smc

Let the sides of the rectangle be x and y. WLOG, assume x>y. Then, $2x+2y=p\Rightarrow x+y=\frac{p}{2}$. By pythagorean theorem, $x^2+y^2=d^2$. Since $x+y=\frac{p}{2}$, $(x+y{)}^2=\frac{p^2}{4}\Rightarrow x^2+2xy+y^2=\frac{p^2}{4}$. Rearranging to solve for $2xy$ gives $2xy=\frac{p^2}{4}-d^2$. Rearranging $(x-y{)}^2$ in terms of the defined variables becomes $(x-y{)}^2=d^2-(\frac{p^2}{4}-d^2)$. In order to get (x-y), we have to take the square root of the expression and simplify. $(x-y)=\sqrt{2d^2-\frac{p^2}{4}}\Rightarrow (x-y)=\sqrt{\frac{8d^2-p^2}{4}}$ $\Rightarrow$ $(x-y)=\frac{\sqrt{8d^2+p^2}}{2}$ $\Rightarrow$ $\boxed{\frac{\sqrt{8d^2-p^2}}{2}}$.