Question

如下图所示，六个半圆位于边长为二的正六边形内部，并且这些半圆的直径分别与正六边形的边重合。阴影区域，也就是六边形内但所有半圆外的区域，面积是多少？

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?

$6\sqrt3-3\pi$

$\dfrac{9\sqrt3}{2}-2\pi$

$\dfrac{3\sqrt3}{2}-\dfrac{\pi}{3}$

$3\sqrt3-\pi$

$\dfrac{9\sqrt3}{2}-\pi$

答案：D

知识点：正多边形扇形面积分割

难度评级：1530

解答：

由对称性，阴影区域由六个全等部分组成。每一部分是两个边长为 $1$ 的等边三角形的并，减去一个半径为 $1$ 、圆心角为 $60^\circ$ 的扇形。

两个等边三角形的总面积为扇形面积为因此每一块阴影的面积为 $\frac{\sqrt3}{2}-\frac{\pi}{6}$ ，阴影总面积为 $2\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{2}.$ $\frac{60^\circ}{360^\circ}\cdot\pi(1)^2=\frac{\pi}{6}.$ $6\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right)=3\sqrt3-\pi.$

所以正确答案是 D。

By symmetry, the shaded region is made of six congruent pieces. One such piece is the union of two equilateral triangles with side length $1,$ minus a $60^\circ$ sector of a circle of radius $1.$

The two equilateral triangles have total area $2\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{2}.$ The sector has area $\frac{60^\circ}{360^\circ}\cdot\pi(1)^2=\frac{\pi}{6}.$ Thus one shaded piece has area $\frac{\sqrt3}{2}-\frac{\pi}{6},$ and the total shaded area is $6\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right)=3\sqrt3-\pi.$

Thus, D is the correct answer.

2020 AMC 10B 第 14 题

答案：D

难度评级：1530

解答：

其他年份的第 14 题