(1)过点C做CE⊥AD,与AD交于点E             （2）∴S平行四边形ABCD= AD×CE 
    ∠B＝30°                                                   =2×6
   平行四边形ABCD                                                                              =12(cm²)
  二推一
推出∠D等于30°
   ∵ CE⊥AD
     CD＝4
    ∴CE=二分之一个DC
           =2（cm)
∴AD与BC之间的距离为2cm