Question

（8分）如圖，在△ABC中，AB=AC，點O為底邊上的中點，以點O為圓心，

1為半徑的半圓與邊AB相切于點D．

（1）判斷直線AC與⊙O的位置關系，并說明理由；

（2）當∠A＝60°時，求圖中陰影部分的面積．

 

Answer 1

解：（1）直線AC與⊙O相切．···································································· 1分

理由是：

連接OD,過點O作OE⊥AC，垂足為點E．

∵⊙O與邊AB相切于點D，

∴OD⊥AB．·························································································· 2分

∵AB=AC，點O為底邊上的中點，

∴AO平分∠BAC····················································································· 3分

又∵OD⊥AB，OE⊥AC

∴OD= OE····························································································· 4分

∴OE是⊙O的半徑．

又∵OE⊥AC，∴直線AC與⊙O相切．·························································· 5分

（2）∵AO平分∠BAC，且∠BAC=60°， ∴∠OAD=∠OAE=30°，

∴∠AOD=∠AOE=60°，

解析:略