Question

如图,在四边形ABCD中,对角线AC,BD交于点E,角BAC等于90度,角CED等于45度,角DC

Answer 1

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