Question

Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points of the adjacent sides of a square of side ‘a’. 

(a) \(\frac{3a^2}{16}\)

(b) \(\frac{3\sqrt3a^2}{16}\)

(c) \(\frac{3a^2(\pi-12)}{4}\)

(d) \(\frac{3\sqrt3a^2}{32}\)

Answer 1

 (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) = \(\frac{\sqrt3a^2}{32}\)