Let's calculate the area of a regular star polygon if we know the distance from the line connecting two adjacent outer vertices to the farthest vertex from this line.

 untranslated portion: (Note: The original text did not contain any additional formatting or line breaks beyond the initial sentence, so the translation maintains this structure.)
The figure is shown below.

![](https://cdn.mathpix.com/cropped/2024_05_02_d858b4eddd3eb6946f1bg-1.jpg?height=485&width=545&top_left_y=213&top_left_x=779)

Given $AB = p, AC = AD = a, AO = m_1, AE = m_2$. Since $\angle AOC = \frac{360^\circ}{10} = 36^\circ$, it follows that $\angle ABC = \angle OCE = 18^\circ$, $\angle ACO = 54^\circ$, and $\angle ACE = 54^\circ - 18^\circ = 36^\circ$.

From $\triangle ABC$, $a = p \tan 18^\circ$,

from $\triangle AOC$, $m_1 = a \tan 54^\circ = p \tan 18^\circ \tan 54^\circ$,

from $\triangle AEC$, $m_2 = a \tan 36^\circ = p \tan 18^\circ \tan 36^\circ$.

The desired area $T$ is clearly five times the area of the kite $OCED$ ($t$). The area of this kite is the difference between the area of $\triangle CDO$ ($t_1$) and the area of $\triangle CDE$ ($t_2$).

But

\[
\begin{aligned}
& t_1 = a m_1 = p^2 \tan^2 18^\circ \tan 54^\circ \\
& t_2 = a m_2 = p^2 \tan^2 18^\circ \tan 36^\circ
\end{aligned}
\]

and thus

\[
T = 5t = 5(t_1 - t_2) = 5p^2 \tan^2 18^\circ (\tan 54^\circ - \tan 36^\circ)
\]

But

\[
\tan \alpha - \tan \beta = \frac{\sin \alpha}{\cos \alpha} - \frac{\sin \beta}{\cos \beta} = \frac{\sin (\alpha - \beta)}{\cos \alpha \cdot \cos \beta}
\]

and thus

\[
\tan 54^\circ - \tan 36^\circ = \frac{\sin 18^\circ}{\cos 54^\circ \cos 36^\circ} = \frac{2 \sin 18^\circ}{2 \sin 36^\circ \cos 36^\circ} = \frac{2 \sin 18^\circ}{\sin 72^\circ} = \frac{2 \sin 18^\circ}{\cos 18^\circ} = 2 \tan 18^\circ
\]

Substituting this value into (1),

\[
T = 10 p^2 \tan^3 18^\circ
\]

If the segment $BC = BD$ is taken as the unit, then $CD$ is nothing other than the side of a regular decagon inscribed in a unit circle, which, as is known, is $\frac{\sqrt{5} - 1}{2}$. Thus, $a = \frac{\sqrt{5} - 1}{4}$, and by the Pythagorean theorem,

\[
p^2 = 1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2 = \frac{10 + 2 \sqrt{5}}{16}
\]

or

\[
p = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}
\]

Therefore,

\[
\tan 18^\circ = \frac{p}{a} = \frac{\sqrt{5} - 1}{\sqrt{10 + 2 \sqrt{5}}} = \frac{\sqrt{25 - 10 \sqrt{5}}}{5}
\]

and thus

\[
\tan^3 18^\circ = \left(\frac{\sqrt{25 - 10 \sqrt{5}}}{5}\right)^3 = \frac{\sqrt{85 - 38 \sqrt{5}}}{5}
\]

Therefore,

\[
T = 10 p^2 \tan^3 18^\circ = 2 \sqrt{85 - 38 \sqrt{5}} p^2 \approx 0.343 p^2
\]

This last result can also be obtained directly without using $\tan 18^\circ$.