设梯形上底为a1,下底为a2,高为（h1+h2）,
三角形AOD面积为4,即a1h1=4*2=8,△BOC的面积为9,及a2h2=9*2=18
由于△AOD和△BOC相似,则,a1/a2=h1/h2=根号下4/9=2/3
及a1=(2/3)a2
梯形面积为△ABC+△BDC-△BOC+△AOD
=2*(1/2)a2(h1+h2)-9+4
=a2h1+a2h2-5
=(3/2)a1h1+18-5
=12+18-5
=25