A circle is inscribed in an equilateral triangle and a square is inscribed in that circle. The ratio of the areas of the triangle and the square is

Answer: Option C

Let the side of equilateral triangle = 'a' and the side of square = 'b'
In circle radius of equilateral $$\Delta =\frac{a}{2\sqrt{3}}$$
$$\begin{array}{rl}& \therefore \text{Diagonal of square}=2\times \frac{a}{2\sqrt{3}}=\frac{a}{\sqrt{3}}\\ & \text{Now, }b=\frac{\text{Diagonal}}{\sqrt{2}}=\frac{a}{\frac{\sqrt{3}}{\sqrt{2}}}=\frac{a}{\sqrt{6}}\\ & \text{Required ratio}=\frac{\frac{\sqrt{3}}{4}{a}^2}{{\left(\frac{a}{\sqrt{6}}\right)}^2}\\ & =\frac{\sqrt{3}}{4}{a}^2\times \frac{6}{a^2}\\ & =\frac{3\sqrt{3}}{2}\Rightarrow 3\sqrt{3}:2\end{array}$$