在等腰RT三角形ABC中,角C=90度,AC=1,过点C做直线L平行AB,F是L上一点,且AB=AF,则线段FC的长为?
人气:203 ℃ 时间:2019-08-18 18:12:01
解答
过点C作CD⊥AB于D,过点F作FE⊥AB于E;则有:CDEF是矩形,可得:FC = DE ,FE = CD ;已知,等腰Rt△ABC中,∠C = 90度,AC = 1 ,且 AB = AF ,可得:FE = CD = AD = √2/2 ,AF = AB = √2 ,AE = √(AF²-FE²) = ...
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