Jirka had three identical rectangles. First, he placed them together as shown in the picture and got a rectangle with a perimeter of $20 \mathrm{~cm}$. Then he placed them together differently and got a rectangle with a different perimeter. What was the perimeter of this rectangle?

(When placing the rectangles together, no two rectangles overlapped and there were no gaps between them.)

(E. Novotná)

![](https://cdn.mathpix.com/cropped/2024_04_17_c6cec8a95bc7df296d45g-1.jpg?height=305&width=211&top_left_y=1401&top_left_x=928)
From the assembled shape, we see that the longer side of each of the three Jirka's rectangles is as long as two shorter sides. The large rectangle in the picture has a perimeter composed of three longer and four shorter sides of the smaller rectangles. Since the sum of the three longer sides is the same as the sum of six shorter ones, the perimeter of the large rectangle is equal to $6+4=10$ shorter sides of the small rectangle. Since the large rectangle has a perimeter of $20 \mathrm{~cm}$, the shorter side of the small rectangle is equal to 20: $10=2(\mathrm{~cm})$. Therefore, the small rectangles have dimensions $2 \mathrm{~cm} \times 4 \mathrm{~cm}$.

The three given rectangles can be arranged according to the given rules in the following additional ways:
![](https://cdn.mathpix.com/cropped/2024_04_17_c6cec8a95bc7df296d45g-1.jpg?height=216&width=1108&top_left_y=2216&top_left_x=474)
a) In the first case, the perimeter of the large rectangle is equal to the sum of 6 shorter and 2 longer sides of the small rectangle, i.e., $6 \cdot 2+2 \cdot 4=20(\mathrm{~cm})$.

b) In the second case, the perimeter of the large rectangle is equal to the sum of 6 longer and 2 shorter sides of the small rectangle, i.e., $6 \cdot 4+2 \cdot 2=28(\mathrm{~cm})$.

In the first case, the perimeter is the same as the perimeter of the rectangle from the problem statement, while in the second case, it is different. The newly formed Jirka's rectangle had a perimeter of $28 \mathrm{~cm}$.

Scoring suggestion. 2 points for discovering that one dimension of the small rectangle is twice as large as the other; 2 points for calculating the dimensions of the small rectangle; 1 point for a rectangle with a different perimeter; 1 point for its perimeter.

Discussion of option a) is not a mandatory part of the solution.