SSC CGL 16 Aug 2015 Shift 1 Paper

AD = 12 $\sqrt{3}$ cm.
AB = 2x cm. (let)
BD = x cm.
From ΔABD,
AD = $\sqrt{\mathit{A}{\mathit{B}}^2-\mathit{B}{\mathit{D}}^2}$
= $\sqrt{{\left(2\mathit{x}\right)}^2-{\mathit{x}}^2}$
= $\sqrt{4{\mathit{x}}^2-{\mathit{x}}^2}$
= $\sqrt{3{\mathit{x}}^2}$ = $\sqrt{3}$ x
∴ $\sqrt{3}$ x = 12 $\sqrt{3}$
⇒ x = 12 cm
∴ AB = 2x = 12 = 24 cm
∴ Area of ΔABC = $\frac{\sqrt{3}}{4}$ × ${\mathrm{side}}^2$
= $\frac{\sqrt{3}}{4}$ × 24 × 24
= 144 $\sqrt{3}$ sq. cm.