∵DO=AO=4
又∵E为DO中点
∴EO=2
∴AE=根号20
连接BF,∠AFB所对弦为直径
∴∠AFB=90°
∵cosA=2根号5/5
∴AF/AB=cosA
∴AF=16根号5/5
∴EF=AF-AE=6根号5/5
AB CD是两条互相垂直的直径,E是OD中点,延长AE交圆于F,AO=4厘米,则EF=________?厘米
AB CD是两条互相垂直的直径,E是OD中点,延长AE交圆于F,AO=4厘米,则EF=________?厘米
数学人气：613 ℃时间：2020-05-09 09:42:55
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