Basic Engineering Mathematics, Fifth Edition

228 Basic Engineering Mathematics

his calculated using Pythagoras’ theorem:

82 =h^2 + 42

from which h=

√

82 − 42 = 6 .928cm

Hence,

Area of one triangle=

1

2

× 8 × 6. 928 = 27 .71cm^2

Area of hexagon= 6 × 27. 71

=166.3cm^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 £16.80perm^2 , assuming 30%

extra carpet is required due to wastage in fitting

L 2.5 m M

K

J

4m

2m

0.6 m

0.6 m

0.8 m

2m 0.8 m

30  60 

3m

3m

2m 3m

I

H

G

F

A

B B

C

E D

Figure 25.27

Area of floor plan

=area of triangleABC+area of semicircle

+area of rectangleCGLM

+area of rectangleCDEF

−area of trapeziumHIJK

TriangleABCis equilateral sinceAB=BC=3m and,

hence, angleB′CB= 60 ◦.

sinB′CB=BB′/ 3

i.e. BB′=3sin60◦= 2 .598m.

Area of triangleABC=

1

2

(AC)(BB′)

=

1

2

( 3 )( 2. 598 )= 3 .897m^2

Area of semicircle =

1

2

πr^2 =

1

2

π( 2. 5 )^2

= 9 .817m^2

Area ofCGLM= 5 × 7 =35m^2

Area ofCDEF= 0. 8 × 3 = 2 .4m^2

Area ofHIJK=

1

2

(KH+IJ)( 0. 8 )

SinceMC=7mthenLG=7m, hence

JI= 7 − 5. 2 = 1 .8m. Hence,

Area of HIJK=

1

2

( 3 + 1. 8 )( 0. 8 )= 1 .92m^2

Total floor area= 3. 897 + 9. 817 + 35 + 2. 4 − 1. 92

= 49 .194m^2

To allow for 30% wastage, amount of carpet required

= 1. 3 × 49. 194 = 63 .95m^2

Cost of carpet at £16.80perm^2

= 63. 95 × 16. 80 =£1074, correct to the nearest pound.

Now try the following Practice Exercise

PracticeExercise 99 Areas of common

shapes (answers on page 351)

1. Calculate the area of a regular octagon if each
   side is 20mm and the width across the flats is
   48 .3mm.


2. Determine the area of a regular hexagon which
   has sides 25mm.


3. A plot of land is in the shape shown in
   Figure 25.28. Determine

20m

20m

20m

30m

10m

20m

30m

20m

15m

40m

15m

20m

Figure 25.28