UPSC CDS 1 2023 Math Paper

Concept:Geometry of an inscribed circle, a rectangle at one corner and a small square at the opposite corner; using Pythagoras theorem and diagonal of a square.Explanation:Let the radius of the inscribed circle be $r$ cm.From the $4$ cm $\times$ $2$ cm rectangle at corner $P$, we have $\mathrm{OD} = r-2$, $\mathrm{CD} = r-4$, and $\mathrm{OC} = r$.In $\triangle \mathrm{OCD}$, by Pythagoras: $r^2 = (r-2)^2 + (r-4)^2$.Simplifying: $r^2 = r^2 -4r+4 + r^2 -8r+16$ $\Rightarrow 0 = r^2 -12r+20$.Solving: $r^2 -12r+20=0$ $\Rightarrow (r-10)(r-2)=0$ $\Rightarrow r=10$ (since $r>4$).The diagonal of the outer square $\mathrm{PQRS}$ is $10\sqrt{2}$ cm.This diagonal consists of the radius $r$ (along $\mathrm{OA}$) plus the diagonal of the small square ($\sqrt{2}\,a$), where $a$ is the side of the small square.Thus $10 + \sqrt{2}\,a = 10\sqrt{2}$ $\Rightarrow \sqrt{2}\,a = 10(\sqrt{2}-1)$ $\Rightarrow a = \frac{10(\sqrt{2}-1)}{\sqrt{2}}$.Area of small square $= a^2 = \frac{100(\sqrt{2}-1)^2}{2} = 50(2+1-2\sqrt{2}) = 50(3-2\sqrt{2})$ cm$^2$.Answer:$50(3-2\sqrt{2})$ square cm.