1.51. 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 $\frac{a}{3}$, the center of which coincides with the center of the triangle.
1.51. The desired area $S=S_{1}-S_{2}+3 S_{3}$, where $S_{1}=\frac{a^{2} \sqrt{3}}{4}-$ area of the triangle, $S_{2}=\frac{\pi a^{2}}{9}$ - area of the circle, $S_{3}$ - area of the segment cut off by the triangle from the circle. The chord of this segment is $\frac{a}{3};$ therefore,

$$
S_{3}=\frac{1}{6} \cdot \frac{\pi a^{2}}{9}-\frac{a^{2} \sqrt{3}}{9 \cdot 4}=\frac{\pi a^{2}}{54}-\frac{a^{2} \sqrt{3}}{36}
$$

Finally, we get

$$
\begin{aligned}
& \qquad=\frac{a^{2} \sqrt{3}}{4}-\frac{\pi a^{2}}{9}+\frac{\pi a^{2}}{18}-\frac{a^{2} \sqrt{3}}{12}=\frac{a^{2}(3 \sqrt{3}-\pi)}{18} \\
& \text { Answer: } \frac{a^{2}(3 \sqrt{3}-\pi)}{18}
\end{aligned}
$$