2012 AMC 10B 第 16 题

先试着解答 2012 AMC 10B 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2012 AMC 10B 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

三个半径为二的圆两两相切。如图,求这三个圆以及它们围成的中间区域的总面积。

Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?

10π+4310\pi+4\sqrt{3}

13π313\pi-\sqrt{3}

12π+312\pi+\sqrt{3}

10π+910\pi+9

13π13\pi

答案:A
知识点:相切圆扇形等边三角形面积分割
难度评级:1630
解答:

连接三个圆心,得到边长为 44 的等边三角形,面积为 434\sqrt3

每个圆中计入的部分是 300300^\circ 扇形,面积为 300360π22=10π3\dfrac{300}{360}\pi\cdot2^2=\dfrac{10\pi}{3}

所以总面积为 310π3+43=10π+433\cdot\dfrac{10\pi}{3}+4\sqrt3=10\pi+4\sqrt3

所以正确答案是 A

Connect the centers of the three circles. This forms an equilateral triangle of side 44, with area 434\sqrt3.

The included part of each circle is a 300300^\circ sector, whose area is 300360π22=10π3\dfrac{300}{360}\pi\cdot2^2=\dfrac{10\pi}{3}.

The total area is 310π3+43=10π+433\cdot\dfrac{10\pi}{3}+4\sqrt3=10\pi+4\sqrt3.

Thus, A is the correct answer.

← 第 15 题#15
完整试卷

其他年份的第 16 题