UPSC CDS I 2024 Math Paper

Concept:The problem involves comparing the area of the shaded region (composed of a small circle and a circular segment) to the area of the non‑shaded region (remainder of the large circle).Explanation:The large circle has centre $\mathrm{O}$ and radius $R = 7\ \mathrm{cm}$.The shaded region has two parts:1. A smaller circle with diameter $\mathrm{OC}$, so its radius $r = \frac{R}{2} = \frac{7}{2}\ \mathrm{cm}$. Area of small circle $= \pi r^{2} = \frac{22}{7} \times \left(\frac{7}{2}\right)^{2} = \frac{77}{2}\ \mathrm{cm}^{2}$.2. A segment of the large circle: area of semicircle $\mathrm{DAB}$ minus area of triangle $\mathrm{DBA}$. Semicircle area $= \frac{1}{2}\pi R^{2} = \frac{1}{2} \times \frac{22}{7} \times 7^{2} = 77\ \mathrm{cm}^{2}$. Triangle area $= \frac{1}{2} \times \mathrm{DB} \times \mathrm{OA} = \frac{1}{2} \times 14 \times 7 = 49\ \mathrm{cm}^{2}$. So, segment area $= 77 - 49 = 28\ \mathrm{cm}^{2}$.Total shaded area $= \frac{77}{2} + 28 = \frac{133}{2}\ \mathrm{cm}^{2}$.Area of the large circle $= \pi R^{2} = \frac{22}{7} \times 7^{2} = 154\ \mathrm{cm}^{2}$.Non‑shaded area $= 154 - \frac{133}{2} = \frac{308 - 133}{2} = \frac{175}{2}\ \mathrm{cm}^{2}$.Required ratio $= \frac{\text{shaded area}}{\text{non-shaded area}} = \frac{133/2}{175/2} = \frac{133}{175} = \frac{19}{25}$.Answer:$\frac{19}{25}$