SSC CHSL 11 Jan 2017 Shift 2 Paper

Let ∠ 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.
⇒ ∠OAD = 30° and OD = 4 cm
In $△\mathrm{OAD}\tan \left(\angle \mathit{O}\mathit{A}\mathit{D}\right)=\frac{\mathit{O}\mathit{D}}{\mathit{O}\mathit{A}}$
⇒ $\mathit{t}\mathit{a}\mathit{n}\left(30\right)=\frac{4}{\mathit{O}\mathit{A}}$
⇒ $\frac{1}{\sqrt{3}}=\frac{4}{\mathit{O}\mathit{A}}$
⇒ $\mathit{O}\mathit{A}=4\sqrt{3}$
Thus, AC $=2\times 4\sqrt{3}=8\sqrt{3}$ = cm
∴ Area of rhombus = $\frac{1}{2}\times$ (product of diagonals)
= $\frac{1}{2}\times \left(\mathit{A}\mathit{C}\right)\times \left(\mathit{B}\mathit{D}\right)$
= $\frac{1}{2}\times 8\sqrt{3}\times 8$
= $32\sqrt{3}\mathit{c}{\mathit{m}}^2$