The length of diagonal of a square is $$9\sqrt2$$ cm. The square is reshaped to form an equilateral triangle. What is the area (in cm2) of largest incircle that can be formed in that triangle?

Diagonal of square = $$9\sqrt2$$ cm

=> Side of square = $$\frac{9\sqrt2}{\sqrt2}=9$$ cm

=> Perimeter of square = Perimeter of equilateral triangle = $$4\times9=36$$ cm

Thus, side of equilateral triangle = $$a=\frac{36}{3}=12$$ cm

Also, inradius (r) of an equilateral triangle = $$\frac{a}{2\sqrt3}$$

=> $$r=\frac{12}{2\sqrt3}=2\sqrt3$$ cm

$$\therefore$$ Area of incircle = $$\pi (r)^2$$

= $$\pi(2\sqrt3)^2=12\pi$$ $$cm^2$$

=> Ans - (C)