Basic Engineering Mathematics, Fifth Edition

Areas of common shapes 223

Area of rectangle=l×b= 7. 0 × 4. 5

= 31 .5cm^2

Perimeter of rectangle= 7 .0cm+ 4 .5cm

+ 7 .0cm+ 4 .5cm

=23.0cm

Problem 4. Calculate the area of the

parallelogram shown in Figure 25.11

21 mm

16 mm

H G

EF

9mm

Figure 25.11

Area of a parallelogram=base×perpendicular height

The perpendicular heighthis not shown in Figure 25.11
but may be found using Pythagoras’ theorem (see
Chapter 21).
From Figure 25.12, 9^2 = 52 +h^2 , from which
h^2 = 92 − 52 = 81 − 25 =56.
Hence, perpendicular height,

h=

√

56 = 7 .48mm.

9mm

E

G

h

F

H

16 mm 5 mm

Figure 25.12

Hence,area of parallelogramEFGH

=16mm× 7 .48mm

=120mm^2.

Problem 5. Calculate the area of the triangle

shown in Figure 25.13

I

K

J

5.68 cm

1.92 cm

Figure 25.13

Area of triangleIJK=

1

2

×base×perpendicular height

=

1

2

×IJ×JK

To findJK, Pythagoras’ theorem is used; i.e.,

5. 682 = 1. 922 +JK^2 ,from which

JK=

√

5. 682 − 1. 922 = 5 .346cm

Hence,area of triangleIJK=

1

2

× 1. 92 × 5. 346

=5.132cm^2.

Problem 6. Calculate the area of the trapezium

shown in Figure 25.14

27.4 mm

8.6 mm

5.5 mm

Figure 25.14

Area of a trapezium=

1

2

×(sum of parallel sides)

×(perpendicular distance

between the parallel sides)

Hence,area of trapeziumLMNO

=

1

2

×( 27. 4 + 8. 6 )× 5. 5

=

1

2

× 36 × 5. 5 =99mm^2