## 7. Area of a Rectangle

A square is divided into eight rectangles as shown in the figure. The sum of the perimeters of these eight rectangles is 288 cm. Determine the area of the highlighted rectangle in square centimeters, if its longer side is three times the length of the shorter side, and the shorter side is four times shorter than the side of the square.

Result: $\mathbf{1 0 8}$

![](https://cdn.mathpix.com/cropped/2024_05_30_d078161e0adaef934ec1g-5.jpg?height=386&width=377&top_left_y=218&top_left_x=1473)
## Solution.

On the diagram, we observe rectangles labeled 1, 2, 3, 4, and 5. Rectangles 1 and 2 share one common side. The sum of the lengths of the right sides of rectangles 1, 2, 3, and 4 is equal to the length of the side of the drawn square. The side of rectangle 1 is part of the side of rectangle 5, and the sum of the lengths of the lower sides of rectangle 1 and rectangle 5 is equal to the length of the side of the drawn square. The sides of the rectangles that are inside the square (colored red, blue, green, and brown on the diagram) are used twice in the sum of the perimeters of the rectangles.

![](https://cdn.mathpix.com/cropped/2024_05_30_d078161e0adaef934ec1g-5.jpg?height=397&width=437&top_left_y=818&top_left_x=1415)
Let \( a \) be the length of the side of the square. Then the perimeter of all rectangles is equal to \( 4a + 2 \cdot 4a = 12a \), from which we get \( a = 24 \, \text{cm} \). The length of the side of the square is \( 24 \, \text{cm} \).

Let \( b \) be the length of the shorter, and \( c \) the length of the longer side of the highlighted rectangle. Then

\[
b = a : 4 = 24 : 4 = 6 \, \text{cm} \quad \text{and} \quad c = b \cdot 3 = 6 \cdot 3 = 18 \, \text{cm}
\]

The lengths of the sides of the highlighted rectangle are \( 6 \, \text{cm} \) and \( 18 \, \text{cm} \), and its area is \( 6 \, \text{cm} \cdot 18 \, \text{cm} = 108 \, \text{cm}^2 \).