Xavier Aptitude Test 2020 Paper

The two circles are symmetric about the diagonal. $NC1 = \frac{2\sqrt{2}}{3} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \frac{2\sqrt{2}}{3}$ The lengths $FC1$, $EC1$ are the radius of the circle which is $\frac{2\text{ km}}{3}$. The length $C1 P$ is the radius of the circle. Because of symmetry $C1 O = C2 O$ and $C1 N = C2 B$ $2\times (C1 N + C1 O) = 2\sqrt{2}$ the length of the diagonal of the square. $C1 N + C1 O = \sqrt{2}$ $C1 O = \sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{\sqrt{2}}{3}$ $OP = \left(\frac{2 - \sqrt{2}}{3}\right)$ The diagonal perpendicularly bisects the line $GH$ Hence $C1OH$ is $90^{\circ}$ $C1H^2 = C1O^2 + OH^2$ $OH = \frac{\sqrt{2}}{3}$ Similarly $OG = \frac{\sqrt{2}}{3}$ $HC1, C1G$ both are equal to $\frac{2}{3}$ each. $H1G$ is $\frac{2\sqrt{2}}{3}$ .$HC1G$ is a right angled triangle with angle $HC1G$ is $90^{\circ}$ Area of the region required is $2\times(\text{Area of segment OGH})$ .'. Area of required region $= 2\left(\frac{90}{360} \cdot \frac{4\pi}{9} - \frac{1}{2} \cdot \frac{2\sqrt{2}}{3} \cdot \frac{\sqrt{2}}{3}\right) = \frac{2(\pi - 2)}{9}$