解：由折叠知：BE = DE,CF = C' F因为ABCD是矩形所以AB = CD ,AD = BC所以△ABE的周长= AE+BE+AB=AE+DE+AB=AD+AB=1+2=3△BC' F 的周长 = BC' + C' F + BF = CD + CF + BF = CD + BC = 1+ 2 = 3
