Question

In Fig. 3, APB and AQO are semicircles, and AO=OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. Use $\pi =\frac{22}{7}$

Answer 1

Let the radius of the semi-circle APB be r.
$\Rightarrow$ The radius of the semi-circle $AQO=\frac{r}{2}$
Now,
Perimeter of the given figure = Length of arc AQO + Length of arc of APB + OB
$\pi \times \frac{\pi }{2}+\pi \times r+r$
$r(\frac{3}{2}\pi +1)$
$r(\frac{3}{2}\times \frac{22}{7}+1)$
$=r\frac{80}{14}$
$r\left(\frac{40}{7}\right)$
$\Rightarrow r\left(\frac{40}{7}\right)=40$
$\Rightarrow r=7cm$
$\therefore$ Area of the shaded region = Area of semi-circle AQO + Area of semi-circle APB
$=\frac{\pi {\left(\frac{7}{2}\right)}^2}{2}+\frac{\pi 7^2}{2}$
$=49\pi (\frac{1}{8}+\frac{1}{2})$
$=96.25c{m}^2$