(8分)如圖,在△ABC中,AB=AC,點(diǎn)O為底邊上的中點(diǎn),以點(diǎn)O為圓心,

1為半徑的半圓與邊AB相切于點(diǎn)D.

(1)判斷直線AC與⊙O的位置關(guān)系,并說明理由;

(2)當(dāng)∠A=60°時(shí),求圖中陰影部分的面積.

 

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

理由是:

連接OD,過點(diǎn)O作OE⊥AC,垂足為點(diǎn)E.

∵⊙O與邊AB相切于點(diǎn)D,

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

∵AB=AC,點(diǎn)O為底邊上的中點(diǎn),

∴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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