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From the statement i have got the area of the square S is 10. Because the square T is inscribed into the square S then why not 45 is the smallest square for T?

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Actually the area of $S$ is $100$ rather than $10$ –  Henry Sep 30 '13 at 7:34

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It rather depends on what inscribed means, but it probably means that the vertices of $T$ lie on the edges of $S$, something like this.

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$S$ is the outer square, while $T$ is the rotated square made up of a smaller square and four grey triangles. Since the area of the grey triangles is equal to the area of the green triangles, you will minimise the area of $T$ if the innermost square has zero area.

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Hint: The vertices of $T$ are on the sides of $S$, so the least possible value for the diagonal of $T$ is $10$ (the length of the sides of $S$).

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