SSC CGL 16 Aug 2015 Shift 2 Paper

The ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding sides.
∴ $\frac{\mathrm{Area}\mathrm{of}\mathit{\Delta }\mathit{P}\mathit{Q}\mathit{R}}{\mathrm{Area}\mathrm{of}\mathit{\Delta }\mathit{A}\mathit{B}\mathit{C}}$ = $\frac{\mathit{P}{\mathit{R}}^2}{\mathit{A}{\mathit{C}}^2}$
⇒ $\frac{{12}^2}{{\mathit{A}\mathit{C}}^2}$ = $\frac{256}{441}$
⇒ Taking square roots of both sides,
⇒ 16 × AC = 12 × 21
⇒ AC = $\frac{12\times 21}{16}$ = $\frac{63}{4}$
= 15.75 cm.