AMC12 2015 B

AMC12 2015 B · Q14

AMC12 2015 B · Q14. It mainly tests Circle theorems, Area & perimeter.

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?

半径为 2 的圆以 $A$ 为圆心。一个边长为 4 的正三角形有一个顶点在 $A$。位于圆内但三角形外的区域面积与位于三角形内但圆外的区域面积之差是多少？

(A) 8 - \pi 8 - \pi

(B) \pi + 2 \pi + 2

(D) 4(\pi - \sqrt{3}) 4(\pi - \sqrt{3})

(E) 2\pi + \frac{\sqrt{3}}{2} 2\pi + \frac{\sqrt{3}}{2}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $x$ equal the area of the circle, $y$ the area of the triangle, and $z$ the area of the overlapped sector. The answer is $(x-z)-(y-z)=x-y$. The area of the circle is $4\pi$ and the area of the triangle is $\frac{\sqrt{3}}{4}\cdot 4^2=4\sqrt{3}$, so the result is $4(\pi-\sqrt{3})$.

答案（D）：设 $x$ 为圆的面积，$y$ 为三角形的面积，$z$ 为重叠扇形的面积。所求为 $(x-z)-(y-z)=x-y$。圆的面积为 $4\pi$，三角形的面积为 $\frac{\sqrt{3}}{4}\cdot 4^2=4\sqrt{3}$，因此结果为 $4(\pi-\sqrt{3})$。

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