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

Solution (By Examveda Team)

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