AMC8 2017

AMC8 2017 · Q25

AMC8 2017 · Q25. It mainly tests Circle theorems, Area & perimeter.

In the figure shown, US and UT are line segments each of length 2, and m$\angle$TUS = 60°. Arcs TR and SR are each one-sixth of a circle with radius 2. What is the area of the region shown?

如图所示，线段 US 和 UT 各长 2，∠TUS = 60°。弧 TR 和弧 SR 各为半径为 2 的圆的六分之一。求图示区域的面积。

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

(B) $4\sqrt{3} - 4\pi/3$ $4\sqrt{3} - 4\pi/3$

(D) $4\sqrt{3} - 2\pi/3$ $4\sqrt{3} - 2\pi/3$

(E) $4 + 4\pi/3$ $4 + 4\pi/3$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The region shown is what remains when two one-sixth sectors of a circle of radius 2 are removed from an equilateral triangle with side length 4. The area of an equilateral triangle with side length $s$ is $\frac{\sqrt{3}}{4}s^2$. Thus the area of the region is $4\sqrt{3} - 2\left(\frac{1}{6}\cdot 4\pi\right) = 4\sqrt{3} - \frac{4\pi}{3}$.

答案 (B)：所示区域是边长为 4 的等边三角形中，去掉两个半径为 2 的六分之一圆扇形后剩下的部分。边长为 $s$ 的等边三角形面积是 $\frac{\sqrt{3}}{4}s^2$。因此该区域的面积为 $4\sqrt{3} - 2\left(\frac{1}{6}\cdot 4\pi\right) = 4\sqrt{3} - \frac{4\pi}{3}$。

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