Question

In the given figure, ABCD is a trapezium of area 24.5 cm² , If AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. [CBSE 2014]

Answer 1

Area of trapezium = $\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)\times \mathrm{AB}$
$$\begin{gathered} \Rightarrow 24.5=\frac{1}{2}\left(10+4\right)\times \mathrm{AB} \\ \Rightarrow \mathrm{AB}=3.5\mathrm{cm} \end{gathered}$$
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
$$\begin{gathered} =24.5-\frac{1}{4}\mathrm{\pi }{\left(\mathrm{AB}\right)}^2 \\ =24.5-\frac{1}{4}\times \frac{22}{7}{\left(3.5\right)}^2 \\ =24.5-9.625 \\ =14.875{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 14.875 cm²