\(\frac{{Diagonal^2}}{2} = Area of square\)

So for large square of side 1 inch

\(\frac{{Diagonal^2}}{2} = 1\) ( area of the large square with side 1 inch)

or \(Diagonal = \sqrt2\)

Now for small square

\(\frac{{Diagonal^2}}{2} = \frac{1}{2}\) ( area of the small square = \(\frac{1}{2}\) the area of larger square)

or \(Diagonal = 1\)

So the diagonal of the larger square is longer than the diagonal of the smaller square = \(\sqrt2 - 1\)
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