3 [ Text problems (other) $\quad]$

Four photographs can be arranged to form three different rectangles (see fig.). The perimeter of one of them is 56 cm. Find the perimeters of the other two rectangles, if the perimeter of a photograph is 20 cm.
![](https://cdn.mathpix.com/cropped/2024_05_06_53798a1e86d520071242g-07.jpg?height=362&width=958&top_left_y=2380&top_left_x=548)
The width and height of the rectangle on the left are twice the width and height of the photograph, respectively. Therefore, its perimeter is twice that of the photograph, which is 40 cm.

The width of the rectangle in the center is four times the width of the photograph, while the height is the same. Therefore, its perimeter is the sum of the perimeter of the photograph and six times its width.

The height of the rectangle on the right is four times the height of the photograph, while the width is the same. Therefore, its perimeter is the sum of the perimeter of the photograph and six times its height.

Let 56 cm be the perimeter of the second rectangle. Since the perimeter of the photograph is 20 cm, then 36 cm is six times its width. Therefore, in this case, the width of the photograph is 6 cm, and its height is 4 cm.

Therefore, the perimeter of the third rectangle is $20 + 6 \cdot 4 = 44$ (cm).

If 56 cm is the perimeter of the third rectangle, then the height and width will switch "roles," so the perimeter of the second rectangle will now be 44 cm.

## Answer

40 cm and 44 cm.