Question

A circle is inscribed in an equailateral triangle of side $a.$ The area of any square inscribed in this circle is

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

Answer 1

Answer: (D)

$\frac{a^2}{6}$

$△=\frac{1}{2}.r(a+b+c)$
$\frac{\sqrt{3}{a}^2}{4}=\frac{1}{2}.3a.r$
$\sqrt{3}{a}^2=6ar$
$r=\frac{a}{2\sqrt{3}}$.
Now considering the square, the diameter of the circle will be its hypotenuse.
Hence
$2(side{)}^2=4r^2$
$2(side{)}^2=4(\frac{a^2}{4\times 3})$
$(side{)}^2=\frac{a^2}{6}$
=Area of square.