basic_engineering_mathematics0

Areas of plane figures 169
(b) Area = πd2
4 = π(15)2 = 225π
= 19.35 cm 2
= 176.7mm 2
4 4
Problem 5. Figure 22.9 shows the gable end of a building.
(c) Circumference, c = 2πr, hence
Determine the area of brickwork in the gable end.
r =
c
2π = 70
2π = 35
π mm
A
( ) 35 2
5m 5m
Area of circle = πr 2 = π = 352
π π
B
C D
= 389.9 mm 2 or 3.899 cm 2
6m
Problem 8. Calculate the areas of the following sectors of
circles:
8m
(a) having radius 6 cm with angle subtended at centre 50 ◦
Fig. 22.9
(b) having diameter 80 mm with angle subtended at centre
107 ◦ 42 ′
(c) having radius 8 cm with angle subtended at centre
The shape is that of a rectangle and a triangle.
1.15 radians.
Area of rectangle = 6 × 8 = 48 m 2
Area of sector of a circle = θ 2
Area of triangle = 1 × base × height.
360 (πr2 )or 1 2 r2 θ (θ in radians).
2
(a) Area of sector
CD = 4m, AD = 5 m, hence AC = 3 m (since it is a 3, 4, 5
triangle).
= 50 50 × π × 36
360 (π62 ) = = 5π = 15.71 cm 2
360
Hence, area of triangle ABD = 1 × 8 × 3 = 12 m2
2
Total area of brickwork = 48 + 12 = 60 m 2
(b) If diameter = 80 mm, then radius, r = 40 mm, and area of
sector
= 107◦ 42 ′ 107 42
Problem 6. Determine the area of the shape shown in
360 (π402 ) =
60
360 (π402 ) = 107.7
360 (π402 )
Fig. 22.10.
= 1504 mm 2 or 15.04 cm 2
27.4 mm
(c) Area of sector = 1 2 r2 θ = 1 2 × 82 × 1.15 = 36.8cm 2
5.5 mm
8.6 mm
Problem 9. A hollow shaft has an outside diameter of
Fig. 22.10
The shape shown is a trapezium.
5.45 cm and an inside diameter of 2.25 cm. Calculate the
cross-sectional area of the shaft.
The cross-sectional area of the shaft is shown by the shaded part
in Fig. 22.11 (often called an annulus).
Area of trapezium = 1 (sum of parallel sides)(perpendicular
2
distance between them)
= 1 (27.4 + 8.6)(5.5)
2
= 1 × 36 × 5.5 = 99 mm2
2
Problem 7. Find the areas of the circles having (a) a radius
d
2.25 cm
of 5 cm, (b) a diameter of 15 mm, (c) a circumference of
d 5.45 cm
70 mm.
Fig. 22.11
Area of a circle = πr 2 or πd2
Area of shaded part = area of large circle − area of small circle
4
(a) Area = πr 2 = π(5) 2 = 25π = 78.54 cm 2
= πD2 − πd2 = π 4 4 4 (D2 − d 2 ) = π 4 (5.452 − 2.25 2 )