UPSC CDS I 2024 Math Paper

Concept:Use the incircle radius of an equilateral triangle to find its side length and area, then subtract the unshaded region composed of circular sectors and triangles.Explanation:Area of the circle is $9\pi$ square units.Radius $r = 3$ units because $\pi r^2 = 9\pi$.Height of the equilateral triangle is $2r = 6$ units.For an equilateral triangle, height $= \frac{\sqrt{3}}{2} \times \text{side}$.So $\frac{\sqrt{3}}{2} a = 6 \Rightarrow a = \frac{12}{\sqrt{3}} = 4\sqrt{3}$ units.Area of equilateral triangle $= \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 48 = 12\sqrt{3}$ square units.Total area of triangle and circle together $= 12\sqrt{3} + 9\pi$ square units.Angle at vertex $A$ is $60^{\circ}$.Each unshaded region consists of a sector of the circle with angle $120^{\circ}$ and a triangle formed by two radii and the chord.Area of one unshaded part $= \frac{1}{2} \times 3 \times 3 \times \sin 120^{\circ} + 9\pi \times \frac{120}{360} = \frac{9\sqrt{3}}{4} + 3\pi$ square units.Total unshaded area $= 2 \times \left( \frac{9\sqrt{3}}{4} + 3\pi \right) = \frac{9\sqrt{3}}{2} + 3\pi$ square units (since $2 \times 3\pi = 6\pi$ but