Question

已知如图,在四边形ABCD中,∠BCD=60°,∠ADC=45°,CA平分∠BCD,AB=AD=2被根号2,求四边形AB

Answer 1

过点A作AE⊥CD,垂足为E,作AF⊥BC,垂足为 F

在Rt△ADE中,∠ADC=45°

∵AD=2√2

∴AE=ED=2

∵AC平分∠BCD=60°

∴∠ACD=∠ACB=30°,AE=AF=2

在Rt△ACE和Rt△ACF中

FC=2√3,CE=2√3

在Rt△AFB中

∵AB=2√2,AF=2

∴BF=2

则四边形ABCD的面积

=S△ACD+S△ABC

=S△ACD+S△AFC-S△ABF

=1/2×（CE+ED)×AE+1/2×AF×FC-1/2×AF×BF

=1/2×（2+2√3）×2+1/2×2×2√3-1/2×2×2

=2+2√3+2√3-2

=4√3