Task B-1.4. In a circle of radius $r$, two parallel chords $\overline{A B}$ and $\overline{C D}$ are drawn on the same side of the center. $\overline{A B}$ is a side of an equilateral triangle inscribed in the circle, and $\overline{C D}$ is a side of a regular hexagon inscribed in the same circle. Determine the perimeter and area of trapezoid $A B C D$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6bd7d8198e5de61c26bbg-03.jpg?height=593&width=562&top_left_y=1422&top_left_x=749)

The side length of a regular hexagon is equal to the radius of the circle in which the hexagon is inscribed, i.e., \( |DC| = r \).

The radius of the circumscribed circle of an equilateral triangle is

\[
r = \frac{|AB| \sqrt{3}}{3} \Rightarrow |AB| = r \sqrt{3}
\]

Let the perpendicular from \( S \) to \( CD \) intersect \( CD \) at \( F \) and \( AB \) at \( E \).

The height of the equilateral triangle is \( |EH| = \frac{|AB| \sqrt{3}}{2} = \frac{3}{2} r \).

The height of the characteristic triangle in the hexagon is \( |SF| = \frac{r \sqrt{3}}{2} \).

The height of the trapezoid is \( |EF| = |SF| - \frac{1}{3} |EH| = \frac{r}{2} (\sqrt{3} - 1) \), so

\[
|AD| = |BC| = \sqrt{\left(\frac{|AB| - |DC|}{2}\right)^2 + |EF|^2} = \sqrt{\left(\frac{r \sqrt{3} - r}{2}\right)^2 + \frac{r^2}{4} (\sqrt{3} - 1)^2} = \frac{r (\sqrt{3} - 1) \sqrt{2}}{2}
\]

The perimeter of the trapezoid is

\[
o = |AB| + |DC| + 2|AD| = r (\sqrt{3} + 1 + \sqrt{6} - \sqrt{2})
\]

The area of the trapezoid is

\[
P = \frac{|AB| + |CD|}{2} \cdot |EF| = \frac{r \sqrt{3} + r}{2} \cdot \frac{r}{2} (\sqrt{3} - 1) = \frac{r^2}{2}
\]