Xavier Aptitude Test 2018 Paper

O is the centre of the bigger circle. Let P be the centre of the smaller circle. The two circles (not shown) intersect at A and B. As ∠AOB = 60°, it follows that
∠AOP = 30°
As OA = 2R, (for the sake of clarity, we use r instead of R) it follows that AM= R (M is the point of intersection of segment AB and line OP) and OM = $\sqrt{3}$R. As PA = $\sqrt{2}$R , it follows that MP = R and ∠APB = 90°.
The common area is the area of the smaller circle minus the shaded area. ...(1)
Shaded area = Ar of quadrant APB + Ar ΔOPA
+ Ar OPB - Ar of sector AOB
= $\frac{\mathit{\pi }{\left(\sqrt{2}\mathit{R}\right)}^2}{4}$ + $\left(\sqrt{3}-1\right){\mathit{R}}^2$ - $\frac{\mathit{\pi }}{6}{\left(2\mathit{R}\right)}^2$
= $\mathit{\pi }{\mathit{R}}^2\left(\frac{1}{2}-\frac{2}{3}\right)$ + $\left(\sqrt{3}-1\right){\mathit{R}}^2$
= $\frac{-\mathit{\pi }{\mathit{R}}^2}{6}$ + $\left(\sqrt{3}-1\right){\mathit{R}}^2$
From (1) and (2),
Common Area = $2\mathit{\pi }{\mathit{R}}^2$ + $\frac{\mathit{\pi }{\mathit{R}}^2}{6}$ + $\left(1-\sqrt{3}\right){\mathit{R}}^2$
= $\left(\frac{13\mathit{\pi }}{6}+1-\sqrt{3}\right){\mathit{R}}^2$