Sainik School Entrance Exam 2020 Solved Paper

(d) As, In a ∥gm PQRS

$\mathrm{PQ}=6\mathrm{cm},\mathrm{QR}=5\mathrm{cm}$
and $\mathrm{PR}=7\mathrm{cm}$
$\therefore \mathrm{RS}=6\mathrm{cm},\mathrm{PS}=5\mathrm{cm}$
Let PM be the perpendicular distance between $\overline{\mathrm{PQ}}$ and $\overline{\mathrm{RS}}$
and $\mathrm{MS}=\mathit{a}\mathrm{cm}$
∴ $\mathrm{MR}=6-\mathit{a}\mathrm{cm}$
In △PMS
${\mathrm{PS}}^2={\mathrm{PM}}^2+{\mathrm{MS}}^2$
$\Rightarrow {\mathrm{PM}}^2={\mathrm{PS}}^2-{\mathrm{MS}}^2$
$=5^2-{\mathit{a}}^2=25-{\mathit{a}}^2\mathrm{cm}$ ...........(i)
In △PMR
${\mathrm{PR}}^2={\mathrm{PM}}^2+{\mathrm{MR}}^2$
$\Rightarrow {\mathrm{PM}}^2={\mathrm{PR}}^2-{\mathrm{MR}}^2$
$=7^2-{\left(6-\mathit{a}\right)}^2=49-\left(6^2+{\mathit{a}}^2-12\mathit{a}\right)$
$=49-36-{\mathit{a}}^2+12\mathit{a}$
$=13-{\mathit{a}}^2+12\mathit{a}$
$\Rightarrow {\mathit{P}\mathit{M}}^2=13-{\mathit{a}}^2+12\mathit{a}$ .............(ii)
From (i) and (ii), we get
$25-{\mathit{a}}^2=13-{\mathit{a}}^2+12\mathit{a}$
$\Rightarrow 25-{\mathit{a}}^2-13+{\mathit{a}}^2-12=0$
$\Rightarrow 12-12\mathit{a}=0$
$\Rightarrow 12\mathit{a}=12\Rightarrow \mathit{a}=1\mathrm{cm}$
∴ from (i), ${\mathrm{PM}}^2=25-{\mathit{a}}^2=25-1=24\mathrm{cm}$
$\Rightarrow \mathrm{PM}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}\mathrm{cm}$
So, the perpendicular distance
$=\mathrm{PM}=2\sqrt{6}\mathrm{cm}$