The parallelogram, the perimeter of which is 28 cm, is divided by diagonals into 4 triangles.

The parallelogram, the perimeter of which is 28 cm, is divided by diagonals into 4 triangles. The difference between the perimeters of the two triangles is 2 cm. Find the sides of the parallelogram.

Let us denote the lengths of the sides of the parallelogram by a and b.
Then a + b + a + b = 28 => 2 * a + 2 * b = 28 => a + b = 14.
The diagonals divide the parallelogram into 4 pairwise equal triangles.
Since the difference between the perimeters of two triangles is 2 cm, then these triangles are not equal.
Let it be 2 triangles highlighted in the figure (see attachment).
It is known that the diagonals of a parallelogram are halved at the intersection point. Another side of the triangles is common (or they are as equal as the cut off parts of the diagonal).
Then it turns out that with the difference in perimeters, two of the three sides of the triangles cancel out, and the difference between the third sides (sides of the parallelogram) is 2.
Hence, let a = b + 2, then:
a + b = 14 => b + 2 + b = 14 => 2b = 12 => b = 6;
a = b + 2 = 6 + 8 = 14.
Answer: the sides of the parallelogram are 6, 8, 6, 8.