## Task B-4.3.

In a water park, there is a large pool in the shape of an octagon that can be circumscribed by a circle with a diameter of 15 meters. It is known that every other side of the pool is 9 meters long, and the remaining four sides are equal but of unknown length. The park owners need to order a tarp to cover the pool. What is the minimum area that this tarp must have, and what is the length of the unknown side of the pool?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-18.jpg?height=479&width=474&top_left_y=660&top_left_x=720)

$$
4 \alpha + 4 \beta = 360^{\circ} \Rightarrow \alpha + \beta = 90^{\circ}
$$

By the cosine rule, we have

$$
\begin{aligned}
& \cos \beta = \frac{7.5^2 + 7.5^2 - 9^2}{2 \cdot 7.5 \cdot 7.5} = \frac{7}{25} \\
& \Rightarrow \sin \beta = \cos \alpha = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \frac{24}{25}
\end{aligned}
$$

The area of the given octagon is four times the area of an isosceles triangle with a base of length 9 and four times the area of an isosceles triangle with a base of length \( x \):

$$
P = 4 \cdot \frac{1}{2} \left( r^2 \sin \alpha + r^2 \sin \beta \right) = 2 r^2 (\cos \beta + \sin \beta) = 2 \cdot 7.5^2 \cdot \left( \frac{7}{25} + \frac{24}{25} \right) = \frac{225}{2} \cdot \frac{31}{25} = \frac{279}{2}.
$$

The length of the unknown side can also be calculated using the cosine rule.

$$
x^2 = 7.5^2 + 7.5^2 - 2 \cdot 7.5 \cdot 7.5 \cdot \cos \alpha = \frac{9}{2} \Rightarrow x = \frac{3 \sqrt{2}}{2} \text{ m}
$$