According to the given information the sides $a$, $b$, $c$ of the described triangle is a primitive Pythagorian triple. Hence there exists two coprime positive integers $s > t$ s.t. $s - t$ is odd and $(1) \;\; (a,b,c) = (2st,s^2-t^2,s^2+t^2)$. The area $A$ and perimeter $P$ of the triangle is $(2) \;\; A = st(s^2 - t^2)$ and $(3) \;\; P = 2s(s + t)$ respectively. The ratio of the area to the perimeter is a perfect square, which according to (2) and (3) means there is a positive integer s.t. ${\textstyle n^2 = \frac{A}{P} = \frac{st(s - t)(s + t)}{2s(s + t)}}$, i.e. $(4) \;\; t(s - t) = 2n^2$. Let $A_{\min}$ be the minimal value of the area $A$. Next set $n=1$. Then $t(s - t) = 2$ by (4), yielding $(s-t,t) = (1,2)$ since $s-t$ is odd. Hence $(s,t) = (3,2)$, yielding $A = np = p = 2s(s + t) = 2 \cdot 3 (3 + 2) = 6 \cdot 5 = 30$. Hence $A_{\min} \leq 30$. Finally assume $A_{\min} < 30$. Then $n > 1$, which combined with (2) and (4) give us $A = st(s - t)(s + t) = 2n^2s(s + t) \geq 2 \cdot 2^2 \cdot 2 (2 + 1) = 16 \cdot 3 = 48$, which is impossible since $A_{\min} < 30$. This contradiction implies the minimal possible area of these triangles is 30.