Task B-2.8. A path $2 \mathrm{~m}$ wide and with an area of $36 \mathrm{~m}^{2}$ has been built around a swimming pool in the shape of a regular hexagon. What is the perimeter of the pool?
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_247a158c2fc9e6452974g-09.jpg?height=767&width=851&top_left_y=1297&top_left_x=583)

$x=2 \text{ m}, P=36 \text{ m}^2$.

The path consists of 6 isosceles trapezoids with bases $a$ and $b$, and height $x$. The area of one trapezoid is $6 \text{ cm}^2$.

$$
P_{t}=\frac{a+b}{2} \cdot x \quad \Rightarrow \quad a+b=6
$$

The area of the trapezoid is the difference in the areas of equilateral triangles with sides $b$ and $a$.

$$
\begin{aligned}
& \frac{b^{2} \sqrt{3}}{4}-\frac{a^{2} \sqrt{3}}{4}=6 \\
& b^{2}-a^{2}=8 \sqrt{3} \\
& (b-a) \cdot(b+a)=8 \sqrt{3} \\
& (b-a) \cdot 6=8 \sqrt{3} \\
& b-a=\frac{4 \sqrt{3}}{3}
\end{aligned}
$$

By solving the system of equations: $a+b=6$ and $b-a=\frac{4 \sqrt{3}}{3}$, we get $a=\frac{9-2 \sqrt{3}}{3} \text{ m}$.

(2 points)

The perimeter of the pool is $6a$.

$$
O=18-4 \sqrt{3} \text{ m}
$$

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

January 24, 2011.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE AND GRADE THAT APPROACH APPROPRIATELY.