jmc

Let the other leg have length $x$. From the area of the triangle, we have $\frac{1}{2}(24)(x)=120$, so $12x=120$ and $x=10$. Let $c$ be the hypotenuse of the triangle. The Pythagorean Theorem gives us $c^2=10^2+24^2=100+576=676$, so $c=26$. Therefore, the perimeter is $10+24+26=\boxed{60}$.