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
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