过B作BE垂直于AD,过C作CF垂直于AD.
那么有:AE=3,AF=4,EF=1,BE=6,CF=8,FD=2
S(ABCD)=S(ABE)+S(BEFC)+S(CFD)
=1/2AE*BE+(BE+CF)*EF/2+1/2FD*CF
=1/2*3*6+(6+8)*1/2+1/2*2*8
=9+7+8
=24
过B作BE垂直于AD,过C作CF垂直于AD.
那么有:AE=3,AF=4,EF=1,BE=6,CF=8,FD=2
S(ABCD)=S(ABE)+S(BEFC)+S(CFD)
=1/2AE*BE+(BE+CF)*EF/2+1/2FD*CF
=1/2*3*6+(6+8)*1/2+1/2*2*8
=9+7+8
=24