2021 Fall AMC 12B Problem 9

For an equilateral triangle, $O$ is also the circumcenter, so $OA = OC = \dfrac{6}{\sqrt3} = 2\sqrt3.$ The central angle $\angle AOC = 120^\circ.$ In triangle $AOC,$ side $AC = 6$ is opposite the $120^\circ$ angle, so the circumradius $R'$ of this triangle satisfies $$2R' = \dfrac{6}{\sin 120^\circ} = 4\sqrt3,$$ giving $R' = 2\sqrt3.$ The area of the circle is $\pi (2\sqrt3)^2 = 12\pi.$ Thus, the correct answer is B.