2003 AMC 10A Problem 19

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 lune equals a triangle plus the small semicircle, minus a sector of the large circle

The chord of length

1

subtends a

60^\circ

arc of the large semicircle

The small semicircle’s diameter is a chord of length

1

in the large circle. Joining its endpoints to the large circle’s center gives an equilateral triangle of side

1

and area

\dfrac{\sqrt{3}}{4}.

The region between the chord and the small arc, taken together with that triangle, has area

\dfrac{\sqrt{3}}{4} + \dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^2 = \dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8}.

Subtracting the

60^\circ

sector of the large circle, of area

\dfrac{1}{6}\pi(1)^2 = \dfrac{\pi}{6},

leaves the lune:

\dfrac{\sqrt{3}}{4} + \dfrac{\pi}{8} - \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{4} - \dfrac{\pi}{24}.

Thus, the correct answer is C.

Problem 19 in other years