Question

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

• A. $\sqrt{3}:\sqrt{2}$
• B. $\sqrt{3}:1$
• C. $3\sqrt{3}:2$
• D. $3\sqrt{2}$

Answer 1

The correct option is C $3\sqrt{3}:2$

Clearly, centre of triangle , circle and square coincides. (Centroid, incenter of equilateral triangle , orthocenter coincides)
Let $a$ be the length of side of an equilateral triangle.
Area of equilateral triangle $A_1=\frac{\sqrt{3}{a}^2}{4}$
We have radius of inscribed circle $r=\frac{\sqrt{3}}{2}a\times \frac{1}{3}=\frac{a}{2\sqrt{3}}$
So, diameter= $\frac{a}{\sqrt{3}}$
We know that the diameter of the inscribed circle is equal to the diagonal of the square.
So, diagonal of square $=\frac{a}{\sqrt{3}}$
Let $x$ be the length of side of square.
Length of diagonal of square $=x\sqrt{2}$
$\Rightarrow x\sqrt{2}=\frac{a}{\sqrt{3}}$
$\Rightarrow x=\frac{a}{\sqrt{6}}$
Area of square $A_2=x^2=\frac{a^2}{6}$
Now,$\frac{A_1}{A_2}=\frac{\frac{\sqrt{3}{a}^2}{4}}{\frac{a^2}{6}}$
$\Rightarrow \frac{A_1}{A_2}=\frac{3\sqrt{3}}{2}$