UPSC CDS 2 2021 Math Paper

Concept:The major segment of a circle is the region larger than the semicircle. Its area is found by subtracting the area of the minor segment from the total area of the circle.Explanation:Given: radius $r = 2.1$ cm, central angle for the minor segment $\theta = 120^{\circ}$.Area of the circle: $\pi r^{2} = \frac{22}{7} \times (2.1)^{2} = 13.86$ cm$^{2}$.Area of the minor segment = area of sector – area of triangle.Sector area: $\frac{\theta}{360^{\circ}} \times \pi r^{2} = \frac{120}{360} \times 13.86 = 4.62$ cm$^{2}$.Triangle area: $\frac{1}{2} r^{2} \sin 120^{\circ} = \frac{1}{2} \times (2.1)^{2} \times \frac{\sqrt{3}}{2} \approx 1.91$ cm$^{2}$.Minor segment area: $4.62 - 1.91 = 2.71$ cm$^{2}$.Area of major segment = area of circle – area of minor segment = $13.86 - 2.71 = 11.15$ cm$^{2}$.Answer:$11.15$ cm$^{2}$ (Option D).