(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)判断直线AC与⊙O的位置关系,并说明理由;(2)当∠A=60°时,求图中阴影部分的-九

题目简介

(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)判断直线AC与⊙O的位置关系,并说明理由;(2)当∠A=60°时,求图中阴影部分的-九

题目详情

(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)判断直线AC与⊙O的位置关系,并说明理由;
(2)当∠A=60°时,求图中阴影部分的面积.
题型:解答题难度:中档来源:不详

答案

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)∵AO平分∠BAC,且∠BAC=60°,∴∠OAD=∠OAE=30°,
∴∠AOD=∠AOE=60°,

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