228 Basic Engineering Mathematics
h is calculated using Pythagoras’ theorem:
8
2
= h
2
+ 4
2
from which h =
8 2 − 4 2 = 6.928 cm
Hence,
Area of one triangle =
1
2
× 8 × 6.928 = 27.71 cm
2
Area of hexagon = 6 × 27.71
= 166.3 cm
2
Problem 19. Figure 25.27 shows a plan of a floor
of a building which is to be carpeted. Calculate the
area of the floor in square metres. Calculate the cost,
correct to the nearest pound, of carpeting the floor
with carpet costing £16.80 per m 2 , assuming 30%
extra carpet is required due to wastage in fitting
2 . 5 m
M
L
K
J
4 m
2 m
0.6 m
0.6 m
0.8 m
2 m
0.8 m
30Њ 60Њ
3 m
3 m
2 m
3 m
I
H
G
F
A
BЈ
B
C
D
E
Figure 25.27
Area of floor plan
= area of triangle ABC + area of semicircle
+ area of rectangle CGLM
+ area of rectangle CDEF
− area of trapezium HIJK
Triangle ABC is equilateral since AB = BC = 3 m and,
hence, angle B CB = 60 ◦ .
sin B
CB = BB
/3
i.e.
BB
= 3 sin60
◦
= 2.598 m.
Area of triangle ABC =
1
2
(AC)(BB
)
=
1
2
(3)(2.598) = 3.897 m
2
Area of semicircle =
1
2
πr
2
=
1
2
π(2.5)
2
= 9.817 m
2
Area of CGLM = 5 × 7 = 35 m
2
Area of CDEF = 0.8 × 3 = 2.4 m
2
Area of HIJK =
1
2
(KH + IJ )(0.8)
Since MC = 7 m then LG = 7 m, hence
JI = 7 − 5.2 = 1.8 m. Hence,
Area of HIJK =
1
2
(3 + 1.8)(0.8) = 1.92 m 2
Total floor area = 3.897 + 9.817 + 35 + 2.4 − 1.92
= 49.194 m
2
To allow for 30% wastage, amount of carpet required
= 1.3 × 49.194 = 63.95 m 2
Cost of carpet at £16.80 per m 2
= 63.95 × 16.80 = £1074, correct to the nearest pound.
Now try the following Practice Exercise
Practice Exercise 99 Areas of common
shapes (answers on page 351)
1. Calculate the area of a regular octagon if each
side is 20 mm and the width across the flats is
48.3 mm.
2. Determine the area of a regular hexagon which
has sides 25 mm.
3. A plot of land is in the shape shown in
Figure 25.28. Determine
20 m
20 m
20 m
30 m
10 m
20 m
3 0 m
20 m
15
m
40 m
15 m
20 m
Figure 25.28
h is calculated using Pythagoras’ theorem:
8
2
= h
2
+ 4
2
from which h =
8 2 − 4 2 = 6.928 cm
Hence,
Area of one triangle =
1
2
× 8 × 6.928 = 27.71 cm
2
Area of hexagon = 6 × 27.71
= 166.3 cm
2
Problem 19. Figure 25.27 shows a plan of a floor
of a building which is to be carpeted. Calculate the
area of the floor in square metres. Calculate the cost,
correct to the nearest pound, of carpeting the floor
with carpet costing £16.80 per m 2 , assuming 30%
extra carpet is required due to wastage in fitting
2 . 5 m
M
L
K
J
4 m
2 m
0.6 m
0.6 m
0.8 m
2 m
0.8 m
30Њ 60Њ
3 m
3 m
2 m
3 m
I
H
G
F
A
BЈ
B
C
D
E
Figure 25.27
Area of floor plan
= area of triangle ABC + area of semicircle
+ area of rectangle CGLM
+ area of rectangle CDEF
− area of trapezium HIJK
Triangle ABC is equilateral since AB = BC = 3 m and,
hence, angle B CB = 60 ◦ .
sin B
CB = BB
/3
i.e.
BB
= 3 sin60
◦
= 2.598 m.
Area of triangle ABC =
1
2
(AC)(BB
)
=
1
2
(3)(2.598) = 3.897 m
2
Area of semicircle =
1
2
πr
2
=
1
2
π(2.5)
2
= 9.817 m
2
Area of CGLM = 5 × 7 = 35 m
2
Area of CDEF = 0.8 × 3 = 2.4 m
2
Area of HIJK =
1
2
(KH + IJ )(0.8)
Since MC = 7 m then LG = 7 m, hence
JI = 7 − 5.2 = 1.8 m. Hence,
Area of HIJK =
1
2
(3 + 1.8)(0.8) = 1.92 m 2
Total floor area = 3.897 + 9.817 + 35 + 2.4 − 1.92
= 49.194 m
2
To allow for 30% wastage, amount of carpet required
= 1.3 × 49.194 = 63.95 m 2
Cost of carpet at £16.80 per m 2
= 63.95 × 16.80 = £1074, correct to the nearest pound.
Now try the following Practice Exercise
Practice Exercise 99 Areas of common
shapes (answers on page 351)
1. Calculate the area of a regular octagon if each
side is 20 mm and the width across the flats is
48.3 mm.
2. Determine the area of a regular hexagon which
has sides 25 mm.
3. A plot of land is in the shape shown in
Figure 25.28. Determine
20 m
20 m
20 m
30 m
10 m
20 m
3 0 m
20 m
15
m
40 m
15 m
20 m
Figure 25.28