ABC is an equilateral triangle with side $$AB=a$$. AO, BO and CO are the angle bisectors of $$angle$$ A, $$angle$$ B and $$angle$$ C respectively. O is the centre of the circle and let radius of circle = $$r$$ and side of square = $$s$$ Also, we know that the angle bisector from the vertex of an equilateral triangle is the perpendicular bisector of the opposite side. => AD is the perpendicular bisector of BC. => $$BD=frac{a}{2}$$ and $$angle$$ OBD = $$frac{1}{2}angle B=frac{1}{2} imes60^circ=30^circ$$ Now, in $$ riangle$$ OBD, => $$tan(30^circ)=frac{OD}{BD}=frac{r}{frac{a}{2}}$$ => $$r=frac{1}{sqrt3} imesfrac{a}{2}=frac{a}{2sqrt3}$$ Now, in right $$ riangle$$ EDG, using Pythagoras theorem => $$(ED)^2=(EG)^2+(GD)^2$$ => $$(2r)^2=(s)^2+(s)^2$$ => $$4r^2=2s^2$$ => $$s^2=2 imes(frac{a}{2sqrt3})^2$$ => $$s^2=frac{a^2}{6}$$ $$ herefore$$ ar($$ riangle$$) ABC : ar(DEFG) = $$(frac{sqrt3}{4}a^2):(s)^2$$ = $$(frac{sqrt3}{4}a^2):(frac{a^2}{6})$$ = $$3sqrt3:2$$ => Ans - (D)