2014 AMC 12A Problem 12

Let the circles have radii $R$ (with the $30^\circ$ arc) and $r$ (with the $60^\circ$ arc). The common chord has length $2R\sin15^\circ=2r\sin30^\circ,$ so $\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}.$ The smaller central angle gives the larger radius, so $R\gt r.$ The area ratio is $$\begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3. \end{gathered}$$ Thus, the correct answer is D.