Question

In fig. $3$, $APB$ and $AQO$ are semicircle, and $AO=OB$. If the perimeter of the figure is $40cm$, find the area of the shaded region $[use\pi =\frac{22}{7}]$

Answer 1

let $Ao=r\Rightarrow$ diameter of $AOQ=r\Rightarrow$ radius $=\frac{r}{2}$ & radius of $AOB=r$

$\begin{array}{ccccc}\Rightarrow & r& +\frac{2\pi \left(\frac{r}{2}\right)}{2}& +\frac{2\pi \left(r\right)}{2}& =40\\ & \uparrow & \uparrow & \uparrow & \\ & OB& AOQ& AOB& \end{array}$

$\Rightarrow r+\frac{\pi r}{2}+\pi r=40$

$\Rightarrow r(1+\frac{3\pi }{2}+\pi r)=40$

$\Rightarrow r=\frac{40}{1+\frac{3\pi }{2}}$

Area $=\frac{\pi {\left(\frac{r}{2}\right)}^2}{2}+\frac{\pi r^2}{2}=\frac{\pi r^2}{2}(\frac{1}{4}+1)$

$=\frac{5}{4}\times \frac{\pi r^2}{2}=\frac{5\pi }{8}\times \frac{1600}{{(2+3\pi )}^2}\times 4$

$=\frac{4000\pi }{{(2+\frac{66}{7})}^2}=\frac{4000\times \frac{22}{7}\times 49}{{\left(80\right)}^2}$

$=96.25c{m}^2$