Why do these parallelograms have the same area 
In both parallelograms 1 and 2, each has a side equal to length l. These side are colinear with two parallel lines. Why do these parallelograms have equal areas?
 A: Consider the following figure, in which the blue region is a single quadrilateral that encloses both of your parallelograms:

This quadrilateral has some area. We don't need to measure the area, just recognize that it is a well-defined area. Call that area $A_1.$
Let's change the color of part of the figure, forming a yellow triangle as shown below:

Now return to the blue quadrilateral and again color a triangular portion yellow:

The two yellow triangles are congruent, which you can confirm in any of various ways.
(You can independently show three pairs of congruent angles and three pairs of congruent sides.)
So if one triangle has area $A_2,$ the other also has area $A_2.$
See below gif:

In the second figure we see that the areas of paralellogram $1$ plus a yellow triangle together add up to the area of the quadrilateral in the first figure:
$$ \operatorname{Area}(\text{parallelogram $1$}) + A_2 = A_1.$$
In the third figure we see that the areas of paralellogram $2$ plus a yellow triangle together add up to the area of the quadrilateral in the first figure:
$$ \operatorname{Area}(\text{parallelogram $2$}) + A_2 = A_1.$$
That is,
$$\operatorname{Area}(\text{parallelogram $1$}) = A_1 - A_2 =
\operatorname{Area}(\text{parallelogram $2$}).$$
A: The simple answer is that the area of a parallelogram is base times altitude. 
A much deeper answer is that you can apply the two-dimensional case of
Cavalieri's principle
It reads as follows
  Suppose two regions in a plane are included between two parallel 
  lines in that plane. If every line parallel to these two lines 
  intersects both regions in line segments of equal length, then 
  the two regions have equal areas.

In the case of your problem, every line parallel to the bases of the two parallelograms will intersect them in lines segments, each with a width of $\ell$.
