Question

梯形ABCD,AD:BC=2:3,画对角线AC和BD相交于O点,三角形ABO的面积是6平方米,求梯形的面积

Answer 1

∵AD//BC

∴△AOD∽△COB,

∴AD/BC=OD/BO=AO/CO=2/3,

S△AOD/S△COB=（AD/BC）^2=(2/3)^2=4/9

∵△AOD和△AOB同高,S△COD=(3/2)S△AOD=6m^2,

∴S△AOD/S△AOB=OD/OB=2/3,

S△AOD=2S△AOB/3=4m^2,

∴S△COB=(9/4)*4=9m^2,

S△AOD/S△COD=AO/CO=2/3

S△COD=(3/2)S△AOD=6m^2,

∴S梯形ABCD=4+6+9+6=25m^2.