Not all solutions of a problem are created equal. There are many ways to show how the size of a parallelogram can be found: one solution is very simple, but not general enough; some are general but complex, and some are general, simple and beautiful.

When looking for a clear geometrical approach to find the area of a parallelogram I first thought about the simplest case, one like this:

The solution is rather easy, once you think about how it compares with the rectangle of base \(b\) and height \(h\):

You can remove the green triangle on the right and put it on the left, building a rectangle of height \(h\) and width \(b\) that has exactly the same area as the original parallelogram: and we know how to compute the area of a rectangle, \(h\times b\).

This is an interesting result: it tells you that you can stretch your parallelogram as much as you want and, as long as you keep \(b\) and \(h\) constant it will have the same area.

As much as we want? Well, we don’t actually know that. It turns out that, as you stretch, you reach a point where the argument above does not hold anymore. What would you say about this parallelogram?

It has the same \(h\) and \(b\) as the previous one, and maybe its area is the same, but we haven’t showed it yet: the argument above does not apply to it, as we cannot cut the triangle on the right an put it on the left to build the rectangle. It just doesn’t fit:

You could of course rotate it, lay it on the long side, and you are in the first case. But at this point the goal is not only to know its area: it is to show that it should be \(h\times b\). So we needed another argument, and an argument I found. But, as it turns out, it was clear but long, not very pretty and therefore not satisfying.

Talking about it over lunch with some friends a couple of better solutions appeared. Jaime Fernández and Ján Morovič quickly came up with a smart way to convert the new problem to the previous one, and then Utpal Sarkar, aka Doetoe, thought of a simple and very elegant solution.

I encourage you to think about it yourself before looking at the solutions, and then read them in order: first the long one, then the smart one, then the elegant (and best) one.

- The long and winding road

The hypothesis is that it should be possible to fit this parallelogram into the \(b\) times \(h\) rectangle, so the area of the parallelogram should be \(b\times h\). One way to go about it is to stretch the rectangle so that it fills the gap:

Now it’s clear that the blue triangle on the right is exactly the same size of the dark green triangle, so we can move it to form a small rectangle:

We have still the light green rectangle to cover, and the blue area to be fit. If the hypothesis was right, and the area of the original parallelogram was the same as the area of the \(b\times h\) rectangle, the blue area should fit exactly in the light green rectangle. One way to see if this is true is to consider these two triangles:

They are identical. Each of them contains one of the areas we want to show that are equal, plus an identical medium triangle (dark green), and an identical small triangle (gray):

So the light green and the blue areas must be equal, and we have been able to fit exactly all the blue into the green.

- Jaime and Ján’s solution

It is smarter and much simpler. Take again the original parallelogram

and cut it like this

We can move the green triangle to the left, like this

We have just constructed one parallelogram that has the exact same area as the original one, but is less slanted. In fact, it is a simple parallelogram for which we already know how to calculate the area.

If the original slant had been bigger we would have required more than one such transformation to reach the simple case; but we’ll always reach it, so we can always reduce the problem to another that we know how to solve. And the solution for this simplest case is \(b\times h\), so we know that the slanted parallelogram is also \(b\times h\).

- Doetoe’s solution

Doetoe was able to come up with an even simpler and more elegant solution by looking, literally, out of the box. Take the original slanted parallelogram

and draw a box around it:

Now just slide the lower right triangle until it finds its match with the other one:

There you have it, so general and beautiful and simple.

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