The diagonals of a rectangle measure $\sqrt{1993} \mathrm{~cm}$. What are its dimensions, knowing that they are integers?
If $a \geq b$ are the lengths of the sides of the rectangle, then by the Pythagorean theorem we have

$$
a^{2}+b^{2}=1993
$$

Since $a^{2} \geq b^{2}$, it follows that

$$
2 a^{2} \geq a^{2}+b^{2}=1993>a^{2}
$$

Therefore,

$$
\sqrt{1993}>a \geq \sqrt{996.5}
$$

Thus, $44 \geq a \geq 32$. Using the fact that $a^{2}-(a-1)^{2}=2 a-1$, we can complete the following table by adding the number $2 a-1$ to the elements in the second column of the row $a-1$ to obtain the element in the second column of the row $a$.

| $a$ | $b^{2}=1993-a^{2}$ |
| :---: | :---: |
| 44 | 57 |
| 43 | 144 |
| 42 | 229 |
| $\vdots$ | $\vdots$ |

Thus, $a=43$ and $b=12$ is a solution.