The rectangle $A B C D$ is divided into five equal rectangles. If the perimeter of $A B C D$ is $20 \mathrm{~cm}$, determine its area.

![](https://cdn.mathpix.com/cropped/2024_05_01_80f7f1185898b39e7876g-04.jpg?height=414&width=499&top_left_y=2032&top_left_x=687)

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Solution

Let $x$ and $y$, with $x < y$, be the dimensions of each of the smaller rectangles. Thus, the perimeter of rectangle $ABCD$ is $5x + 4y = 20 \text{ cm}$. Additionally, comparing the dimensions of the sides of the smaller rectangles, we have $3x = 2y$. Therefore, $20 = 5x + 6x$ and $x = 20 / 11 \text{ cm}$. Consequently, $y = \frac{3x}{2} = \frac{30}{11} \text{ cm}$ and

$$
A_{ABCD} = 3x(x + y) = 3 \cdot \frac{20}{11} \cdot \frac{50}{11} = \frac{3000}{121} \text{ cm}^2
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_80f7f1185898b39e7876g-05.jpg?height=459&width=534&top_left_y=770&top_left_x=864)