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?

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

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

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

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

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

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.