From four photographs, three different rectangles can be formed (see figure). The perimeter of one of them is 56 cm. Find the perimeters of the other two rectangles, if the perimeter of the photograph is 20 cm.
![](https://cdn.mathpix.com/cropped/2024_05_06_297c289048f559686628g-20.jpg?height=366&width=944&top_left_y=1018&top_left_x=562)
The width and height of the rectangle on the left image are twice the width and height of the photo, respectively. Therefore, its perimeter is twice the perimeter of the photo, which is 40 cm.

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

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

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