Problem 5. Given a rectangle with sides $a \mathrm{~cm}$ and $b \mathrm{~cm}$. If the side with length $a \mathrm{~cm}$ is extended by $b \mathrm{~cm}$ and the side with length $b \mathrm{~cm}$ is extended by $a \mathrm{~cm}$, a square with area $100 \mathrm{~cm}^{2}$ is obtained. If the lengths of the sides of the given rectangle are natural numbers, determine the rectangle with the smallest area.
Solution. The area of the obtained square is $P_{\kappa B}=(a+b)^{2}=100 \mathrm{~cm}^{2}$, so $a+b=10 \mathrm{~cm}$. Since the lengths of the sides of the rectangle are natural numbers, for $a$ and $b$ the following pairs of values are possible:

$$
\begin{aligned}
& a=1, b=9 ; a=2, b=8 ; \quad a=3, b=7 ; \quad a=4, b=6 ; a=5, b=5 \\
& a=6, b=4 ; \quad a=7, b=3 ; \quad a=8, b=2 \quad \text { or } a=9, b=1
\end{aligned}
$$

The rectangle with the smallest area $P_{n p}=a b$, is obtained for $a=1, b=9$ or $a=9, b=1$ and is $P_{n p}=9 \mathrm{~cm}^{2}$.

## VI section