Question

如图,折叠长方形的一边AD,使点D落在BC边上的点F处,已知AB=8,BC=10 求DE的长

Answer 1

F是从D点折过来的,因此△AEF与△AED全等.

AF = AD = BC = 10,DE = EF.

Rt△ABF中,斜边AF = 10,AB = 8,因此BF = 6.CF = BC - BF = 4.

△CEF为直角三角形,CE = 根号下(EF²-CF²).又DE+CE = CD = AB = 10,因此有

10 - DE = 根号下(DE²-4²)

DE = 5.8