Let's determine those right-angled triangles whose side lengths are integers, and the measure of their perimeter and area are equal.
I. solution. Let the sides of the triangle be $a \leq b$ (there are no sides of 0 units or negative sides) |
| -2 | $\overline{-4}$ | 2 | $\overline{0}$ |  |
| -4 | -2 | 0 | $\overline{2}$ |  |
| -8 | -1 | -4 | 3 |  |

The sides of the triangle are $5,12,13$ or $6,8,10$ units long.

II. solution. Let the sides of the triangle be $a \leq b < c$ (there are no sides of 0 units or negative sides). For the case $a+b > 0$, $\left.(a+b)^{2}>a^{2}+b^{2},\right)$

$$
\frac{a b}{2} \leq a+b+a+b \leq 4 b
$$

Thus, $\frac{a b}{2} \leq 4 b$, which means $a \leq 8$. Therefore, the smallest term of the Pythagorean triple we are looking for cannot be greater than 8. All such triples are:

| $a$ | $b$ | $c$ | $\frac{a b}{2}$ | $a+b+c$ |
| :---: | :---: | :---: | :---: | :---: |
| 3 | 4 | 5 | 6 | 12 |
| 5 | 12 | 13 | 30 | 30 |
| 6 | 8 | 10 | 24 | 24 |
| 7 | 24 | 25 | 84 | 56 |
| 8 | 15 | 17 | 60 | 40 |

Among these, two suitable triples are: $5,12,13$ and $6,8,10$.