SSC CHSL 16 Apr 2021 Shift 2 Paper

Given, sides of a triangle are $96\mathit{m},110\mathit{m}$ and $146\mathit{m}$, respectively.
Since, ${\left(96\right)}^2+{\left(110\right)}^2={\left(146\right)}^2$
$\therefore$ It proves Pythagoras theorem and it is a right angle triangle with hypotenuse $146\mathit{m}$.
Let the sides of reactangle be $3\mathit{x}\mathit{m}$ and $2\mathit{x}\mathit{m}$, respectively.
According to the question,
Area of rectangle $=$ Area of triangle
$3\mathit{x}\times 2\mathit{x}=\frac{1}{2}\times 96\times 110=5280$
$\Rightarrow {\mathit{x}}^2=880$
$\Rightarrow \mathit{x}=\sqrt{880}=4\sqrt{55}\mathit{m}$
$\therefore$ Perimeter of the rectangular park
$=2\left(3\mathit{x}+2\mathit{x}\right)$
$=2\times 5\mathit{x}=10\mathit{x}$
$=10\times 4\sqrt{55}$
$=40\sqrt{55}\mathit{m}$