The flag in the figure is formed by a rectangle $40 \mathrm{~cm} \times 20 \mathrm{~cm}$ and has two colors (white and gray). The white part is composed of two stripes of the same width that cross each other and are perpendicular to the sides of the rectangle. The gray part is the area of the rectangle that was not covered by the stripes. What should be the width of the stripes so that the white area is equal to the gray area?

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-23.jpg?height=386&width=576&top_left_y=780&top_left_x=820)

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Solution

Let $x$ be the width of the stripes. Thus, one of the stripes measures $x$ by $40 \mathrm{~cm}$, while the other measures $x$ by $20 \mathrm{~cm}$. If we add the areas of the two stripes, we will be counting their intersection twice and, therefore, will need to subtract it. Since the area of the stripes should be half the area of the flag, we have:

$$
\begin{aligned}
40 x+20 x-x^{2} & =\frac{40 \cdot 20}{2} \\
60 x-x^{2} & =400 \\
x^{2}-60 x+400 & =0 \\
x & =\frac{60 \pm \sqrt{60^{2}-4 \cdot 1 \cdot 400}}{2} \\
& =\frac{60 \pm \sqrt{2000}}{2} \\
& =30 \pm 10 \sqrt{5}
\end{aligned}
$$

Since $0<x<20$, the width of the stripes is $x=(30-10 \sqrt{5}) \mathrm{cm}$.