Question

圆 $A, B$ 和 $C$ 的半径都为一。圆 $A$ 和圆 $B$ 有一个切点。圆 $C$ 与 $\overline{AB}$ 的中点相切。圆 $C$ 内但圆 $A$ 和圆 $B$ 外的面积是多少？

Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

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

$\dfrac{\pi}{2}$

$2$

$\dfrac{3\pi}{4}$

$1+\dfrac{\pi}{2}$

答案：C

知识点：圆面积扇形面积分割

难度评级：1790

解答：

所求区域面积等于圆 $C$ 的面积减去它与圆 $A$ 和圆 $B$ 的重叠区域面积。

由图可知，一个重叠区域的一半可由一个四分之一圆扇形减去一个直角三角形得到：

这个面积为 $\dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}.$

共需减去四块这样的区域，所以最后的面积为 $\pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2.$

所以正确答案是 C。

The area of this region is the area of circle $C$ minus the area of the overlapping regions with $A$ and $B.$

From the diagram, we can find the area of half of one of the overlapping regions by finding the area of the sector and subtracting the area of the triangle.

This area is then $\dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}.$

There are four of these that we must subtract, which leaves us with a final answer of $\pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2.$

Thus, C is the correct answer.

2011 AMC 10A 第 18 题

答案：C

难度评级：1790

解答：

其他年份的第 18 题