + 9.36 ø 122 square units.
B
100 5 AB
2
6
10 5 AB
A
8
} }
Because ABCD is a parallelogram, AB > CD and
} }
BC > DA. So, CD 5 10. Next extend a vertical line
}
}
up from GD to meet BC. Because BAD > BCD,
} }
AB > CD, and two more corresponding angles of
the triangles are congruent (by Alternate Interior and
Suppementary Angles Theorems), you know the triangle
}
}
}
made by AB and parts of AD and BF is congruent to
}
}
}
the triangle made by CD and parts of BC and AD by
AAS. You can now say GH 5 8 because you have
corresponding parts of congruent triangles. Now draw a
}
horizontal line connecting C to BF. You know the base is
the length of FG 1 GH 5 8 1 2 5 11 and the height is
16 2 9 5 7. Use the Pythagorean Theorem to ﬁnd BC.
Problem Solving
1
1
34. A 5 } h(b1 1 b2) 5 } (35)(70 1 79) 5 2607.5 in.2
2
2
The area of the glass in the wind shield is
2607.5 square inches.
1
1
35. A 5 } d1d2 5 }(8)(5) 5 20 mm2
2
2
The area of the logo is 20 square millimeters.
Geometry
Worked-Out Solution Key
341
Chapter 11,
continued
1
A 5 }2 d1d2
39. The kite was cut along the diagonal and then the other
1
432 5 }2 (36)(d2)
24 5 d2
The length of the other diagonal is 24 inches.
37. a. The two polygons are a trapezoid and a right triangle.
b. Area of ﬁeld 5 Area of triangle 1 Area of trapezoid
1
1
Area of ﬁeld 5 }2 bh 1 }2 h(b1 1 b2)
1
5 103,967.5
Theorem 11.5.
1
1
40. The area of nPRS is }b1h and the area of nPQR is }b2h.
2
2
1
2
1
2
The area of the playing ﬁeld is about 103,968 square
feet or about 11,552 square yards.
1
2
41. By the SSS Congruence Postulate, nPQR > nPSR.
1
1
1
The area of nPQR is }2d2 }2d1 5 }4d1d2. So, the area of
1
2
kite PQRS is twice the area of nPQR, or
21 }4 2d1d2 5 }2d1d2.
1
2
1
Diagram
Area, A
2
4
3
6
Rhombus
number, n
original kite
rhombus
isosceles
triangle
right triangle
many different triangles
many different
kites
many different
4
Diagram
Area, A
you can go through the same process with a rhombus,
}b1h 1 }b2h 5 }h(b1 1 b2).
103,968 ft2 1 yd2
}+}
5 11,552 yd2
1
9 ft2
Rhombus
number, n
1
The area of trapezoid PQRS is equal to
ø 103,968 ft2
1
1
1
1
Rhombus
number, n
have A 5 (d1)1 }2 d2 2, which simpliﬁes to }2d1d 2. Because
the formula will also simplify to }2 d1d2, which is
Area of ﬁeld 5 }2 (315)(322) 1 }2(179)(145 1 450)
38. a.
diagonal to make 1 big triangle and 2 smaller triangles.
The resulting ﬁgure, once the smaller triangle was
moved, was a rectangle. The formula for the area of a
rectangle is b + h. In this case, the base was the ﬁrst
diagonal, and the height was half of the other diagonal.
Substituting those values into the area formula, you
8
5
b. Yes, you can make isosceles and right triangles by
moving the vertical diagonal down all the way and to
the middle or down all the way and to the left or right
all the way.
Diagram
1
1
c. Area of original kite: }d1d2 5 }(4)(6) 5 12 units2
2
2
1
1
Area, A
10
Area of rhombus: }2d1d2 5 }2(4)(6) 5 12 units2
1
1
b. The area is twice the rhombus number n. The area of
the nth rhombus is A n 5 2n.
c. The length of the other diangonal is 2n for the nth
1
1
rhombus. A 5 }2 d1d2 5 }2(2)(2n) 5 2n. The rule for
area in this part is the same as the rule for area
in part (b).
342
Geometry
Worked-Out Solution Key
Area of isosceles triangle: }2 bh 5 }2(6)(4) 5 12 units2
1
1
Area of right triangle: }2 bh 5 }2(6)(4) 5 12 units2
1
1
Area of different kiles: }2d1d2 5 }2(4)(6) 5 12 units2
All of the areas are equal, 12 square units. The lengths
of the diagonals are not being changed, just moved
around. Although the shapes made by connecting the
diagonals will be different, the area will remain the
same.
36.
Chapter 11,
continued
Lesson 11.3
43. Looking at the trapezoid:
b1 5 a, b2 5 b1 and h 5 a 1 b
11.3 Guided Practice (pp. 738–739)
1
1
A 5 }2 h(b1 1 b2) 5 }2(a 1 b)(a 1 b)
1
1
a
4
16
1. Ratio of perimeters: } 5 } 5 }
3
b
12
1
5 }2 (a2 1 2ab 1 b2) 5 }2a2 1 ab 1 }2b2
Looking at the triangles that make up the trapezoid:
1
2
1
2
1
2
1
2
A 5 }(a)(b) 1 }(a)(b) 1 }(c)(c) 5 ab 1 }c2
1
2
1
2
1
2
64(9) 5 16(Area of nDEF)
36 5 Area of nDEF
16
1
2
20
2. Ratio of area: }
36
21 }2a2 1 }2 b2 2 5 21 }2 c2 2
1
576 5 16(Area of nDEF)
. The area of nDEF is 36 square feet.
is }
9
}a2 1 }b2 5 }c2
1
1
}
}
2Ï5
Ï20
Ï36
}
Ï5
Ratio of sides: }
} 5 } 5 }
6
3
a2 1 b2 5 c2
}
Ï5
.
The ratio of their corresponding side lengths is }
3
Mixed Review for TAKS
3. Step 1 Find the ratio of the perimeters. Use the same
44. C;
units for both lengths in the ratio.
0.25 in. j dimension in blueprint
Scale 5 }
1 ft j actual dimension
Perimeter of Rectangle I
Perimeter of Rectangle II
0.25 in. 2
0.0625 in.2
Let x 5 the area of the actual kitchen
5.5 in.2
x
66 in.
(Periemter of Rectangle I)2
(Perimeter of Rectangle II)
12
20
88 5 x
1
400
}2 5 }
The area of the actual kitchen is 88 square feet.
Step 3 Find the area of Rectangle I.
45. G;
Area of Rectangle I
Area of Rectangle II
1
400
Area of Rectangle I
(35)(20)
1
400
0
1
}} 5 }
(23)2 2 1
02 2 1
12 2 1
}} 5 }
5921
5021
5121
58
5 21
50
2(23)2 2 1
2(0)2 2 1
2(1)2 2 1
5 2(9) 2 1
5 2(0) 2 1
5 2(1) 2 1
5 18 2 1
5021
5221
23
x
Area of Rectangle I + 400 5 700
Area of Rectangle I 5 1.75 ft2
2
144 in.
Step 4 1.75 ft2 + }
5 252 in.2
2
1 ft
1
The ratio of the area is }
and the area of
400
5 17
5 21
51
2(23)2 1 1
2(0)2 1 1
2(1)2 1 1
5 29 1 1
5 20 1 1
5 21 1 1
11.3 Exercises (pp. 740–743)
5 28
5011
50
Skill Practice
51
y 5 2x 2 1 1
Area of Rectangle I
Area of Rectangle II
}}}2 5 }}
5.5 5 0.0625x
y 5 2x 2 1 1
1
Step 2 Find the ratio of the areas of the two rectangles.
0.0625 in.2
1 ft
y 5 2x 2 2 1
5.5 ft
5}
5}
5}
20
110 ft
110 ft
}5}
2
y 5 x2 2 1
66 in.
2(35 ft) 1 2(20 ft)
}} 5 }}
j area in blueprint
5}
(Scale)2 5 1 }
1 ft 2
j actual area
1 ft2
16
9
64
Area of nDEF
The ratio of the area of nABC to the area of nDEF
}a2 1 ab 1 }b2 5 ab 1 }c2
1
2
16
}} 5 }
Because the areas are equal, you can set the area of the
trapezoid and the combined area of the triangles equal to
each other.
1
2
42
3
a2
b
Ratio of areas: }2 5 }2 5 }
9
Rectangle I is 252 square inches.
2(23)2 1 1
2(0)2 1 1
2(1)2 1 1
5 2(9) 1 1
5 2(0) 1 1
5 2(1) 1 1
5 18 1 1
5011
5211
5 19
51
53
E
B
5
A
4
10
3
C
D
8
6
F
The function y 5 2x 2 1 best represents the mapping
shown.
2
Geometry
Worked-Out Solution Key
343
Chapter 11,
continued
The side lengths of nABC are each multiplied by the
same scale factor to get the lengths of each side of
nDEF. AB has corresponding side length DE, BC has
corresponding side length EF, and CA has corresponding
side length FD. Each side in n ABC is multiplied by 2 to
get the length of its corresponding side in nDEF.
2. You don’t need to know the value of n to ﬁnd of the ratio
of the perimeters or the ratio of the areas of the polygons,
because you know the ratio of the side lengths, 3 : 4. This
ratio is also the ratio of the perimeters. According to
Theorem 11.7, the ratio of the areas is the square of the
ratio of the perimeters 32 : 42, or 9 : 16.
12. C; Ratio of areas 5 18 : 24
}
}
}
}
2Ï6
Ï6
}
1 }
}
Ï3
3Ï 12
6Ï3
3Ï2 Ï6
}
} + }
} 5 } 5 } 5 } or Ï 3 : 2
12
12
2
2
13. Ratio of areas 5 7 : 281 : 4
}
}
Ratio of lengths 5 Ï 1 : Ï4 5 1 : 2
1
2
4
XY
}5}
8 5 XY
The length of XY is 8 centimeters.
14. Ratio of areas 5 198 : 88 5 9 : 4
Ratio of
perimeters
Ratio of areas
6 : 11
6 :11
36 : 121
XY 5 15
25 : 81
The length of XY is 15 inches.
}
20 : 36 5 5 : 9
5:9
4.
}
}
Ratio of
corresponding
side lengths
3.
}
Ratio of lengths 5 Ï 18 : Ï24 5 3Ï 2 : 2Ï6 5
}
Ratio of lengths 5 Ï 9 : Ï4 5 3 : 2
3
2
XY
10
}5}
15. The ratio of areas is 1 : 4, so the ratio of lengths
5. The ratio of the perimeters is 1 : 3 and the ratio of the
2
2
areas is 1 : 3 , or 1 : 9.
12
ZY
The area of the blue triangle is 18 square feet.
6. The ratio of the perimeters is 15 : 20, or 3 : 4, and the ratio
of the areas is 32 : 42, or 9 : 16.
So, the ratio of corresponding lengths is 3 : 7 and the
ratio of areas is 32 : 72, or 9 : 49.
9
49
248
x
9
49
}5}
The area of the red quadrilateral is 135 square centimeters.
7. The ratio of the perimeters is 7 : 9 and the ratio of the
areas is 72 : 92, or 49 : 81.
12,152 5 9x
1350 ø x
The area of regular pentagon VWXYZ is about
1350 square centimeters.
49
81
}5}
1
1
17. Area of MNPQ 5 }d1d2 5 }(14)(25) 5 175 ft2
2
2
x ø 127
The area of the red hexagon is about 127 square inches.
8. The ratio of the perimeters is 5 : 3 and the ratio of the
areas is 52 : 32, or 25 : 9.
Ratio of areas 5 28 : 175 5 4 : 25
}
}
Ratio of lengths 5 Ï 4 : Ï25 5 2 : 5
Shorter diagonal of RSTU
14
2
5
}} 5 }
25
9
}5}
Shorter diagonal of RSTU 5 5.6 ft
14.4 5 x
The area of the blue parallelogram is 14.4 square yards.
9. Ratio of areas 5 49 : 16
}
}
Ratio of lengths 5 Ï49 : Ï16 5 7 : 4
The ratio of the lengths of corresponding sides is 7 : 4.
10. Ratio of areas 5 16 : 21
}
}
Ratio of lengths 5 Ï16 : Ï121 5 4 : 11
The ratio of the lengths of corresponding sides is 4 : 11.
11. Ratio of areas 5 121 : 144
}
}
Ratio of lengths 5 Ï121 : Ï 144 5 11 : 22
The ratio of the lengths of corresponding sides is 11 : 12.
Geometry
Worked-Out Solution Key
Longer diagonal of MNPQ
25
2
5
}} 5 }
Longer diagonal of MNPQ 5 10 ft
The area of MNPQ is 175 square feet. The lengths of the
diagonals are 10 feet and 5.6 feet.
18. Case 3, Case 1, Case 2
}
In Case 3, the enlargement is Ï 5 , or approximately
2.24. Because 2.24 is less than the enlargement of 3
(Case 1) and 2.24 and 3 are both less than an enlargement
of 4 (Case 2), the order of enlargement from smallest to
largest is Case 3, Case 1, Case 2.
}} 5 }
x 5 135
344
5(12 cm)
Perimeter of QRSTU
3
16. }} 5 } 5 }
7
140 cm
Perimeter of VWXYZ
Area of QRSTU
Area of VWXYZ
9
16
}5}
40
x
1
2
ZY 5 24
18 5 x
x
210
}
}5}
1
2
}5}
9
x
x
240
}
is Ï1 : Ï4 or 1 : 2. You need to use the 1 : 2 ratio.
Chapter 11,
continued
19. Doubling the side length of a square never doubles
the area. Doubling the side length of a square always
20. Two similar octagons sometimes have the same perimeter.
The perimeters will be the same only when the octagons
are congruent.
The corresponding lengths are UV and VW.
}
Ratio of lengths: Ï3 : 3
}
Ratio of areas: (Ï3 )2 : 32 5 3 : 9 5 1 : 3
25. a. Ratio of area nAGB to nCGE 5 9 : 25
}
5 3:5
3 7.5
3 9
3
4.5
21. ratio of lengths: } 5 }, } 5 }, } 5 }
4 10
4 12
4
6
Set up proportions of corresponding sides to the ratio
of lengths.
The ratio of lengths is 3 : 4, so the ratio of areas is 32 : 42,
or 9 : 16.
9
Area of nABC
You can use the proportion }} 5 }
to solve
16
Area of nDEF
for the area of nDEF.
22. First, you have to ﬁnd the width of ABCD.
P 5 2* 1 2w
}5}
AG
CG
3
5
}5}
GB
GE
3
5
}5}
AG
10
3
5
}5}
GB
15
3
5
AG 5 6
GB 5 9
Look at n AGF and nBGC. AGF and BGC are
right angles, so AGF > BGC.  GFA >  GBC
because they are alternate interior angles. So, n AGF
, nBGC by the AA Similarity Postulate. Find the
}
length of GF using a ratio.
84 5 2(24) 1 2w
36 5 2w
18 5 w
The width of ABCD is 18 feet, or
GF
GC
AG
BG
GF
10
6
9
}5}
1 yard
3 feet
18 feet + } 5 6 yards
}5}
Ratio of width 5 6 : 3 5 2 : 1
Ratio of areas 5 22 : 12 5 4 : 1
20
GF 5 }
3
The ratio of the area ABCD to the area of DEFG is 4 : 1.
23. D and N are right angles, so D > N by the Right
GE 5 GF 1 FE
Angles Congruence Theorem. F > M is given. So,
nDEF , nNLM by the AA Similarity Postulate.
20
1 FE
15 5 }
3
Use the area formula to ﬁnd NL.
}
Ratio of lengths of nAGB to nCGE 5 Ï9 : Ï25
25
3
} 5 FE
1
A 5 }2 bh
20
1
2
294 5 }(NL)(21)
28 5 NL
ABCD is a parallelogram. DAC > BCA because
they are alternate interior angles. So, n ADC , nCBA
by the AA Similarity Postulate. Find the area of
nCBA, whose height is BG and base length is
AC 5 AG 1 GC.
Find ML so that you can ﬁnd the ratio of corresponding
lengths.
(MN)2 1 (NL)2 5 (ML)2
212 1 282 5 (ML)2
1
35 5 ML
The ratio of the area of nADC to nCBA is 72 : 72,
or 1 : 1.
ratio of lengths: 10 : 35 5 2 : 7
24. TUY > UVW because they are corresponding angles.
UTY > VUW because they are corresponding angles.
So, nTUY , nUVW by the AA Similarity Postulate.
}
You have to ﬁnd the length of VW to ﬁnd the ratio of
lengths.
}
Ï3
1
A 5 }2bh 5 }2(6 1 10)(9) 5 72
1225 5 (ML)2
ratio of areas: 22 : 72 5 4 : 49
25
So, AG 5 6, GB 5 9, GF 5 }
, and FE 5 }
.
3
3
Problem Solving
26. ratio of longest sides 5 3 : 5
ratio of areas + 32 : 52 5 9 : 25
Area of old banner
Area of new banner
9
25
3(1)
Area of new banner
9
25
}} 5 }
}} 5 }
tan 308 5 }
VW
}
Ï3
tan 308
1
VW 5 }
8}3 5 Area of new banner
VW 5 3
The area of the new banner will be 8}3 square feet.
1
Geometry
Worked-Out Solution Key
345
Chapter 11,
continued
27. ratio of areas 5 360 : 250 5 36 : 25
}
Ratio of lengths of nQRS to nABC: 1 : 2
}
ratio of lengths 5 Ï36 : Ï 25 5 6 : 5
Ratio of areas of nQRS to nABC: 1 : 4
distance on new patio
distance on similar patio
The area of the smaller triangle is }4 the area of the
6
5
1
}} 5 }
distance on new patio
12.5
larger triangle. The smaller triangle has an area of
6
5
}} 5 }
1
4
}(18) 5 4.5 square units.
distance on new patio 5 15
The distance on the new patio is 15 feet.
x
y
I
28. B;
II
a
b
ratio of lengths 5 90 : 60 5 3 : 2
Rectangle I , Rectangle II
ratio of areas 5 32 : 22 5 9 : 4
AI 5 ax, AII 5 by
pounds for baseball diamond
pounds for softball diamond
9
4
Because the rectangles are similar, the ratios of
20
pounds for softball diamond
9
4
corresponding sides are proportional. So, } 5 }y.
}}} 5 }
}}} 5 }
9 ø pounds for softball diamond
You need about 9 pounds for the softball diamond.
a
a
b
a2
b
So, the ratio of corresponding sides is a : b, and the ratio
of their areas is a2 : b2.
y
B
31. The graph is visually midleading because although the
2 13
5
amount of science ﬁction books read is only half the
amount of mysteries read, the graph makes it seem like
the amount of science ﬁction books read is a quarter of
the mysteries read. The student could redraw the graph
by using bars of the same width for the same categories.
1
C
9
1
1
that into the previous equation.
Ratio of areas 5 }b + } 5 }2.
A
x
a
b
ax
a x
x
a
Ratio of areas 5 }
5 }b + }y. Because } 5 }y, substitute
by
b
x
1
A 5 }2 bh 5 }2 (9)(4) 5 18 square units
013 014
3
}
midpoint of AB: S 1 }
,}
5 S 1 }2, 2 2
2
2 2
3
319 410
}
midpoint of BC: Q 1 }
,}
5 Q(6, 2)
2
2 2
}
019 010
9
,}
5 S }2, 2
midpoint of AC: R }
2
2
1
2
1
2
AI
B
P II
R
1
Î1 6 2
C
}
9 2
1 (2 2 0)2 5 Ï2.25 1 4
2
}
2
}
5 Ï 6.25
5 2.5
Î1 2
QS 5 Î1 6 2
}}
9
2
}
}
}
3 2
1 (0 2 2)2 5 Ï9 1 4 5 Ï13
2
}
2
}}
}
3 2
1 (2 2 2)2 5 Ï 20.25 1 0 5 4.5
2
}
2
}
BC
AC
5
2Ï 13
9
AB
} 5 } 5 2, } 5 }
} 5 2, } 5 } 5 2
2.5
4.5
QR
RS
QS
Ï13
Because all of the ratios of corresponding lengths are
equal, the triangles are similar.
346
9
6
}
Ï9
Ï6
3
Ï6
}
3Ï6
6
}
Ï6
2
C
x
}}
RS 5
6
33. a. BFC > DFE because they are vertical angles.
1
QR 5
3
6
}5}
} 5 }
} 5 } 5 }
S
A
II
}5}
A II
PI
y
I
Geometry
Worked-Out Solution Key
BCD > DEB because they intercept BD . So,
nBFC , nDFE by the AA Similarity Postulate. Also,
CAD > EAB because they are the same angle. So,
nCAD , nEAB by the AA Similarity Postulate.
Ratio of lengths 5 10 : 9
Ratio of areas 5 102 : 92 5 100 : 81
c. One way: Use AB + AC 5 AD + AE to solve for the
}
length of DE.
9 + (9 1 11) 5 10 + (10 1 DE)
180 5 100 1 10 DE
80 5 10 DE
8 5 DE
b. A(0, 0), B(3, 4), C(9, 0)
Chapter 11,
continued
Another way: Use the ratios of corresponding sides of
the similar triangles.
Area of ﬁgure 5 Area of rectangle 1 Area of triangle
1
5 102 square units
9
10
}5}
10 1 DE
20
1
1
3. A 5 }h(b1 1 b2) 5 }(4)(3 1 6.5) 5 19 square units
2
2
10(10 1 DE) 5 180
4. d1 5 4 1 4 5 8 units, d2 5 5 1 8 5 13 units
1
1
A 5 }2 d1d2 5 }2 (8)(13) 5 52 square units
10 1 DE 5 18
DE 5 8
}
}
34. a. You are given JL 5 KL, so JK > KL. Because n JKL
is in the corner of a cube, L is a right angle. So,
n JKL is an isosceles right triangle. D is a right
angle because it is the corner of a cube. A cube has
} }
equal sides, which means MP > NP and n MNP is
an isosceles right triangle. Because both n JKL and
n MNP are isosceles right triangles, their side lengths
are proportional. So, n JKL , n MNP by the SAS
Similarity Theorem.
b. Ratio of lengths of triangles: 1 : 3
Ratio of areas of triangles: 12 : 32 5 1 : 9
5. The ratio of the lengths of two similar heptagons is 7 : 20,
so the ratio of their perimeters is also 7 : 20. The ratio of
their areas is 72 : 202, or 49 : 400.
6. The ratio of the areas is 1200 : 48, or 25 : 1. The ratio of
}
50
XY
5
1
}5}
10 5 XY
XY is 10 feet.
1. A 5 *w, where x 5 width and 1.5x 5 length
1.44(15 + 10) 5 1.5x + x
144 5 x2
c. The ratio of the area of n JKL to a face of the cube is
1 : 18. The area of the pentagon JKQRS is the area of
a face of the cube with the triangle n JKL subtracted
from it. So, the ratio of the area of n JKL to the area of
pentagon JKQRS is 1 : (18 2 1), or 1 : 17.
Mixed Review for TAKS
35. C;
x11
The algebraic tiles show that
x 2 x 2 2 5 (x 1 1)(x 2 2).
}
the lengths is Ï25 : Ï1 , or 5 : 1.
Problem Solving Workshop 11.3 (p. 744)
Because the area of the face PMTN is twice the
area n MNP, the scale factor of n JKL to the area
of one face of the cube is 1 : 9(2) 5 1 : 18.
1
5 bh 1 }2ba 5 (12)(6) 1 }2(12)(5)
AB
}5}
AE
AC
12 5 x
The length of the third pan is 1.5(12) 5 18 inches.
1
1
2. Area of PQRS 5 }h(b1 1 b2) 5 }(6)(9 1 12) 5 63
2
2
4
28
} of the area of PQRS. So,
The area of WXYZ is }
5
63} 9
Ï4
2
each side in WXYZ is }
} , or } the size of each
3
Ï9
corresponding side in PQRS. So,
2
So, the solutions to the equation
x 2 2 x 5 2, or x2 2 x 2 2 5 0, are
x1150
x 5 21
and
x2250
and
x52
x22
1
1
2
2
2
2
2
2
WZ 5 }3 PS 5 }3(12) 5 8 units.
3. ﬁrst square: A 5 s2
second square: A 5 2s2
Each side of a square is the square root of the area,
}
1
1
4. Area of nABC 5 }bh 5 }(8)(5) 5 20 cm2
2
2
1. A 5 bh; h 5 3b, A 5 108 ft2
108 5 b(3b)
11.25
The area of nDEF is }
, or 0.5625 of the area of
20
108 5 3b2
}
nABC. So, each side in nDEF is Ï 0.5625 , or 0.75 the
size of each corresponding side in nABC.
36 5 b2
65b
The base is 6 feet and the height is 3(6) 5 18 feet.
2. Use the Pythagorean Theorem to ﬁnd the height of the
triangle.
a55
DF 5 0.75(5) 5 3.75 cm
EF 5 0.75(8) 5 6 cm
DE2 1 EF2 5 DF2
(3.75)2 1 62 5 DF2
a2 1 122 5 132
a2 5 25
}
so the second length is Ï2s2 5 sÏ2 .
Quiz 11.1–11.3 (p. 743)
a
13
12
50.0625 5 DF2
7.08 ø DF
The length of DE is 3.75 cm and the length of DF is
Geometry
Worked-Out Solution Key
347
Chapter 11,
continued
Mixed Review for TEKS (p. 745)
4. G;
Ratio of costs 5 1.5 : 2.25 5 2 : 3
1. C;
Area of green band
Area of orange band
2
11
23
Ratio of side lengths 5 12 : 18 5 2 : 3
121
529
}} 5 }2 5 }
Ratio of areas 5 122 : 182 5 22 : 32 5 4 : 9
Let x 5 area enclosed by the orange band.
121
4
}5}
529
x
Ratio of perimeters 5 4(12) : 4(18) 5 2 : 3
Area is not proportional to the cost per tile.
5. D;
2116 5 121x
The area, A, of a rhombus is found by using the formula
17.5 ø x
1
A 5 }2 (d1)(d2) where d1 and d2 are the lengths of its
So, about 17.5 2 4 5 13.5 square feet of area are
enclosed by the orange band but outside the green band.
diagonals. If each diagonal is multiplied by the same
number n, the lengths of the diagonals become nd1 and
2. H;
Perimeter of pentagon LMNPQ:
nd2, and the area becomes A 5 }2 (nd1)(nd2) 5 n21 }2 d1d2 2.
5 1 6 1 13 1 14 1 4 5 42 mm
So, the area changes by a factor of n2.
1
Using the Pythagorean Theorem,
}
}
MQ 5 Ï5 2 4 5 Ï25 2 16 5 Ï9 5 3 mm
2
2
NQ 5 Ï142 2 132 5 Ï196 2 169 5 Ï 27 ø 5.2 mm
1
1250 5 }2(5)(x 1 3x)
Area of pentagon LMNPQ:
1250 5 2.5(4x)
Area of nLMQ 1 Area of nMNQ 1 Area of nNPQ
1250 5 10x
}
1
}
1
}
1
ø }2(3)(4) 1 }2(3)(5.2) 1 }2 (13)(5.2)
5 6 1 7.8 1 33.8
5 47.6 mm2
Let x 5 the perimeter of a pentagon that is similar to
pentagon LMNPQ but whose area is 190.28 square
millimeters.
190.28 mm2
47.6 mm
190.28
47.6
B
x
A
5
C
3x
D
125 5 x
}
The length of CD is 3(125) 5 375 units.
} }
7. Quadrilateral EFGH is a kite, so FH > EG.
Because nEFH
is an
isosceles right triangle, by Theorem
}
}
7.8, FH 5 5Ï2 + Ï2 5 5 + 2 5 10.
mFGJ 5 mHGJ 5 308, so mEFJ 5 mEHJ 5
1
2
}(3608 2 908 2 608 2 608 2 308 2 308) 5 458,
x 2 mm2
(42 mm)
}
ø }2
2
}
5Ï2
Ï2
so EJ 5 FJ 5 }
} 5 5
2
x
1764
}ø}
JG
tan 608 5 }
FJ
335,653.92 ø 47.6x 2
7051.6 ø x2
JG
tan 608 5 }
5
84 ø x
A concave pentagon similar to pentagon LMNPQ but
whose area is 190.28 square millimeters has a perimeter
that is closest to 84 millimeters.
3. C;
Area of rectangle 5 bh 5 (15)(9.5) 5 142.5 ft2
1
1
5 + tan 608 5 JG
8.66 ø JG
EG 5 EJ 1 JG ø 5 1 8.66 5 13.66
Using the formula for area, A, of a kite with diagonal
lengths d1 and d2:
1
1
1
Area of rhombus 5 }2 d1d2 5 }2(9.5)(15) 5 71.25 ft2
A 5 }2d1d2 5 }2(FH)(EG) ø }2(10)(13.66) ø 68
Area for marigolds 5 Area of rectangle
2 Area of rhombus
The area of EFGH, to the nearest square unit, is
68 square units.
Area for marigolds 5 142.5 2 71.25 5 71.25 ft2
Total cost 5 cost for marigolds 1 cost for asters
5 \$.30(71.25) 1 \$.40(71.25) ø \$49.88
Walter will spend about \$49.88 on ﬂowers.
348
1
A 5 }2 h(b11 b2)
6.
Geometry
Worked-Out Solution Key
}
1
Chapter 11,
continued
Lesson 11.4
7.
11.4 Guided Practice (pp. 747–749)
14 m 5 r
1. Circle with diameter of 5 inches:
8. C 5 :d 5 :(14) ø 43.98 units
C 5 :d 5 :(5) ø 15.71
9. C 5 2:r 5 2:(3 1 2) 5 2:(5) ø 31.42 units
The circumference is about 15.71 inches.
10. C 5 :d 5 :(10 4 2) 5 :(5) ø 15.71 units
C C
Circle with circumference of 17 feet: C 5 :d
408
mAB
11. Arc length of AB 5 } + 2:r 5 } + 2:(6)
3608
3608
17 5 :d
5.41 ø d
ø 4.19 m
C C
The diameter is about 5.41 feet.
1208
mAB
12. Arc length of AB 5 } + 2:r 5 } + 2:(14)
3608
3608
2. Circumference: C 5 :d 5 :(28) ø 87.96 in.
Distance traveled 5 Number of revolutions +
circumference
ø 29.32 cm
1 ft
12 in.
500 ft 5 Number of revolutions + 87.96 in. + }
500 ft 5 Number of revolutions + 7.33 ft
The tire makes about 68 revolutions while traveling
500 feet.
758
3. Arc length of DQ 5 } + :(9) ø 5.89 yards
3608
C
C
Arc length of LM
mLM
4. }} 5 }
C
3608
3608 2 608
C
C
1508
3608
}
3780 5 300:r
1
1
6. Distance 5 2(84.39) 1 2 + } + 2: + 44.02
2
2
Distance ø 445.4 meters
The runner on the blue path travels about 445.4 meters.
11.4 Exercises (pp. 749–752)
Skill Practice
C C
mRSQ
2108
20. Arc length of RSQ 5 } + 2:r 5 } + 2:(8)
3608
3608
C
C
ø 29.32 ft
C
CB
mA
8.73
}5}
Arc length of AB
mAB
21. }} 5 }
2:r
3608
2:(10)
3608
C
508 ø mC
AB
8.73
3608 + }
F
5 mAB
20:
Arc length of AB
mAB
1. }} 5 }
2:r
3608
2. The arc measure is the number degrees of a circle the
arc is bounded by. The arc length is the part of the
circumference of the circle that the arc occupies.
3. C 5 2:r 5 2:(6) ø 37.70 in.
4. C 5 :d 5 :(17) ø 53.41 cm
63 5 2:r
}
ø 20.94 ft
4.01 ft ø r
10.03 ft ø r
3008
C
CS 5 1508 (from Exercise 17)
18. mR
C
CS 1 mSCQ 5 1508 1 608 5 2108
SQ 5 mR
mR
C
QR
1508
CR 5 m3608
19. Arc length of Q
+ 2:r 5
+ 2:(8)
3608
}5}
C 5 2:r
}
mQR 5 1508
mEF
EF
5. Arc length of } 5 }
2:r
3608
5.
}
mQPR 5 }
5}
5 1508
2
2
3
4
81.68 m 5 C
C
C
C
C
RS 5 3608 2 608
mQ
C
RS 5 3008
mQ
mC
QRS
3008
C
16. Arc length of Q
RS 5
+ 2:r 5
+ 2:(8)
3608
3608
15. mQRS 5 3608 2 mQS
17. QPR > RPS
}5}
10.5 ft
2:r
14. Two arcs from different circles have the same length only
ø 41.89 ft
2708
61.26 m
}5}
C
3608
61.26 m
C
C C
458
mAB
13. Arc length of AB 5 } + :d 5 } + :(8) ø 3.14 ft
3608
3608
if the circles have the same circumference.
68 ø Number of revolutions
C
C 5 2:r
28: 5 2:r
6. C 5 :d 5 :(5) 5 5: in.
C
C
Arc length of CD
mCD
22. }} 5 }
C
3608
7.5
C
768
3608
7.5
C
19
90
}5}
}5}
35.53 units ø C
Geometry
Worked-Out Solution Key
349
Chapter 11,
Arc length C
LM
CM
mL
23. }} 5 }
38.95
2:r
2608
3608
38.95
2:r
13
18
C
AB
28.978 ø mC
2.86
2:(4Ï 2 )
3608 + }
} 5 mAB
}5}
701.1 5 26:r
8.58 units ø r
24. P 5 2 lengths 1 circumference of the circle
5 2(13) 1 :(6) ø 44.85 units
1
25. P 5 2 lengths 1 } (circumference of the circle)
2
1
5 2(6) 1 }2 (2:(3)) ø 21.42 units
}
26. x 1 y 5 16 has a radius of Ï 16 5 4.
2
C 5 2:r 5 2:(4) 5 8:
5.09
2
0.8
4.2
2.05
4Ï2
mAB
458
608
21.498
1838
908
28.988
4
2.09
0.3
13.41
3.22
2.86
C
C
Length of
AB
C 5 2:r 5 2:(3) 5 6:
is twice as large when the radius is doubled.
C 5 :d
}F
5r
C
2:
}F
5d
When C 5 26::
When C 5 26::
26 :
}5r
2:
26 :
}5d
:
13 5 r
26 5 d
C
30. When mAB 5 458 and length of AB 5 4:
C
CF 5 x8:r
Arc length of E
908
}
x8:r
908
32. A;
C C
1408 1 mC
YZ 5 1808
CZ 5 408
mT
}
mXY 1 mYZ 5 1808 because XZ is the diameter of the
circle.
C
When r 5 2 and mAB 5 608:
608
3608
The diameter is 6, so the radius is 6 4 2 5 3.
C m3608YCZ
Arc length of AB 5 } + 2:(2) ø 2.09
C
0.3
2:(0.8)
C
mAB
3608
C
CB
21.498 ø mA
CB 5 1838:
When r 5 4.2 and mA
CB 5 1838
Arc length of A
+ 2:(4.2) ø 13.41
3608
CB 5 908 and length of ACB 5 3.22:
When mA
0.3
3608 + } 5 mAB
2:(0.8)
}
908
3608
3.22 5 } + 2:r
2.05 ø r
408
3608
2
Arc length of YZ 5 } + 2:r 5 } + 2:(3) 5 }3 :
When r 5 0.8 and length of AB 5 0.3:
}5}
C
1808
is twice as large when the measure of EF is doubled.
458
3608
C
C
x8:r
Because } is twice as large as }, the length of EF
4 5 } + 2:r
5.09 ø r
C
b. If you double the measure of EF :
2x8
+ 2:r
Arc length of EF 5 }
360
C
:
C
C
C
}
C 5 2:r 5 2:(3Ï 2 ) 5 6Ï2 :
29. C 5 2:r
C
x8
a. double the radius: Arc length of EF 5 } + 2:(2r)
3608
x8:r
Arc length of EF 5 }
908
x8:r
x8:r
Because } is twice as large as }, the length of EF
908
1808
}
}
28. x2 1 y2 5 18 has a radius of Ï 18 5 3Ï 2 .
}
C
x8:r
31. When mEF 5 x8 and r 5 r: Arc length of EF 5 }
1808
}
2
}
27. (x 1 2) 1 ( y 2 3) 5 9 has a radius of Ï 9 5 3.
2
C
mAB
2.86
}
} 5 }
3608
2:(4Ï2 )
}5}
2
C
}
When r 5 4Ï 2 and length of AB 5 2.86:
3608
33.
6 cm r
12 cm
8 cm
16 cm
A rhombus has diagonals that bisect each other. The
rhombus can also be split into 4 congruent triangles.
Look at one such triangle.
16
12
5 8 and }
5 6 2. Use the
1 with legs }
2
2
Pythagorean Theorem to ﬁnd the side length.
62 1 82 5 s2
100 5 s2
10 5 s
6 cm r
s
8 cm
Now look at the radius of the circle and break the
original right triangle into two smaller right triangles.
350
Geometry
Worked-Out Solution Key
2:r
continued
Chapter 11,
continued
Use the Pythagorean Theorem and a system of equations
to ﬁnd the length of the radius.
r2 1 (10 2 x)2 5 64
2 r2 1 x2
5 36
d
10 2 x
r
6 cm
r
8 cm
100 2 20x 1 x 2 x 5 28
2
72 5 20x
3.6 5 x
Now substitute 3.6 for x to ﬁnd r.
r2 1 (3.6)2 5 36
r 1 12.96 5 36
r2 5 23.04
r 5 4.8
C 5 2:r 5 2:(4.8) ø 30.16
The circumference of a circle inscribed in a rhombus
with diagonals that are 12 centimeters and 16 centimeters
34. Because you know the measure of the shaded red angle
is 308 and the arc length is 2, you can set up a proportion
for the circumference of the blue circle.
C
mblue
3608
}} 5 }
In Exercise 31, you found that when you double the
the blue circle is double that of the red circle, the arc
length is twice as big, or 2(2) 5 4. Because the angle
measure of the blue angle and the shaded red angle are
the same, mblue 5 308. So, substituting those values in
the proportion, you get:
C
4
C blue
62 1 62 5 d2
72 5 d2
d
6
}
6Ï2 5 d
}
6
The circumference of the circle is about 26.66 units.
37. Find the circumference of the tire:
C 5 :d 5 :(8 in.) ø 25.133 in.
You have the radius, so you can now ﬁnd the
circumference.
C
To ﬁnd the diameter, use the Pythagorean Theorem.
C 5 :d 5 :(6Ï2 ) ø 26.66
2
Arc length blue
C blue
6
x
(10 2 x)2 2 x2 5 28
2
36.
308
3608
}5}
48 5 C blue
So, the circumference of the blue circle is 48 units.
Problem Solving
35. The rope length, 21 feet 8 inches, represents
length of path 5 number of rotations + circumference
ø 87 + 25.133 5 2186.571
The length of the path is about 2187 inches long.
38. a. The chain touches about half of each sprocket and has
9
inches. Find half of
an additional two lengths of 6}
16
each sprocket’s circumference, add them together and
to that.
16 2
9
larger sprocket: C 5 :d 5 :1 6}8 2 ø 19.24
1
So, half of the circumference of the larger sprocket is
1
smaller sprocket: C 5 :d 5 :1 1}
ø 4.52
16 2
7
So, half of the circumference of the smaller sprocket is
1
length of chain 5 9.62 1 2.26 1 21 6}
ø 25
16 2
9
The length of the chain is about 25 inches.
b. The chain touches about half of each sprocket, so only
half of the teeth on each sprocket are gripping the
chain at any given time.
1
Half of the teeth on the larger sprocket: }2(76) 5 38
the circumferencec of the tree. If you divide the
circumference by :, you will get the diameter of
the tree.
Half of the teeth on the smaller sprocket: }2 (15) 5 7.5
2
21 feet 8 inches 5 21}3 feet
2
21}3 feet 4 : ø 6.9
There are about 46 teeth gripping the chain at any
given time.
The diameter of the tree is about 7 feet.
1
Total teeth 5 38 1 7.5 5 45.5
39. Because *1{{*2, 1 > 2 by the Alternate Interior
Angles Theorem. So, m1 5 7.28.
distance from Alexandria to Syene
C
m1
3608
575
C
7.28
3608
}}} 5 }
}5}
28,750 ø C
The Earth’s circumference is about 28,750 miles.
Geometry
Worked-Out Solution Key
351
Chapter 11,
continued
40. Because each arc is a half-circle, the measure of each arc
is 1808.
m each arc
3608
length of each arc 5 } + 2:r
x
1
y 5 214x 2 11
1808
length of each arc 5 } + 2:r
3608
214(1) 2 11
214(3) 2 11
5 214 2 11
5 242 2 11
5 225
5 253
4(1) 2 1
4(3)2 2 1
5 4(1) 2 1
5 4(9) 2 1
5421
5 36 2 1
53
5 35
22(1)3 1 1
22(3)3 1 1
5 22(1) 1 1
5 22(27) 1 1
5 22 1 1
5 254 1 1
5 21
5 253
22(1) 1 1
22(3) 1 1
5 22 1 1
5 26 1 1
5 21
5 25
2
length of each arc 5 :r
Because there are 4 arcs, the sum of the arc lengths
is 4:r.
y 5 4x2 2 1
m each arc
41. 8 segments: length of each arc 5 } + :d where
3608
d5r
y 5 22x 3 1 1
1808
length of each arc 5 } + :r
3608
1
length of each arc 5 }2:r
Because there are 8 arcs, the sum of all of their arc
y 5 22x 1 1
lengths is 81 }2:r 2 or 4:r.
1
m each arc
3608
16 segments: lengths of each arc 5 } + :d where
1
d 5 }2r
The equation that best describes the functional
relationship between x and y is y 5 22x3 1 1.
11.4 Extension (p. 754)
1 2
1808
3608
1
length of each arc 5 }4:r
3
1
2
length of each arc 5 } + : }r
1. The equator and the longitude lines are great circles.
Because there are 16 arcs, the sum of all their arc lengths
The latitude lines are not great circles. Earth’s center is
the center for the equator and lines of longitude. Earth’s
center is not the center for lines of latitude.
1
The sum of the arc lengths for 8 segments is 4:r, the
sum for the arc lengths for 16 segments is 4:r, and the
sum of the arc legnths for n segments is 4:r. The length
will be the same, you just have to allocate the radius
differently according to how many segments you have.
More than 1 line
through endpoints of
diameter on a sphere
Mixed Review for TAKS
42. C;
22
21
214(22) 2 11
214(21) 2 11
5 28 2 11
5 14 2 11
5 17
53
x
y 5 214x 2 11
4(22) 2 1
4(21)2 21
5 4(4) 2 1
5 4(1) 2 1
5 16 2 1
5421
5 15
53
22(22)3 1 1
22(21)3 1 1
2
y 5 4x2 2 1
y 5 22x 3 1 1
y 5 22x 1 1
352
Geometry
Worked-Out Solution Key
5 22(28) 1 1 5 22(21) 1 1
5 16 1 1
5211
5 17
53
22(22) 1 1
22(21) 1 1
5411
5211
55
53
Only 1 line through 2
points not endpoints
of diameter
3. If two lines intersect, then their intersection is exactly
2 points.
C
C
x
20:
1208
3608
Arc length of AB
mAB
4. }} 5 }
2:r
3608
1
3
}F
5}5}
2
x 5 6}3:
2
1
The distance are 6}3 : and 13}3:.
C
Arc length of ACB
2:r
C
mACB
3608
}} 5 }
x
20:
3608 2 1208
3608
2408
3608
2
3
}F
5}5}5}
1
x 5 13}3:
is 1 16}4: 2r, or 4:r.
Chapter 11,
CB mC
Arc length of A
AB
5. }} 5 }
2:r
continued
6. Yes, you can ﬁnd the measure of the intercepted arc if
3608
908
3608
x
16:
you know the area and radius of a sector of the circle.
The formula for the area of a sector is
1
4
}F
5}5}
measure of arc
3608
Area of sector 5 }} + :r2. If you solve for
x 5 4:
C
Arc length of ACB
2:r
C
the measure of the arc, you get:
mACB
3608
}} 5 }
x
16:
3608 2 908
3608
2708
3608
Area of circle
:r
3
4
}F
5}5}5}
3608 + Area of circle
:r
}}
5 measure of arc
2
x 5 12:
11.5 Exercises (pp. 758–761)
The distances are 4: and 12:.
C
C
Skill Practice
Arc length of AB
mAB
6. }} 5 }
2:r
3608
1. A sector of a circle is the region bounded by two radii of
the circle and their intercepted arc.
1408
7
x
}F
5}5}
30:
18
3608
2. Doubling the arc measure of a sector in a given circle
2
will double its area. When you double the arc measure,
you make the sector twice as big, which means that the
area also doubles.
x 5 11}3 :
C
Arc length of ACB
2:r
C
mACB
3608
}} 5 }
x
30:
3. A 5 :r2 5 : + 52 5 25: ø 78.54
3608 2 1408
3608
2208
3608
11
18
}F
5}5}5}
1
x 5 18}3 :
2
The area of the circle is exactly 25: square inches, which
4. The radius of a circle is half of the diameter, so the radius
1
is }2(16) 5 8 ft.
1
The distances are 11}3: and 18}3:.
A 5 :r2 5 : + 82 5 64: ø 201.06
The area of the circle is exactly 64: square feet, which is
Lesson 11.5
11.5 Guided Practice (pp. 756-757)
1. A 5 :r2 5 :(14)2 ø 615.75
measure of arc
3608
}}
5 }}
2
5. The radius of a circle is half of the diameter, so the radius
1
is }2(23) 5 11.5 cm.
The area of (D is about 615.75 square feet.
A 5 :r2 5 :(11.5)2 5 132.25: ø 415.48
C
mFE
2. Area of red sector 5 } + :r2
3608
The area of the circle is exactly 132.25: square
centimeters, which is about 415.48 square centimeters.
1208
3608
5 } + : + 142 ø 205.25
6. A 5 :r2 5 : + (1.5)2 5 2.25: ø 7.07
The area of the circle is exactly 2.25: square kilometers,
which is about 7.07 square kilometers.
The area of the red sector is about 205.25 square feet.
C
mFGE
3. Area of the blue sector 5 } + :r2
3608
7.
3608 2 1208
3608
5 } + : + 142 ø 410.50
C
mFG
4. Area of sector FJG 5 } + Area of (H
3608
858
3608
The area of (H is 907.92 square centimeters.
5. Area of ﬁgure 5 Area of semicircle 1 Area of triangle
1
2
5 } + : + 1 }2 2 1 }2(7)(7)
3608
7 2
A 5 :r2
380 5 :r2
154
:
} 5 r2
380
:
7.00 ø r
11.00 ø r
7 meters.
11 meters.
A 5 :r2
676: 5 :r2
907.92 5 Area of (H
1808
9.
8.
154 5 :r2
} 5 r2
The area of the blue sector is about 410.50 square feet.
214.37 5 } + Area of (H
A 5 :r2
1
5 6.125: 1 24.5 ø 43.74
676 :
:
} 5 r2
26 5 r
The radius is 26 centimeters, so the diameter is
52 centimeters.
The area of the ﬁgure is about 43.74 square meters.
Geometry
Worked-Out Solution Key
353
Chapter 11,
continued
10. The area of sector XZY should be divided by the area of
the circle, not 3608. Also, the right side should be the
C
mJK
16. Area of sector JK 5 } + :r2
3608
898
3608
758
3608
12.36 5 } + :r2
measure of the sector divided by 3608, or }.
50
:
758
3608
} 5 r2
3.99 ø r
n 5 10
The area of sector XZY is 10 square feet.
C
mED
11. Area of small sector 5 } + :r2
3608
17. Area of shaded region 5 Area of square
2 2(Area of semicircle)
1 1808
3608
608
5 } + : + 102 ø 52.36
3608
5 6(6) 2 2 }(: + 32)
C
5 36 2 9: ø 7.73
mEGD
3608
Area of large sector 5 } + :r2
3608 2 608
3608
5 } + : + 102 ø 261.80
The areas of the small and large sectors are about 52.36
square inches and 261.80 square inches, respectively.
7.73 square meters.
18. Area of shaded region 5 Area of trapezoid
2 Area of semicircle
C
1
5 }2(8)(16 1 20)
mED
12. Area of small sector 5 } + :r2
3608
1 1808
3608
2 } + (: + 42)
3608 2 2568
3608
5 } + : + 142
C
118.87 square inches.
Area of large sector 5 } + :r2
19. A;
2568
3608
5 } + : + 142 ø 437.87
Area of the putting green 5 Area of square
1 Area of sector of circle
The areas of the small and large sectors are about
177.88 square centimeters and 437.87 square
centimeters, respectively.
1 Area of rectangle
908
1 3608
C
5 3.52 1 } + (:(3.5)2)
mDE
13. Area of small sector 5 } + :r2
3608
C
mEGD
Area of large sector 5 } + :r2
3608
3608 2 1378
3608
5 } + : + 282 ø 1525.70
The areas of the small and large sectors are about 937.31
square meters and 1525.70 square meters, respectively.
C
mJK
14. Area of sector JK 5 } + Area of (M
3608
1658
38.51 5 } + Area of (M
3608
1 7(3.5)
The area of the putting green is about 46 square feet.
A 5 :r2
20.
260.67 5 :r2
260.67
:
} 5 r2
9.11 ø r
21. C 5 2:r ø 2:(9.11) ø 57.24
The circumference of (M is about 57.24 inches.
C
CL
mK
42 5 } + 260.67
mKL
22. Area of sector KML 5 } + Area of (M
3608
84.02 ø Area of (M
The area of (M is about 84.02 square meters.
C
mJLK
15. Area of sector KLJ 5 } + Area of (M
3608
3608 2 508
56.87 5 } + Area of (M
3608
66.04 ø Area of (M
The area of (M is about 66.04 square centimeters.
354
Geometry
Worked-Out Solution Key
2
ø 12.25 1 9.62 1 24.5 ø 46.37
1378
5 } + : + 282
3608
ø 937.31
2
5 144 2 8: ø 118.87
ø 177.88
mEGD
3608
2
3608
C
C
588 ø mKL
n
48
}5}
Chapter 11,
continued
C
23. Perimeter of blue region 5 arc length of KNL
30. Area of shaded region 5 Area of largest circle
2 Area of second largest circle
C
1 Area of third largest circle
mKNL
3608
3608 2 588
ø } + 2:(9.11) 1 2(9.11)
3608
5 } + 2:r 1 2r
2 Area of smallest circle
5 : + (2 1 2 1 2 1 2)2
2 : + (2 1 2 1 2)2
ø 48.02: 1 18.22 ø 66.24
1 : + (2 1 2)2 2 :(2)2
The perimeter of the blue region is about 66.24 inches.
C
5 64: 2 36: 1 16:
mKL
24. Arc length of KL 5 } + 2:r
3608
C
2 4: ø 125.66
588
ø } + 2:(9.11) ø 9.22
3608
31. Both triangles are right triangles by Theorem 10.9.
The length of KL is about 9.22 inches.
C
25. Perimeter of red region 5 length of KL 1 2(radius)
ø 9.22 1 2(9.11) ø 27.44
32 1 42 5 d2
The perimeter of the red region is about 27.44 inches.
26. Use the Pythagorean Theorem to ﬁnd the radius of the
circle and base and height of the triangle.
5 2
}
5Ï 7
r5}
2
Area of shaded region 5 Area of circle 2 Area of triangle
}
1 5Ï2 2
2
}
1
1 5Ï2
5:+ }
2 }2 }
2
2
2
C
1088
3608
5 } + : + (4 1 4)2 5 19.2:
26.77 square inches.
Area of the blue region 5 :r2 5 : + 42 5 16:
1C
5 2 } + :r2
3608
2
1808 2 1098
5 2 } + : + (5.2)2
3608
1
Area of the yellow region 5 Area of (P
2 Area of red region
2 Area of blue region
2
ø 2(16.75) ø 33.51
28. Area of shaded region 5 Area of square
2 4(Area of 1 circle)
2 4(: + (5)2)
5 400 2 100: ø 85.84
29. Area of shaded region 5 Area of bigger semicircle
2 Area of smaller semicircle
1808
F 3608
G
1808
2F
+ (: + 17 ) G
3608
5 } + (: + (17 1 17)2
}
7.63 square meters.
mRS
32. Area of the red region 5 } + :r2
3608
ø 12.5: 2 12.5 ø 26.77
5 20
1
5 6.25: 2 12 ø 7.63
r
2
55d
5 : 1 }2 2 2 21 }2(3)(4) 2
5 in.
r
25 5 d2
Area of shaded region 5 Area of circle + 2(Area of triangles)
r2 1 r2 5 52
2r 5 25
Use the Pythagorean Theorem to ﬁnd the length of the
diameter of the circle
2
5 : + (4 1 4)2 2 19.2:
2 16: 5 28.8:
The area of the red region is 19.2: square units, the area
of the blue region is 16: square units, and the area of the
yellow region is 28.8: square units.
33. Rewrite of the Perimeters of Similar Polygons Theorem:
For any two circles, the ratio of their circumferences is
equal to the ratio of their radii. Rewrite of the Area of
Similar Polygons Theorem: For any two circles, if the
length of their radii is in the ratio of a : b, then the ratio of
their areas is a2 : b2. All circles are similar, so you do not
need to add that the circles must be similar.
34. These sectors are not similar, therefore the student
shouldn’t have used Theorem 11.7 (Areas of Similar
Polygons). The correct ratio of the arcs of the sectors
bounded by these areas is 2 : 1.
5 578: 2 144.5: ø 1361.88
1361.88 square centimeters.
Geometry
Worked-Out Solution Key
355
Chapter 11,
continued
C
Now substitute 4 5 r to ﬁnd mFG .
35.
C
CG
mF
10
}5}
mFG
10 5 } + 2:(4)
3608
r
s
x
2(:)4
3608
C
CH ø 143.28.
FD 5 mE
So, mC
143.28 ø mFG
radius of large circle 5 r, radius of small circle 5 x,
side of square 5 s
Use the Pythagorean Theorem to ﬁnd the radius of the
small circle in terms of r.
r2 1 r2 5 s2
12 2
r
Problem Solving
r2
2
x 5r 2}
2
r
1
x2 5 }2r2
Î
1
37. The radius for the eye of the hurricane is }(20) 5 10
2
s
2
miles.
x
A 5 :r2 5 : + 102 ø 314.16
}
The area of land underneath the eye is about
314.16 square miles.
1
x 5 }2r2
}
rÏ2
38. A 5 :r2
x5}
2
Area of large circle
:r2
:r2
:r2
1
2
}} 5 }2 5 }
5}
5 }1 5 }1
}
1
Area of small circle
:x
rÏ 2 2
}:r2
}
: }
2
2
2
1 2
The ratio of the area of the large circle to the area of the
small circle is 2 : 1.
36. Let r be the radius of the smaller circle.
CG
F
C m3608
Length of FG 5 } + 2:r
C
C
C
C
C
mFG
3608
mFG
5
}5}
:r
3608
mEH
Length of EH 5 } + 2:r
3608
mEH
30 5 } + 2:(8 1 r)
3608
mEH
15
}5}
:(8 1 r)
3608
10 5 } + 2:r
C
C C
Because mFG 5 mEH , set their equations equal to each
other to ﬁnd the length of the radius.
5
:r
13,656 5 :r2
13,656
:
}5 r2
210.08 ø r
The distance you walk around the edge is the
circumference of the pond.
C 5 2:r ø 2:(210.08) ø 1319.97
39. a. A circle graph is appropriate because the data is
b. percentage + (3608) 5 central angle
Bus: 0.65 + (3608) 5 2348
Walk: 0.25 + (3608) 5 908
Other: 0.10 + (3608) 5 368
The central angle for the bus is 2348, the central angle
for walking is 908, and the central angle for other
modes of transportation is 368.
15
:(8 1 r)
}5}
5:(8 1 r) 5 15:r
Walk
25%
Bus
65%
8 1 r 5 3r
Other
10%
8 5 2r
45r
Measure of arc
c. Area of sector 5 }} + :r 2
3608
2348
3608
Bus: } + :(2)2 ø 8.17
908
3608
Walk: } + :(2)2 5 :
356
Geometry
Worked-Out Solution Key
2
143.28
3608
160 square meters.
s
}
rÏ 2 2
} 1 x2 5 r2
143.28
3608
5 } + : + (8 1 4)2 2 } + : + 42
ø 160
r
2r2 5 s2
}
rÏ 2 5 s
Area of shaded region 5 Area of larger sector
2 Area of smaller sector
Chapter 11,
continued
368
3608
Other: } + :(2)2 ø 1.26
c.
The area of the bus sector is about 8.17 square inches,
the area of the walking sector is : square inches, and
the area of the sector for other modes of transportation
40. The area of a tortilla with a 6 inch diameter is
6 2
: }2 5 9:. The area of a tortilla with a 12 inch diameter
12 2
is : }
5 36:. Because the 12 inch tortilla is 4 times
2
1 2
1 2
larger, you need 4 times as much dough. So, you need
41 }4 2 5 1 cup of dough to make a tortilla with a 12 inch
1
Circumference of
10
14
1 2: + }
two 10-inch pizzas 5 21 2 + : + }
22 1
22
and one 14-inch
5 34: ø 106.81
pizza
You should buy four 10-inch pizzas because the
circumference is greatest, which means more crust.
43. a. The height is equal to the radius, the base is equal to
1
1
old “a” 5 }2 (: + 82) 1 }2 (12 1 9)(10 1 16) ø 374
new “a” 5 :(14)2 1 3(22) ø 682
The area of the old “a” is about 374 square
millimeters, the area of the new “a” is about 682
square millimeters, and the percentage increase in
b. area of old “a”: 75.5(66) 5 4983
area of new “a”: 85(76) 5 6460
6460 2 4983
ø 0.296 ø 30%
percent increase 5 }
4983
Sample answer: No; the percent increase of the area of
“a” is about 30%, which is much less than the percent
increase of the interior area.
Circumference of
14
5 28: ø 87.96
5 21 2 + : + }
22
two 14-inch pizzas
half of the circumference and the area is the base times
the height.
diameter.
1
Circumference of
10
5 41 2 + : + }
5 40: ø 125.66
22
four 10-inch pizzas
12 2
42. Area of 12-inch diameter pizza: : } 5 36: ø 113.10
2
1 2
Because a 12-inch diameter pizza has an area of about
113 square inches and feeds you and two friends, each
person eats about 113 4 3 5 38 square inches of pizza.
Area of 10-inch diameter pizza: :1 }
5 25: ø 78.54
22
10 2
h 5 r, b 5 }2(2:r) 5 :r, r + :r 5 :r 2
b. Sample answer: When you cut the circle into 16
congruent sectors, you are not losing area. When
you rearrange the 16 pieces of the circle to make a
parallelogram, you can determine that the area is :r 2.
Because no area is lost when you make the circle into
a parallelogram, the area of the circle will be the same
as the area of the parallelogram, or :r 2.
44. Let black diameter 5 x, blue diameter 5 y, and red
diameter 5 z. Using Pythagorean Theorem, you can say
x2 1 y2 5 z2.
1
Area of triangle 5 }2 xy
1808
3608
1
y 2
:y2
2
Area of semicircle y 5 } + : + 1 }2 2 5 }
8
3608
1808
1808
3608
:x2
x
Area of semicircle x 5 } + 1 : + 1 }2 22 2 5 }
8
Area of non-shaded areas 5 Area of semicircle z
2 Area of triangle
1 2
14 2
5 49: ø 153.94
Area of 14-inch diameter pizza: : }
2
If you want to feed yourself and 7 friends, you need
about 8(38) 5 304 square inches of pizza.
:z2
z
Area of semicircle z 5 } + 1 : + 1 }2 22 2 5 }
8
:z2
1
2 }2 xy
5}
8
Area of shaded crescent 5 Area of semicircle y
1 Area of semicircle x
a. In order to have 304 square inches of pizza you could
buy four 10-inch pizzas, two 14-inch pizzas, or two
10-inch pizzas and one 14-inch pizza.
:y2
:x2
1 :z
2
1 :z
2
2
1
1}
2 }
2 }2xy
5}
8
8
8
Price of four 10-inch pizzas 5 4(\$6.99) 5 \$27.96
:(y 1 x
2
2
)
2
Price of two 14-inch pizzas 5 2(\$12.99) 5 \$25.98
1
5}
2 }
2 }2 xy
8
8
Price of two 10-inch pizzas and one 14-inch pizza
5 2(\$6.99) 1 \$12.99 5 \$26.97
5}
2}
1 }2 xy 5 }2 xy
8
8
To spend as little money as possible, buy two
14-inch pizzas.
b. You should buy two 10-inch pizzas and one 14-inch
:z2
1
:z2
1
1
1
So, because }2xy 5 }2xy, the sum of the area of the two
shaded crescents equals the area of the triangle.
pizza. Two 10-inch pizzas and one 14-inch pizza is
three pizzas total, the amount of pizza you need, and is
cheaper than than three 14-inch pizzas.
Geometry
Worked-Out Solution Key
357
Chapter 11,
continued
6. Area of shaded region 5 Area of circle
Mixed Review for TAKS
2 Area of rhombus
45. A;
5 : + 1 }2 2 2 }2(3)(6)
6 2
Red socks 1 Blue socks 5 Total number of socks
x 1 y 5 28
1
5 9: 2 9 ø 19.27
Blue socks 5 Red socks 1 4
19.27 square centimeters.
y5x14
x 1 (x 1 4) 5 28
Lesson 11.6
2x 1 4 5 28
2x 5 24
11.6 Guided Practice (pp. 762–764)
} }
1. The center of the polygon is P, a radius is PY or PX,
}
the apothem is PQ, and the central angle is XPY.
x 5 12
y5x14
y 5 12 1 4
3608
1
2. mXPY 5 } 5 908, mXPQ 5 } mXPY 5 458,
4
2
y 5 16
Because x 5 12 and y 5 16, the solution (x, y) of the
system is (12, 16).
and mPXQ 5 1808 2 908 2 458 5 458
3. Use the Pythagorean Theorem to ﬁnd the base of
the triangle.
46. F;
The rate of change is the slope. Choose two points on the
graph, for example (0, 1) and (2, 22).
8
6.5
y2 2 y1
23
22 2 1
5}
5 21.5
m5}
x2 2 x1 5 }
220
2
s
The rate of change is 21.5.
s 1 6.52 5 82
2
s2 5 21.75
Quiz 11.4–11.5 (p. 761)
1 2
C
C
Arc length of GE
mGE
2. }} 5 }
C
3608
36
C
1
The perimeter of the regular pentagon is about 46.6 units
and the area is about 151.5 square units.
1028
3608
4. A decagon has 10 sides, so P 5 10(7) 5 70 units.
127.06 ø C
The circumference of (F is about 127.06 inches.
Arc length of C
JK
mJC
K
3. }} 5 }
3608
29
2:r
1
A 5 }2 aP ø }2 (6.5)(46.6) 5 151.45
}5}
2:r
s ø 4.66
The base of the triangle is about 4.66, so the length of
one side of the pentagon is about 2(4.66) 5 9.32. A
pentagon has 5 sides, so P ø 5(9.32) 5 46.6 units.
Because each of the 10 triangles that make up the polygon
is isosceles, each apothem bisects the side length and
central angle. Each side length is 7 units, so the bisected
1
side length is }2 (7) 5 3.5 units. Each central angle is
658
3608
}5}
3608
1
5 368, so the bisected angle is }2 (368) 5 188.
is }
10
10,440 5 130:r
To find the length of the apothem a use the sine function
with one of the bisected isosceles triangles.
25.56 ø r
4. Area of shaded region 5 Area of larger circle
2 Area of smaller circle
2
5 968: ø 3041.06
2
638
5 2 } + : + (8.7)2 ø 83.23
3608
83.23 square inches.
358
Geometry
Worked-Out Solution Key
3.5
a ø 10.77
meters.
1
a
tan 188 5 }
a
5 : + 33 2 : + 11
2
3.5
188
1
1
A 5 }2 aP ø }2 (10.77)(70) 5 376.95
The perimeter for the regular decagon is 70 units and the
area is about 377.0 square units.
C C
The length of C
788
mAB
14
1. Arc length of AB 5 } + 2:r 5 } + 2: } ø 9.53
2
3608
3608
Chapter 11,
continued
3608
5. The measure of the central angle is } 5 1208. The
3
a 2 1 3.422 5 102
a2 5 88.3036
bisected angle is 608. To find the side length of the
triangle, use trigonometric ratios.
a ø 9.4
1
1
Area 5 }2aP ø }2 (9.4)(61.56) ø 289.3
608
5
The area of the regular nonagon is about 289.3 square
units.
s
s
tan 608 5 }5
16. Use the Pythagoream Theorem to find the length of a
side of the regular heptagon.
5 + tan 608 5 s
}
5Ï 3 5 s
}
}
The regular triangle has side length 2(5Ï3 ) 5 10Ï3 .
}
2.77
2.5
}
So, the perimeter is P 5 3s 5 3(10Ï3 ) 5 30Ï3 .
1
1
}
A 5 }2 aP 5 }2 (5)(30Ï 3 ) ø 129.9 square units.
}
The perimeter for the regular triangle is 30Ï3 ø 52.0
units and the area is about 129.9 square units.
s
s2 1 2.52 5 2.772
s2 5 1.4229
s ø 1.19
6. You can use a special right triangle to solve Exercise 5.
The length of the base of the triangle is about 1.19,
so the length of one side of the polygon is about
2(1.19) 5 2.38.
11.6 Exercises (p. 765–768)
Skill Practice
1
1. F
2.  AFE
3. 6.8 units
4. 5.5 units
5. To find the measure of a central angle of a regular
polygon with n sides, divide 3608 by the number of
sides n of the polygon.
3608
6. } 5 368
10
3608
7. } 5 208
18
3608
8. } 5 158
24
3608
9. } ø 51.48
7
1
A 5 }2a + ns ø }2 (2.5)(7)(2.38) ø 20.8
The area of the regular polygon is about 20.8 square
units.
17. 7.5 is not the length of one side of the hexagon, it is the
length of half of one side (or the base of the triangle). 7.5
should be doubled to get the actual side of the hexagon.
1
3608
8
10. GJH is a central angle. So mGJH 5 } 5 458.
}
11. JK is an apothem, which makes it an altitude of isosceles
1
A 5 }2 a + ns ø }2 (13)(6)(15) 5 585 units2
18. B; The measure of the central angle of a dodecagon
3608
1
5 308. The bisected angle is }2 (308) 5 158. You
is }
12
know the side length is 8, so the base of the triangle is
}
}(8) 5 4. Use a trigonomic ratio to find the apothem.
1
tan 158 5 }a
nGJH. So, JK bisects GJH and
mGJK 5 }2 mGJH 5 22.58.
12. mKGJ 5 1808 2 908 2 22.58 5 67.58
13. EJH is 3 central angles combined.
So mEJH 5 3(458) 5 1358.
14. P 5 3(12) 5 36 units
}
1
1
A 5 }2 aP 5 }2 (2Ï 3 )(36) ø 62.4
The area of the regular triangle is about 62.4 square
units.
15. P 5 9(6.84) 5 61.56 units. Use the Pythagorean
Theorem to find the apothem a.
1
2
4
4
tan 158
a5}
3608
19. The measure of the central angle is } 5 458. The
8
1
bisected angle is }2(458) 5 22.58. Use trigonometric
ratios to find the apothom a and the side length s of
the triangle.
s
sin 22.58 5 }
20
20 + sin 22.58 5 s
a
cos 22.58 5 }
20
20 + cos 22.58 5 a
10
a
10
22.58
20 a
20
3.42
6.84
s
Geometry
Worked-Out Solution Key
359
Chapter 11,
continued
The regular octagon has side length
2(20 + sin 22.58) 5 40 sin 22.58.
23. You need to know the apothem and side length to find
the area of the regular square. You can use the methods
of special right triangles or trigonometry to find the
apothem length. The measure of the central angle is
P 5 8(40 + sin 22.58) ø 122.5 units
1
1
A 5 }2 aP 5 }2 (20 + cos 22.58)(122.5) ø 1131.8 units2
The perimeter of the regular octagon is about 122.5 units
and the area is about 1131.8 square units.
3608
20. The measure of the central angle is } 5 728. The
5
1
}
bisected angle is 2 (728) 5 368. To find the side length
3608
4
1
2
} 5 908 and the bisected angle is } (908) 5 458.
To find the length of the apothem and side length, use a
special triangle
458
458
14
14
x
of the pentagon, use a trigonometric ratio.
x 2
458
}
xÏ 2 5 14
14
Ï2
}
}
14Ï2
Ï2
Ï2
}
x5}
} + }
} 5 } 5 7Ï 2
2
s
s
}
}
tan 368 5 }
4.1
The regular square has side length s 5 2(7Ï2 ) 5 14Ï2 .
4.1 + tan 368 5 s
So, the area is A 5 }2 a + ns 5 }2 (7Ï2 )(4)(14Ï 2 )
1
The regular pentagon has side length 5 2(4.1 + tan 368)
5 8.2 + tan 368. So, the perimeter is 5(8.2 + tan 368)
1
1
ø 29.8 units, and the area is A 5 }2aP 5 }2(4.1)(29.8)
ø 61.1 square units.
}
5 392 square units.
24. You need to know the apothem to find the area of
the regular hexagon. You can use the methods of the
Pythagorean Theorem, special triangles, or trigonometry.
3608
3608
7
21. The measure of the central angle is } ø 51.48. The
1
}
1
bisected angle is }2 (51.4) ø 25.78. The side of the
1
heptagon is 9, so the base of the triangle is }2(9) 5 4.5.
To find the apothem use a trigonometric ratio.
The measure of the central angle is }
5 608 and the
6
1
bisected angle is }2(608) 5 308. The side of the hexagon
1
is 10, so the base of the triangle is }2(10) 5 5. To find the
apothem, use a special triangle.
4.5
tan 25.78 5 }
a
25.78
308
10
a ø 9.35
a
2x
a
x
5
So, the perimeter is 7(a) 5 63 units, and the area is
1
1
A 5 }2aP 5 }2 (9.35)(63) ø 294.5 square units.
22. There is enough information to find the area. The
3608
5 408. The bisected
measure of the central angle is }
9
1
}
angle is 2 (408) 5 208. Because the perimeter is 18 inches
and a nonagon has 9 sides, each side length is
18 4 9 5 2 inches. The side length of the nonagon is
}
So, x 5 5. The apothem is 5Ï3 .
1
}
1
The area is A 5 }2a + ns 5 }2 (5Ï3 )(6)(10) ø 259.8 square
units.
25. You need to know the side length to find the area of the
decagon. You can use the methods of the Pythagorean
Theorem or trigonometry. To find the side length, use the
Pythagorean Theorem.
1
2 inches, so the base of the triangle is }2 (2) 5 1 inch.
To find the apothem, use a trigonometric ratio.
x 3
608
608
4.5
308
x2 1 82 5 8.42
8.4
x2 5 6.56
8
}
x 5 Ï6.56
208
x
a
}
The regular decagon has side length s 5 2Ï656 .
So, the area is
1
1
tan 208 5 }a ø 2.8
1
1
The area is A 5 }2 aP 5 }2 (2.8)(18) ø 24.8 square inches.
360
Geometry
Worked-Out Solution Key
1
}
A 5 }2 a + ns 5 }2 (8)(10)(2Ï 6.56 ) ø 204.9 square units.
1
4.1
x
368
Chapter 11,
continued
26. Area of unshaded region 5 Area of circle
29. The measure of the central angle is 608, so the bisected
1
angle is }2(608) 5 308. Use a special triangle to find the
2 Area of square
5 : + 142 2 392
radius of the circle and base of the smaller triangle.
5 196: 2 392 ø 223.8
223.8 square units.
3608
27. The measure of the central angle is } 5 728 and the
5
1
measure of the bisected angle is }2 (728) 5 368. The side
length of the pentagon is 12, so the base of the triangle
1
608
r
2x
b
308
2 3
The base of the larger triangle is 2(2) 5 4.
Area of shaded region 5 Area of sector
2 Area of triangle
C
F
of the apothem and the radius of the circle.
368
x
So, the base b of the triangle is 2 and the radius r is
2(2) 5 4.
is }2 (12) 5 6. Use trigonometric ratios to find the length
r
608
308
x 3
m arc
3608
1
2 }2(base of larger n)(a)
5 } + :r 2
a
F
608
3608
G
}
1
5 } + : + 42 2 }2(4)(2Ï3 )
6
6
tan 368 5 }a
6
sin 368
ø 1.4 square units
6
tan 368
r5}
a5}
}
30. The circle has a radius of Ï 25 5 5. The central angle of
3608
5 728 and the bisected angle is
the pentagon is }
5
1
}(728) 5 368. Use trigonometric ratios to find the side
2
Area of shaded region 5 Area of circle
2 Area of pentagon
5 : + r 2 2 1 }2 a + ns 2
1
1
length and apothem of the pentagon.
y
2
2
6
5:+ }
sin 368
F
1
}
8
5 }3: 2 4Ï3
6
sin 368 5 }r
G
6
tan 368
2
1
2 }2 } (5)(12)
G
2
22
368
(4, 22)
x
5
5
ø 79.6 square units
a
b
3608
28. The measure of the central angle is } 5 1208 and the
3
1
measure of the bisected angle is }2 (1208) 5 608. Use a
special triangle to find the apothem and size length of
the triangle.
a
8
x
608
308
2x
308
x 3
b
So, x 5 4. The apothem
a is 4 and the side length b of
}
the triangle is 4Ï 3 .
The length
of}a side of the equilateral triangle is
}
2(4Ï 3 ) 5 8Ï3 .
Area of shaded region 5 Area of circle 2 Area of triangle
1
5 : + r 2 2 }2 + a + ns
sin 368 5 }5
5 cos 368 5 a
5 sin 368 5 b
The regular pentagon has side length
2b 5 2(5 + sin 368) 5 10 + sin 368. So, the area is
1
a 608
b
cos 368 5 }5
1
A 5 }2 a + ns 5 }2 (5 + cos 368)(5)(10 + sin 368)
ø 59.4 square units.
31. True. Because the radius is fixed and the circle around
the n-gons also stays the same, more and more of the
circle gets covered up as n gets larger.
32. True. Because the hypotenuse of the right triangle
represents the radius and the leg represents the apothem,
the apothem must always be less than the radius.
33. False. The radius can be equal to the side length, like in a
regular hexagon.
}
1
5 : + 8 2 }2 (4)(3)(8Ï3 )
2
}
5 64: 2 48Ï3
ø 117.9 square units
Geometry
Worked-Out Solution Key
361
Chapter 11,
continued
3608
Pentagon: The central angle measures }
5 728 and the
s
34.
308
s
h
x 3
1
bisected angle measures }2 (728) 5 368. Use a
2x
trigonometric ratio to find the apothem.
608
x
s
2
368
b or s
a
Use a special triangle to find the height of the triangle.
}
bÏ 3
b
.
So, x 5 }2 , s 5 b, and h 5 }
2
4
4
tan 368 5 }a
Because b 5 s, substitute that into the height you just
}
}
bÏ3
sÏ 3
1
found: h 5 }
5}
. Use the formula A 5 }2 bh to find
2
2
}
}
1 sÏ3 2
1
4
tan 368
a5}
Ï3 s 2
So, the area of the pentagon is
5}
.
the area in terms of s. A 5 }2 (s) }
4
2
A 5 }2a + ns 5 }2 1 } 2(5)(8) ø 110 square
1
1
4
tan 368
units.
Square: The area of the square is A 5 s2 5 82 5 64
square units.
a
a
Triangle: Look at the smaller triangle used for the
hexagon. The height of the triangle is the
apothem of the smaller triangle used for the
hexagon. So, the
In an equilateral triangle, each altitude creates two
congruent triangles. So, each altitude is also a median.
1
Looking at the figure the apothem is }3 the altitude of
the equilateral triangle. You found the height of the
}
1
1
1 2
}
5 16Ï3 square units.
}
Ï3 s2
1
+3+s5}
square units.
A 5 }2 }3 }
4
2
square units.
35. Shaded area 5 Area of hexagon 2 Area of pentagon
1 Area of square
Problem Solving
2 Area of triangle
36.
3608
5 608 and the
Hexagon: The central angle measures }
6
1
bisected angle measures }2(608) 5 308. The side
length of the hexagon is 8, so the base of the
1
triangle is }2 (8) 5 4. Use a special triangle
to find the apothem.
308
2x
a
608
308
x 3
3608
a. The measure of the central angle is } 5 608 and the
6
1
bisected angle measures }2 (608) 5 308. Use a special
8 in. 308
2x 308 x 3
a
0 8
6
608
x
x
}
}
So, x 5 4. The apothem is 4Ï3 inches.
So, x 5 4 and a 5 4Ï3 .
b. A regular hexagon is made up of 6 equilateral
So, the area of the hexagon is
1
8 in.
triangle to ﬁnd the apothem.
608
4
1
}
}
Shaded area 5 96Ï 3 2 110 1 64 2 16Ï 3
ø 92.6
}
}
1 1 sÏ3 2 2
}
1
area 5 A 5 }2 bh 5 }2 (8)(4Ï 3 )
}
sÏ 3
1 sÏ 3
, so the apothem is }3 }
.
triangle above, }
2
2
1
Substituting into the formula A 5 }2a + ns, you get
}
A 5 }2a + ns 5 }2 (4Ï 3 )(6)(8) 5 96Ï 3 square units.
triangles, so the side length of the hexagon is the same
as the radius of the hexagon, or 8. So, the perimeter is
P 5 6(8) 5 48 inches and the area is
1
1
}
A 5 }2 aP 5 }2 (4Ï3 )(48) ø 166 square inches.
362
Geometry
Worked-Out Solution Key
a
Chapter 11,
continued
37. The apothem of the octagon is 1 1 0.2 5 1.2 centimeters.
3608
5 458
The measure of the central angle is }
8
1
and the bisected angle measures }2 (458) 5 22.58.
Use a trigonometric function to ﬁnd the side length of
the octagon.
b
39. A regular pentagon has 5 sides of equal length. So, the
side length of the smaller pentagon is 15 4 5 5 3 inches.
3608
The measure of the central angle is }
5 728 and the
5
1
bisected angle measures }2(728) 5 368. Because the side
length of the smaller pentagon is 3 inches, the base of the
1
triangle is }2(3) 5 1.5 inches. Use a trigonometric ratio
to ﬁnd the apothem.
1.2 cm
368
a
22.58
b
tan 22.58 5 }
1.2
1.5 in.
1.2 + tan 22.58 5 b
1.5
tan 368 5 }
a
The regular octagon has side length
1.5
tan 368
a5}
s 5 2b 5 2(1.2 + tan 22.58) 5 2.4 + tan 22.58.
Area of silver border 5 Area of octagon 2 Area of circle
1
5 }2 a + ns 2 :r 2
1
5 }2 (1.2)(8)(2.4 + tan 22.58) 2 :(1)2
ø 1.6 square centimeters
The apothem is 1.2 centimeters, the area of the octagon
is about 4.8 square centimeters, and the area of the silver
border is about 1.6 square centimeters.
38. Sample prediction: The pentagon will have the greatest
area and the circle will have the smallest area.
13 2
a. Area of circle 5 :r 2 5 : } ø 132.7 in.2
2
1 2
So, the area of the small trivet is
1
1
1 tan1.5368 2
A 5 }2 aP 5 }2 } (15) ø 15.5 square inches.
area of small trivet
area of large trivet
(perimeter of small trivet)2
}} 5 }}
2
(perimeter of large trivet)
152
15.5
}} 5 }2
area of large trivet
25
43.0 ø area of large trivet
The area of the small trivet is about 15.5 square inches
and the area of the large trivet is about 43.0 square inches.
40. a.
A
B
1
1
b. Area of triangle 5 } bh 5 } (18)(15) 5 135 in.2
2
2
3608
c. The measure of the central angle is } 5 728 and the
5
1
}
bisected angle measures 2 (728) 5 368. The side length
of the pentagon is 9 inches, so the base of the triangle
1 in.
1
is }2 (9) 5 4.5 inches. Use a trigonomic ratio to ﬁnd
the apothem.
368
3608
b. The measure of the central angle is } 5 608 and the
6
1
bisected angle measures }2 (608) 5 308. Because the
a
side length is 1 inch, the base of the triangle is
1
2
}(1) 5 0.5 inch. Use a trigonometric ratio to ﬁnd
4.5 in.
the apothem.
4.5
tan 368 5 }
a
4.5
tan 368
a5}
0 8
3
1
a
2
1
1 4.5
So, the area is A 5 }2 a + ns 5 }2 } (5)(9)
tan 368
ø 139.4 in.
2
The area of the circle is about 132.7 square inches, the
area of the triangle is 135 square inches, and the area
of the pentagon is about 139.4 square inches.
.5in.
0
0.5
tan 308 5 }
a
0.5
tan 308
a5}
Geometry
Worked-Out Solution Key
363
Chapter 11,
continued
So, the area of the hexagon is
1
1
1 tan0.5308 2
A 5 }2 a + ns 5 }2 } (6)(1) ø 2.6 square inches.
Area of shaded region 5 Area of circle
2 Area of hexagon
ø : + 12 2 2.6
3
2
ø 0.54 square inch
The area of the hexagon is about 2.6 square inches and
}
c. Sample answer: Draw AB with length 1 inch. Open
compass to 1 inch and draw a circle with that radius.
Using the same compass setting, mark off equal parts
along the circle. Connect every other intersection
or connect 2 consecutive points with the center of
the circle.
F
G
1
5 2F }2 (Ï 3 )(2 1 (2 1 2)) G
1
Area of 2 isosceles trapezoids 5 2 }2 h(b1 1 b2)
}
5 2F 3Ï3 G
}
}
5 6Ï 3
41. Let x be the side length of a regular hexagon. A hexagon
3
has 6 central angles, so 6 traingles are formed. The
3608
measure of a central angle is }
5 608. The bisected
6
1
3
angle has a measure of }2 (608) 5 308. Draw one of the
6 triangles formed by the central angles and find the side
length s and an expression in terms of a for x.
}
308
a
3
3
3
1
x
2
x
1
2
}
2
}x
a
cos 308 5 }s
tan 308 5 a
a
s5}
cos 308
43. Because PC is a radius, then P is the circumcenter of
}
n ABC. Let E be the midpoint of AB. Because P is the
}
}
circumcenter, CE and BD are perpendicular bisectors and
medians of n ABC. By the Concurrency of Medians of a
2a + tan 308 5 x
}
1 Ï3 2
a
2
Ï3
1}
2 2
}
2
2aÏ 3
3
2
Ï3
5a+}
}
2
Triangle Theorem, CP 5 }3 CE and BP 5 }3 BD.
1
}
}
So, DP 5 }3 BD. Because CE and BD are perpendicular
2a }
5x
3
5}
}
2
bisectors, BP 5 CP. So, CP 5 }3BP and 2DP 5 }3 BD.
}5x
So, radius CP is twice the apothem DP.
}
2aÏ3
5}
3
2.6
44. a. The average distance across a cell is }
5
The side length x of the hexagon is equal to the distance
s from the center of the hexagon to a vertex. So, each
of the 6 triangles is an equilateral triangle.
42. Sample answer: a regular hexagon can be broken into
6 equilateral triangles, 2 isosceles trapezoids, and
3 parallelograms.
1
1
}
}
Area of hexagon 5 }2 a + ns 5 }2 (Ï 3 )(6)(2) 5 6Ï3
364
F G
1
}
}
}
5 6Ï 3
Geometry
Worked-Out Solution Key
1
1
b. The apothem is } diameter 5 } (0.52)
2
2
5 0.26 centimeter. The measure of the central angle
3600
5 608 and the bisected angle measures
is }
6
1
2
} (608) 5 308. Use a trigonometric ratio to ﬁnd the
length of the base of the triangle.
b
1
5 6 1 }2 2(2)(Ï3 )
5 6[Ï 3 ]
5 0.52 centimeter.
tan 308 5 }
0.26
1
Area of 6 equilateral triangles 5 6 }2 bh
364
}
Area of 3 parallelograms 5 3(bh) 5 3(2)(Ï 3 ) 5 6Ï3
2
0.26 + tan 308 5 b
0.26 cm
308
b
s
2
Chapter 11,
continued
The length of a side of the hexagon is
2b 5 2(0.26 + tan 308) 5 0.52 + tan 308 ø
0.3 centimeter. So, the area of the cell is
1
regular hexagon:
10cm
1
A 5 }2 a + ns 5 }2 (0.26)(6)(0.3)
ø 0.234 square centimeter.
c. 100 cells + 0.234 square centimeter/cell 5
23.4 square centimeters
(1 decimeter)2
23.4 square centimeters + }}2
(10 centimeter)
5
tan 308 5 }a
5
tan 308
5 0.234 square decimeter
a5}
d. From part (c), 100 cells have an area of about
308
0.234 square decimeter. To ﬁnd how many cells are in
1 square decimeter, divide 100 by 0.234. So, there are
approximately 427 cells per square decimeter.
45. a. triangle:
a
10 cm
5 cm
A 5 }2aP 5 }21 } 2(60) ø 259.8 cm2
1
20 cm
20 cm
h
1
5
tan 308
regular decagon:
608
10 cm
20 cm
6cm
h
tan 608 5 }
10
10 + tan 608 5 h
1
2
3
1
2
A 5 } bh 5 } (20)(10 + tan 608) ø 173.2 cm2
tan 188 5 }a
3
tan 188
a5}
square:
188
a
15 cm
6 cm
3 cm
15 cm
A 5 s2 5 152 5 225 cm2
regular pentagon:
12cm
1
1 tan3188 2
1
A 5 }2 aP 5 }2 } (60) ø 277
The area of the equilateral triangle is about 173.2 square
centimeters. The area of the square is 225 square
centimeters. The area of the regular pentagon is about
247.7 square centimeters. The area of the regular
hexagon is about 259.8 square centimeters. The area of
the regular decagon is about 277 square centimeters.
The area increases as the number of sides increase.
6
a
tan 368 5 }
b. 60-gon: The central angle for a 60-gon measures
3608
1
} 5 68, the bisected angle measures } (68) 5 38,
60
2
6
tan 368
a5}
and the side length is 60 4 60 5 1 centimeter.
Because the side length is 1 centimeter, the base of
368
1
a
the triangle is }2 (1) 5 0.5 centimeter. Use a
trigonometric ratio to ﬁnd the apothem.
6 cm
12 cm
1
1 tan6368 2
1
A 5 }2 aP 5 }2 } (60) ø 247.7 cm2
Geometry
Worked-Out Solution Key
365
Chapter 11,
continued
46. The inscribed n-gon would have a central angle of
3608
1808
1 3608
} and a bisected angle of } } , or }. When
n
2 n
n
38
1
a
2
broken down into reference triangles, the radius of the
polygon would equal the radius of the circle.
1808
n
0.5 cm
1 cm
r
Not drawn to scale
0.5
tan 38 5 }
a
a
b
0.5
tan 38
a5}
1
2
1
1 0.5
The area of a 60-gon is A 5 }2 aP 5 }2 } (60)
tan 38
ø 286.2 square centimeters.
120-gon: The central angle for 120-gon measures
3608
128
1
2
} 5 38, the bisected angle measures } (38) 5 1.58,
and the side length is 60 4 120 5 0.5 centimeter.
Because the side length is 0.5 centimeter, the base
1
of the triangle is }2 (0.5) 5 0.25 centimeter. Use a
trigonometric ratio to find the apothem.
1 1808 2
1 1808 2
b
cos }
5 }r
n
r + sin }
5b
n
r + cos 1 }
5a
n 2
sin }
5 }r
n
1 1808 2
a
180
The side length of the inscribed n-gon is
1
1808
2
s 5 2b 5 2r + sin }
.
n
So, the area of the inscribed n-gon is
1
1
1
1 1808 2 2 1
1 1808 2 2
(n) 2r + sin }
A 5 }2 a + ns 5 }2 r + cos }
n
n
1 1808 2
1 1808 2
5 r2n cos }
sin }
.
n
n
1.58
The circumscribed n-gon would have a central angle of
a
1
3608
n
1 3608
2 n
1808
n
broken down into reference triangles the apothem would
equal the radius of the circle.
0.25 cm
1808
n
Not drawn to scale
0.25
tan 1.58 5 }
a
r
0.25
tan 1.58
a5}
b
1
2
1
1 0.25
The area of a 120-gon is A 5 }2 aP 5 }2 } (60)
tan 1.58
ø 286.4 square centimeters.
1 1808 2
b
tan }
5 }r
n
1 1808 2
r + tan }
5b
n
A
300
The side length of the circumscribed n-gon is
Area
240
1 1808 2
s 5 2b 5 2r + tan }
. So, the area of the
n
180
circumscribed n-gon is
120
1
0
0
2
4
6
8
10
n
Number of sides of polygon
A circle will have the greatest area.
C 5 2:r
1
1 1808 2 2
1 1808 2
The area between the two polygons is equal to the area
of the inscribed polygon subtracted from the area of
the circumscribed polygon. So, the area between the
polygons is
1 1808 2
1 1808 2
1 1808 2
1 1808 2
1 1808 2 G
60 5 2:r
r2n tan }
2 r2n cos }
sin }
, which factors
n
n
n
30
:
to r 2n tan }
2 cos }
sin }
.
n
n
n
}5r
30 2
A 5 :r 2 5 : 1 }
5}
ø 286.5 cm2
:
:2
The area of the circle will be about 286.5 square
centimeters.
366
1
A 5 }2 a + ns 5 }2 (r)(n) 2r + tan }
5 r 2n tan }
.
n
n
60
Geometry
Worked-Out Solution Key
900
F
1 1808 2
0.5 cm
c.
2
} and a bisected angle of } } , or }. When
Chapter 11,
continued
}
50 0 16 1 26
47. C;
The given coordinates of nABC are:
50 Þ 42
A(23, 22)
The triangle is not a right triangle.
B(23, 2)
Choice D: C(6, 22)
AB 5 {2 2 (22){ 5 {4{ 5 4
BC 5 Ï(6 2 (23)]2 1 (22 2 2)2
}}}
Choice A: C(26, 22)
}
}}}
5 Ï 92 1 (24)2
}}
5 Ï 81 1 16
BC 5 Ï[26 2 (23)]2 1 (22 2 2)2
}
5 Ï (23)2 1 (24)2
}
5 Ï 97
}
5 Ï 9 1 16
AC 5 {6 2 (23){ 5 {9{ 5 9
}
5 Ï 25
}
(Ï97 )2 0 42 1 92
55
97 0 16 1 81
AC 5 {26 2 (23){ 5 {23{ 5 3
97 5 97 52 0 32 1 42
The triangle is a right triangle.
25 0 9 1 16
The triangle is not a right triangle if vertex C has
coordinates (2, 23).
25 5 25 The triangle is a right triangle.
48. J;
Choice B: C(21, 0)
}}}
BC 5 Ï[21 2 (23)]2 1 (0 2 2)2
}
5 Ï 22 1 (22)2
}
5 Ï4 1 4
}
5 Ï8
}
5 2Ï 2
}}}
AC 5 Ï[21 2 (23)]2 1 [0 2 (22)]2
}
}
(5Ï2 )2 0 42 1 (Ï26 )2
Mixed Review for TAKS
2
}
675 5 b
}
So, the function m 5 675 2 50w can be used to describe
the remaining balance, m, after w withdrawals.
5 Ï4 1 4
5 Ï8
}
5 2Ï 2
}
}
4 0 (2Ï 2 )2 1 (2Ï2 )2
2
m 5 b 2 50w
625 5 b 2 50(1)
5 Ï2 1 2
2
Each time the number of withdrawals increases by 1, the
remaining balance decreases by \$50. So, there is a linear
relationship between the number of withdrawals and
remaining balance whose rate of change is 2\$50. This
relationship can be written as an equation in the form
m 5 b 2 50w, where m is the remaining balance and w
is the number of withdrawals. Substitute values from the
table to ﬁnd b.
16 0 8 1 8
1. The regular n-gons approach the shape of a circle as the
value of n gets very large. The more sides there are, the
smaller the central angle is, which makes the shape
more circular.
16 5 16 The triangle is a right triangle.
Choice C: C(2, 23)
2. As n gets very large, the perimeter approaches 2:. The
1808
radius is 1 and as n increases, the value of n + sin }
n
}}}
BC 5 Ï[2 2 (23)]2 1 (23 2 2)3
1
}
5 Ï 52 1 (25)2
2
approaches :, which suggests that if the number of sides
were inﬁnitely large, the circumference would be 2:r.
}
5 Ï 25 1 25
}
3. As n gets very large, the areas appraoch :. The radius
5 Ï 50
is 1, so r 2 5 1. As n increases, the value of
}
5 5Ï 2
1 1808 2
1 1808 2
}}}
+ cos }
approaches :, which suggests
n + sin }
n
n
}
that if the number of sides were inﬁnitely large, the area
would be :r 2.
AC 5 Ï[2 2 (23)] 1 [23 2 (22)]
2
5 Ï 52 1 (21)2
2
}
5 Ï 25 1 1
}
5 Ï 26
Lesson 11.7
Investigating Geometry Activity 11.7 (p. 770)
Geometry
Worked-Out Solution Key
367
Chapter 11,
continued
2. Geometric probabilities are determined by comparing
3. As the number of tosses increases, the experimental
probability gets closer to the theoretical probability.
11.7 Guided Practice (pp. 772–773)
}
Length of RT
{21 2 (22){
}
1. P(point is on RT ) 5 }}
} 5 }} Length of PQ
{5 2 (25){
1
5}
; 0.1, 10%
10
}
{4 2 (21){
Length of TS
5
}
2. P(Point is on TS ) 5 }}
5}
} 5 }
Length of PQ
{5 2 (25){ 10
1
2
5 }; 0.5; 50%
}
Length of PJ
{21 2 (25){
4
}
3. P(Point is on PT ) 5 }}
5}
} 5 }} 10
Length of PQ
{5 2 (25){
2
5 }5 ; 0.4, 40%
}
Length of RQ
{5 2 (22){
}
4. P(Point is on RQ ) 5 }}
} 5 }
Length of PQ
{5 2 (25){
7
5}
; 0.7, 70%
10
5. The longest you can wait is 8:49 2 8:43 5 6 minutes.
P(You get to the station by 8:58)
geometric measures. A probability, found by dividing
the number of favorable outcomes by the total number
of possible outcomes, deals with events, not geometric
measures. In a geometric probablity, you divide the
“favorable measure” by the total measure.
}
}
3. P(Point k is on AD ) 5 }}
}
Length of AE
{3 2 (212){
15
5 }}
5 }
{12 2 (212){ 24
5
5 }8; 0.625; 62.5%
}
Length of BC
{23 2 (26){
}
4. P(Point k is on BC) 5 }
} 5 }}
Length of AE
{12 2 (212){
3
1
5}
5 }8; 0.125; 12.5%
24
}
Length of DE
{12 2 3{
}
5. P(Point k is on DE) 5 }}
} 5 }}
Length of AE
{12 2 (212){
9
3
5}
5 }8 ; 0.375; 37.5%
24
}
Length of AE
{12 2 (212){
}
6. P(Point k is on AE ) 5 }
} 5 }}
Length of AE
{12 2 (212){
24
5}
5 1; 1.0; 100%
24
Favorable waiting time
6
1
5 }}
5}
5 }2
12
maximum waiting time
3
5
7. AD 1 DE 5 AE, so } 1 } 5 1. The sum of their
8
8
The probability that you will get to the station by
1
8:58 is }2 , or 50%.
probabilities will add up to 1, or 100%.
}
5 :r 2 2 s 2 5 : + 22 2 (2Ï2 )2
Area of large white circle 1 Area of smaller black circle
2 Area of smaller white circle
5 :F
(8 1 8 1 8 1 8 1 8)2 2 :(8 1 8 1 8 1 8)2
1 :(8 1 8 1 8)2 2 :(8 1 8)2
5 4: 2 8
P(Point is in shaded region) 5 }}
Area of entire ﬁgure
5 1600: 2 1024: 1 576: 2 256: 5 896:
4: 2 8
Area of black region
P(arrow lands in black region) 5 }}
Area of target
896:
56
14
5}
5}
5}
1600:
100
25
The probability that the arrow lands in the black region
5}
ø 0.36 or 36%.
4:
The probability that a randomly chosen point lies in the
Area of smaller triangle
9. P(Point is in shaded region) 5 }}
Area of larger triangle
14
or 56%.
is }
25
5
7. Area of woods 5 Total area of park 2 Area of ﬁeld
21
1
5}
5 }4
84
5 58.5 2 30 5 28.5
The probability that a randomly chosen point lies in the
Area of woods
285
19
28.5
ø}
5}
5}
P(ball in woods) 5 }}
585
39
Total area of park
58.5
shaded region is }4, or 25%.
19
The probability that the ball is in the ﬁeld is about }
, or
39
48.7%.
11.7 Exercises (pp. 774–777)
Skill Practice
1. If an event cannot occur, its probability is 0. If an event is
certain to occur, its probability is 1.
368
1
2
}
1
}(12)(14)
2
}(6)(7)
Geometry
Worked-Out Solution Key
1
10. Area of shaded region 5 Area of trapezoid
2 Area of rectangle
1
5 }2 h(b1 1 b2) 2 b1h
1
5 }2 (5)(8 1 20) 2 (8)(5)
5 70 2 40 5 30
6. Area of black region 5 Area of larger black circle 2
8. Area of shaded region 5 Area of circle 2 Area of square
Chapter 11,
continued
P(Point is in shaded region) 5 }}
Area of trapezoid
30
20. Use The Pythagorean Theorem to ﬁnd the height of the
triangle.
3
5}
5 }7
70
The probability that a randomly chosen point lies in the
h
5
3
11. In the numerator, the area of the unshaded rectangle at
the bottom should be subtracted. This rectangle has a
height of 7 2 5 5 2.
1
10(7) 2 }2:(5)2 2 2(10)
50 2 12.5:
70
h2 5 16
h54
}} 5 } ø 0.153
10(7)
4
32 1 h2 5 52
Find the area of the entire ﬁgure.
The probability that a randomly chosen point in the
Area of entire ﬁgure 5 Area of shaded region
1 Area of trapezoid
1
12. Sample answer: The area of the north side of the island is
Area of north side of island
13. P(location is on north side) 5 }}
Total area of island
31.5
315
63
5}
5}
5}
64
640
128
1
5 }2b: ht 1 }2h2(b1 1 b2)
about 31.5 square units. The area of the south side of the
island is about 32.5 square units. The area of the whole
island is about 64 square units.
1
1
5 }2(3)(4) 1 }2 (3)(5 1 7)
5 6 1 18 5 24
Write a ratio of the areas to ﬁnd the probability.
P(Point is in shaded region) 5 }}
Area of entire ﬁgure
The probability that a randomly chosen location on the
63
island lies on the north side is }
128
Area of south side of island
14. P(location is on south side) 5 }}
Total area of island
32.5
325
65
5}
5}
5}
64
640
128
The probability that a randomly chosen location on the
65
island lies on the south side is }
128
15. The shaded triangle is similar to the whole triangle by
1
6
5 }2
the AA Similarity Postulate. The ratio of sides is }
12
1
7
5 }2. Because you know the ratio of sides, you
and }
14
can use the Area of Similar Polygons Theorem to ﬁnd the
desired probability. The ratio of areas is 12 : 22 5 1 : 4,
which is the desired probability.
16. x 2 6a1
5
1a2x 2 3a5
4a2xa8
2axa4
2
P(2axa4) 5 }7
x
}q7
18.
2
xq14
P(xq14) 5 0
3xa27
The probability that a randomly chosen point in the
1
ﬁgure lies in the shaded region is }4, or 25%.
21. Set up a ratio of side lengths to ﬁnd the base length of
the smaller triangle.
base of smaller triangle
base of larger triangle
height of smaller triangle
height of larger triangle
base of smaller triangle
14
(12 2 8)
12
}} 5 }}
}} 5 }
14
base of smaller triangle 5 }
3
Find the area of the shaded region.
Area of shaded region 5 Area of larger triangle
2 Area of smaller triangle
1
P(xa7) 5 }7
19.
1
1
5 }2 bihi 2 }2bshs
xa7
17.
6
5}
5 }4
24
5 }2 (14)(12) 2 }2 1 4}3 2(4)
1
1
28
2
224
5}
5 84 2 }
3
3
Write a ratio of the areas to ﬁnd the probability.
P(Point lies in shaded region) 5 }}
Area of larger triangle
224
}
8
3
}
5
5}
84
9
The probability that a randomly chosen point in the
8
xa9
P(xa9) 5 1
Geometry
Worked-Out Solution Key
369
Chapter 11,
continued
22. Use a special triangle to ﬁnd the radius of the circle.
458
458
30 ø r
Area of circle: A 5 :r 2 5 :(30)2 5 900:
x
r
x 2
458
C 5 2:r
25.
188.5 5 2:r
x
8
458
The measure of the central angle of a regular hexagon
3608
}
So, r 5 8Ï 2 . The base of the triangle formed by the radii
is 8 1 8 5 16.
1
is }
5 608. The bisected angle measures }2 (608) 5 308.
6
Use a 308-608-908 triangle to find the apothem a of the
hexagon and base b of the triangle.
Find the area of the shaded region.
Area of shaded region 5 Area of sector
2 Area of triangle
308
a
30
x 3
C
m arc
1
3608
}
908
1
5 } + : + (8Ï 2 )2 2 }2(16)(8)
3608
308
2x
5 } + :r 2 2 }2 bh
5 32: 2 64
Write a ratio of the areas to ﬁnd the probability.
608
x
b
}
So, a 5 15Ï3 and b 5 15. The side length of the
hexagon is s 5 2b 5 2(15) 5 30. The area of the
hexagon is
1
1
}
A 5 }2a + ns 5 }2 (15Ï 3 )(6)(30) 5 1350Ï3 .
32: 2 64
: (8Ï 2 )
P(Point is in the hexagon) 5 }}
Area of circle
32: 2 64
5}
5}
}
2
128:
Area of hexagon
}
1350Ï 3
ø 0.091
5}
ø 0.827
900:
Area of U 2 Area of A
Area not in A
5 }}
P(x not in A) 5 }
Total Area
Area of U
C
Circumference of arc
24. P(Point is an arc) 5 }}
circumference of cirlce
5
C
m arc
3608
}
} + 2:r
2:r
808
2
5 } 5 }9
3608
The probability that a randomly chosen point in the circle
lies in the inscribed regular hexagon is about 0.827,
or 82.7%.
26. Area of circle 5 :r 2
The measure of the central angle for a regular octagon
3608
5 458 and the bisected angle measures
is }
8
1
2
} (458) 5 22.58. Use trigonometric ratios to find the
apothem and side length.
C
22.58
m arc
} + :r 2
Area of sector
3608
}}
}
P(Point is in sector) 5 Area of circle 5
:r 2
808
3608
a
r
2
5 } 5 }9
The probability that a randomly chosen point on the
2
circle lies on the arc is }9 , or about 22.2%. The probability
that a randomly chosen point in the circle lies in the
2
b
sin 22.58 5 }r
b
cos 22.58 5 }r
a
r + sin 22.58 5 b
r + cos 22.58 5 a
sector is }9 , or about 22.2%. The probabilities do not
The side length of the regular octagon is
depend on the radius because the circumference and the
area of a circle end up canceling with the values in the
denominator.
A 5 }2a + ns 5 }2 (r + cos 22.5)(8)(2r + sin 22.58)
s 5 2b 5 2r + sin 22.58. So, the area of the octagon is
1
1
5 8r2 + cos 22.58 + sin 22.58
Area of octagon
P(Point is in polygon) 5 }}
Area of circle
8r 2 cos 22.58 + sin 22.58
:r
5 }}
ø 0.90
2
The probability that a randomly chosen point in the circle
lies in the inscribed polygon is about 0.90, or 90%.
370
Geometry
Worked-Out Solution Key
The probability that a randomly chosen point in the
23. D;
}
P(Point lies in shaded region) 5 }}
Area of circle
Chapter 11,
continued
27. Because points A and B are end points of a diameter
of (D, the angle they form when connected to point
C will always be a right angle, by Theorem 10.9. So,
the probability that nABC is a right triangle is 100%.
mCAB will be less than or equal to 458 half the time,
so the probability that mCABa458 is 50%.
28.
Problem Solving
Area of inner square
62
30. P(Dart hits inner square) 5 }} 5 }2
Total area of board
18
36
1
5}
5 }9
324
Area outside inner square but inside circle
y
5 Area of circle 2 Area of inner square
2 62 5 81: 2 36
5 : + 1}
22
18 2
1
21
P(Dart hits outside inner square but inside circle)
x
Area outside inner square but inside circle
5 }}}}
Total area of board
b1 5 {3 2 2{ 5 1, b2 5 {3 2 0{ 5 3,
81: 2 36
and h 5 {2 2 0{ 5 2
5}
ø 0.674
324
1
1
Area of trapezoid 5 }2 h(b1 1 b2) 5 }2 (2)(1 1 3) 5 4
}
x 1 y q4 is the equation of a circle with r 5 Ï4 5 2.
2
2
1
8
} of the circle is in the shaded region, so the area of the
1
1
shaded circle is }8 (:(2)2) 5 }2:.
P(the point (x, y) in the solution region is in x 2 1 y 2q4)
Area of trapezoid 2 Area of shaded circle
5 }}}}
Area of trapezoid
1
4 2 }2:
5}
ø 0.607 or 60.7%
4
The probability that a point (x, y) in the solution region is
in x 2 1 y 2q4 is about 60.7%.
29. An expression for each step is
stepn 2 1 1 0.5(1 2 step n 2 1), which simplifies to
0.5 stepn 2 1 1 0.5, where n is the current step number.
Step 1: 0.5 painted
Step 2: 0.5(0.5) 1 0.5 5 0.75 5 75%
Step 3: 0.5(0.75) 1 0.5 5 0.875 5 87.5%
Step 4: 0.5(0.875) 1 0.5 5 0.9375 5 93.75%
Step 5: 0.5(0.9375) 1 0.5 5 0.96875 5 96.875%
Step 6: 0.5(0.96875) 1 0.5 5 0.984375 5 98.4375%
Step 7: 0.5(0.984375 1 0.5 5 0.9921875 5 99.21875%
Step 8: 0.5(0.9921875) 1 0.5 5 0.99609375
5 99.609375%
Step 9: 0.5(0.99609375) 1 0.5 5 0.998046875
5 99.8046875%
Step 10: 0.5(0.998046875) 1 0.5 5 0.9990234375
5 99.90234375%
Step 11: 0.5(0.9990234375) 1 0.5 5 0.99951171875
5 99.951171875%
After 11 steps, the painted portion of the stick is greater
than 99.95%. So, nq11 steps.
1
The probability that it hits inside the inner square is }9, or
about 11.1%. The probability that it hits outside the inner
square but inside the circle is about 0.674 or 67.4%.
wait time
4
2
31. a. P(bus waiting) 5 }} 5 } 5 }
5
10
time between arrivals
The probability that there is a bus waiting when a
2
passenger arrives at a random time is }5 , or 40%.
time between arrivals 2 waiting time
b. P(bus not waiting) 5 }}}
time between arrivals
10 2 4
6
3
5}
5}
5 }5
10
10
The probability that there is not a bus waiting when a
3
passenger arrives at a random time is }5 , or 60%.
amount of time before lunch
4
32. P(ﬁre drill for lunch) 5 }}} 5 }
7
total amount of time at school
4
The probability that the ﬁre drill begins before lunch is }7,
33. P(You miss the call)
amount of overlap while practicing drums
5 }}}}
Interval of time for phone call
{7:10 2 7:00{
5 }}
{8:00 2 7:00{
10 minutes
1
5}
5 }6
60 minutes
1
The probability that you missed your friend’s call is }6 ,
34. a. First count all the whole squares. There are about 64
whole squares. Next make groups of partially covered
squares, so the combined area of each group is about
1 square unit. There are about 14 more squares, which
makes a total of 64 1 14 5 78 squares. Each side of
each square is 2 kilometers, so the area of each square
is 22 5 4 square kilometers. 78 squares + 4 km2/square
5 312 square kilometers. The area of the planned
landing region was about 312 square kilometers.
Geometry
Worked-Out Solution Key
371
Chapter 11,
continued
Area of crater
b. P(Beagle 2 landing in crater) 5 }}
Area of landing region
2
1 2
1
: + }2
}
5
ø 0.0025
312
The probability that Beagle 2 landed in the crater is
37. a. From Exercise 30, the probability that one dart hits the
1
yellow square is }9 . Because the throws are
independent, you multiply the probabilities.
1 2
[P(Dart hits yellow square)]2 5 1 }9 2 5 }
81
1
The probability that both darts hit the yellow square
1
is }
81
b. P(Dart hits outside the circle)
D
182 2 : + 92
18
608
F
B
324 2 81:
5}
5}
2
324
3 cm
E
Because the throws are independent, you multiply the
probabilities.
P(Dart hits yellow square)
1
C
608
3608
mDE
3608
Area of sector DEF 5 } + :r 2 5 } + : + 32
5 1.5:
Area of (C 5 :r 2 5 : + (1.5)2 5 2.25:
ø 0.024
The probability that the ﬁrst dart hits the yellow square
and the second hits outside the circle is about 2.4%.
c. From Exercise 30. The probability that one dart hits
C
mAB
3608
608
5 } + :(1.5)2 5 0.375:
3608
Area of sector ABC 5 } + :r 2
Area of sector
Smaller circle: P(Point in sector) 5 }}
Area of circle
0.375:
5}
2.25:
inside the circle but outside the yellow square is
about 0.674. Because the throws are independent, you
multiply the probabilities.
[P(Dart hits inside the circle but outside the yellow
square)]2 ø (0.674)(0.674) ø 0.454
The probability that both darts hit inside the circle but
outside the yellow square is about 45.4%.
38. P(Part of bird call erased) 5 Time of silence
1 Time of bird call
1
Time of erased data
1 }}
Total time of tape
Area of sector
Larger circle: P(Point in sector) 5 }}
Area of circle
8 min 1 5 min 1 10 min
1.5:
5 }}
60 min
1
5}
60
5}
9:
23
The probability of a randomly selected point being in
the sector stays the same when the central angle stays
the same and the radius of the circle doubles. As the
radius doubles, the area of the sector and entire circle
quadruples, but they are still praportional by the Areas
of Similar Polygons Theorem. Because they are still
praportional, the probabilities remain equal.
36. P(both pieces are at least 1 in.) is the same as
1 2 P(one piece is less than 1 inch) because both pieces
may not be less than 1 inch. The probability of producing
1
1
2
a piece less than 1 inch is }3 . So, 1 2 }3 5 }3 ø 66.7% is
the probability we want to find.
372
Geometry
Worked-Out Solution Key
324 2 81:
+ P(Dart hits outside the circle) 5 }9 + }
324
Area of (F 5 :r2 5 : + 32 5 9:
P(All of the bird call erased)
Time of bird call 1 Time of erased data
5 }}}
Total time of tape
5 1 10
15
1
5}
5}
5 }4 or 25%
60
60
The probability that part of the bird call was erased
23
, or about 38.3%. The probability that all of the bird
is }
60
1
call was erased was }4 , or 25%.
C 608
1.5 cm
Area of outer square 2 Area of circle
5 }}}
Area of outer square
A
Chapter 11,
continued
3. P(point is in shaded region) 5 }}
Area of large circle
Mixed Review for TAKS
3 2
9
122
4
: 1 }2 2
}:
: s2
4
5 }2 5 }
5}
100
10 2
:*
}:
: }
39. B;
b 5 1500(1.02)
t
t 5 18 months 5 1.5 years
9
5}
100
b 5 1500(1.02)
1.5
b ø 1500(1.0301495)
The probability that a randomly chosen point in the
b ø 1545.22
9
, or 9%.
ﬁgure lies in the shaded region is }
100
Kyle’s balance after 18 months is \$1545.22.
40. G;
If the slope of line g is increased it will become steeper.
If the x-intercept remains the same, the y-intercept will
move down in order for g to become steeper. So, the
y-intercept will decrease.
Quiz 11.6–11.7 (p. 777)
4. P(Point is in shaded region) 5 }}
Area of rectangle
Area of rectangle 2 Area of trapezoid
5 }}}
Area of rectangle
1
1
bh 2 }2ht(b1 1 b2)
8(5) 2 }2 (2)(3 1 5)
32
4
}}
}}
5
5
5}5}
bh
1. The central angle of the regular pentagon measures
3608
1
} 5 728 and the bisected angle measures } (728) 5 368.
5
2
8(5)
40
5
The probability that a randomly chosen point in the
4
ﬁgure lies in the shaded region is }5, or 80%.
Mixed Review for TEKS (p. 778)
Use a trigonometric ratio to ﬁnd the apothem a.
1. B;
a 368 17 cm
Probability that a coin will land in the dish:
6 2
:1 }2 2
: (3)2
9:
Area of the dish
}}} 5 }2 5 }2 5 }
100:
Area of the bottom of the jar
20
:
(10)
}
:1 2 2
a
cos 368 5 }
17
17 + cos 368 5 a
1
1
A 5 }2 a + ns 5 }2 (17 + cos 368)(5)(20) ø 687.66
The area of the regular pentagon is about 687.7 square
centimeters.
2. The central angle of the regular octagon measures
3608
} 5 458 and the bisected angle measures
8
1
} (458) 5 22.58. Use a trigonometric ratio to ﬁnd the
2
The expected number of prizes won after 400 coins are
dropped into the jar is 400 times the probability that a
coin will land in the dish. So, the expression that gives
.
this expected number is 4001 }
100: 2
9:
2. J;
Small mirror:
a
apothem a and base b.
22.58
a
0.5 m
Each small mirror is a regular hexagon (the number of
sides, n, is 6). So, the measure of a central angle is
25 m
3608
6
} 5 608. Use a special triangle to ﬁnd a, the length
of the apothem.
b
b
25
sin 22.58 5 }
25 + sin 22.58 5 b
1
2
a
25
cos 22.58 5 }
25 + cos 22.58 5 a
(608) 5 308
x 3 308 2x
a
The side length of the octagon is s 5 2b 5 50 + sin 22.58.
1
1
A 5 }2 a + ns 5 }2(25 + cos 22.5)(8)(50 + sin 22.5)
ø 1767.8
The area of the regular octagon is about 1767.8 square
meters.
608
x
0.25 m
0.5 m
}
So, a 5 0.25(Ï3 ) ø 0.433 meter and the area of a small
1
1
2
mirror is }2a + ns ø } (0.433)(6)(0.5) ø 0.6495 square
meter. Because the primary mirror is made of 91 small
mirrors, the area of the primary mirror, to the nearest
tenth of a square meter, is 91(0.6495) ø 59.1 square
meters.
Geometry
Worked-Out Solution Key
373
Chapter 11,
continued
Chapter 11 Review (pp. 780–783)
3. D;
The triangle is}an isosceles right triangle whose leg
6Ï 2
Ï2
1. A sector of a circle is the region bounded by two radii of
lengths are }
} 5 6 units, so the area of the triangle is
1
2
}(6)(6) 5 18 square units. Because P and Q are
congruent semicircles with diameters of 6 units, the area
6 2
5 4.5: square units,
of P, }2:1 }2 2 5 }2:(3)2 5 }
2
equals the area of Q. The area of the semicircle R is
1
1
9:
}
}
18:
1
1 6Ï 2 2
}: } 5 }:(3Ï 2 )2 5 } 5 9: square units, so
2
1
2
2
2
2
a circle and their intercepted arc.
2. Either pair of parallel sides can be used as the bases
of a parallelogram and the height is the perpendicular
distance between them.
}
3. An apothem of the square is XZ.
}
4. A radius of the square is XY.
5. A 5 bh 5 10(6) 5 60 square units
6. Use the Pythagorean Theorem to ﬁnd the base of the
triangle
the area of P plus the area of Q equals the area of R. The
area of the entire ﬁgure is 18 1 4.5: 1 4.5: 1 9: 5
18 1 18: square units, NOT 18: square units.
b
32
68
4. J;
1308
The area of the fan shown is A 5 } + : + 62 ø 40.84
3608
square inches.
322 1 b2 5 682
b2 5 3600
Choice F: Angle measure 1008, radius 6.25 in.:
b 5 60
1008
3608
A 5 } + : + 6.252 ø 34.09 in.2
Choice G: Angle measure 1058, radius 6.25 in.:
1058
3608
A 5 } + : + 6.252 ø 35.79 in.2
1
1
A 5 }2 bh 5 }2 (32)(60) 5 960 square units
7. Height of triangle: 40 2 16 5 24
A 5 Area of square 1 Area of triangle
1
Choice H: Angle measure 1408, radius 5.75 in.:
5 s2 1 }2 bh
1408
3608
A 5 } + : + 5.752 ø 40.39 in.2
1
Choice J: Angle measure 1558, radius 5.75 in.:
1558
3608
A 5 } + : + 5.752 ø 44.72 in.2
5 162 1 }2 (16)(24) 5 448 square units
8. A 5 147 in.2 and h 5 1.5b
A fan with the angle measure and radius given in
choice J, 155 degrees and 5.75 inches, has a greater area
than the fan shown, so it will do a better job of cooling
than the one shown.
1
147 5 }2 b(1.5b)
147 5 0.75b2
3608
5. The central angle measures } 5 728 and the bisected
5
1
angle measures }2 (728) 5 368. Use a trigonometric ratio
196 5 b2
14 5 b
The base of the triangle is 14 inches and the height is
1.5(14) 5 21 inches.
to find the side length of the pentagon.
b
tan 368 5 }7
7 + tan 368 5 b
7
9.
368
The side length of the
pentagon is s 5 2b 5 14 + tan 368.
P
b
Area of shaded region 5 Area of pentagon
2 Area of circle
1
5 }2 a + ns 2 :r 2
1
5 }2 (7)(5)(14 + tan 368) 2 :(7)2
5 245 + tan 368 2 49:
ø 24.1
374
Geometry
Worked-Out Solution Key
y
L
1
N
M
1
x
b 5 LM 5 {6 2 2{ 5 4 and h 5 {4 2 2{ 5 2
A 5 bh 5 (4)(2) 5 8
The area of the parallelogram is 8 square units.
1
A 5 }2 bh
Chapter 11,
10.
continued
C
Q
S
x
5.5(3608) 5 358(C)
56.57 5 C
T
The circumference of (F is about 56.57 centimeters.
d1 5 QS 5 {21 2 (23){ 5 2 and
d2 5 RT 5 {3 2 (22){ 5 5
1
17.
1
A 5 }2 d1d2 5 }2 (2)(5) 5 5
y
D
Arc length of GH 5 }
1
x
G
F
b1 5 GF 5 {3 2 1{ 5 2, b2 5 DE 5 {5 2 (21){ 5 6,
C
mTWU
18. Area of sector TWU 5 } + :r 2
3608
and h 5 {4 2 (22){ 5 6
1
3608
C 1158(26:)
3608
Arc length of C
GH ø 26.09
The length of G
E
21
C mC
GH
3608
Arc length of C
GH
1158
}} 5 }
Arc length of GH
2:r
}} 5 }
2:(13)
The area of the kite is 5 square units.
11.
358
3608
5.5
C
}5}
1
25
C
Arc length of GH
mGH
16. }} 5 }
C
3608
y
R
2408
368
5 } + : + 92 ø 169.65
1
A 5 }2 h(b1 1 b2) 5 }2 (6)(2 1 6) 5 24
inches.
The area of the trapezoid is 24 square units.
19. Area of blue shaded region 5 Area of rectangle
12. Ratio of sides 5 3 : 4
Ratio of perimeters 5 3 : 4 and ratio of areas 5 3 : 4
2
2 Area of semicircle
2
5 9 : 16
Area of red
Area of blue
9
16
4.5
Area of blue
9
16
1
5 bh 2 }2 :r 2
}5}
1
}5}
5 24 2 2: ø 17.72
8 5 Area of blue
The ratio of the perimeters is 3 : 4, the ratio of the areas is
9 : 16, and the area of the blue triangle is 8 square feet.
13. Ratio of sides 5 10 : 13
inches.
C
mRQ
20. Area of red sector 5 } + :r 2
3608
Ratio of perimeters 5 10 : 13 and ratio of areas
508
3608
27.93 5 } + : + r 2
5 102 : 132 5 100 : 169
Area of red
Area of blue
100
169
}5}
100
90
}5}
169
Area of blue
152.1 5 Area of blue
The ratio of the perimeters is 10 : 13, the ratio of the areas
is 100 : 169, and the area of the blue quadrilateral is 152.1
square centimeters.
14. The ratio of the lengths of corresponding sides is
}
}
Ï 144 : Ï 49 , or 12 : 7.
15.
4 2
5 6(4) 2 }2 :1 }2 2
8ør
Find the measure of the major arc.
C
C
mRQ 5 50, so mRTQ 5 3608 2 508 5 3108.
C
mRTQ
3608
Area of blue sector 5 } + :r 2
3108
3608
ø } + :(8)2 ø 173.14
feet.
C 5 :d
94.24 5 :d
94.24
:
}5d
30 ø d
The diameter of (F is about 30 feet.
Geometry
Worked-Out Solution Key
375
Chapter 11,
continued
3608
21. The central angle is } 5 458 and the bisected angle
8
1
is }2 (458) 5 22.58. Use a trigonometric ratio to ﬁnd the
side length.
22.58
25. P(Point is in shaded area)
Area of triangle
5 }}}
Area of semicircle 1 Area of triangle
1
2
}}
1
6 2
1
}: + } 1 } (6)(15)
2
2
2
} (6)(15)
5
1 2
6 in.
45
5}
ø 0.761
4.5: 1 45
The probability that a randomly chosen point in the
b
26. P(Point is in shaded area)
b
tan 22.58 5 }6
Area of rectangle 1 Area of semicircle
5 }}}
Area of square
6 + tan 22.58 5 b
The side length s 5 2b 5 12 + tan 22.58
An octagon has 8 sides, so the perimeter is
P 5 8(12 + tan 22.58) 5 96 + tan 22.58 ø 39.8.
1
1
A 5 }2 a + ns 5 }2 (6) + 8(12 tan 22.58) ø119.3
The perimeter of the platter is about 39.8 inches and
the area is about 119.3 square inches.
22. The pentagon’s side length is 20 centimeters, so the
1
length of the base of the triangle is }2 (20) 5 10
centimeters. Use the Pythagorean Theorem to ﬁnd
the apothem.
41 }2 2 1 }2 + : + 1 }2 2
4 2
1
4
4
The probability that a randomly chosen point in the
Chapter 11 Test (p. 784)
1. A 5 bh 5 7(4.7) 5 32.9
The area of the parallelogram is 32.9 square centimeters.
2. Use the Pythagorean Theorem to ﬁnd the base of the
triangle.
13 ft
5 ft
17 cm
8 1 2:
5 }}
5}
ø 0.893
2
16
a
b
5 1 b 5 132
2
10 cm
b 5 12
a 2 5 189
1
1
A 5 }2 bh 5 }2 (12)(5) 5 30
}
a 5 Ï189
1
}
A 5 }2 a + ns 5 }2 (Ï189 )(5)(20) ø 687.39
The area of the jigsaw puzzle is about 687.4 square
centimeters.
}
length of AB {2 2 (22){
4
}
23. P(k is on AB) 5 }
5}
} 5 }
length of AC
{5 2 (22){ 7
}
4
The probability that k is on AB is }7 , or about 57.1%.
24. P(Point is in shaded region) 5 }}
Area of semicircle
5
258
12 2
2
3608
}}
1
12 2
}+:+ }
2
2
1 2
1 2
}+:+ }
258
3608
}
5}
ø 0.139
1
}
2
The probability that a randomly chosen point in the
376
Geometry
Worked-Out Solution Key
The area of the triangle is 30 square feet.
3. b2 5 18 cm 1 9 cm 5 27 cm
1
1
A 5 }2 h(b1 1 b2) 5 }2 (10)(18 1 27) 5 225
The area of the trapezoid is 225 square centimeters
1
1
4. A 5 } h(b1 1 b2) 5 }(9)(8 1 15) 5 103.5
2
2
The area of the trapezoid is 103.5 square meters.
1
1
5. A 5 } d1d2 5 } (32)(40) 5 640
2
2
The area of the rhambus is 640 square inches.
1
1
6. A 5 } d1d2 5 } (41)(67) 5 1373.5
2
2
The area of the kite is 1373.5 square centimeters.
b2 5 144
102 1 a2 5 172
1
2
Chapter 11,
continued
C
mRPS
16. Area of sector RPS 5 } + :r 2
3608
1148
36 5 } + :r 2
3608
7. b 5 3h and A 5 108 in2
A 5 bh
108 5 3(h)(h)
108 5 3h2
6.02 ø r
36 5 h2
65h
17. A hexagon has 6 sides. The perimeter is 18 inches, so the
The height of the parallelogram is 6 inches and the base
is 3(6) 5 18 inches.
8. Ratio of perimeters 5 40 : 16 5 5 : 2
side length is 18 4 6 5 3 inches. The central angle
3608
1
5 608 and the bisected angle is }2 (608) 5 308.
is }
6
Use a special triangle to find the apothem.
9. Ratio of corresponding side lengths
5 ratio of perimeters 5 5 : 2
10. Ratio of areas 5 52 : 22 5 25 : 4
a
C C
1088
mAB
11. Arc length of AB 5 } + 2:r 5 } + 2:(17)
360
3608
C
ø 32.04
Arc length of C
EHD
C
HD
mE
12. }} 5 }
608
1.5 in.
The apothemis about 1.5Ï 3 ø 2.6 inches.
1
}
1
A 5 }2a + ns ø }2 (1.5Ï3 )(6)(3) ø 23.38
2(area of red semicircle)
109.71 ø C
The circumference of (F is about 109.71 inches.
C
Arc length of GH
mGH
13. }} 5 }
2:r
3608
608
x
5 }}}
Area of square 1 2(Area of seimicircle)
64(3608) 5 C(2108)
1
10 2
1
2 }2 + : + 1 }
22
2
25:
5 }}
5}
ø 0.44
2
100 1 25:
1
1 22
10
1
102 1 2 }2 + : + }
2
The probability that a randomly chosen point in the
figure lies in the red region is about 44%.
mC
GH
35
}5}
2: (27)
2x
18. P(Point is in red region)
2108
3608
}5}
C
308
The area of the tile is about 23.4 square inches.
3608
64
C
x 3
}
The length of AB is about 32.04 centimeters.
C
308
3608
C
CH
74.278 ø mG
The measure of G
14. Find the measure of the major arc.
C
TR 5 3608 2 1058 5 2558
mQ
35
3608 + } 5 mGH
2: (27)
C
m QTR
3608
Area of sector QTR 5 } + :r 2
2558
3608
5 } + : + 82 ø 142.42
inches.
C
mLM
15. Area of shaded sector  NM 5 } + Area of (N
3608
688
3608
49 5 } + Area of (N
259.41 ø Area of (N
The area of (N is about 259.41 square meters.
19. P(Point is in blue region)
Area of square 2 2(Area of red semicircle)
5 }}}}
Area of square 1 2(Area of semicircle)
10
1
2 100 2 25:
5 }}
5}
ø 0.12
2
100 1 25:
10
1
102 1 21 }2 + : + 1 }
22 2
2
1
102 2 2 }2 + : + 1 }
22
The probability that a randomly chosen point in the
figure lies in the blue region is about 12%.
Chapter 11 Algebra Review (p. 785)
1 hr
1. 90 min + } 5 1.5 hour
60 min
d 5 rt when d 5 14.25 and t 5 1.5
14.25 5 r(1.5)
14.25
1.5
}5r
(2) 5 19
d 5 rt 5 1 }
1.5 2
14.25
(2). You can bike
The algebraic model is d 5 1 }
1.5 2
14.25
19 miles in 2 hours.
Geometry
Worked-Out Solution Key
377
Chapter 11,
continued
2. 25% off means she paid 75% of the original price for
the jacket.
TAKS Practice (pp. 788–789)
1. A;
Let j represent the original price of the jacket.
12 1 0.75j 5 39
0.75j 5 27
Number of successes
Experimental probability 5 }}
Number of trials
312
5
1
5}
5 }4. The experimental
5 }}
21513151312
20
j 5 36
The algebraic model is 12 1 0.75j 5 39. The original
cost of the jacket is \$36.
3. 29.50 1 0.25m 5 32.75 where m is the number of
1
probability of rolling a number greater than 4 is }4.
2. G;
2
1
Theoretical probability 5 }6 5 }3
0.25m 5 3.25
The theoretical probabilty of rolling a number greater
m 5 13
1
The algebraic model is 29.50 1 0.25m 5 32.75.
4. 7.6(20) 1 12.1rq400 where r is the number of minutes
spent running
12.1rq248
rq20.5
The algebraic model is 7.6(20) 1 12.1rq400. Jaime
needs to run 20.5 minutes or more to meet his goal.
5. If the value of the car decreases 10% each year, it keeps
90% (or 0.9) of its value per year.
than 4 is }3 .
3. C;
Because there are an even number of tiles in the bag and
each is labeled with a different number, half of the tiles
are labeled with numbers that are greater than the median
and half of the tiles are labeled with numbers that are
less than the median. So, the probability of randomly
choosing a tile whose number is less than the median
1
of all the numbers in the bag is }2.
4. H;
18,000(0.9) 5 A
5
The mode of {1, 2, 4, 3, 1, 3, 5, 3} is 3.
18,000(0.59049) 5 A
5. D;
10,628.82 5 A
Write the data in ascending order.
35, 40, 40, 40, 45, 50, 55, 55, 60
420
2
5 46}3
Mean 5 }
9
Median 5 45
student tickets sold and a 5 number of adult tickets sold.
Mode 5 40
Range 5 60 2 35 5 25
5(a 1 62) 1 8a 5 2065
The median is between the mode and the mean.
6. 5s 1 8a 5 2065 and s 5 a 1 62 where s 5 number of
5a 1 310 1 8a 5 2065
6. H;
13a 5 1755
Including the score of 4 on his next quiz, in ascending
a 5 135
s 5 a 1 62 5 135 1 62 5 197
4, 4, 6, 7, 8, 10
The algebraic models are 5s 1 8a 5 2065 and
s 5 a 1 62. There were 197 student tickets sold
Mean 5 }
5 6.5
6
Median 5 }
5}
5 6.5
2
2
Mode 5 4
Range 5 10 2 4 5 6
7. 0 5 216t 2 1 47t 1 6
247 6 Ï47 2 4(216 + 6)
2 + (216)
2
617
7. B;
}
247 6 Ï 2209 1 384
232
}}} 5 }}
ø 3.06 or 20.12
The algebraic model is 0 5 216t 2 1 47t 1 6. It takes
about 3.06 seconds for the tennis ball to reach the
ground.
The measure of a central angle of the regular hexagon
3608
5 608. Use a special triangle to ﬁnd a, the length
is }
6
of the apothem.
1
2
(608) 5 308
x 3 308 2x
a
3 in.
608
x
6 in.
The} apothem, a, of the regular hexagon is
3Ï3 ø 5.2 inches.
378
Geometry
Worked-Out Solution Key
13
The mode will be the lowest measure.
}}
39
The algebraic model is 18,000(0.9)5 5 A. The value of
the car after 5 years is \$10,628.82.
Chapter 11,
continued
8. G;
13. B;
If the coefﬁcient of x2 in y 5 x2 2 1 is increased to 3, the
graph has points with y-coordinates that are greater than
their original values. The graph below illustrates
the situation.
x
1 in.
} 5 } j dimension on model
1150 ft j dimension on actual building
100 ft
1150 5 100x
11.5 5 x
y
The height of the model will be 11.5 inches.
(2, 11)
9. B;
Because the length of the base is (x 1 c) feet and the
height is x feet, the equation that best represents the area
6
y 5 x2 2 1
1
2
A of the triangle is A 5 }x(x 1 c).
y 5 3x 2 2 1
10. G;
Of the 14,691 kilometers of Brazil’s land border, about
23%, or about 0.23(14,691) 5 3378.93 kilometers are
shared with Bolivia. Of the remaining border, which is
6%, or about 0.06(11,312.07) ø 678.72 kilometers are
shared with French Guiana. So, to the nearest hundred
kilometers, about 700 kilometers of Brazil’s border are
shared with French Guiana.
(1, 2)
(2, 3)
(1, 0)
(0, 21)
4 x
The graph of y 5 3x2 2 1 is a vertical stretch of the
graph of y 5 x 2 2 1. So, the parabola will be narrower.
14.
2x 1 3y 5 24
2x 1 5y 5 217.5
l
2x 1 3y 5 24
l
22x 1 10y 5 235
13y 5 239
y 5 23
2x 1 5(23) 5 217.5
11. B;
2x 2 15 5 217.5
(x, y) l (1.5x, 1.5y)
2x 5 22.5
M(1, 2) l M9(1.5, 3)
If rectangle KLMN is dilated by a factor of 1.5 with the
origin as the center of dilation, the coordinates of M9 are
(1.5, 3).
x 5 2.5
The x-coordinate of the solution of the system of linear
equations is 2.5.
12. H;
1
2
}x 2 y 5 4
1
2y 5 2}2 x 1 4
1
y 5 }2 x 2 4
1
2
}m 5 21
m 5 22
1
A line perpendicular to }2 x 2 y 5 4 has a slope of 22.
Equation of the line that passes through (23, 2) and is
1
perpendicular to }2x 2 y 5 4:
y 5 mx 1 b
2 5 22(23) 1 b
24 5 b
y 5 22x 2 4
Geometry
Worked-Out Solution Key
379
```
Fly UP