Question

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2,$ as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

$\dfrac{1}{6}\pi - \dfrac{\sqrt{3}}{4}$

$\dfrac{\sqrt{3}}{4} - \dfrac{1}{12}\pi$

$\dfrac{\sqrt{3}}{4} - \dfrac{1}{24}\pi$

$\dfrac{\sqrt{3}}{4} + \dfrac{1}{24}\pi$

$\dfrac{\sqrt{3}}{4} + \dfrac{1}{12}\pi$

Solution:

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.

2003 AMC 12A Problem 15

Solution:

Problem 15 in Other Years