Task B-3.7. While setting the table for dinner, mother came up with a task to get Matko to help her. It is necessary to arrange six identical coasters in the shape of a rectangle along the edge of a round table with a radius of $8 \mathrm{dm}$. The width of each coaster is $2 \mathrm{dm}$, and the length is $x \mathrm{dm}$. Two vertices of each coaster, which are the endpoints of the side with length $x$, are located on the very edge of the table. Each two adjacent coasters share a common vertex - the one that is not on the edge of the table. Matko needs to calculate the length of the coaster $x$. What number should be obtained?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9c6c142e457d4763ad17g-14.jpg?height=738&width=764&top_left_y=1905&top_left_x=648)

Image - can be without the marked radii $AO$ and $OD$.

The inner vertices of all rectangles determine a regular hexagon, so the measure of angle $\angle CDO$ is $60^{\circ}$.

Then the measure of angle $\angle ADO$ is $90^{\circ} + 60^{\circ} = 150^{\circ}$. Apply the cosine rule to triangle $ADO$,

$$
8^{2} = 2^{2} + x^{2} - 2 \cdot 2 \cdot x \cdot \cos 150^{\circ}
$$

Then $64 = 4 + x^{2} + 2x \sqrt{3}$,

i.e., $x^{2} + 2x \sqrt{3} - 60 = 0$.

The positive solution to this equation is $x = -\sqrt{3} + 3 \sqrt{7} \text{ dm}$.

## COUNTY COMPETITION IN MATHEMATICS

## 4th grade - high school - B variant

18th February 2014.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO EVALUATE THAT APPROACH APPROPRIATELY.