TS ICET 23 May 2018 Shift 1 Paper

Let the side of the square PQRS and that of the second square be ${\mathit{s}}_1$ and S respectively.
The area of the second square = twice area of PQRS$\Rightarrow {\mathit{S}}^2=2{\left({\mathit{s}}_1\right)}^2=\mathit{S}\sqrt{2}{\mathit{s}}_1$
The diagonal of PQRS$=\sqrt{2}{\mathit{s}}_1=\mathit{a}+\mathit{b}$
Hence $\mathit{S}=\mathit{a}+\mathit{b}$
Perimeter of the second square$=4\mathit{S}=4\left(\mathit{a}+\mathit{b}\right)$