已知圆O的直径AB=10,弦CD⊥AB于点M,若OM:OA=3:5,则AC的长为多少?
问题描述：
已知圆O的直径AB=10,弦CD⊥AB于点M,若OM:OA=3:5,则AC的长为多少?
答
因为AB = 10,所以 OA = OB = 51.若M在OA中间OM = 3/5OA = 3,MC = 根号（OC^2 - OM^2) = 4AM = OA - OM =2,所以 AC = 根号（AM^2 + CM^2) = 2根号52.若M在OB中间则AM = OA + OM = 8AC = 根号（AM^2 + CM^2) = 4根号5...