One of the internal angle of a rhombus is 60° and length of its shorter diagonal is 8 cm. What is the area of the rhombus?

Let $$\angle$$ A = 60° and BD be the shorter diagonal = 8 cm

The diagonals of a rhombus bisect each other at right angle and also bisect the angles of rhombus.

=> $$\angle$$ OAD = 30° and OD = 4 cm

In $$\triangle$$ OAD, $$tan(\angle OAD) = \frac{OD}{OA}$$

=> $$tan(30) = \frac{4}{OA}$$

=> $$\frac{1}{\sqrt{3}} = \frac{4}{OA}$$

=> $$OA = 4\sqrt{3}$$

Thus, AC = $$2 \times 4\sqrt{3} = 8\sqrt{3}$$ cm

$$\therefore$$ Area of rhombus = $$\frac{1}{2} \times $$ (product of diagonals)

= $$\frac{1}{2} \times (AC) \times (BD)$$

= $$\frac{1}{2} \times 8\sqrt{3} \times 8$$

= $$32\sqrt{3}$$ $$cm^2$$

=> Ans - (D)