做DF垂直于AC,交AC于F点，根据题有 △ABE,△DFE为等腰直角三解形，△CFD为直角三角形，CD为△CFD的斜边 ∵BE=2√2, ∴AB=AE=2 ∵DE=√2，∴DF=EF=1，又∵∠DCE=30度，∴CD=2DF=2，同时求得CF=√3 ∴AC=AE+EF+CF = 2 + 1 + √3=3+√3 ∴Sabcd = S△abc + S△acd = AC*AB/2 + AC*DF/2 =AC*(AB+DF)/2 =(3+√3)(2+1)/2 = 3(3+√3)/2