UPSC CDS 2 2021 Math 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:Radius of the circle $r = 2.1$ cm and central angle $\theta = 120^{\circ}$.Area of sector $= \dfrac{\theta}{360^{\circ}} \times \pi r^2 = \dfrac{120}{360} \times \dfrac{22}{7} \times (2.1)^2$.Simplify: $\dfrac{1}{3} \times \dfrac{22}{7} \times 4.41 = \dfrac{22 \times 4.41}{21} = \dfrac{97.02}{21} = 4.62$ cm$^{2}$.Area of triangle $= \dfrac{1}{2} \times r^2 \times \sin\theta = \dfrac{1}{2} \times (2.1)^2 \times \sin 120^{\circ}$.$\sin 120^{\circ} = \dfrac{\sqrt{3}}{2} \approx 0.866$.So triangle area $= \dfrac{1}{2} \times 4.41 \times 0.866 = 2.205 \times 0.866 \approx 1.91$ cm$^{2}$.Area of minor segment $=$ area of sector $-$ area of triangle.Segment area $= 4.62 - 1.91 = 2.71$ cm$^{2}$.Answer:The approximate area of the minor segment is $2.71$ cm$^{2}$, which corresponds to option A.