Question

In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. [CBSE 2015]

Answer 1

Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB
$$\begin{gathered} \Rightarrow 40=\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{\mathrm{AO}}{2}\right)+\frac{1}{2}\times 2\mathrm{\pi }\left(\mathrm{OB}\right)+\mathrm{OB} \\ \Rightarrow 40=\frac{11}{7}\mathrm{AO}+\frac{22}{7}\mathrm{OB}+\mathrm{OB} \\ \Rightarrow 40=\frac{11}{7}\mathrm{OB}+\frac{22}{7}\mathrm{OB}+\mathrm{OB}\left[\because \mathrm{AO}=\mathrm{OB}\right] \\ \Rightarrow 40=\frac{40}{7}\mathrm{OB} \\ \Rightarrow \mathrm{OB}=7\mathrm{cm} \end{gathered}$$
Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB
$$\begin{gathered} =\frac{1}{2}\mathrm{\pi }{\left(\frac{7}{2}\right)}^2+\frac{1}{2}\mathrm{\pi }{\left(7\right)}^2 \\ =\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^2+\frac{1}{2}\times \frac{22}{7}\times {\left(7\right)}^2 \\ =96.25{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of the shaded portion is 96.25 cm².