Question

（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.（1）判斷直線AC與⊙O的位置關(guān)系，并說明理由；

 2.（2）當(dāng)∠A＝60°時(shí)，求圖中陰影部分的面積．

 

Answer 1

 

1.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.（2）∵AO平分∠BAC，且∠BAC=60°，∴∠OAD=∠OAE=30°，

∴∠AOD=∠AOE=60°，

在Rt△OAD中，∵tan∠OAD = ，∴AD==，同理可得AE=

∴S_{四邊形ADOE} =×OD×AD×2=×1××2=························································· 6分

又∵S_扇形形ODE==π·························································································· 7分

∴S_陰影= S_{四邊形ADOE} －S_扇形形ODE=－π．······································································· 8分

 

解析:略