Area

Consider the figure above. The area of the bigger part is area of the circle minus the area of the segment.

$\cos \phi=\dfrac{2}{6}$ $\implies$ $\phi=\cos^{-1}\left(\dfrac{2}{6}\right) \approx70.529$ $\implies$ $2\phi=2(70.529) \approx 141.058$

$A_{bigger~part}=A_{circle}-A_{segement}=\pi(6^2)-\left[\dfrac{141.058}{360}\pi (6^2)-\dfrac{1}{2}(6^2)(\sin~141.058)\right] \approx$ $\boxed{80.01}$