五边形ABCDE中,已知AB=CD=AE=BC+DE=2,∠ABC=∠AED=90°,求五边形ABCDE的面积.

连结AC,将ΔABC绕点A旋转,使AB与AE重合,设C点落在点F处.
则AF = AC,DF = EF + DE = 2 = CD,故ΔADF≌ΔADC.
由于AE⊥DF,故S(ΔADF) = AE * DF / 2 = 2.S(ΔADC) = S(ΔADF) = 2.
故S(五边形ABCDE) = S(四边形AFDC) = 4.