This step can be counter-intuitive, so the first step should be to use right-isosceles triangles. This is because these can be used to show that they still form a rectangle, and the formula is still valid. Once the students are proficient in calculating the areas of isosceles triangles, they should be asked to cut out two identical, non-right isosceles triangles -- of some specific length and width -- and calculate their combined area using the formula given above. This should be fairly straightforward at this point, though most students will probably not understand why they were asked to cut out two identical triangles.

The teacher should then draw a parallelogram on the board, and mark the lengths of each side -- these lengths should be identical to those of the triangles the students were just asked to cut out in paper. Now, the teacher should ask the students what the area of the parallelogram is.

In most classrooms, this will be a key point...

Some students will recognize that it is the exact same calculation they were already required to do, but others will not. It is important to ask the students who grasp the notion quickly how they came to their conclusion: the parallelogram is simply two isosceles triangles.
At this point, the teacher can ask the students to cut out a new parallelogram of different dimensions, and ask them to calculate the area. Now the teacher can define a parallelogram as any four-sided shape in which opposite sides run parallel to each other; but, it can also be defined, in terms of area, as the combination of two, identical isosceles triangles. It should then be made clear that the formula for any parallelogram, including rectangles, is simply length times width; and it should be explained that all of the triangles yet encountered were simply half of a parallelogram; thus, explaining the factor of 2 in the triangle formula.