D
设AE=x,由折叠可知,EC=x,BE=4-x,
在Rt△ABE中,AB 2+BE 2=AE 2,即3 2+(4-x) 2=x 2,
解得:x=
;
由折叠可知∠AEF=∠CEF,由AD∥BC得∠CEF=∠AFE,
∴∠AEF=∠AFE,即AE="AF="
;
∴S△AEF=
×AF×AB=
×
×3=
.故选D
D
设AE=x,由折叠可知,EC=x,BE=4-x,
在Rt△ABE中,AB 2+BE 2=AE 2,即3 2+(4-x) 2=x 2,
解得:x=
;
由折叠可知∠AEF=∠CEF,由AD∥BC得∠CEF=∠AFE,
∴∠AEF=∠AFE,即AE="AF="
;
∴S△AEF=
×AF×AB=
×
×3=
.故选D