Xavier Aptitude Test 2021 Paper

We know that radius of circle $\mathrm{P}$ is $2\mathrm{cm}$
Length of $\mathrm{MO}=1\mathrm{cm}$
radius of circle $\mathrm{R}=\mathrm{AO}=\sqrt{2^2-1^2}=\sqrt{3}$
Area of overlap between the circle $\mathrm{R}$ and $\mathrm{P}$(shaded region) $=\mathrm{semi}\mathrm{circle}\mathrm{area}\mathrm{of}\mathrm{R}+\mathrm{area}\mathrm{of}\mathrm{segment}\mathrm{ANBOA}$
$\mathrm{Area}\mathrm{of}\mathrm{segment}\mathrm{ANBOA}=\mathrm{area}\mathrm{of}\mathrm{sector}\mathrm{ANBM}-\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{AMB}$
$\frac{120}{360\mathit{\pi }}{2}^2-\frac{1}{2}\left(1\right)\left(2\sqrt{3}\right)$
$\frac{4\mathit{\pi }}{3}-\sqrt{3}$
Area of overlap between circle $\mathrm{R}$ and $\mathrm{P}=\frac{\mathit{\pi }{\left(\sqrt{3}\right)}^2}{2}+\frac{4\mathit{\pi }}{3}-\sqrt{3}$
$=\frac{17\mathit{\pi }}{6}-\sqrt{3}$