如图,连接AF,作GH⊥AE于点H,则有AE=EF=HG=4,FG=2,AH=2, ∵AG==2,AF==4, ∴AF2=AD2+DF2=(AG+GD)2+FD2=AG2+GD2+2AG•GD+FD2,GD2+FD2=FG2 ∴AF2=AG2+2AG•GD+FG2∴32=20+2×2×GD+4, ∴GD=,FD=, ∵∠BAE+∠AEB=90°=∠FEC+∠AEB, ∴∠BAE=∠FEC, ∵∠B=∠C=90°,AE=EF, ∴△ABE≌△ECF(AAS), ∴AB=CE,CF=BE, ∵BC=BE+CE=AD=AG+GD=2+, ∴AB+FC=2+, ∴矩形ABCD的周长=AB+BC+AD+CD=2BC+AB+CF+DF =2++2++2++=8. 故答案为,8.
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