2. In a circle with radius 1, the area of the figure formed by all chords starting from point $A$ and having a length not less than the side length $a$ of the inscribed equilateral $\triangle ABC$ is ( ).
(A) $\frac{\pi}{2}+\frac{\sqrt{2}}{3}$
(B) $\frac{\pi}{3}+\frac{\sqrt{2}}{2}$
(C) $\frac{\pi}{2}+\frac{\sqrt{3}}{3}$
(D) $\frac{\pi}{3}+\frac{\sqrt{3}}{2}$
2.D.

From the given conditions, it is easy to know that the figure formed by these chords is precisely the equilateral $\triangle A B C$ and its corresponding segments. Let the center of $\triangle A B C$ be $O$, then the area of the small sector $\overparen{B O C}$ is $\frac{\pi}{3}$. And
$$
S_{\triangle A O B}=S_{\triangle A O C}=\frac{1}{2} \times 1^{2} \times \sin 120^{\circ}=\frac{\sqrt{3}}{4} .
$$

Therefore, the area of the required figure is $\frac{\sqrt{3}}{2}+\frac{\pi}{3}$.