Question

矩形 $PQRS$ 在平面内，且 $PQ = RS = 2$ 、 $QR = SP = 6$ 。矩形先绕 $R$ 顺时针旋转 $90^\circ$ ，再绕第一次旋转后点 $S$ 移到的位置顺时针旋转 $90^\circ$ 。点 $P$ 经过的路径长度是多少？

Rectangle $PQRS$ lies in a plane with $PQ = RS = 2$ and $QR = SP = 6.$ The rectangle is rotated $90^\circ$ clockwise about $R,$ then rotated $90^\circ$ clockwise about the point that $S$ moved to after the first rotation. What is the length of the path traveled by point $P?$

$\left(2\sqrt{3} + \sqrt{5}\right)\pi$

$6\pi$

$\left(3 + \sqrt{10}\right)\pi$

$\left(\sqrt{3} + 2\sqrt{5}\right)\pi$

$2\sqrt{10}\pi$

答案：C

知识点：变换弧勾股定理

难度评级：1840

解答：

第一次旋转中， $P$ 绕 $R$ 走四分之一圆，半径为 $PR = \sqrt{2^2 + 6^2} = 2\sqrt{10}$ 。弧长为 $\dfrac{1}{4}\left(2\pi \cdot 2\sqrt{10}\right) = \sqrt{10}\,\pi$ 。

第二次旋转中， $P$ 绕 $S$ 的新位置走四分之一圆，半径为 $6$ 。弧长为 $\dfrac{1}{4}(2\pi \cdot 6) = 3\pi$ 。

总路径长度为 $\left(3 + \sqrt{10}\right)\pi$ 。

所以正确答案是 C。

In the first rotation, $P$ moves on a quarter circle about $R$ with radius $PR = \sqrt{2^2 + 6^2} = 2\sqrt{10}.$ The arc length is $\dfrac{1}{4}\left(2\pi \cdot 2\sqrt{10}\right) = \sqrt{10}\,\pi.$

In the second rotation, $P$ moves on a quarter circle about the new position of $S$ with radius $6.$ The arc length is $\dfrac{1}{4}(2\pi \cdot 6) = 3\pi.$

The total path length is $\left(3 + \sqrt{10}\right)\pi.$

Thus, the correct answer is C.

2008 AMC 10A 第 19 题

答案：C

难度评级：1840

解答：

其他年份的第 19 题