Question

如图,梯形ABCD中,AD平行于BC,AC,BC交于点O,S△AOD=4,S△=9,则S△AOD= ______.

Answer 1

过点O作OE⊥AD,延长EO交BC于点F

因为AD∥BC

所以OF ⊥BC

由AD∥BC,得∠OAD=∠OCB,∠ODA=∠OBC

所以△AOD∽△COB

所以AD/BC=OE/OF=√(S△AOD/S△BOC)=2/3

所以BC=1.5AD,OF=1.5OE

所以 S梯形ABCD=1/2(AD=BC)/*EF

=0.5**2.5AD*2.5OE

=2.5*2.5*(0.5*AD*OE)

=6.25* S△AOD

=6.25*4

=25