Right triangle ABC .

Let the two sides of the triangle be $a$ and $b$ . Let $c$ be the hyotenuse. Then $a^2 + b^2 = c^2 ..... (1)$

Its Given that $\frac{1}{2}ab = a + b + c = 24 ..... (2)$ $\therefore a + b = 24 - c$ $(a + b)^2 = (24 - c)^2$ $a^2 + b^2 +2ab = 24^2 - 2.24.c + c^2$ $c^2 +24.4 = 576 - 48c + c^2$
$48c = 576 - 96$ $48c = 480$ $c = 10$

Since the perimeter is a whole number, we should first consider that all sides are integers. Since the area of the right triangle is 24, the 2 sides (excluding the hypotenuse) must multiply to be 48, since 24=1/2 * ab, ab=48. Considering the prime factorization of 48, 2^4 * 3, we have to create two factors of 48 that would allow a familiar pythagorean triple. In this example, 2*3 and 2^3 (6 and 8) do multiply to 48, and would have a hypotenuse of 10. 6 + 8 + 10 = 24, so the perimeter condition is also satisfied.