226 Basic Engineering Mathematics
11. Determine the area of an equilateral triangle
of side 10.0 cm.
12. If paving slabs are produced in 250 mm by
250 mm squares, determine the number of
slabs required to cover an area of 2 m 2 .
Here are some further worked problems on finding the
areas of common shapes.
Problem 11. Find the area of a circle having a
radius of 5 cm
Area of circle = πr
2
= π(5)
2
= 25π = 78.54 cm
2
Problem 12. Find the area of a circle having a
diameter of 15 mm
Area of circle =
πd 2
4
=
π(15) 2
4
=
225π
4
= 176.7 mm
2
Problem 13. Find the area of a circle having a
circumference of 70 mm
Circumference, c = 2πr, hence
radius, r =
c
2π
=
70
2π
=
35
π
mm
Area of circle = πr
2
= π
35
π
2
=
35 2
π
= 389.9 mm
2 or 3.899 cm
2
Problem 14. Calculate the area of the sector of a
circle having radius 6 cm with angle subtended at
centre 50 ◦
Area of sector =
θ 2
360
(πr
2
) =
50
360
(π6
2
)
=
50 × π × 36
360
= 15.71 cm
2
Problem 15. Calculate the area of the sector of a
circle having diameter 80 mm with angle subtended
at centre 107 ◦ 42
If diameter = 80 mm then radius, r = 40 mm, and
area of sector =
107 ◦ 42
360
(π40
2
) =
107
42
60
360
(π40
2
)
=
107.7
360
(π40
2
)
= 1504 mm
2 or 15.04 cm
2
Problem 16. A hollow shaft has an outside
diameter of 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 Figure 25.22 (often called an annulus).
d 5
2.25 cm
D 5 5.45 cm
Figure 25.22
Area of shaded part = area of large circle – area of
small circle
=
π D 2
4
−
πd 2
4
=
π
4
(D
2
− d
2
)
=
π
4
(5.45
2
− 2.25
2
)
= 19.35 cm
2
Now try the following Practice Exercise
Practice Exercise 98 Areas of common
shapes (answers on page 351)
1. A rectangular garden measures 40 m by 15 m.
A 1 m flower border is made round the two
shorter sides and one long side. A circular
swimming pool of diameter 8 m is constructed
11. Determine the area of an equilateral triangle
of side 10.0 cm.
12. If paving slabs are produced in 250 mm by
250 mm squares, determine the number of
slabs required to cover an area of 2 m 2 .
Here are some further worked problems on finding the
areas of common shapes.
Problem 11. Find the area of a circle having a
radius of 5 cm
Area of circle = πr
2
= π(5)
2
= 25π = 78.54 cm
2
Problem 12. Find the area of a circle having a
diameter of 15 mm
Area of circle =
πd 2
4
=
π(15) 2
4
=
225π
4
= 176.7 mm
2
Problem 13. Find the area of a circle having a
circumference of 70 mm
Circumference, c = 2πr, hence
radius, r =
c
2π
=
70
2π
=
35
π
mm
Area of circle = πr
2
= π
35
π
2
=
35 2
π
= 389.9 mm
2 or 3.899 cm
2
Problem 14. Calculate the area of the sector of a
circle having radius 6 cm with angle subtended at
centre 50 ◦
Area of sector =
θ 2
360
(πr
2
) =
50
360
(π6
2
)
=
50 × π × 36
360
= 15.71 cm
2
Problem 15. Calculate the area of the sector of a
circle having diameter 80 mm with angle subtended
at centre 107 ◦ 42
If diameter = 80 mm then radius, r = 40 mm, and
area of sector =
107 ◦ 42
360
(π40
2
) =
107
42
60
360
(π40
2
)
=
107.7
360
(π40
2
)
= 1504 mm
2 or 15.04 cm
2
Problem 16. A hollow shaft has an outside
diameter of 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 Figure 25.22 (often called an annulus).
d 5
2.25 cm
D 5 5.45 cm
Figure 25.22
Area of shaded part = area of large circle – area of
small circle
=
π D 2
4
−
πd 2
4
=
π
4
(D
2
− d
2
)
=
π
4
(5.45
2
− 2.25
2
)
= 19.35 cm
2
Now try the following Practice Exercise
Practice Exercise 98 Areas of common
shapes (answers on page 351)
1. A rectangular garden measures 40 m by 15 m.
A 1 m flower border is made round the two
shorter sides and one long side. A circular
swimming pool of diameter 8 m is constructed