Smaller diagonal of a rhombus is equal to length of its sides. If length of each side is 4 cm, then what is the area $$(in cm^2)$$ of an equilateral triangle with side equal to the bigger diagonal of the rhombus?

Let bigger diagonal of rhombus = BD = $$2x$$ cm and smaller diagonal = AC = 4 cm

Diagonals of a rhombus bisect each other at right angle.

=> OC = 2 cm and OD = $$x$$ cm

In $$\triangle$$ OCD,

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

=> $$(OD)^2=(4)^2-(2)^2$$

=> $$(OD)^2=16-4=12$$

=> $$OD=\sqrt{12}=2\sqrt3$$ cm

Thus, side of equilateral triangle = bigger diagonal = $$2\times2\sqrt3=4\sqrt3$$ cm

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

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

= $$\frac{\sqrt3}{4}\times48=12\sqrt3$$ $$cm^2$$

=> Ans - (D)