2003 AMC 12A Problem 15

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

The small diameter is a chord of length

1

in the large radius-

1

circle, spanning a

60^\circ

arc

Lune

=

(equilateral triangle

+

small semicircle)

-

the

60^\circ

sector of the large circle

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.

Problem 15 in other years