1.15. The side of an equilateral triangle inscribed in a circle is equal to $a$. Calculate the area of the segment cut off by it.
### 1.15. Area of Segment \( A \cap B \)

is the difference between the areas of sector \( A O B \) and triangle \( A O B \) (Fig. 1.13). We find \( S_{\text{sect } A O B} = \frac{\pi R^2}{3} \), \( S_{\triangle A O B} = \frac{1}{2} a \cdot \frac{R}{2} = \frac{a R}{4} \), hence \( S = \frac{\pi R^2}{3} - \frac{a R}{4} \). Since \( R = \frac{a}{\sqrt{3}} \), then

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-420.jpg?height=485&width=450&top_left_y=1367&top_left_x=690)

Fig. 1.13

\[
S = \frac{\pi a^2}{9} - \frac{a^2}{4 \sqrt{3}} = a^2 \left( \frac{\pi}{9} - \frac{\sqrt{3}}{12} \right) = \frac{a^2 (4 \pi - 3 \sqrt{3})}{36}
\]

Answer: \( S = \frac{a^2 (4 \pi - 3 \sqrt{3})}{36} \).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-421.jpg?height=344&width=356&top_left_y=474&top_left_x=45)

Fig. 1.14