Determine all triangles such that the lengths of the three sides and its area are given by four consecutive natural numbers.
1. Denote \( S, x, y, z \) as the area and the three sides of the triangle, respectively. Assume \( x < y < z \).
2. Let \( a, a+1, a+2, a+3 \) be the four consecutive natural numbers. We need to find \( x, y, z \) and \( S \) such that they match these four numbers.
3. If \( a = 1 \), the four numbers are \( 1, 2, 3, 4 \). Clearly, there is no triangle with sides \( 1, 2, 3 \) or any permutation of these numbers that has an area of 4.
4. If \( a = 2 \), the four numbers are \( 2, 3, 4, 5 \). Similarly, there is no triangle with sides \( 2, 3, 4 \) or any permutation of these numbers that has an area of 5.
5. Consider \( a \geq 3 \). Then we have \( x \geq 3 \).

Using Heron's formula for the area of a triangle:
\[
S = \frac{1}{4} \sqrt{(x+y+z)(x+y-z)(y+z-x)(z+x-y)}
\]

6. Since \( S = a+2 \) or \( S = a+3 \), we need to check both cases.

### Case 1: \( S = a+2 \)
7. Assume \( x = a, y = a+1, z = a+3 \). Then:
\[
S = a+2 = \frac{1}{4} \sqrt{(3a+4)(a-2)(a+4)(a+2)}
\]
8. Simplify the expression inside the square root:
\[
(3a+4)(a-2)(a+4)(a+2)
\]
9. Solving this equation for positive integer \( a \) shows that there is no solution.

### Case 2: \( S = a+3 \)
10. Assume \( x = a, y = a+1, z = a+2 \). Then:
\[
S = a+3 = \frac{1}{4} \sqrt{(3a+3)(a-1)(a+3)(a+1)}
\]
11. Simplify the expression inside the square root:
\[
(3a+3)(a-1)(a+3)(a+1)
\]
12. Solving this equation for positive integer \( a \):
\[
a+3 = \frac{1}{4} \sqrt{(3a+3)(a-1)(a+3)(a+1)}
\]
13. Testing \( a = 3 \):
\[
3+3 = \frac{1}{4} \sqrt{(3 \cdot 3 + 3)(3-1)(3+3)(3+1)}
\]
\[
6 = \frac{1}{4} \sqrt{12 \cdot 2 \cdot 6 \cdot 4}
\]
\[
6 = \frac{1}{4} \sqrt{576}
\]
\[
6 = \frac{1}{4} \cdot 24
\]
\[
6 = 6
\]
14. Therefore, \( a = 3 \) is a valid solution, and the sides of the triangle are \( 3, 4, 5 \).

The final answer is \( \boxed{3, 4, 5} \).