Question

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

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

Answer 1

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

We know that the area of a triangle can be written asWe know that the area of a triangle can be written as $\frac{a\times b\times c}{4R}$, where a, b and c are the three sides and R is the circum-radius of the triangle.,

So, in case of an equilateral triangle, the area of the triangle is $\frac{a^3}{4R}$.

Also the area of an equilateral triangle is given by $\frac{\sqrt{3}{a}^2}{4}$.

Equating the two, we get

$R=\frac{a}{\sqrt{3}}$

So, the in-radius, r = $R=\frac{a}{\sqrt{3}}$.

Hence, the diameter of the inscribed circle is $\frac{a}{3}$.

The diameter of the inscribed circle is equal to the diagonal of the square.
Let each side of the square be x.
So the diagonal of the square will be ($\frac{1}{\sqrt{2}x}$).

Equating the diameter of the circle with the diagonal of the square we get, x = ($\frac{a}{\sqrt{6}}$).

Therefore, the area of the square = $x^2=\frac{a^2}{6}$

Then the ratio of area of triangle to area of square=$\frac{\sqrt{3}{a}^2}{4}:\frac{a^2}{6}::3\sqrt{3}:2$