If a circle is inscribed in an equilateral triangle of side o, then area of the square inscribed in the circle is

$\begin{array}{l}s=\frac{1}{2}(a+b+c)=\frac{3a}{2}\\ \mathrm{\Delta }=\frac{\sqrt{3}{a}^2}{4}\therefore r=\frac{\mathrm{\Delta }}{s}=\frac{a}{2\sqrt{3}}\end{array}$
If r is the length of a side of the square inscribed
in incircle of the triangle then
$\begin{array}{l}{x}^2+x^2=(\text{ diameter }{)}^2=(2r{)}^2\\ \Rightarrow x^2=2r^2=\frac{a^2}{6}\\ \Rightarrow \text{ Area of square }=\frac{a^2}{6}\end{array}$