In a regular dodecagon $A B C D E F G H I J K L$ inscribed in a circle with radius $6 \mathrm{~cm}$, determine the perimeter of the pentagon $A C F H K$.

(K. Pazourek)

![](https://cdn.mathpix.com/cropped/2024_04_17_aa8865afb54b9c300d40g-4.jpg?height=565&width=577&top_left_y=948&top_left_x=745)

Idea. Divide the pentagon into triangles with a common vertex at the center of the circumscribed circle and determine their properties.
First, let's calculate the size of the central angle of the regular dodecagon \(A B C D E F G H I J K L\), i.e., the angle with its vertex at the center \(S\) of the circumscribed circle and its sides passing through adjacent vertices. The sum of all twelve central angles is \(360^\circ\), so the size of one central angle is \(360^\circ : 12 = 30^\circ\).

![](https://cdn.mathpix.com/cropped/2024_04_17_aa8865afb54b9c300d40g-4.jpg?height=574&width=577&top_left_y=1917&top_left_x=745)

The sizes of the central angles of the pentagon \(A C F H K\) are:

\[
\begin{gathered}
|\angle A S C| = |\angle F S H| = |\angle K S A| = 2 \cdot 30^\circ = 60^\circ, \\
|\angle C S F| = |\angle H S K| = 3 \cdot 30^\circ = 90^\circ.
\end{gathered}
\]

Triangles \(A S C\), \(F S H\), and \(K S A\) are thus equilateral, and triangles \(C S F\) and \(H S K\) are isosceles right triangles. Therefore,

\[
\begin{gathered}
|A C| = |F H| = |K A| = r = 6 \text{ cm}, \\
|C F| = |H K| = r \sqrt{2} = 6 \sqrt{2} \text{ cm}.
\end{gathered}
\]

The perimeter of the pentagon \(A C F H K\) is thus \(3 \cdot 6 + 2 \cdot 6 \sqrt{2} = 18 + 12 \sqrt{2} \approx 34.97\) (cm).