A rectangle has a perimeter of $124 \mathrm{~cm}$. The perimeter of the rhombus determined by the midpoints of the sides is $100 \mathrm{~cm}$. What are the lengths of the sides of the rectangle?
If one side of the rectangle is $x$, then the other side is $62-x$, and according to the problem,

$$
\left(\frac{x}{2}\right)^{2}+\left(\frac{62-x}{2}\right)^{2}=\left(\frac{100}{4}\right)^{2}
$$

or

$$
x^{2}-62 x+672=0
$$

from which

$$
x_{1}=48 \mathrm{~cm}, x_{2}=14 \mathrm{~cm}
$$

Therefore, the other side of the rectangle is $14 \mathrm{~cm}$, or $48 \mathrm{~cm}$.

(Jeczuskó Imre, Eperjes.)

Number of solutions: 41.