Sainik School Entrance Exam 2020 Solved Paper

(a) Given, perimeter of a square $=40\mathrm{cm}$
$\Rightarrow 4\mathit{a}=40\mathrm{m}$
$\Rightarrow \mathit{a}=10\mathrm{m}\left(\mathit{a}=\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{square}\right)$
and, perimeter of a rectangle $=40\mathrm{m}$
$\Rightarrow 2\left(\mathit{l}+\mathit{b}\right)=40\mathrm{m}$
$\Rightarrow \mathit{l}+\mathit{b}=20\mathrm{m}$ .................(i)
Now, Area (square) - area (rectangle) $=9{\mathrm{m}}^2$
${\mathit{a}}^2-\mathit{l}\mathit{b}=9$ .............$\left[\mathit{a}=10\mathrm{m}\right]$
$\Rightarrow {10}^2-\mathit{l}\mathit{b}=9$
$\Rightarrow \mathit{l}\mathit{b}=100-9=91$
$\because {\left(\mathit{l}-\mathit{b}\right)}^2={\left(\mathit{l}+\mathit{b}\right)}^2-4\mathit{l}\mathit{b}$
$\therefore {\left(\mathit{l}-\mathit{b}\right)}^2={\left(20\right)}^2-4\left(91\right)$
$\Rightarrow {\left(\mathit{l}-\mathit{b}\right)}^2=400-364=36$
$\Rightarrow {\left(\mathit{l}-\mathit{b}\right)}^2=6^2$
$\Rightarrow \mathit{l}-\mathit{b}=6\mathrm{m}$ ................(ii)
Adding (i) and (ii), we get
$2\mathit{l}=26\Rightarrow \mathit{l}=13\mathit{n}$
from (i), $\mathit{b}=20-\mathit{l}=20-13=7\mathrm{m}$
So, the possible dimentions of the rectangle is
$\mathit{l}=13\mathrm{m},\mathit{b}=7\mathrm{m}$