Question

1如图,矩形ABCD中,AB=3cm,AD=6 cm,点E为AB边上的任意一点,四边形EFGB也是矩形,且EF=2BE,

Answer 1

1、设BE=x,AE=3-x,EF=2x,

S△AEF=2x*(3-x)/2=3x-x^2,

S矩形FEBG=2x^2,

S△ABC=3*6/2=9(cm^2),

S△AGC=(6+2x)*x/2=3x+x^2,

S△AFC=S△AEF+S矩形FEBG+S△ABC-S△AGC=3x-x^2+2x^2+9-(3x+x^2)=9cm^2.

2、S矩形ABCD=AB*BC=24cm^2,

S△PHD+S△PBF=4*4/2=8cm^2,

S△PBE+S△PDG=(4-3)*6/2=3cm^2,

S四边形PFCD=S矩形ABCD-（S△PHD+S△PBF）-（S△PBE+S△PDG）-S四边形AEPH

=24-8-3-5=8cm^2.