AMC10 2012 B

AMC10 2012 B · Q16

AMC10 2012 B · Q16. It mainly tests Circle theorems, Area & perimeter.

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?

三个半径为2的圆互切。圆与它们所围区域的总面积是多少，如图所示？

(A) $10\pi + 4\sqrt{3}$ $10\pi + 4\sqrt{3}$

(B) $13\pi - \sqrt{3}$ $13\pi - \sqrt{3}$

(D) $10\pi + 9$ $10\pi + 9$

(E) $13\pi$ $13\pi$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Connect the centers of the three circles to form an equilateral\ntriangle with side length 4. Then the requested area is equal to the area of this\ntriangle and a $300^\circ$ sector of each circle. The equilateral triangle has base 4 and\naltitude $2\sqrt{3}$, so its area is\n\n$\frac{1}{2} \cdot 4 \cdot 2\sqrt{3} = 4\sqrt{3}$.\n\nThe area of each sector is $\frac{300}{360} \cdot \pi \cdot 2^2 = \frac{10\pi}{3}$. Hence the total area is $3 \cdot \frac{10\pi}{3} + 4\sqrt{3} = 10\pi + 4\sqrt{3}$.

连接三个圆心，形成边长为4的等边三角形。所求面积等于该三角形面积加上每个圆的$300^\circ$扇形面积。等边三角形底边为4，高为$2\sqrt{3}$，面积为$\frac{1}{2} \cdot 4 \cdot 2\sqrt{3} = 4\sqrt{3}$。每个扇形面积为$\frac{300}{360} \cdot \pi \cdot 2^2 = \frac{10\pi}{3}$。因此总面积为$3 \cdot \frac{10\pi}{3} + 4\sqrt{3} = 10\pi + 4\sqrt{3}$。

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