2004 AMC 10B Problem 25

Let the large circles have centers $A$ and $B,$ let $C$ be the center of the small circle, and let $D$ be a point where the two large circles meet. Then $\triangle ACD$ is right with $AC = 1$ and $AD = 2,$ so $CD = \sqrt3,$ $\angle CAD = 60^\circ,$ and its area is $\dfrac{\sqrt3}{2}.$ One quarter of the shaded region equals the $60^\circ$ sector of the radius-$2$ circle (area $\dfrac{2\pi}{3}$) minus $\triangle ACD$ (area $\dfrac{\sqrt3}{2}$) minus a quarter of the small circle (area $\dfrac{\pi}{4}$), giving $\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} - \dfrac{\sqrt3}{2}.$ Multiplying by $4,$ the shaded area is $\dfrac{5\pi}{3} - 2\sqrt3.$ Thus, the correct answer is B.