Three equal circles are described with vertices of the triangles as centres. If the radius of each circle is r, the sum of areas of the portions of the circles intercepted in a triangle is :

Answer: Option D

Solution

Answer: Option D

Solution:

We have :
Required area :
$$\begin{array}{rl}& =\frac{\pi r^2{\theta }_1}{360}+\frac{\pi r^2{\theta }_2}{360}+\frac{\pi r^2{\theta }_3}{360}\\ & =\frac{\pi r^2}{360}\left({\theta }_1+{\theta }_2+{\theta }_3\right)\\ & =\frac{\pi r^2\times 180}{360}\left[\because {\theta }_1+{\theta }_2+{\theta }_3=180^{\circ }\right]\\ & =\frac{\pi r^2}{2}Or=\frac{1}{2}\pi r^2\end{array}$$