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 :

Solution (By Examveda Team)

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