UPSC CDS II 2024 Math Paper

Concept:The circle passing through the vertices of three identical cones placed in contact is the circumcircle of an equilateral triangle formed by those vertices.Explanation:The base radius of each cone is $3$ cm.When placed on their bases, each cone touches the other two.The centers of the three bases form an equilateral triangle with side length equal to twice the base radius.Side length $= 2 \times 3 = 6$ cm.Since the cones are identical, the vertices lie directly above the base centers at the same height.Thus, the vertices also form an equilateral triangle with side length $s = 6$ cm.A unique circle passes through these three vertices — it is the circumcircle of the equilateral triangle.For an equilateral triangle, circumradius $R = \frac{s}{\sqrt{3}}$.Here, $R = \frac{6}{\sqrt{3}} = 2\sqrt{3}$ cm.Area of the circle $= \pi R^2 = \pi \times (2\sqrt{3})^2 = \pi \times 4 \times 3 = 12\pi$ square cm.Answer:$12\pi$ square cm (Option D).