Question

In figure $7.43$, $A$ is the centre of the circle. $\angle ABC={45}^o$ and $AC=7\sqrt{2}cm$. Find the area of segment $BXC$.

Answer 1

Radius of circle $r=7\sqrt{2}cm$

$\angle ABC=\angle ACB={45}^o(\because AB=AC)$

$\therefore$ $△ABC$ $\Rightarrow$ ${180}^o-{45}^o-{45}^o={90}^o$

So, the angle of sector $\theta ={90}^o$

W.K.T Area of segment $=\frac{{\pi }^2}{2}(\frac{\pi \theta }{180}-\sin \theta )=\frac{{\left(7\sqrt{2}\right)}^2}{2}(\frac{\pi \times {90}^o}{{180}^o}-\sin \left({90}^o\right))=49(\frac{\pi }{2}-1){cm}^2$

Hence Area of segment is $49(\frac{\pi }{2}-1){cm}^2$