SSC CHSL 15 Nov 2015 Shift 1 Paper

Let length of rectangle = $\mathit{l}\mathrm{m}$ and breadth = $\mathit{b}\mathrm{m}$
⇒ Diagonal of rectangle, $\mathit{d}={\mathit{l}}^2+{\mathit{b}}^2$
⇒ Area = $\mathit{l}\mathit{b}=120$ -----(i)
and Perimeter = $2\left(\mathit{l}+\mathit{b}\right)=46$
⇒ $\mathit{l}+\mathit{b}=\frac{46}{2}=23$
Squaring both sides, we get
⇒ ${\left(\mathit{l}+\mathit{b}\right)}^2={\left(23\right)}^2$
⇒${\mathit{l}}^2+{\mathit{b}}^2+2\mathit{l}\mathit{b}=529$
⇒ ${\mathit{l}}^2+{\mathit{b}}^2+2\left(120\right)=529$ [Using equation (i)]
⇒ ${\mathit{l}}^2+{\mathit{b}}^2=529-240=289$
Taking square root on both sides
⇒ ${\mathit{l}}^2+{\mathit{b}}^2=\sqrt{289}=17$
∴ Leanth of diagonal = 17m