Section 6–4
◆
Exercise 3
163
The Circle
Quadrilaterals
◆
1. Find the area of the parallelogram in Fig. 6–43.
2. Find the area of the rhombus in Fig. 6–44.
3. Find the area of the trapezoid in Fig. 6–45.
16.8
25.2
FIGURE 6–43
100
20.2
60°
60°
100
FIGURE 6–44
45°
28.6
FIGURE 6–45
4. Find the cost, to the nearest dollar, of applying roofing membrane to a flat roof, 21.5 m
long and 15.2 m wide, at $12.40/m2.
5. What will be the cost of flagging a sidewalk 104 m long and 2.2 m wide, at $13.50 per
square metre?
6. How many 25-cm-square tiles will cover a floor 16 m by 4 m?
7. What will it cost to carpet a floor, 6.25 m by 7.18 m, at $7.75 per square metre?
8. How many rolls of paper, each 8.00 yd. long and 18.0 in. wide, will paper the sides
of a room 16.0 ft. by 14.0 ft. and 10.0 ft. high, deducting 124 sq. ft. for doors and
windows?
9. What is the cost of plastering the walls and ceiling of a room 13 m long, 12 m wide,
and 7.5 m high, at $12.50 per square metre, allowing 135 m2 for doors, windows, and
baseboard?
10. What will it cost, to the nearest cent, to cement the floor of a cellar 12.6 m long and 9.2 m
wide, at $6.25 per square metre?
11. Find the cost of lining a topless rectangular tank 68 in. long, 54 in. wide, and 48 in. deep,
with zinc, weighing 5.2 lb. per square foot, at $1.55 per pound installed.
6–4
The Circle
Circumference and Area
For a circle of diameter d and radius r (Fig. 6–46), the following formulas apply:
Circle of
Radius r and
Diameter d
␲ ⬵ 3.1416
Circumference 2r d
113
d 2
Area r 2 4
114
Work to the nearest dollar in
this exercise, unless otherwise
asked.
164
Chapter 6
◆◆◆
◆
Geometry
Example 13: The circumference and the area of a 25.70-cm-radius circle are
C 2 (25.70) 161.5 cm
and
A (25.70)2 2075 cm2
◆◆◆
Radian Measure
A central angle (Fig. 6–46) is one whose vertex is at the centre of the circle. An arc is a portion
of the circle. If an arc is laid off along a circle with a length equal to the radius of the circle, the
central angle subtended by this arc is defined as one radian.
Center
We cover radian measure, and
compute areas of sectors and
segments, in Chapter 16.
Di
am
ete
rd
us r
Radi
Central
angle Arc, S
Sector
FIGURE 6–46 A circle.
Other Parts of a Circle
A sector (Fig. 6–47) is the region bounded by two radii and one of the arcs that they
intercept. A chord is a line segment connecting two points on the circle. A segment is
the region bounded by a chord and one of the arcs that it intercepts. When a circle has
two intersecting chords, the following theorem applies:
Chord
Segment
FIGURE 6–47
Intersecting
Chords
If two chords in a circle intersect, the product of the
parts of one chord is equal to the product of the parts
of the other chord.
121
ab cd
10
.1
In Fig. 6–48, you can see that a 16.3, b x, c 10.1, and d 12.5
◆◆◆
x
16.3
Example 14: Find the length x in Fig. 6–48.
12
.5
Solution: In Fig. 6–48, we can see that a 16.3, b x, c 10.1, and d 12.5. By
statement 121,
16.3x 10.1(12.5)
10.1(12.5)
x 7.75
16.3
FIGURE 6–48
Tangents and Secants
A tangent (Fig. 6–49) is a straight line that touches a circle in just one point. A secant
cuts the circle in two points. A chord is that portion of a secant lying within the circle.
We have two very useful theorems about tangents. First:
A
P
Tan
gen
C
◆◆◆
t
O
B
Secant
FIGURE 6–49 Tangent and
secant.
D
Tangents
to a Circle
A tangent is perpendicular to the radius drawn through
the point of contact.
Thus in Fig. 6–49, angle OPB is a right angle.
119
Section 6–4
◆
165
The Circle
115 cm
O
P
Q
340 cm
◆◆◆
FIGURE 6–50
Example 15: Find the distance OP in Fig. 6–50.
Solution: By statement 119, we know that angle PQO is a right angle. Also,
PQ 340 115 225 cm
So, by the Pythagorean theorem,
OP (115)2 (225)2 253 cm
◆◆◆
The second theorem follows:
Tangents
to a Circle
Two tangents drawn to a circle from a point outside the
circle are equal, and they make equal angles with a line
from the point to the centre of the circle.
120
B
Thus, in Fig. 6–51,
AB AD
and
BAC CAD
C
A
D
Semicircle
FIGURE 6–51
Any angle inscribed in a semicircle is a right angle.
Example 16: Find the distance x in Fig. 6–52.
θ
Exercise 4
◆
x
◆◆◆
The Circle
1. In a park is a circular fountain whose basin is 22.5 m in circumference. What is the
diameter of the basin?
2. How much larger is the side of a square circumscribing a circle 155 cm in diameter than a
square inscribed in the same circle?
3. The area of the bottom of a circular pan is 707 cm2. What is its diameter?
5
x (250) (148) 201
14
2
2
8
Solution: By statement 118, we know the largest angle, is 90. Then, by Eq. 145,
12
◆◆◆
118
FIGURE 6–52
166
Chapter 6
By cutting from both sides, one
can fell a tree whose diameter is
twice the length of the chainsaw
bar.
◆
Geometry
4. Find the diameter of a circular solar pond that has an area of 125 m2.
5. What is the circumference of a circular lake 33.0 m in diameter?
6. The radius of a circle is 5.00 m. Find the diameter of another circle containing 4 times the
area of the first.
7. The distance around a circular park is 2.50 km. How many hectares does it contain?
8. A woodcutter uses a tape measure and finds the circumference of a tree to be
248 cm. Assuming the tree to be circular, what length of chainsaw bar is needed to fell the
tree?
9. Find the distance x in Fig. 6–53.
10. In Fig. 6–54, distance OP is 8.65 units. Find the distance PQ.
11. Figure 6–55 shows a semicircle with a diameter of 105. Find distance PQ.
4.12
Q
x
Q
3.16
FIGURE 6–53
48.0 cm
52.0 cm
FIGURE 6–56
95.
0
P
O
6.35 diameter
P
FIGURE 6–54
FIGURE 6–55
R
12. What must be the diameter d of a cylindrical piston so that a pressure of
865 kPa on its circular end will result in a total force of 16 200 N?
13. Find the length of belt needed to connect the three 15.0-cm-diameter
pulleys in Fig. 6–56. Hint: The total curved portion of the belt is equal to
the circumference of one pulley.
14. Seven cables of equal diameter are contained within a circular conduit,
as in Fig. 6–57. If the inside diameter of the conduit is 25.6 cm, find the
cross-sectional area not occupied by the cables.
15. A 60 screw thread is measured by placing three wires on the thread and
measuring the distance T, as in Fig. 6–58. Find the distance D if the wire
diameters are 0.1000 in. and the distance T is 1.500 in., assuming that
the root of the thread is a sharp V shape.
16. Figure 6–59 shows two of the supports for a hemispherical dome. Find
the length of girder AB.
60°
.9
T = 1.500 in.
12
D
m
A
B
22.8 m
0.1000 in. diameter
FIGURE 6–57
FIGURE 6–58 Measuring a screw
thread “over wires.”
FIGURE 6–59
◆
Section 6–5
167
Volumes and Areas of Solids
17. A certain car tire is 78.5 cm in diameter. How far will the car move forward with one revolution of the wheel?
18. Archimedes claimed that the area of a circle is equal to the area of a triangle that has an
altitude equal to the radius of the circle and a base equal to the circumference of the circle.
Use the formulas of this chapter to show that this is true.
6–5
Volumes and Areas of Solids
Volume
The volume of a solid is a measure of the space it occupies or encloses. It is measured in cubic
units (m3, cm3, cu. ft., etc.) or, usually for liquids, in litres or gallons. One of our main tasks in
this chapter is to compute the volumes of various solids.
Area
We speak about three different kinds of areas. Surface area will be the total area of the surface
of a solid, including the ends, or bases. The lateral area, which will be defined later for each
solid, does not include the areas of the bases. The cross-sectional area is the area of the plane
figure obtained when we slice the solid in a specified way.
The formulas for the areas and volumes of some common solids are given in Fig. 6–60.
a
a
a
Base
122
Surface area = 6a 2
123
Cube
h
h
Volume = lwh
Rectangular
parallelepiped
l
w
Volume = a3
Surface area = 2(lw + hw + lh)
125
Any cylinder
or prism
Volume = (area of base)(altitude)
126
Right
cylinder
or prism
Lateral area = (perimeter of base)(altitude)
(not incl. bases)
127
Volume =
2r
4
πr 3
3
Any cone
or pyramid
h
h
s
A1
s
A2
h
Frustum
FIGURE 6–60
128
Sphere
Surface area = 4πr 2
s
124
Right circular
cone or
regular pyramid
Lateral area =
Any cone
or pyramid
Volume =
Right circular
cone or
Lateral area =
regular pyramid
Some solids.
h
(area
3
Volume =
s
(sum
2
s
2
129
of base)
130
(perimeter of base)
131
h
(A1
3
+ A2 + A1A2)
132
of base perimeters) = 2s (P1 + P2) 133