Question

（8分）如图，在△ABC中，AB=AC，点O为底边上的中点，以点O为圆心，1为半径的半圆与边AB相切于点D．

 1.（1）判断直线AC与⊙O的位置关系，并说明理由；

 2.（2）当∠A＝60°时，求图中阴影部分的面积．

 

Answer 1

 

1.解：（1）直线AC与⊙O相切．···················································································· 1分

理由是：

连接OD,过点O作OE⊥AC，垂足为点E．

∵⊙O与边AB相切于点D，

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

∵AB=AC，点O为底边上的中点，

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

又∵OD⊥AB，OE⊥AC

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

∴OE是⊙O的半径．

又∵OE⊥AC，∴直线AC与⊙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_{四边形ADOE} =×OD×AD×2=×1××2=························································· 6分

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

∴S_阴影= S_{四边形ADOE} －S_扇形形ODE=－π．······································································· 8分

 

解析:略