smc

We will proceed by coordinate bashing. Call the first leg $2a$, and the second leg $2b$ (We are using the double of a variable to avoid any fractions) Notice that we want to find $\sqrt{(2a{)}^2+(2b{)}^2}$ Two equations can be written for the two medians: $a^2+4b^2=40$ and $4a^2+b^2=25$. Add them together and we get $5a^2+5b^2=65$, Dividing by 5 gives $x^2+y^2=13$ Multiplying this by 4 gives $4x^2+4y^2=52⟹(2x{)}^2+(2y{)}^2=52$, just what we need to find the hypotenuse. Recall that he hypotenuse is $\sqrt{(2a{)}^2+(2b{)}^2}$. The value inside the radical is equal to $52$, so the hypotenuse is equal to $\sqrt{52}=\boxed{2\sqrt{13}}$