Find all Pythagorean triangles whose areas are equal to their perimeters.

We need to find all Pythagorean triple $x,\ y,\ z$ such that

$$\begin{equation} x^2+y^2=z^2 \end{equation}$$

and

$$\begin{equation} \dfrac{xy}{2}=x+y+z \end{equation}$$

From $(1)$ and $(2)$, we obtain that

$${\color{#c34632}\begin{align*} x^2+y^2&=(x+y-\dfrac{xy}{2})^2\\ \cancel{x^2+y^2}&=\cancel{x^2+y^2}-\dfrac{x^2y^2}{4}+2xy-x^2y-xy^2\\ 0&=\dfrac{xy}{4}(xy+8-4y-4x)\\ 8&= (x-4)(y-4) \end{align*}}$$

So that, only possible Pythagorean triples are

$${\color{#4257b2}6 ,\ 8 ,\ 10}$$

$${\color{#4257b2} 5,\ 12,\ 13}$$