(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.

 

【答案】

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

 

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