UPSC CDS 2 2022 Math Paper

Concept:The major segment area is found by subtracting the minor segment area from the total area of the circle.Explanation:Let the chord length be $l$ and the radius be $r$.The chord subtends a right angle at the centre (implied by the given formula for minor segment).In $\triangle OAB$, using Pythagoras theorem: $l^2 = r^2 + r^2 = 2r^2$, so $r = \frac{l}{\sqrt{2}}$.Area of circle $= \pi r^2 = \pi \cdot \frac{l^2}{2} = \frac{\pi l^2}{2}$.Area of minor segment (for angle $\theta = 90^\circ$): $A_{\text{minor}} = \frac{1}{2}\left(\frac{\pi}{2} - \sin 90^\circ\right) r^2 = \frac{1}{2}\left(\frac{\pi}{2} - 1\right) \cdot \frac{l^2}{2} = \frac{l^2}{4}\left(\frac{\pi}{2} - 1\right)$.Now, area of major segment $A_{\text{major}} = \text{Area of circle} - A_{\text{minor}} = \frac{\pi l^2}{2} - \frac{l^2}{4}\left(\frac{\pi}{2} - 1\right)$.Simplify: $A_{\text{major}} = \frac{\pi l^2}{2} - \frac{\pi l^2}{8} + \frac{l^2}{4} = \frac{4\pi l^2 - \pi l^2}{8} + \frac{l^2}{4} = \frac{3\pi l^2}{8} + \frac{l^2}{4} = \frac{l^2}{4}\left(\frac{3\pi}{2} + 1\right)$.Thus the required area is $\frac{l^2}{4}\left(\frac{3\pi}{2} + 1\right)$.Answer:$\frac{l^{2}}{4}\left(\frac{3\pi}{2}+1\right)$ (Option A).