(d) \(\frac{3\sqrt3a^2}{32}\)
Let AB = a be the side of the outermost square.
Then AG = AH = \(\frac{a}{2}\)
⇒ GH = \(\sqrt{\frac{a^2}{4}+\frac{a^2}{4}}\) = \(\frac{a}{\sqrt2}\)
∴ Diameter of circle = \(\frac{a}{\sqrt2}\)
⇒ Radius of circle = \(\frac{a}{2\sqrt2}\)
If O is the centre of the circle, then ∠POQ = 120º.
∴ Area of ΔPOQ = \(\frac{1}{2}\) x PO x PQ x sin 120º
= \(\frac{1}{2}\) x \(\frac{a}{2\sqrt2}\) x \(\frac{a}{2\sqrt2}\) x \(\frac{\sqrt3}{2}\) = \(\frac{\sqrt3a^2}{32}\)
∴ Area of ΔPQR = 3 (Area of ΔPOQ) = \(\frac{\sqrt3a^2}{32}\)