UPSC CDS I 2019 Math Paper

Concept:The common area is the overlap of two quarter-circle sectors inside a square, each centered at opposite vertices.Explanation:Consider a square with side length $a$.Label the vertices as $A$, $B$, $C$, $D$ in order.Take $A$ and $C$ as the two opposite vertices.Draw a quarter‑circle arc from $B$ to $D$ with centre $A$ and radius $a$.Draw another quarter‑circle arc from $B$ to $D$ with centre $C$ and radius $a$.These two arcs intersect at $B$ and $D$.The overlapping region is bounded by both arcs and is symmetric about chord $BD$.Compute the area of one segment (say between arc $BD$ and chord $BD$ for the sector with centre $A$):Area of sector $ABD$ (quarter‑circle) $= \frac{1}{4}\pi a^{2}$.Area of $\triangle ABD$ (right‑angled isosceles) $= \frac{1}{2} a \times a = \frac{a^{2}}{2}$.Area of one segment $= \frac{1}{4}\pi a^{2} - \frac{a^{2}}{2} = \frac{a^{2}}{2}\left(\frac{\pi}{2} - 1\right)$.The same segment appears on the other side (from centre $C$).Thus the common area (the lens‑shaped overlap) is twice that segment:$= 2 \times \frac{a^{2}}{2}\left(\frac{\pi}{2} - 1\right) = a^{2}\left(\frac{\pi}{2} - 1\right)$.Answer:$a^{2}\left(\frac{\pi }{2}-1\right)$ (Option C).