AMC10 2012 A

AMC10 2012 A · Q18

AMC10 2012 A · Q18. It mainly tests Area & perimeter, Polygons.

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?

图中的闭合曲线由9个全等的圆弧组成，每个圆弧长度为$\frac{2\pi}{3}$，对应圆心是边长为2的正六边形的顶点之一。求该曲线围成的面积。

(A) 2\pi + 6 2\pi + 6

(B) 2\pi + 4\sqrt{3} 2\pi + 4\sqrt{3}

(D) 2\pi + 3\sqrt{3} + 2 2\pi + 3\sqrt{3} + 2

(E) \pi + 6\sqrt{3} \pi + 6\sqrt{3}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The labeled circular sectors in the figure each have the same area because they are all $\frac{2\pi}{3}$-sectors of a circle of radius 1. Therefore the area enclosed by the curve is equal to the area of a circle of radius 1 plus the area of a regular hexagon of side 2. Because the regular hexagon can be partitioned into 6 congruent equilateral triangles of side 2, it follows that the required area is $$ \pi + 6\left(\frac{\sqrt{3}}{4}\cdot 2^2\right)=\pi+6\sqrt{3}. $$

答案（E）：图中标出的圆扇形面积都相同，因为它们都是半径为 1 的圆的 $\frac{2\pi}{3}$ 扇形。因此，曲线所围成的面积等于半径为 1 的圆的面积加上边长为 2 的正六边形的面积。由于正六边形可以分割成 6 个边长为 2 的全等正三角形，所以所求面积为 $$ \pi + 6\left(\frac{\sqrt{3}}{4}\cdot 2^2\right)=\pi+6\sqrt{3}. $$

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