10.300. The side of a regular triangle is equal to $a$. Determine the area of the part of the triangle that lies outside a circle of radius $a / 3$, the center of which coincides with the center of the triangle.
## Solution.

The desired area $S=S_{1}-S_{2}+3 S_{3}$, where $S_{1}=\frac{a^{2} \sqrt{3}}{4}$ is the area of the triangle, $S_{2}=\frac{\pi a^{2}}{9}$ is the area of the circle, and $S_{3}$ is the area of the segment cut off by the triangle from the circle. The chord of this segment is $\frac{a}{3}$; therefore, $S_{3}=\frac{1}{6} \cdot \frac{\pi a^{2}}{9}-\frac{a^{2} \sqrt{3}}{9 \cdot 4}=\frac{\pi a^{2}}{54}-\frac{a^{2} \sqrt{3}}{36}$.

We obtain that $S=\frac{a^{2} \sqrt{3}}{4}-\frac{\pi a^{2}}{9}+\frac{\pi a^{2}}{18}-\frac{a^{2} \sqrt{3}}{12}=\frac{a^{2}(3 \sqrt{3}-\pi)}{18}$.

Answer: $\frac{a^{2}(3 \sqrt{3}-\pi)}{18}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0780.jpg?height=272&width=440&top_left_y=108&top_left_x=238)

Fig. 10.91 .1

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0780.jpg?height=300&width=442&top_left_y=76&top_left_x=752)

Fig. 10.91 .2