Problem 3. A can of paint is enough to paint a piece of cardboard in the shape of a rectangle where one side is three times larger than the other. If we make a new rectangle from the piece of cardboard, shortening the longer side by $18 \mathrm{~cm}$ and extending the other side by $8 \mathrm{~cm}$, the same amount of paint will be used for painting. Determine the perimeter of the new rectangle.
Solution. Let's denote the initial rectangle as $A B C D$, where $\overline{A B}=a$ and $\overline{B C}=b$. From the condition of the problem, we have $a=3 b$. If we denote the new rectangle as $A E F G$, then

$$
\overline{A E}=\overline{A B}=3 b-18 \text { and } \overline{E F}=\overline{B C}+8=b+8
$$

Since we will use the same amount of paint for coloring, we have

$$
P_{A B C D}=P_{A E F G}
$$

i.e.

$$
3 b^{2}=(3 b-18)(b+8)
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_a1a5b9bf3a5d296219b1g-09.jpg?height=353&width=483&top_left_y=727&top_left_x=999)

which simplifies to

$$
3 b^{2}=3 b^{2}+24 b-18 b-144
$$

From the last equation, we have $b=24 \mathrm{~cm}$.

Accordingly,

$$
L_{A E F G}=2(3 \cdot 24-18+24+8)=172 \mathrm{~cm}
$$