Sum of the areas of $2$ squares is $544$ m$^2$. If the difference of their perimeter is $32$ m, find the sides of two squares.

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Step 1: Let $a$ and $b$ be the sides of two squares, with $a > b$. Then $a^2 + b^2 = 544$ and $4a - 4b = 32 \Rightarrow a - b = 8 \Rightarrow a = b + 8$.
Step 2: $(b + 8)^2 + b^2 = 544 \Rightarrow 2b^2 + 16b + 64 = 544 \Rightarrow b^2 + 8b - 240 = 0$
Step 3: $(b + 20)(b - 12) = 0 \Rightarrow b = 12$ (rejecting negative value)
Step 4: $b = 12$ m, $a = 12 + 8 = 20$ m

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