如图,正方形ABCD中,AD=1,G为CD中点,E为BC上任一点,FE//AG交AB与F,设BE=x,四边形AEFG面积
2个回答

EF//AG,则:∠BFE=∠BAG=∠DGA;

又∠B=∠D=90度,则:⊿EBF∽⊿ADG,BF/BE=DG/AD.

点G为CD中点,则BF/BE=DG/AD=1/2,则BF=(1/2)BE=0.5X.

S四边形AFEG=S正方形ABCD-S⊿EBF-S⊿ECG-S⊿ADG

即:y=1-(1/2)X*0.5X-(1/2)*(1-X)*0.5- (1/2)*1*0.5

化简得:y=(-1/4)X^2+(1/4)X+1/2.(0