Problem 3. A square is divided into five equal rectangles by four parallel lines. The sum of the perimeters of these rectangles is 96 cm. Determine the perimeter of the square.
Solution. First method. If we denote the lengths of the sides of the five identical rectangular partitions by $a$ and $b$, then the sum of the perimeters of the partitioned rectangles will be equal to $5 \cdot 2(a+b)=10 a+10 b$. Furthermore, from the condition of the problem, it follows that $a=5 b$, so $2 a=2 \cdot 5 b=10 b$. Therefore, the sum of the perimeters of the five partitioned rectangles is

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-04.jpg?height=394&width=407&top_left_y=978&top_left_x=1075)
equal to $10 a+10 b=10 a+2 a=12 a$. Therefore, $12 a=96$, i.e., $a=96: 12=8 \mathrm{~cm}$. Finally, the perimeter of the original square is equal to $4 a=4 \cdot 8=32 \mathrm{~cm}$.

Second method. From the condition of the problem, it follows that each of the partitioned segments participates twice in the sum of the perimeters of the partitioned rectangles (once in the perimeter of the upper and once in the perimeter of the lower rectangle). If we denote the length of the side of the square by $a$, then the perimeter of the square is $L=4 a$, and the sum of the perimeters

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-04.jpg?height=385&width=418&top_left_y=1504&top_left_x=1065)
of the partitioned rectangles is $L^{\prime}=L+4 \cdot 2 a=L+2 \cdot 4 a=L+2 L=3 L$. Therefore, $3 L=96$, from which we get $L=32 \mathrm{~cm}$.

Note. The second method of solving can also be applied when the partitioned rectangles are not identical, in which case it is again obtained that the perimeter of the original square is $L=32 \mathrm{~cm}$.