UPSC CDS 2 2022 Math Paper

Concept:The inscribed circle (incircle) of an equilateral triangle touches all three sides, and its radius has a fixed relation to the side length.Explanation:Step 1: Write the given side length $a = 2\sqrt{3}\,\text{cm}$.Step 2: For an equilateral triangle, the radius $r$ of the incircle is $r = \frac{a}{2\sqrt{3}}$. Substituting $a$: $r = \frac{2\sqrt{3}}{2\sqrt{3}} = 1\,\text{cm}$.Step 3: Compute the area of the equilateral triangle: $A_{triangle} = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \times (2\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 4 \times 3 = 3\sqrt{3}\,\text{cm}^2$.Step 4: Compute the area of the inscribed circle: $A_{\text{circle}} = \pi r^2 = \pi (1)^2 = \pi\,\text{cm}^2$.Step 5: The remaining portion of the triangle (the region outside the circle) is the difference: $A_{triangle} - A_{\text{circle}} = 3\sqrt{3} - \pi\,\text{cm}^2$.Answer:$(3\sqrt{3} - \pi)\,\text{cm}^2$, which matches option B.