Question

In the given figure, ABCD is a rectangle of dimensions 21 cm × 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure. [CBSE 2017]

Answer 1

Area of the shaded region

= Area of the rectangle − Area of the semicircle

$$\begin{gathered} =21\times 14-\left\{\frac{1}{2}\times \mathrm{\pi }\times {\left(\frac{14}{2}\right)}^2\right\} \\ =294-\left\{\frac{1}{2}\times \frac{22}{7}\times 7\times 7\right\} \\ =294-77 \\ =217{\text{cm}}^2 \end{gathered}$$

Therefore, area of shaded region is 217 cm².

Length of the boundary (or perimeter) of the shaded region

$$\begin{gathered} =\mathrm{AB}+\mathrm{AD}+\mathrm{DC}+\mathrm{arc}\mathrm{BC} \\ =\mathrm{AB}+\mathrm{AD}+\mathrm{DC}+\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{14}{2}\right) \\ =21+14+21+\frac{22}{7}\times 7 \\ =56+22 \\ =78\mathrm{cm} \end{gathered}$$

Therefore, the perimeter of the shaded region is 78 cm.