A rhombus has a perimeter of $100 \mathrm{~cm}$, and the difference between its diagonals is $34 \mathrm{~cm}$. What are the lengths of the diagonals?
If one diagonal is $x$; then the other is $x+34$ and

$$
\left(\frac{x}{2}\right)^{2}+\left(\frac{x+34}{2}\right)^{2}=\left(\frac{100}{4}\right)^{2}
$$

or

$$
x^{2}+34 x-672=0
$$

from which

$$
x_{1}=14, \quad x_{2}=-48
$$

14, So one diagonal of the rhombus is $14 \mathrm{~cm}$, the other is $48 \mathrm{~cm}$.

(Pál Sárközy, Pannonhalma.)

Number of solutions: 24.