Question

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

Answer 1

In the given figure, ABCD is the trapezium.



Draw a line BE parallel to AD.

In ∆BCE,
The sides of the triangle are of length 15 m, 13 m and 14 m.
∴ Semi-perimeter of the triangle is
$s=\frac{15+13+14}{2}=\frac{42}{2}=21\mathrm{m}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆BCE=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-15\right)\left(21-13\right)\left(21-14\right)} \\ =\sqrt{21\left(6\right)\left(8\right)\left(7\right)} \\ =84{\mathrm{m}}^2...\left(1\right) \end{gathered}$$

Also,
Area of ∆BCE = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$
$$\begin{gathered} \Rightarrow 84=\frac{1}{2}\times 14\times \mathrm{Height} \\ \Rightarrow 84=7\times \mathrm{Height} \\ \Rightarrow \mathrm{Height}=\frac{84}{7} \\ \Rightarrow \mathrm{Height}=12\mathrm{m} \end{gathered}$$

∴ Height of ∆BCE = Height of the parallelogram ABED = 12 m

Now,
Area of the parallelogram ABED = Base × Height
= 11 × 12
= 132 m² ...(2)

∴ Area of the trapezium = Area of the parallelogram ABED + Area of the triangle BCE
= 132 + 84
= 216 m²

Hence, the area of a trapezium is 216 m².