2013 AMC 10A 第 20 题

Question

一个单位正方形绕其中心旋转

45^\circ

。正方形内部扫过区域的面积是多少？

A unit square is rotated

45^\circ

about its center. What is the area of the region swept out by the interior of the square?

1 - \dfrac{\sqrt2}{2} + \dfrac{\pi}{4}

\dfrac{1}{2} + \dfrac{\pi}{4}

2 - \sqrt2 + \dfrac{\pi}{4}

\dfrac{\sqrt2}{2} + \dfrac{\pi}{4}

1 + \dfrac{\sqrt2}{4} + \dfrac{\pi}{8}

只看扫过区域的一个象限。

Work with one quadrant of the swept region

把该象限分解成一个扇形和两个直角三角形部分。

Decompose that quadrant into a sector and two right-triangle pieces

考虑扫过区域的四分之一，最后乘以

4

。

其中扇形圆心角为

45^\circ

，半径为

\frac{\sqrt2}{2}

，面积为

\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}

。

该四分之一区域中的两个直角三角形面积分别为

\frac12\cdot\frac12\cdot\frac12=\frac18

和

\frac12\left(\frac{\sqrt2-1}{2}\right)^2

。

把四分之一区域的面积和乘以

4

，得到

2-\sqrt2+\frac{\pi}{4}

。

所以正确答案是 C。

Consider one quarter of the swept region and multiply its area by

4

.

The sector has angle

45^\circ

and radius

\frac{\sqrt2}{2}

, so its area is

\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}

.

The two right-triangle pieces in that quarter have areas

\frac12\cdot\frac12\cdot\frac12=\frac18

and

\frac12\left(\frac{\sqrt2-1}{2}\right)^2

.

Multiplying the quarter-area sum by

4

gives

2-\sqrt2+\frac{\pi}{4}

.

Thus, C is the correct answer.

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