Question

A circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in the circle is:

• A. $\frac{a^2}{4}$
• B. $\frac{a^2}{6}$
• C. $\frac{a^2}{9}$
• D. $\frac{2a^2}{3}$

Answer 1

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

Area of Triangle $=△=\frac{\sqrt{3}}{4}{a}^2$

and from Above formula we can see that ,

$△=\frac{1}{2}r\ast (a+a+a)=\frac{1}{2}\ast r\ast 3a\Rightarrow r=\frac{2}{3a}△........................\left(1\right)$

Putting value of $△$ in equation (1)

we get,

$r=\frac{2\sqrt{3}{a}^2}{4\ast 3a}=\frac{a}{2\sqrt{3}}$

now, we know that Diameter of the incircle will be the Diagonal of the Square

inscribed in the circle of side $b\left(Let\right)$

so, length of diagonal of square will be $=b\sqrt{2}$

$2r=\frac{a}{\sqrt{3}}=b\sqrt{2}\Rightarrow b=\frac{a}{\sqrt{6}}$

With the help of $b$ we can find the Area of the Square as

${△}_{square}=b^2=\frac{a^2}{6}$