Problem 5. The figure on the right consists of three squares. The vertices of the middle square are the midpoints of the sides of square $ABCD$. The sum of the perimeters of the smallest and the largest square is $60 \mathrm{~cm}$ and the perimeter of the smallest square is four times smaller

![](https://cdn.mathpix.com/cropped/2024_06_05_b353751bdd87cc544500g-06.jpg?height=249&width=255&top_left_y=981&top_left_x=1209)
than the perimeter of the largest square. What is the area of the shaded part of the figure, i.e., the area of the figure bounded by the middle and the smallest square?
Solution. Let $M, N, K, L$ be the vertices of the middle square. Since the perimeter of the smallest square is four times smaller than the perimeter of the largest square, it follows that the perimeter of the smallest square is $60: 5=12 \mathrm{~cm}$, and that of the largest square is $12 \cdot 4=48 \mathrm{~cm}$. The length of the side of the smallest square is

![](https://cdn.mathpix.com/cropped/2024_06_05_b353751bdd87cc544500g-06.jpg?height=273&width=274&top_left_y=1380&top_left_x=1206)
$12: 4=3 \mathrm{~cm}$ and its area is $3^{2}=9 \mathrm{~cm}^{2}$. The length of the side of the largest square is $48: 4=12 \mathrm{~cm}$ and its area is $12^{2}=144 \mathrm{~cm}^{2}$. The area of the right triangle $L A M$ is $\frac{6 \cdot 6}{2}=18 \mathrm{~cm}^{2}$, which is the same as the area of each of the other three triangles. The area of the middle square $M N K L$ is equal to the difference between the area of the largest triangle and the sum of the areas of the four triangles, i.e., it is equal to $144-4 \cdot 18=72 \mathrm{~cm}^{2}$. It follows that the area of the shaded part is $72-9=63 \mathrm{~cm}^{2}$.

## VII Section