jmc

Let the leg lengths of the right triangle be $a$ and $b.$ It follows that $\frac{ab}{2}=2(a+b).$ Expanding and moving all the terms to the left hand side, $ab-4a-4b=0.$ Adding 16 to both sides allows us to factor: $$a(b-4)-4(b-4)=(a-4)(b-4)=16.$$ From this point, the pairs $(a,b)$ that provide different areas are $(5,20),$ $(6,12),$ and $(8,8),$ and the sum of the possible areas is $50+36+32=\boxed{118}$.