Xavier Aptitude Test 2020 Paper

The two circles are symmetric about the diagonal.
$\mathrm{NC}1=\frac{2\sqrt{2}}{3}=\sqrt{{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2}=\frac{2\sqrt{2}}{3}$
The lengths $\mathrm{FC}1$, $\mathrm{EC}1$ are the radius of the circle which is $\frac{2\mathit{k}\mathit{m}}{3}$.
The length $\mathrm{C}1\mathrm{P}$ is the radius of the circle.
Because of symmetry $\mathrm{C}1\mathrm{O}=\mathrm{C}2\mathrm{O}$ and $\mathrm{C}1\mathrm{N}=\mathrm{C}2\mathrm{B}$
$2\times \left(\mathrm{C}1\mathrm{N}+\mathrm{C}1\mathrm{O}\right)=2\sqrt{2}$ the length of the diagonal of the square.
$\mathrm{C}1\mathrm{N}+\mathrm{C}1\mathrm{O}=\sqrt{2}$
$\mathrm{C}1\mathrm{O}=\sqrt{2}-\frac{2\sqrt{2}}{3}=\frac{\sqrt{2}}{3}$
$\mathrm{OP}=\left(\frac{2-\sqrt{2}}{3}\right)$
The diagonal perpendicularly bisects the line $\mathrm{GH}$
Hence $\mathrm{C}1\mathrm{OH}$ is $90\mathrm{degrees}$ $\mathit{C}1{\mathrm{H}}^2=\mathit{C}1{\mathrm{O}}^2+{\mathrm{OH}}^2$
$\mathrm{OH}=\frac{\sqrt{2}}{3}$
Similarly $\mathrm{OG}=\frac{\sqrt{2}}{3}$
$\mathrm{HC}1,\mathrm{C}1\mathrm{G}$ both are equal to $\frac{2}{3}$ each.
$\mathrm{H}1\mathrm{G}$ is $\frac{2\sqrt{2}}{3}$ .$\mathrm{HC}1\mathrm{G}$ is a right angled triangle with angle $\mathrm{HC}1\mathrm{G}$ is $90\mathrm{degrees}$
Area of the region required is $2\times \left(\mathrm{Area}\mathrm{of}\mathrm{segment}\mathrm{OGH}\right)$
.'. Area of required region $=2\left(\frac{90}{360}.4\frac{\mathit{\pi }}{9}-\frac{1}{2}.\frac{2\sqrt{2}}{3}.\frac{\sqrt{2}}{3}\right)=\frac{2\left(\mathit{\pi }-2\right)}{9}$