Petr had a rectangle with a width of $2 \mathrm{~cm}$ and an unknown length. Radka had a rectangle with a width of $2 \mathrm{~cm}$, whose length was equal to the perimeter of Petr's rectangle. When they placed the rectangles together by their widths, they obtained a new rectangle with a perimeter of $63 \mathrm{~cm}$.

Determine the area of Petr's rectangle.

(K. Pazourek)

Hint. Express the perimeters of the rectangles using the length of Petr's rectangle.
Let the unknown length of Petr's rectangle be denoted as $d$. The perimeter of Petr's rectangle was $4+2d(\mathrm{~cm})$, which was the length of Rad's rectangle.

![](https://cdn.mathpix.com/cropped/2024_04_17_d9ed5968656390921398g-1.jpg?height=231&width=1382&top_left_y=2146&top_left_x=340)

The combined rectangle had a perimeter of $6d+12=63(\mathrm{~cm})$. From this, we get $6d=63-12=51(\mathrm{~cm})$, and thus $d=51:6=17:2(\mathrm{~cm})$.

Petr's rectangle had an area of $2 \cdot d=17\left(\mathrm{~cm}^{2}\right)$.

Note. For any rectangle with sides 2 and $a$, the relationship between its area $(S=2a)$ and perimeter $(o=4+2a)$ can be expressed as $S=o-4$.

From the previous illustration, it is clear that the area of Rad's rectangle is $8 \mathrm{~cm}^{2}$ larger than twice the area of Petr's rectangle, which we denote as $P$. Therefore, the area of the combined rectangle is

$$
3P+8=63-4
$$

From this, we get $3P=63-4-8=51$, and thus $P=17\left(\mathrm{~cm}^{2}\right)$.