Question

如图，一个直径为 $1$ 的半圆位于一个直径为 $2$ 的半圆上方。小半圆内部且大半圆外部的阴影区域称为月牙形。求这个月牙形的面积。

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.

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

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

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

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

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

答案：C

知识点：扇形面积分割等边三角形

难度评级：1630

小提示：

小半圆的直径是大圆中长为 $1$ 的弦，对应半径为 $1$ 的圆中的 $60^\circ$ 圆弧。

The small diameter is a chord of length $1$ in the large radius- $1$ circle, spanning a $60^\circ$ arc

大提示：

月牙面积 $=$ （等边三角形 $+$ 小半圆） $-$ 大圆的 $60^\circ$ 扇形。

Lune $=$ (equilateral triangle $+$ small semicircle) $-$ the $60^\circ$ sector of the large circle

解答：

小半圆的直径是大圆半径 $1$ 中长为 $1$ 的弦，所以它在圆心处张成 $60^\circ$ 。

由该弦和小圆弧围成的区域等于一个面积 $\dfrac{\sqrt3}{4}$ 的等边三角形加上小半圆面积 $\dfrac12\pi\left(\dfrac12\right)^2=\dfrac{\pi}{8}$ 。

减去大圆的 $60^\circ$ 扇形面积 $\dfrac16\pi(1)^2=\dfrac{\pi}{6}$ ，得月牙面积 $\dfrac{\sqrt3}{4}+\dfrac{\pi}{8}-\dfrac{\pi}{6}=\dfrac{\sqrt3}{4}-\dfrac{\pi}{24}\text{。}$

所以正确答案是 C。

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 第 15 题

答案：C

难度评级：1630

小提示：

大提示：

解答：

其他年份的第 15 题