Question

From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet. [CBSE 2011]

Answer 1

Since, BFEC is a quarter of a circle.
Hence, BC = EC = 3.5 cm
Now, DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of shaded region = Area of the trapezium ABCD − Area of the quadrant BFEC
$$\begin{gathered} =\frac{1}{2}\times \left(\mathrm{AB}+\mathrm{DC}\right)\times \mathrm{BC}-\frac{1}{4}\times \mathrm{\pi }{\left(\mathrm{EC}\right)}^2 \\ =\frac{1}{2}\times \left(3.5+5.5\right)\times 3.5-\frac{1}{4}\times \frac{22}{7}\times {\left(3.5\right)}^2 \\ =6.125{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of the shaded region is 6.125 cm² .