SSC CGL 24 Aug 2021 Shift 2 Paper

$\mathit{P}\mathit{Q}\mathit{R}$ is a triangle, right angle at $\mathit{Q}$. $\mathit{P}\mathit{N}$ is median divide $\mathit{Q}\mathit{R}$ in equal parts $($ let $\mathit{x})$Similarly, RM is median divide
$\mathit{P}\mathit{Q}$ in equal parts let $\mathit{y}$
In $△\mathit{P}\mathit{Q}\mathit{N}$,
$\Rightarrow {\left(\mathit{P}\mathit{Q}\right)}^2+{\left(\mathit{Q}\mathit{N}\right)}^2={\left(\mathit{P}\mathit{N}\right)}^2$
$\Rightarrow {\left(2\mathit{y}\right)}^2+{\left(\mathit{x}\right)}^2={\left(9\right)}^2$
$\Rightarrow 4{\mathit{y}}^2+{\mathit{x}}^2=81$ . . . (i)
In $△\mathit{M}\mathit{Q}\mathit{R}$,
$\Rightarrow {\left(\mathit{M}\mathit{Q}\right)}^2+{\left(\mathit{Q}\mathit{R}\right)}^2={\left(\mathit{M}\mathit{R}\right)}^2$
$\Rightarrow {\left(\mathit{y}\right)}^2+{\left(2\mathit{x}\right)}^2={\left(7\right)}^2$
$\Rightarrow {\mathit{y}}^2+4{\mathit{x}}^2=49$ . . . (ii)
In $△\mathit{P}\mathit{Q}\mathit{R}$,
$\Rightarrow {\left(2\mathit{y}\right)}^2+{\left(2\mathit{x}\right)}^2=\mathit{P}{\mathit{R}}^2$
$\Rightarrow 4{\mathit{y}}^2+4{\mathit{x}}^2=\mathit{P}{\mathit{R}}^2$ . . . (iii)
$△\mathit{M}\mathit{Q}\mathit{N}={\mathit{y}}^2+{\mathit{x}}^2=9$ . . . (iv)
From Eq. (i) + Eq. (ii)
$\Rightarrow 4{\mathit{x}}^2+4{\mathit{y}}^2+{\mathit{x}}^2+{\mathit{y}}^2=130$
$\left[\therefore {\mathit{x}}^2+{\mathit{y}}^2=9\right]$
$\Rightarrow 4{\mathit{x}}^2+4{\mathit{y}}^2=130-9$
$\Rightarrow =\mathit{P}{\mathit{R}}^2=121$
$\mathit{P}\mathit{R}=11\mathit{c}\mathit{m}.$