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Consider a square of side length ‘a’

Using Pythagoras theorem on $\Delta \,\text{QRS}$,

\[\begin{align}

& \text{S}{{\text{R}}^{\text{2}}}\,\text{+}\,\text{Q}{{\text{R}}^{\text{2}}}\,\text{=}\,\text{Q}{{\text{S}}^{\text{2}}} \\

& {{\text{a}}^{\text{2}}}\,\text{+}\,{{\text{a}}^{\text{2}}}\,\text{=}\,\text{Q}{{\text{S}}^{\text{2}}}

\end{align}\]

$\therefore $ length of diagonal $\text{QS}\,\text{=}\,\sqrt{\text{2}}\text{a}$

Now, according to the question, the diagonal of square

\[\begin{align}

& \text{ABCD}\,\text{=}\,\text{2 }\!\!\times\!\!\text{ }\,\text{diagonal}\,\text{of PQRS} \\

& \text{=}\,\text{2}\,\times \,\sqrt{2}a \\

& =\,2\sqrt{2}a.

\end{align}\]

Let us apply the Pythagoras theorem again on $\Delta \,\text{BCD}$

\[\text{C}{{\text{D}}^{\text{2}}}\,\text{+}\,\text{B}{{\text{C}}^{\text{2}}}\,\text{=}\,\text{BD}\]

But $\text{BD}\,\text{=}\,\text{2}\sqrt{2}\text{a}$, as derived before

\[\begin{align}

& \therefore \,\text{C}{{\text{D}}^{\text{2}}}\,\text{+}\,\text{B}{{\text{C}}^{\text{2}}}\,\text{=}\,\text{8}{{\text{a}}^{\text{2}}} \\

& \text{But}\,\text{CD}\,\text{=}\,\text{BC}\,\text{=}\,\text{b}\,\text{(New square side)} \\

& \text{2}{{\text{b}}^{2}}\,=\,8{{\text{a}}^{2}} \\

& \text{b}\,\text{=}\,\text{2a}

\end{align}\]

So the new square has a side that’s double the length of the old square.

Let’s finally compare areas

$\begin{align}

& \text{Area}\,\text{of}\,\text{PQRS}\,\text{=}\,{{\text{a}}^{\text{2}}} \\

& \text{Area}\,\text{of}\,\text{ABCD}\,\text{=}\,{{\text{b}}^{\text{2}}}\,\text{=}\,{{\text{(2a)}}^{\text{2}}}\,\text{=}\,\text{4}{{\text{a}}^{\text{2}}} \\

\end{align}$

\[\therefore \,\dfrac{\text{Area}\,\text{of}\,\text{ABCD}}{\text{Area}\,\text{of}\,\text{PQRS}}\,\text{=}\,\dfrac{\text{4}{{\text{a}}^{\text{2}}}}{{{\text{a}}^{\text{2}}}}\,\text{=}\,\text{4}\]

Hence

$\begin{align}

& \text{If}\,\text{a}\,\to \,\text{la}\,\text{then}\,\text{area}\,\to \,{{\text{l}}^{\text{2}}}\,\text{area} \\

& \text{If}\,\text{a}\,\to \,\dfrac{\text{l}}{\text{l}}\text{a}\,\text{then}\,\text{area}\,\to \,\dfrac{\text{l}}{{{\text{l}}^{\text{2}}}}\,\text{area} \\

\end{align}$