Question

正方形 $ABCD$ 绕其中心顺时针旋转 $20^\circ$ ，得到正方形 $EFGH$ ，如下图所示。求 $\angle EAB$ 的度数。

Square $ABCD$ is rotated $20^\circ$ clockwise about its center to obtain square $EFGH,$ as shown below. What is the degree measure of $\angle EAB?$

$24^\circ$

$35^\circ$

$30^\circ$

$32^\circ$

$20^\circ$

答案：B

知识点：导角等腰三角形变换

难度评级：1310

解答：

设 $O$ 为共同中心。旋转把 $A$ 送到 $E$ ，所以 $OA = OE$ ，且 $\angle AOE = 20^\circ$ 。三角形 $OAE$ 是等腰三角形，底角为 $\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ$ 。正方形对角线 $AC$ 平分 $A$ 点的直角，所以 $\angle OAB = 45^\circ$ 。两角相减，得到 $\angle EAB = 80^\circ - 45^\circ = 35^\circ$ 。所以正确答案是 B。

Let $O$ be the shared center. The rotation carries $A$ to $E,$ so $OA = OE$ and $\angle AOE = 20^\circ.$ That makes triangle $OAE$ isosceles, with base angles $\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ.$ The diagonal $AC$ splits the right angle at $A,$ so $\angle OAB = 45^\circ.$ Subtracting, $\angle EAB = 80^\circ - 45^\circ = 35^\circ.$ Thus, B is the correct answer.

2023 AMC 10B 第 7 题

答案：B

难度评级：1310

解答：

其他年份的第 7 题