如图:
做AF∥BC交CD于F, EF⊥BC,
∵∠C=45,∴ EF=CF, ∠AFD=∠C=45
∴AD=DF=2
∴AF=2√2
∴BE=AF=2√2
∴CE=BC-BE=4-2√2,
∴CF=CE*√2=(4-2√2)*√2=4(√2-1)
∴ AB=EF=CE =4-2√2,
∴CD=CF+DF=
S(ABCD)=S△ABC+S△ACD
=AB*BC/2+ AD*CD/2
=(4-2√2)*4/2+2*(4(√2-1)+2)/2
=2(4-2√2) +(4(√2-1)+2)
=8-4√2+4√2-2
=6