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Someone posted online a primary school competition problem. I solved it using the Cayley-Menger determinant (finding the answer is 169/2 after getting the length of AD being 13) but it probably isn't the intended solution. I cannot find a "primary school" way to solve it. The problem is to find the area of the square given the following:

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The angles $\angle ABC=90^o,\angle BCD=90^o$.

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By the Pythagorean theorem (the diagram can be rotated such that the interior segments are all horizontal/vertical), the diagonal $DA$ has length $\sqrt{12^2+5^2}=13$, so the square side length is $\frac{13}{\sqrt2}$ and the area $\frac{169}2$.

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