SSC CHSL 12 Oct 2020 Shift 2 Paper

$\mathrm{\Delta }\mathrm{PQR}$ is an isoceles triangle and $\angle \mathrm{PQR}=90°$,
In right angle $\mathrm{\Delta }\mathrm{PQR}$, by Pythagoras theorem
⇒ ${\mathrm{PR}}^2={\mathrm{PQ}}^2+{\mathrm{QR}}^2$
⇒ ${\mathrm{PR}}^2={\left(10\right)}^2+{\left(10\right)}^2$ $\left(\because \mathrm{PQ}=\mathrm{QR}=10\mathrm{cm}\right)$
⇒ ${\mathrm{PR}}^2=200$
⇒ $\mathrm{PR}=\sqrt{200}=10\sqrt{2}\mathrm{cm}$
Now, we know, that perpendicular in an isosceles triangle bisect the third side.
∴ QL is perpendicular to PR.
∴ $\mathrm{LR}=\frac{10\sqrt{2}}{2}=5\sqrt{2}\mathrm{cm}$
Now, in right angle $\mathrm{\Delta }\mathrm{QLR}$, by Pythagoras theorem.
⇒ ${\mathrm{QL}}^2={\mathrm{QR}}^2-{\mathrm{LR}}^2$
⇒ ${\mathrm{QL}}^2={\left(10\right)}^2-{\left(5\sqrt{2}\right)}^2$
⇒ ${\mathrm{QL}}^2=100-50$
⇒ ${\mathrm{QL}}^2=50$
$\mathrm{QL}=\sqrt{50}=5\sqrt{2}\mathrm{cm}$