jmc

Let the side length of the larger square be $x$ and the side length of the smaller square be $y$. We are told $x^2+y^2=65$ and $x^2-y^2=33$. Adding these two equations gives $2x^2=98$, so $x^2=49$. Since $x$ must be positive, we have $x=7$. Substituting this into either equation above gives us $y^2=16$. Since $y$ must be positive, we have $y=4$. The perimeter of the larger square is $4x$ and that of the smaller square is $4y$, so the sum of their perimeters is $4x+4y=4(x+y)=\boxed{44}$.