UPSC CDS Math Area and Perimeter Questions

Let ABCD be the square inscribed in the semi-circle with center 0 , and side CD the diameter of the semi-circle. Let the point $M$ be the mid-point of $AB$. Now, join $OB$ and $OM$ to get the right $\triangle OMB$ with $OB$ as the hypotenuse. Therefore, $OM^{2}+MB^{2}=OB^{2}$ $OM=$ side of the square inscribed in the semi-circle, $MB=$ half of the side; and $OB=$ radius Let the side of the square be x. $x^{2}+\left(\;\frac{x}{2}\right)^{2}=r^{2}$ $x^{2}=\;\frac{4r^{2}}{5}$ Hence, the area of the semi-circle $=\;\frac{4r^{2}}{5}$ Now diagonal of the square inscribed in the circle $=2r$ Therefore, its area $=\;\frac{(2r)^{2}}{2}=2r^{2}$ Area of the square $=\;\frac{\;\text{ diagonal }\;{}^{2}}{2}$ Hence, the required ratio $=\;\frac{4r^{2}}{5}:2r^{2}=\;\frac{2}{5}:1=2:5$