The diagonals of a rectangle measure $\sqrt{1993} \mathrm{~cm}$. What are the dimensions of the rectangle, knowing that they are integers?
Let $a$ and $b$ be the lengths of the sides of the rectangle. Assuming $a \leq b$, we have $b^{2}<a^{2}+b^{2} \leq 2 b^{2}$, since $a^{2} \leq b^{2}$. The diagonals of the rectangle measure $\sqrt{1993}$, so by the Pythagorean Theorem, $a^{2}+b^{2}=1993$. Thus, $b^{2}<1993 \leq 2 b^{2}$, or $996.5 \leq b^{2}<1993$. Therefore,

$$
31<\sqrt{996.5} \leq b<\sqrt{1993}<45
$$

Since $b$ is an integer, $32 \leq b \leq 44$. Additionally, $a \leq b$ and since $a$ is also an integer, $1993-b^{2}=a^{2}$ must be a perfect square.

To calculate the values of $a^{2}=1993-b^{2}$, with $b$ ranging from 32 to 44, we first calculate $1993-32^{2}=969$ and observe that

$$
\begin{aligned}
1993-33^{2} & =1993-(32+1)^{2}=\left(1993-32^{2}\right)-(2 \times 32+1) \\
& =969-65=904
\end{aligned}
$$

and, in general, $1993-(n+1)^{2}=\left(1993-n^{2}\right)-(2 n+1)$. Thus, it is easy to obtain the following table.

| $b$ | $a^{2}$ | $b$ | $a^{2}$ | $b$ | $a^{2}$ | $b$ | $a^{2}$ | $b$ | $a^{2}$ | $b$ | $a^{2}$ | $b$ | $a^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 32 | 969 | 33 | 904 | 34 | 837 | 35 | 768 | 36 | 697 | 37 | 624 | 38 | 549 |
| 39 | 472 | 40 | 393 | 41 | 312 | 42 | 229 | 43 | 144 | 44 | 57 |  |  |

Since 144 is the only perfect square in the list, the only solution is a rectangle with sides measuring 43 and $12 \text{ cm}$.