Question

The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72${}^{\circ }$.

Answer 1

Given that, circumference of a circle is $8cm$ and the central angle of a sector is ${72}^0$

Now, circumference of a circle $=2\pi r$

$\Rightarrow 8=2\pi r$

$\therefore r=\frac{8\times 7}{22\times 2}[\because \pi =\frac{22}{7}]$

$=\frac{14}{11}cm$

We know that, area of sector $=\frac{\theta }{360}\times \left(\pi r^2\right)$

$=\frac{72}{360}\times \frac{22}{7}\times \frac{14}{11}\times \frac{14}{11}$

$=1.02c{m}^2$

Hence, the required area of the sector is $1.02c{m}^2$.