(c) \(\frac{a^2}{6}.\)
If ‘a’ is length of the side of ΔABC, then
Area of ΔABC = \(\frac{\sqrt3}{4}\,a^2\)
semi-perimeter of ΔABC = \(\frac{3a}{2}\)
∴ Radius of in-circle = \(\frac{\text{Area}}{\text{semi-perimeter}}\) = \(\frac{\sqrt3}{4}\,a^2\) x \(\frac{2}{3a}\) = \(\frac{a}{2\sqrt3}\)
∴ Diagonal of square PQRS = Diameter of incircle.
= 2 x \(\frac{a}{2\sqrt3}\) = \(\frac{a}{\sqrt3}\)
∴ Area of squre = \(\frac{(\text{diagonal})^2}{2}\) = \(\frac{\big(\frac{a}{\sqrt3}\big)^2}{2}\) = \(\frac{a^2}{6}.\)