# In the given figure, ABCD is a trapezium in which AD||BC,

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.

Solution:

$\angle \mathrm{ABC}=90^{\circ}$

From the right $\Delta \mathrm{ABC}$, we have:

$\mathrm{AB}^{2}=\left(\mathrm{AC}^{2}-\mathrm{BC}^{2}\right)$

$\Rightarrow \mathrm{AB}^{2}=\left\{\left(41^{2}\right)-\left(40^{2}\right)\right\}$

$\Rightarrow \mathrm{AB}^{2}=(1681-1600)$

$\Rightarrow \mathrm{AB}^{2}=81$

$\Rightarrow \mathrm{AB}=\sqrt{81}$

$\Rightarrow \mathrm{AB}=9 \mathrm{~cm}$

$\therefore$ Length $\mathrm{AB}=9 \mathrm{~cm}$

Now,

Area of the trapezium $=\left\{\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AB}\right\}$

$=\left(\frac{1}{2} \times(16+40) \times 9\right) \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 56 \times 9\right) \mathrm{cm}^{2}$

$=(28 \times 9) \mathrm{cm}^{2}$

$=252 \mathrm{~cm}^{2}$

Hence, the area of the trapezium is $252 \mathrm{~cm}^{2}$.