UPSC CDS 2 2021 Math Paper

Concept:The triangle sides satisfy $6^2 + 8^2 = 10^2$, so it is a right triangle with $\angle A = 90^\circ$.Radii of circles centered at vertices are found from the distances between centers: $AB = r_A + r_B = 6$, $BC = r_B + r_C = 10$, $CA = r_C + r_A = 8$.Solving gives $r_A = 2$ cm, $r_B = 4$ cm, $r_C = 6$ cm.Explanation:Area of sector at A: $\displaystyle P = \frac{90^\circ}{360^\circ} \cdot \pi \cdot (r_A)^2 = \frac{1}{4} \pi \cdot 4 = \pi$ cm$^2$. So statement 1 is correct.Let $\angle B = \theta$. Then $\angle C = 90^\circ - \theta$ (since $\angle A = 90^\circ$).Area of sector at B: $\displaystyle Q = \frac{\theta}{360^\circ} \cdot \pi \cdot (r_B)^2 = \frac{\theta}{360} \cdot \pi \cdot 16$.Area of sector at C: $\displaystyle R = \frac{90^\circ - \theta}{360^\circ} \cdot \pi \cdot (r_C)^2 = \frac{90 - \theta}{360} \cdot \pi \cdot 36$.Compute $9Q + 4R$:$9Q + 4R = 9 \cdot \frac{\theta}{360} \pi \cdot 16 + 4 \cdot \frac{90 - \theta}{360} \pi \cdot 36$$= \frac{\pi}{360} \left(144\theta + 144(90 - \theta)\right) = \frac{\pi}{360} \cdot 144 \cdot 90$$= \frac{144 \cdot 90}{360} \pi = \frac{144}{4} \pi = 36\pi$ cm$^2$.Thus statement 2 is also correct.Answer:Both statements 1 and 2 are correct, so option C (Both 1 and 2) is the answer.