Question

Find the area of a trapezium whose parallel sides are 11 ma and 25 m long, and the nonparallel sides are 15 m and 13 m long.

Answer 1

Draw $DE\parallel BC$ and DL perpendicular to AB.
The opposite sides of quadrilateral DEBC are parallel. Hence, DEBC is a parallelogram.
∴ DE = BC = 13 m
Also,
$AE=(AB-EB)=(AB-DC)=(25-11)=14\mathrm{m}$
For $∆DAE$:
Let:
AE = a =14 m
DE = b = 13 m
DA = c =15 m

Thus, we have:

$$\begin{gathered} s=\frac{a+b+c}{2} \\ s=\frac{14+13+15}{2}=21\mathrm{m} \end{gathered}$$

Area of $∆DAE=\sqrt{s(s-a)(s-b)(s-c)}$
$$\begin{gathered} =\sqrt{21\times (21-14)\times (21-13)\times (21-15)} \\ =\sqrt{21\times 7\times 8\times 6} \\ =\sqrt{7056} \\ =84{\mathrm{m}}^2 \end{gathered}$$

Area of $∆DAE=\frac{1}{2}\times AE\times DL$
$$\begin{gathered} \Rightarrow 84=\frac{1}{2}\times 14\times DL \\ \Rightarrow \frac{84\times 2}{14}=DL \\ \Rightarrow DL=12\mathrm{m} \end{gathered}$$

Area of trapezium = $\frac{1}{2}\times \left(\mathrm{Sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \left(\mathrm{Distance}\mathrm{between}\mathrm{them}\right)$
$$\begin{gathered} =\frac{1}{2}\times (11+25)\times 12 \\ =\frac{1}{2}\times 36\times 12 \\ =216{\mathrm{m}}^2 \end{gathered}$$