imc

Let the legs of the triangle have lengths $a,b$. Then, by the Pythagorean Theorem, the length of the hypotenuse is $\sqrt{a^2+b^2}$, and the area of the triangle is $\frac{1}{2}ab$. So we have the two equations $$\begin{gathered} a+b+\sqrt{a^2+b^2}=32 \\ \frac{1}{2}ab=20 \end{gathered}$$ Re-arranging the first equation and squaring, $$\begin{gathered} \sqrt{a^2+b^2}=32-(a+b) \\ {a}^2+b^2=32^2-64(a+b)+(a+b{)}^2 \\ {a}^2+b^2+64(a+b)=a^2+b^2+2ab+32^2 \\ a+b=\frac{2ab+32^2}{64} \end{gathered}$$ From $(2)$ we have $2ab=80$, so $a+b=\frac{80+32^2}{64}=\frac{69}{4}.$ The length of the hypotenuse is $p-a-b=32-\frac{69}{4}=\frac{59}{4}\boxed{\frac{59}{4}}$.