2003 AMC 10A 第 19 题

Question

一个直径为

1

的半圆位于一个直径为

2

的半圆顶部，如图所示。位于较小半圆内部且位于较大半圆外部的阴影区域称为弓月形。求这个弓月形的面积。

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

这个弓月形等于一个三角形加小半圆，再减去大圆的一个扇形。

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

长度为

1

的弦在大半圆所在圆中对应

60^\circ

的弧。

The chord of length

1

subtends a

60^\circ

arc of the large semicircle

小半圆的直径是大圆中的一条长为

1

的弦。将这条弦的两端与大圆圆心相连，得到边长为

1

的等边三角形，面积为

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

。

弦与小弧之间的区域连同这个三角形，面积为

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

。

减去大圆中

60^\circ

扇形面积

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

，得到弓月形面积为

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

。

所以正确答案是 C。

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.

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