[ Tiling with domino bones and tiles $]$ [ Text problems (other). $]$

A square $8 \times 8$ was cut into squares $2 \times 2$ and rectangles $1 \times 4$. As a result, the total length of the cuts was 54.

How many figures of each type were obtained?
In an $8 \times 8-64$ square, there are 64 cells, and each of the resulting figures has 4 cells. Therefore, there are 16 figures in total.

Let's find the sum of the perimeters of all the resulting figures. Since each cut boundary is part of the perimeter of two figures, we add to the perimeter of the square twice the length of the cuts: $32 + 2 \cdot 54 = 140$.

The perimeter of a $2 \times 2$ square is 8, and the perimeter of a rectangle is 2 more. If all 16 figures were squares, their total perimeter would be $16 \cdot 8 = 128$, which is 12 less than the actual total. To increase the total perimeter by 12, 6 squares need to be replaced with rectangles. Therefore, there were 6 rectangles and 10 squares.

## Answer

10 squares and 6 rectangles.