SSC CHSL 19 Mar 2020 Shift 2 Paper

Given, Let the base and height of the isosceles right angle triangle be $\mathit{a}\mathrm{cm}$.
∴ Area of the triangle
$=\frac{1}{2}\times$ base $\times$ height
$=\frac{1}{2}\times \mathit{a}\times \mathit{a}$
$\frac{1}{2}\times {\mathit{a}}^2=121$ $[\because$ given area $]$
${\mathit{a}}^2=121\times 2$
⇒ $\mathit{a}=11\sqrt{2}$
$\ln \mathrm{\Delta }\mathrm{ABC},{\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2$ $[\because$ by Pythagoras theorem]
${\mathrm{AC}}^2={\mathit{a}}^2+{\mathit{a}}^2=2{\mathit{a}}^2$
${\mathrm{AC}}^2=2\times {\left(11\sqrt{2}\right)}^2$
$=2\times 242=484$
$\mathrm{AC}=\sqrt{484}=22\mathrm{cm}$.