2019 AMC 10A Problem 16

We know $\triangle ABC$ and $\triangle ABO$ are equilateral triangles. We get that $OC = 2\sqrt{3}$ using special right triangles to find the altitudes of the triangles. The radius of the larger circle is therefore $2\sqrt{3} + 1,$ since there is the extra unit radius after $\overline{OC}.$ The area of the larger circle is $$(2\sqrt{3} + 1)^2\pi = (13 + 4\sqrt{3})\pi.$$ The area of all the inner circles is $13\pi.$ The area of the shaded region is $$(13 + 4\sqrt{3})\pi - 13\pi = 4\pi\sqrt{3}.$$ Thus, A is the correct answer.