Question

In the given figure, ABCD is a trapezium in which AD||BC, ∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.

Answer 1

$$\begin{gathered} \angle \mathrm{ABC}=90° \\ \mathrm{From}\mathrm{the}\mathrm{right}∆\mathrm{ABC},\mathrm{we}\mathrm{have}: \\ {\mathrm{AB}}^2=\left({\mathrm{AC}}^2-{\mathrm{BC}}^2\right) \\ \Rightarrow {\mathrm{AB}}^2=\left\{\left({41}^2\right)-\left({40}^2\right)\right\} \\ \Rightarrow {\mathrm{AB}}^2=\left(1681-1600\right) \\ \Rightarrow {\mathrm{AB}}^2=81 \\ \Rightarrow \mathrm{AB}=\sqrt{81} \\ \Rightarrow \mathrm{AB}=9\mathrm{cm} \\ \therefore \mathrm{Length}\mathrm{AB}=9\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{trapezium}=\left\{\frac{1}{2}\times \left(\mathrm{AD}+\mathrm{BC}\right)\times \mathrm{AB}\right\} \end{gathered}$$
$$\begin{gathered} =\left(\frac{1}{2}\times \left(16+40\right)\times 9\right){\mathrm{cm}}^2 \\ =\left(\frac{1}{2}\times 56\times 9\right){\mathrm{cm}}^2 \\ =\left(28\times 9\right){\mathrm{cm}}^2 \\ =252{\mathrm{cm}}^2 \end{gathered}$$
$\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{trapezium}\mathrm{is}252{\mathrm{cm}}^2.$