UPSC NDA I 2023 Mathematics Solved Paper

Concept:The area of a minor segment is found by subtracting the area of the corresponding triangle from the area of the sector.Explanation:We consider a quarter circle of radius $a$.The area of the quarter circle is $\frac{\pi a^2}{4}$.The triangle formed is right‑angled with legs $a$ and $a$, so its area is $\frac{1}{2} \times a \times a = \frac{a^2}{2}$.Subtract the triangle area from the quarter‑circle area to get the minor segment: $\frac{\pi a^2}{4} - \frac{a^2}{2} = \frac{\pi a^2}{4} - \frac{2a^2}{4} = \frac{(\pi - 2)a^2}{4}$.Answer:$\frac{(\pi - 2)a^2}{4}$, which corresponds to option A.