Question

【题目】如图，0°＜∠BAC＜90°，点A1，A3，A5…在边AB上，点 A2，A4，A6…在边AC上，且满足如下规律：A1A2⊥A2A3， A2A3⊥A3A4，A3A4⊥A4A5，…，若AA1=A1A2=A2A3=1，则A11A12的长度为（）

A. B. C. D.

Answer 1

【答案】D

【解析】由已知可得∠A=∠AA2A1=∠A2A4A3=∠A4A6A5，由等角对等边得到AA3=A3A4，AA5=A5A6．由勾股定理得到A1A3=，从而得到AA3=A3A4=，同理可得A3A5=，AA5=A5A6=，依次类推即可得到结论．

∵AA1=A1A2，∴∠A=∠AA2A1．

∵A1A2⊥A2A3， A2A3⊥A3A4，A3A4⊥A4A5，∴A1A2∥A3A4∥A5A6，∴∠AA2A1=∠A2A4A3=∠A4A6A5，∴∠A=∠AA2A1=∠A2A4A3=∠A4A6A5，∴AA3=A3A4，AA5=A5A6．

∵A1A2=A2A3=1，∴A1A3=，∴AA3=A3A4=，∴A3A5=，∴AA5=A5A6==，同理可得：AA7=A7A7=，AA11=A11A12==．

故选D．