∵△AOD和△AOB的面积分别为9和12,∵S△AOD:S△AOB=OD:OB=9:12,
∴OD:OB=3:4,
∵AD∥BC,
∴△OAD∽△COB,
∴
| S△AOD |
| S△BOC |
| OD |
| OB |
| 3 |
| 4 |
| 9 |
| 16 |
∴S△BOC=16,
∵S△ABC=S△DBC,
∴S△COD=S△AOB=12,
∴梯形ABCD的面积是:S△AOD+S△AOB+S△BOC+S△COD=9+12+16+12=49.
故答案为:49.
∵△AOD和△AOB的面积分别为9和12,| S△AOD |
| S△BOC |
| OD |
| OB |
| 3 |
| 4 |
| 9 |
| 16 |