SSC CGL 26 Oct 2014 Shift 1 Paper

Side of triangle = x cm
∴ Its area = $\frac{\sqrt{3}{\mathit{x}}^2}{4}$ sq. cm.
case II,
Side = (x – 2) cm
Area = $\frac{\sqrt{3}}{4}{\left(\mathit{x}-2\right)}^2$ sq. cm.
∴ $\frac{\sqrt{3}}{4}{\mathit{x}}^2-\frac{\sqrt{3}}{4}{\left(\mathit{x}-2\right)}^2$ = 4 $\sqrt{3}$
⇒ $\frac{\sqrt{3}}{4}\left({\mathit{x}}^2-{\left(\mathit{x}-2\right)}^2\right)$ = $4\sqrt{3}$
⇒ ${\mathit{x}}^2-{\mathit{x}}^2$ + 4x - 4 = $\frac{4\sqrt{3}\times 4}{\sqrt{3}}$
⇒ 4x - 4 = 16
⇒ 4x = 16 + 4 = 20
⇒ x = $\frac{20}{4}$ = 5 cm.