Libor divided a rectangle into two smaller rectangles with one line. The perimeter of the large rectangle is $76 \mathrm{~cm}$, and the perimeters of the smaller rectangles are $40 \mathrm{~cm}$ and $52 \mathrm{~cm}$. Determine the dimensions of the large rectangle.

(L. Šimünek)

![](https://cdn.mathpix.com/cropped/2024_04_17_ca6f62942b009cd67ccfg-2.jpg?height=214&width=736&top_left_y=453&top_left_x=660)
Let the common side of the two smaller rectangles be denoted as $a$, and these two rectangles are redrawn separated from each other.
![](https://cdn.mathpix.com/cropped/2024_04_17_ca6f62942b009cd67ccfg-2.jpg?height=214&width=872&top_left_y=891&top_left_x=593)

The sum of the perimeters of the two smaller rectangles is $40+52=92(\mathrm{~cm})$, which is $16 \mathrm{~cm}$ more than the perimeter of the original rectangle, since $92-76=16$. These $16 \mathrm{~cm}$ correspond to two sides $a$, so the side $a$ has a length of $16: 2=8(\mathrm{~cm})$.

The same length applies to two sides of the original rectangle, the sum of the lengths of its remaining two sides is $76-2 \cdot 8=60(\mathrm{~cm})$, each of them is therefore $60: 2=30(\mathrm{~cm})$ long. The dimensions of the original large rectangle are $30 \mathrm{~cm}$ and $8 \mathrm{~cm}$.

Evaluation. 4 points for the shorter side of the rectangle (2 points for justification); 2 points for the longer side of the rectangle (1 point for justification).