Question

在梯形ABCD中,AD平行BC,对角线的交点为O,若AD:BC=4:7,梯形ABCD的面积为121 则AOD,AOB,A

Answer 1

AD‖BC,

△AOD∽△COB,

S△AOD/S△COB=（AD/BC）^2=16/49,

设S△AOD=A,

S△COB=49A/16,

S△ABO/S△AOD=BO/OD,（二三角形等高）,

而BO/OD=BC/AD=7/4,

S△ABO=7S△AOD/4=7A/4,

S△COD/S△BOC=OD/BO=AD/BC=4/7,

S△COD=4S△BOC/7=（49A/16）*4/7=7A/4,

A+49A/16+7A/4+7A/4=121,

121A/16=121,

A=16,

S△AOD=16,

S△AOB=16*7/4=28,

S△ABC=S△AOB+S△BOC=28+16*49/16=77.