Three, (15 points) The side lengths of a right-angled triangle are all integers, and its area is numerically equal to its perimeter. Determine the lengths of the three sides of this right-angled triangle.

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Three, let $a$ and $b$ be the lengths of the two legs of a right triangle, then the length of the hypotenuse $c = \sqrt{a^2 + b^2}$. Since $a$, $b$, and $c$ are all positive integers, $a \neq b$. Without loss of generality, assume $a > b$. According to the problem, we have
$$
a + b + \sqrt{a^2 + b^2} = \frac{ab}{2}.
$$

Squaring both sides and simplifying, we get $\frac{a^2 b^2}{4} - a^2 b - ab^2 + 2ab = 0$, which simplifies to $ab - 4a - 4b + 8 = 0$.
Thus, $(a-4)(b-4) = 8 = 1 \times 8 = 2 \times 4$.
Since $a$ and $b$ are positive integers and $a > b$, we have
$$
\left\{\begin{array}{l}
a - 4 = 8, \\
b - 4 = 1
\end{array} \text { or } \left\{\begin{array}{l}
a - 4 = 4, \\
b - 4 = 2
\end{array}\right.\right.
$$

Solving these, we get $a = 12, b = 5, c = 13$; or $a = 8, b = 6, c = 10$.
Therefore, the lengths of the sides of the right triangle are $(12, 5, 13)$ or $(8, 6, 10)$.