10.339. The center of an equilateral triangle with a side length of 6 cm coincides with the center of a circle with a radius of 2 cm. Determine the area of the part of the triangle that lies outside this circle.
Solution.

Let $O$ be the center of the equilateral $\triangle ABC$ (Fig. 10.124), $AB=6$ cm. The radius $r_{1}=2$ cm of the circle mentioned in the problem satisfies the inequality $r<r_{1}<R$, where $r=\sqrt{3}$ cm is the radius of the inscribed circle, and $R=2\sqrt{3}$ cm is the radius of the circumscribed circle of $\triangle ABC$, so the circle intersects the sides of $\triangle ABC$. $D$ and $E$ are the points of intersection of the circle with side $AB$ of the triangle, and $K$ is the midpoint of $DE$. Then $OK \perp DE$, $OK=r=\sqrt{3}$ cm, $OD=r_{1}=2$ cm. From $\triangle OKD\left(\angle OKD=90^{\circ}\right): \cos \angle DOK=\frac{KO}{OD}=\frac{\sqrt{3}}{2}$, $\angle DOK=30^{\circ}$. $\angle DOE=2 \angle DOK=60^{\circ}$. Therefore, $\triangle DOE$ is equilateral and its area $S_{1}=\sqrt{3} \text{ cm}^2$. The desired area $S=S_{\triangle ABC}-S_{2}$, where $S_{2}$ is the area of the part of the circle inside $\triangle ABC$. $S_{2}=S_{\text{sector}}-3S_{3}$, where $S_{3}$ is the area of the segment bounded by the chord $DE$.

$$
\begin{aligned}
& S_{3}=S_{\text{sector}}-S_{1}=\frac{4\pi}{6}-\sqrt{3}=\frac{2\pi-3\sqrt{3}}{3} \text{ cm}^2 \\
& S_{2}=4\pi-3 \frac{2\pi-3\sqrt{3}}{3}=2\pi+3\sqrt{3} \text{ cm}^2 \\
& S=S_{\triangle ABC}-S_{2}=9\sqrt{3}-(2\pi+3\sqrt{3})=2(3\sqrt{3}-\pi) \text{ cm}^2 \\
& \text{Answer: } 2(3\sqrt{3}-\pi) \text{ cm}^2
\end{aligned}
$$