10.147. In a parallelogram with a perimeter of 32 cm, the diagonals are drawn. The difference between the perimeters of two adjacent triangles is 8 cm. Find the lengths of the sides of the parallelogram.
Solution.

Let the sides of the parallelogram be denoted by $a$ and $b$. Then its perimeter is $p=2(a+b)=32$. The perimeter of one triangle is $p_{1}=b+\frac{d_{1}}{2}+\frac{d_{2}}{2}$, and the second is $p_{2}=a+\frac{d_{1}}{2}+\frac{d_{2}}{2}$, where $d_{1}, d_{2}$ are the diagonals of the parallelogram. The difference $p_{1}-p_{2}=b-a=8$. We obtain the system: $\left\{\begin{array}{l}a+b=16, \\ b-a=8 .\end{array}\right.$ Solving it, we find $a=4, b=12$.

Answer: 12 cm, 4 cm.