The sum of the areas of two squares is 640 m^2. If the difference in their perimeters is 64 m, find the sides of the two squares.

Let the sides of the two squares be \(\displaystyle x \mathrm{~m}\) and \(\displaystyle y \mathrm{~m}(x>y)\) \[x^{2}+y^{2}=640 \] and \(\displaystyle 4 x-4 y=64 \Rightarrow y=x-16\) \[\therefore x^{2}+(x-16)^{2}=640 \] \(\displaystyle \Rightarrow x^{2}-16 x-192=0\) \[\Rightarrow(x-24)(x+8)=0 \] \(\displaystyle \therefore x=24\) \(\displaystyle x=-8\) (Rejected) \[\Rightarrow y=24-16=8 \] Hence the sides of the two squares are $\displaystyle 24$ m and $\displaystyle 8$ m