已知:如图,将长方形ABCD沿着AE折叠,使得点D落在BC边上点F处,AB=8,AD=10,求EC的长 勾股定理

2024-12-15 18:57:03
推荐回答(1个)
回答1:

你的意思是只能用勾股定理?那就这样:
AF = AD = 10
FB = √(AF^2-AB^2) = √(10^2-8^2) = 6
FC = 10-6 = 4
EF = ED = 8-EC

EC^2 + FC^2 = EF^2
EC^2 + 4^2 = (8-EC)^2
EC^2 + 16 = 64 -2*8*EC +EC^2
16*EC = 48
EC=3

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以下是之前答的,不删了,供参考:
AF = AD = 10
FB = √(AF^2-AB^2) = √(10^2-8^2) = 6

∵ ∠AFE = 90°
∴ △ECF ∽ △FBA
∴ EC/FB = FC/AB
EC/6 = (10-6)/8
EC=3