Question

Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and non-parallel sides are 15 cm and 13 cm.

Answer 1

We will divide the trapezium into a triangle and a parallelogram.

Difference in the lengths of parallel sides$=25-11=14\mathrm{cm}$
We can represent this in the following figure:

Trapezium ABCD is divided into parallelogram AECD and triangle CEB.

1. Consider triangle CEB.

In triangle CEB, we have:
$EB=25-11=14\mathrm{cm}$

Using Hero's theorem, we will first evaluate the semiperimeter of triangle CEB and then evaluate its area.

Semiperimeter, $s=\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\left(15+13+14\right)=\frac{42}{2}=21\mathrm{cm}$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}CEB=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-15\right)\left(21-13\right)\left(21-14\right)} \\ =\sqrt{21\times 6\times 8\times 7} \\ =\sqrt{7056} \\ =84{\mathrm{cm}}^2 \end{gathered}$$

Also,

Area of triangle CEB$=\frac{1}{2}\left(\mathrm{Base}\times \mathrm{Height}\right)$

Height of triangle CEB$=\frac{\mathrm{Area}\times 2}{\mathrm{Base}}=\frac{84\times 2}{14}=12\mathrm{cm}$

2. Consider parallelogram AECD.

​​Area of parallelogram AECD$=\mathrm{Height}\times \mathrm{Base}=AE\times CF=12\times 11=132{\mathrm{cm}}^2$

Area of trapezium ABCD$=\mathrm{Ar}\left(∆BEC\right)+\mathrm{Ar}\left(\mathrm{parallelogram}AECD\right)=132+84=216{\mathrm{cm}}^2$