AMC12 2003 A

AMC12 2003 A · Q15

AMC12 2003 A · Q15. It mainly tests Circle theorems, Area & perimeter.

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.

如图所示，一个直径为 $1$ 的半圆放在一个直径为 $2$ 的半圆的顶部。小半圆内部且大半圆外部的阴影区域称为月牙形（lune）。求该月牙形的面积。

(A) \frac{1}{6}\pi - \frac{\sqrt{3}}{4} \frac{1}{6}\pi - \frac{\sqrt{3}}{4}

(B) \frac{\sqrt{3}}{4} - \frac{1}{12}\pi \frac{\sqrt{3}}{4} - \frac{1}{12}\pi

(D) \frac{\sqrt{3}}{4} + \frac{1}{24}\pi \frac{\sqrt{3}}{4} + \frac{1}{24}\pi

(E) \frac{\sqrt{3}}{4} + \frac{1}{12}\pi \frac{\sqrt{3}}{4} + \frac{1}{12}\pi

Answer

Correct choice: (C)

正确答案：（C）

Solution

The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$. The area of the smaller semicircle is $[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi$. Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures $60^\circ$. The area of the $60^\circ$ sector of the larger semicircle is $[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi$. The area of the triangle is $[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$. So the shaded area is $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$. We have thus solved the problem.

阴影面积 $[A]$ 等于小半圆的面积 $[A+B]$ 减去大圆的一个扇形面积 $[B+C]$，再加上由大半圆的两条半径与小半圆的直径构成的三角形面积 $[C]$。小半圆面积为 $[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi$。由于大半圆的半径等于小半圆的直径，该三角形为正三角形，且对应扇形的圆心角为 $60^\circ$。大半圆中 $60^\circ$ 扇形的面积为 $[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi$。三角形面积为 $[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$。因此阴影面积为 $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$。

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