Theory:

If \(2\) figures have the same common side(base), and the vertices(or vertex) opposite to the common side of each figure lie on the same straight line, which is parallel to the common side, then they are said to be on the same base and between the same parallels
Consider the given figure.
 
3.png
 
This figure shows that the trapezium \(PQAB\) and the parallelogram \(ABCD\) have the same base \(AB\). Also, we can see that the vertices \(P\) and \(Q\) of the trapezium \(PQAB\) lie on the line \(PD\), which is opposite to the base \(AB\) and the vertices \(C\) and \(D\) of the parallelogram \(ABCD\) lie on the line \(PD\), which is opposite to the base \(AB\). Thus, we can say that the vertices lie on the line \(PD\) parallel to \(AB\).
 
Therefore, the trapezium \(PQAB\) and the parallelogram \(ABCD\) lie on the same common base \(AB\) and between the same parallels \(AB\) and \(PD\).
 
Also, consider the below figure.
 
4.png
 
This figure shows that the parallelogram \(ABCD\) and the triangle \(DOC\) have the same base \(DC\). And, the vertex \(O\) of the triangle does not lie opposite side to the base \(DC\). Therefore, we cannot say that the \(ABCD\) and \(DOC\) lie between the same parallels \(DC\) and \(AB\).
 
Now, consider the below figure.
 
5.png
 
This figure shows that the trapezium \(ABCD\) and the triangle \(PQR\) do not have the common base and these figures do not lie between the same parallels.