AMC10 2003 A

AMC10 2003 A · Q19

AMC10 2003 A · Q19. 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.

直径为 1 的半圆位于直径为 2 的半圆顶部，如图所示。小半圆内、大半圆外的阴影区域称为月形区。求此月形区的面积。

(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

(C) First note that the area of the region determined by the triangle topped by the semicircle of diameter 1 is $\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^{2}=\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi.$ The area of the lune results from subtracting from this the area of the sector of the larger semicircle, $\dfrac{1}{6}\pi(1)^{2}=\dfrac{1}{6}\pi.$ So the area of the lune is $\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi-\dfrac{1}{6}\pi=\dfrac{\sqrt{3}}{4}-\dfrac{1}{24}\pi.$

（C）首先注意，由边长为 1 的半圆（直径为 1）覆盖在三角形上所确定的区域面积为 $\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\pi\left(\dfrac{1}{2}\right)^{2}=\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi.$ 月牙形（lune）的面积是从该面积中减去较大半圆的扇形面积得到的，即 $\dfrac{1}{6}\pi(1)^{2}=\dfrac{1}{6}\pi.$ 因此月牙形的面积为 $\dfrac{\sqrt{3}}{4}+\dfrac{1}{8}\pi-\dfrac{1}{6}\pi=\dfrac{\sqrt{3}}{4}-\dfrac{1}{24}\pi.$

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