2003 AMC 12A Problem 15

The small semicircle’s diameter is a chord of length $1$ in the large circle of radius $1,$ so it subtends a $60^\circ$ angle at the center. The region bounded by that chord and the small arc is an equilateral triangle of area $\dfrac{\sqrt3}{4}$ topped by the small semicircle of area $\dfrac12\pi\left(\dfrac12\right)^2=\dfrac{\pi}{8}.$ Subtracting the $60^\circ$ sector of the large circle, $\dfrac16\pi(1)^2=\dfrac{\pi}{6},$ gives the lune area $$\dfrac{\sqrt3}{4}+\dfrac{\pi}{8}-\dfrac{\pi}{6}=\dfrac{\sqrt3}{4}-\dfrac{\pi}{24}.$$ Thus, the correct answer is C.