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

Let side of square = 'a'

Diameter of circle = Diagonal of square = $$a\sqrt{2}$$

Radius of circle = $$\frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$$

Let side of triangle = s

Length of iInradius of triangle = $$\frac{s}{2\sqrt{3}} = \frac{a}{\sqrt{2}}$$

$$s = a\sqrt{6}$$

Area of triangle = $$\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(a\sqrt{6})^2 = \frac{3\sqrt{3}a^2}{2}$$

Area of square = $$a^2$$

Required ratio = $$\frac{3\sqrt{3}a^2}{2}: a^2 = 3\sqrt{3}: 2$$