check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.
[tex]\bf \textit{area of a sector of a circle}\\\\
A_x=\cfrac{\theta \pi r^2}{360}\quad
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=6\\
\theta =60
\end{cases}\implies A_x=\cfrac{60\cdot \pi \cdot 6^2}{360}\implies \boxed{A_x=6\pi} \\\\
-------------------------------\\\\[/tex]
[tex]\bf \textit{area of a segment of a circle}\\\\
A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right]
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=6\\
\theta =120
\end{cases}[/tex]
[tex]\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o ) \right]
\\\\\\
A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\
-------------------------------\\\\
\textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}[/tex]
