Assume for the moment that $a \leq b \leq c$. So
$$a^2 + b^2 = c^2. \quad (1)$$
The condition on the area entails $ab = 2(a + b + c)$ which implies that
$$4c^2 = (ab)^2 - 4ab(a + b) + 4(a^2 + b^2 + 2ab). \quad (2)$$
Equations (1) and (2) imply that
$$ab - 4(a + b) + 8 = 0$$
or
$$ab - 4(a + b) + 16 = 8.$$
Thus
$$(a - 4)(b - 4) = 8. \quad (3)$$
Now, there are only two ways to factorise $8$ in positive integers. Thus, either $(a, b) = (5, 12)$ or $(a, b) = (6, 8)$. Thus, there are (up to permutations) two triples that satisfy the given conditions namely, $(5, 12, 13)$ and $(6, 8, 10)$. $\square$