A regular hendecagon has an area of $67,707 \mathrm{~cm}^{2}$; what is the area of the annulus formed by the circles inscribed in and circumscribed around the hendecagon?





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Let the area of the polygon be $T$, the area of the annulus be $t$, and one side of the polygon be $a$.

$$
t=\pi\left(R^{2}-r^{2}\right)
$$

and since

$$
R^{2}=r^{2}+\frac{a^{2}}{4}
$$

therefore

$$
t=\frac{\pi a^{2}}{4}
$$

but

from which

$$
T=\frac{n a^{2}}{4} \operatorname{ctg} \frac{180^{\circ}}{n}
$$

$$
\frac{a^{2}}{4}=\frac{T}{n} \operatorname{tg} \frac{180^{\circ}}{n}
$$

substituting this into (1):

$$
t=\frac{T \pi}{n} \operatorname{tg} \frac{180^{\circ}}{n}
$$

Substituting the given values:

$$
t=3.468 \mathrm{~cm}^{2}
$$

(Lajos Kertész.)

The problem was also solved by: Bojedain F., Dénes A., Détshy K., Devecis M., Goldschmidt Á., Goldziher K., Kallos M., Laczkó E., Lukhaub Gy., Poos M., Probst E., Schiffer H., Spitzer Ö., Spitzer S., Szabó I., Szabó K., Weisz J.