Section 2.6 Problem Solving in Geometry
32. The sum of the measures of the three angles of a
triangle is 180°.
3 40 180
xxx
33.
4 (3 4) (2 5) 180
9 9 180
xx x
x
34.
45180
10 180
xxx
x
35. Let x = the measure of the smallest angle.
Let 2x = the measure of the second angle.
Let x + 20 = the measure of the third angle.
2 ( 20) 180
xxx
36. Let x = the measure of the smallest angle.
Let 3x = the measure of the second angle.
Let x + 30 = the measure of the third angle.
3 30 180
xxx
37. If the measure of an angle is 58
, the measure of its
complement is 90
− 58
= 32
.
41. If the measure of an angle is 132
, the measure of
its supplement is
180 132 48
.
44. If the measure of an angle is 179.5°, the measure of
its supplement is 180° − 179.5° = 0.5°
of its complement, so the equation is
(90 ) 60xx
.
Step 4 Solve this equation
90 60
xx
90° 75° = 15°, and 75° is 60° more
than 15°.
Chapter 2 Linear Equations and Inequalities in One Variable
46. Step 1 Let x = the measure of the angle.
Step 2 Then 90 – x = the measure of its
complement.
Step 3 The angle’s measure is 78° less than that of
47. Step 1 Let x = the measure of the angle.
Step 2 Then 180 – x = the measure of its
supplement.
Step 3 The angle’s measure is three times that of its
48. Step 1 Let x = the measure of the angle.
Step 2 Then 180 – x = the measure of its
supplement.
Step 3 The angle’s measure is 16° more than three
so the proposed solution checks.
49. Step 1 Let x = the measure of the angle.
Step 2 Let 180 − x = the measure of its supplement,
Section 2.6 Problem Solving in Geometry
50. Step 1 Let x = the measure of the angle.
Step 2 Let 180 − x = the measure of its supplement,
and, 90 − x = the measure of its
complement.
Step 3 The measure of the angle’s supplement is
10° more than three times that of its
51. Divide the shape into two rectangles.
entire figure bottom rectangle side rectangle
entire figure
38 49 3
AA A
A
52. Divide the shape into a triangle and a rectangle.
13 m
13 m
10 m
10 m
15 m
5 m
24 m
entire figure rectangle triangle
entire figure
1
2
1
10 6 3 10 3
2
1
60 3 7
AAA
Alwbh
Chapter 2 Linear Equations and Inequalities in One Variable
55. Subtract the volume of the three hollow portions
from the volume of the whole rectangular solid.
cement block rectangular solid hollow
3
VV V
56. Subtract the volume of the smaller cylinder from the
volume of the larger cylinder.
shaded larger cylinder smaller cylinder
22
VV V
Rh rh
ππ
57. The area of the office is
20 ft 16 ft 2
320 ft
. Use
a proportion to determine how much of the yearly
electric bill is deductible.
Let x = the amount of the electric bill that is deductible.
320
x
58. a. The area of the lot is
2
500 ft 200 ft 100,000 ft .
The area of the house is
2
100 ft 60 ft 6000 ft .
The area of the shed is
2
20 ft 20 ft 400 ft .
The area of the driveway is
59. The radius of the large pizza is
114
2
= 7 inches,
and the radius of the medium pizza is
22
22
(3.5 in.)
12.25 in. 38.465 in.
Ar
ππ
For each pizza, find the price per inch by dividing
60. The radius of the large pizza is
116
2
inches = 8
inches, and the radius of each small pizza is
110
2
inches = 5 inches.
Large pizza:
158
2
in. . Since the price of one large pizza is the
same as the price of two small pizzas and the large
pizza has the greater area, the large pizza is the
better buy. (Because the prices are the same, it is
not necessary to find the prices per square inch in
this case.)
61. The area of the larger circle is
Section 2.6 Problem Solving in Geometry
62. The area of the rectangular portion of the floor is
(60 ft)(40 ft) = 2400
2
ft .
Since the radius of each semicircle is 20 ft and the
two semicircles together make one circle, the area
63. To find the perimeter of the entire window, first
find the perimeter of the lower rectangular portion.
This is the bottom and two sides of the window,
which is 3 ft + 6 ft + 6 ft = 15 ft. Next, find the
64. The circumference of the garden is
2 (30 ft) = 60 ft.
ππ
Since 6 in. = 1 ft.
2, the number of plants needed is
60 2 60 120 377.
1
2
π
ππ
To the nearest whole number, 377 plants are
needed.
65. First, find the volume of water when the reservoir
was full.
50 0 20 30,000Vlwh
66. The volume of the foundation is (4 yd)(3 yd). (2 yd)
=
3
24 yd . Since each truck holds 6
3
yd of dirt,
24 4
in. and
(2.5) 6 37.5 117.75Vrh
ππ π
.
The volume of the second can is 117.75 in
2
. Since
the cans are the same price, the can with the greater
volume is the better buy. Choose the can with the
diameter of 6 inches and height of 5 inches.
amount of dirt that had to be removed, is
3400,000
π
33
1,200,000 m 3,769,900 m .
π
69. Find the volume of a cylinder with radius 3 feet and
height 2 feet 4 inches.
2 ft 4 in = 1
23 feet = 7
3 feet
2
2
77
(3) 9 21 65.94
33
Vrh
π
πππ
The volume of the tank is approximately 65.94 ft
3
.
This is a little over 1 ft
3
smaller than 67 ft
3
so it is
Chapter 2 Linear Equations and Inequalities in One Variable
81. does not make sense; Explanations will vary.
Sample explanation: If the radius is doubled, the
area is multiplied by 4.
2
radius
x
A
r
π
82. makes sense
83. true
87. Area of smaller deck
2
(8 ft)(10 ) 80 ft .
Area of larger deck
2
(12 ft)(15 ) 180 ft .
Find the ratio of the areas.
88. Consider the following diagram:
14
30
3
3
3
3
The area of the outer rectangle (pool plus path) is
89. Let x = the radius of the original sphere.
Let 2x = the radius of the larger sphere.
Find the ratio of the volumes of the two spheres.
90. If the length, width, and height of a rectangular
solid are each multiplied by 10, the volume will be
multiplied by
10 10 10 1000.
The volume of the
car will be 1000 times that of the model.
410
35
x
x
The angle of inclination is
35 .
or
22
Pb Pb
ss
93.
713
xx
22852
228 52
32852
328285228
324
324
33
xx
xx xx
x
x
x
x
22
222
33 3 30 0 0
Section 2.7 Solving Linear Inequalities
96.
475
4(6) 7 5
y
97.
2( 3) 5 8( 1)
26588
7688
xxx
xxx
xx
2.7 Check Points
1. a.
b.
3.
69
66 96
3
x
x
x
The solution set is
,3
or
3.xx
4. 8274
xx
5. a. 12
4
x
The solution set is
,8 or
8.xx
b. 618
x
6.
5317
533173
520
520
55
4
y
y
y
y
y
1
x
The solution set is
1,
or
1.xx
Chapter 2 Linear Equations and Inequalities in One Variable
8.
2( 3) 1 3( 2) 14xx
2613614
xx
9. 4( 2) 4 15xx
48415
4484415
815,false
xx
xx xx
There is no solution or
.
11. Let xyour grade on the final examination.
82 74 78 80
5
234 2 80
xx
x
12. Let x the number of people you invite to the
picnic.
2.
(2, )
3.
bc
4.
bc
5.
bc
2.7 Exercise Set
1. x > 5
Section 2.7 Solving Linear Inequalities
7.
4.5x
11.
13x
12.
20x
13.
,3
17.
,0
20.
,5
5
x
,5
23.
410
44104
6
x
x
x
,6
26.
30
3
y
y
3,
Chapter 2 Linear Equations and Inequalities in One Variable
28.
29 2
229
7
xx
xx
x
,7
30.
38211
32118
19
xx
xx
x
,19
32.
8973
87 39
6
xx
xx
x
6,
34.
15
36
51
63
7
6
x
x
x
82
77 17
88 28
47
88
3
y
y
y
36.
13
34
31
43
94
12 12
5
12
y
y
y
y
Section 2.7 Solving Linear Inequalities
38.
12 17 20 13
12 13 20 17
3
y
y
y
40.
13
2
1
223
2
6
x
x
x
6,
41.
2
3
x
42.
1
4
441
4
x
x
43.
420
420
4
x
x
321
33
7
x
x
7,
46.
756
756
77
8
x
x
x
5
x
5,
Chapter 2 Linear Equations and Inequalities in One Variable
49.
315
x
50.
721
721
77
3
x
x
x
3,
53.
1
42
1
24 2
2
81
y
y
y
1
4
55.
4
114
x
x
22
5
x
5,
Section 2.7 Solving Linear Inequalities
59.
3318
333183
x
x
60.
8412
844124
816
x
x
x
61.
37 17
37 3173
714
x
x
x
62.
53 20
53 5 205
315
x
x
x
63.
233
23333
26
x
x
x
64.
3145
31414514
x
x
65.
51
5515
4
x
x
x
66.
33
3333
6
x
x
x
25 6
356
35565
xxxx
x
x
3
Chapter 2 Linear Equations and Inequalities in One Variable
68.
6246
624 464
xx
xxxx
69.
25511
2555115
3511
yy
yyy y
y
70.
4792
479 929
572
57727
yy
yyyy
y
y
71.
32 1 9
639
63393
612
y
y
y
y
72.
42 1 12
8412
y
y
73.
31521
33521
3221
322 212
xx
xx
xx
xxxx
4636
463363
66
66 66
xx
xxxx
x
x
8358
835585
338
33383
xx
xxxx
x
x
Section 2.7 Solving Linear Inequalities
76.
72 4 512
72 8510
15 2 5 10
yy
yy
yy
77.
21
3
2212
3
x
x
78.
31
4
3313
4
x
x
79.
1141
2
3
x
x
80.
15
2
1151
2
x
x
81.
444 5
44420
4444204
4416
xx
xx
xx
xx
Chapter 2 Linear Equations and Inequalities in One Variable
83. 37
37
xx
xxxx
84. 410
410
xx
xxx x
85.
77 2
7714
xx
xx
86.
313 2
3136
xx
xx
87.
2321
xx
88.
54510
520510
xx
xx
inequality is true for all real numbers. The solution
set is
is a real numberxx or
,.
54
54 44
0
xx
xx xx
x
63
30
6333
xx
x
xx xx
3
33
xba
2
22
xba
or
mm
yb yb
xx
mm
Section 2.7 Solving Linear Inequalities
94.
ymxb
ybmxbb
95. x is between −2 and 2, so 2.x
96. x is between 3 and 3, so 3.x
97. x is less than −2 or greater than 2, so 2.x
103. harass
104. cemetery, accommodation, harass
105. a.
0.4 16
0.4 30 16
px
106. a.
0.4 16
0.4 20 16
816
px
b. 0.4 16
0.8 0.4 16
px
x
3
86 88
3390
3
86 88 270
174 270
174 174 270 174
x
x
x
x
86 88 240
174 240
174 174 240 174
66
x
x
x
x
If you get less than a 66 on the final exam, your
88 78 86
4480
4
88 78 86 320
252 320
x
x
x