题目内容
如图,已知AB为⊙O的直径,CD是弦,AB⊥CD于E,OF⊥AC于F,BE=OF.
(1)求证:OF∥BC;
(2)求证:△AFO≌△CEB;
(3)若EB=5cm,CD=
cm,设OE=
,
求
值及阴影部分的面积.
(1)证明:∵AB为⊙O的直径,
∴∠ACB=90°························· 1分
∴AC⊥BC··························· 2分
又∵OF⊥AC
∴OF∥BC··························· 3分
(2)证明:∵AB⊥CD
∴
=
················· 4分
∴∠CAB=∠BCD························ 5分
又∵∠AFO=∠CEB=90°,OF=BE,
∴△AFO≌△CEB························· 6分
(3)∵AB⊥CD
∴CE=
CD=
cm.
在直角△OCE中,OC=OB=
(cm),
根据勾股定理可得:![]()
解得:
···························· 7分
∴tan∠COE=![]()
∴∠COE=60°···························· 8分
∴∠COD=120°,
∴扇形COD的面积是:
cm2
△COD的面积是:
CD•OE=
cm2··········· 9分
∴阴影部分的面积是:
cm2.·············· 10分
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