Smaller diagonal of a rhombus is equal to length of its sides. If length of each side is 6 cm, then what is the area $$(in cm^2)$$ of an equilateral triangle whose side is equal to the bigger diagonal of the rhombus?

Let bigger diagonal of rhombus = BD = $$2x$$ cm and smaller diagonal = AC = 6 cm

Diagonals of a rhombus bisect each other at right angle.

=> OC = 3 cm and OD = $$x$$ cm

In $$\triangle$$ OCD,

=> $$(OD)^2=(CD)^2-(OC)^2$$

=> $$(OD)^2=(6)^2-(3)^2$$

=> $$(OD)^2=36-9=27$$

=> $$OD=\sqrt{27}=3\sqrt3$$ cm

Thus, side of equilateral triangle = bigger diagonal = $$2\times3\sqrt3=6\sqrt3$$ cm

$$\therefore$$ Area of equilateral triangle = $$\frac{\sqrt3}{4}a^2$$

= $$\frac{\sqrt3}{4}\times(6\sqrt3)^2$$

= $$\frac{\sqrt3}{4}\times108=27\sqrt3$$ $$cm^2$$

=> Ans - (B)