2015 AMC 12B 第 14 题

Question

一个半径为

2

的圆以

A

为圆心。一个边长为

4

的等边三角形有一个顶点在

A

。位于圆内但在三角形外的区域面积，与位于三角形内但在圆外的区域面积之差是多少？

A circle of radius

2

is centered at

A.

An equilateral triangle with side

4

has a vertex at

A.

What is the difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?

8 - \pi

\pi + 2

2\pi - \dfrac{\sqrt2}{2}

4(\pi - \sqrt3)

2\pi + \dfrac{\sqrt3}{2}

答案：D

知识点：面积分割圆面积等边三角形

难度评级：1740

小提示：

设

z

为重叠面积；所求差为

(\text{圆} - z) - (\text{三角形} - z)

Let

z

be the overlap area; the requested difference is

(\text{circle} - z) - (\text{triangle} - z)

大提示：

这等于圆面积减去三角形面积；使用

\pi r^2

和

\dfrac{\sqrt3}{4}s^2

。

That equals circle area minus triangle area; use

\pi r^2

and

\dfrac{\sqrt3}{4}s^2

解答：

设

z

为圆和三角形的公共面积。所求差为

\begin{aligned} &(\text{圆} - z) - (\text{三角形} - z) \\ &= \text{圆} - \text{三角形} \end{aligned}\text{。}

圆的面积为

\pi \cdot 2^2 = 4\pi

，等边三角形的面积为

\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3

。差为

4\pi - 4\sqrt3 = 4(\pi - \sqrt3)

。

因此，正确选项是 D。

Let

z

be the area shared by the circle and triangle. The requested difference is

\begin{aligned} &(\text{circle} - z) - (\text{triangle} - z) \\ &= \text{circle} - \text{triangle}. \end{aligned}

The circle has area

\pi \cdot 2^2 = 4\pi,

and the equilateral triangle has area

\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3.

The difference is

4\pi - 4\sqrt3 = 4(\pi - \sqrt3).

Thus, the correct answer is D.

2015 AMC 12B 第 14 题

答案：D

难度评级：1740

小提示：

大提示：

解答：

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