Question

A circle is inscribed in an equilateral triangle of side $a$. The area (in sq. units) of any square inscribed in this circle is

• A. $\frac{a^2}{12}$
• B. $\frac{a^2}{6}$
• C. $\frac{a^2}{8}$
• D. $\frac{a^2}{15}$

Answer 1

The correct option is B $\frac{a^2}{6}$

Given that $ABC$ is an equilateral triangle of sides $a$ and $r$ is the radius of the circle inscribed in it.
In $\Delta ABC,$
Area of $\Delta ABC,\Delta =\frac{\sqrt{3}}{4}{a}^2$
Semi-perimeter of $\Delta ABC,s=\frac{3a}{2}$
$\therefore$ Radius of in-circle, $r=\frac{\Delta }{s}=\frac{\frac{\sqrt{3}}{4}{a}^2}{\frac{3a}{2}}=\frac{a}{2\sqrt{3}}$
Diagonal of square $PQRS=2r=2\times \frac{a}{2\sqrt{3}}=\frac{a}{\sqrt{3}}$
$\therefore$ Area of square $=\frac{{\text{diagonal}}^2}{2}=\frac{{\left(\frac{a}{\sqrt{3}}\right)}^2}{2}=\frac{a^2}{6}$

$\therefore$ Area of square =$\left(\frac{a^2}{6}\right)$square units.