Golenev-Kumuzoova et al.

Yura drew a rectangle (along the cells) on graph paper and painted a picture on it. After that, he drew a frame one cell wide around the picture (see fig.). It turned out that the area of the picture is equal to the area of the frame. What could be the dimensions of Yura's picture?

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-34.jpg?height=432&width=357&top_left_y=2288&top_left_x=862)
Obviously, the width of the picture is more than one cell. Let's draw another frame one cell wide inside the picture (see fig.). Then, in the small frame, as in the large one, there will be four corner cells (they are shaded), and each side will be two cells shorter. Therefore, there will be 8 cells less in the small frame than in the large one (these cells are hatched). This means that the rectangle formed inside the small frame will consist of 8 cells. It is obvious that a rectangle with an area of 8 cells can have dimensions $2 \times 4$ or $1 \times 8$. Hence, the answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-35.jpg?height=651&width=535&top_left_y=292&top_left_x=753)

## Answer

$3 \times 10$ or $4 \times 6$ cells.