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<div id="content">
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<a href="sp-es.html">En castellano</a>.
See also
<a href="../size-triangle/st-en.html">What is the size of a triangle?</a>
<a href="../six-times-nine/stn-en.html">Why is six times nine the same as nine times six?</a>
<div id="outline-container-1" class="outline-2">
<h2 id="sec-1">What is the size of a parallelogram? </h2>
<div class="outline-text-2" id="text-1">
<p>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
<div id="outline-container-1.1" class="outline-3">
<h3 id="sec-1.1">The simple case </h3>
<div class="outline-text-3" id="text-1.1">
<p>When looking for a clear geometrical approach to find the
area of a parallelogram I first thought about the simplest
case, one like this:
<img src="img/paral-1.png" class="clean-center" alt="img/paral-1.png" />
The solution is rather easy, once you think about how it
compares with the rectangle of base <img src="ltxpng/sp-en_f46f0eb142f1e88fce09b363508e32381833ebbc.png"/> and height <img src="ltxpng/sp-en_627f79684ded3e16a85d44bf8bbd232ea90c2b77.png"/>:
<img src="img/paral-2.png" class="clean-center" alt="img/paral-2.png" />
You can remove the green triangle on the right and put it on
the left, building a rectangle of height <img src="ltxpng/sp-en_627f79684ded3e16a85d44bf8bbd232ea90c2b77.png"/> and width <img src="ltxpng/sp-en_f46f0eb142f1e88fce09b363508e32381833ebbc.png"/>
that has exactly the same area as the original parallelogram:
and we know <a href="six-times-nine.html">how to compute</a> the area of a rectangle, <img src="ltxpng/sp-en_1e232eb254f72548cb90a43de4e0100871e0fa5b.png"/>.
<img src="img/paral-3.png" class="clean-center" alt="img/paral-3.png" />
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 <img src="ltxpng/sp-en_f46f0eb142f1e88fce09b363508e32381833ebbc.png"/> and <img src="ltxpng/sp-en_627f79684ded3e16a85d44bf8bbd232ea90c2b77.png"/> 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?
<img src="img/paral-4.png" class="clean-center" alt="img/paral-4.png" />
It has the same <img src="ltxpng/sp-en_627f79684ded3e16a85d44bf8bbd232ea90c2b77.png"/> and <img src="ltxpng/sp-en_f46f0eb142f1e88fce09b363508e32381833ebbc.png"/> 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:
<img src="img/paral-5.png" class="clean-center" alt="img/paral-5.png" />
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
<img src="ltxpng/sp-en_1e232eb254f72548cb90a43de4e0100871e0fa5b.png"/>. 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.
<div id="outline-container-1.2" class="outline-3">
<h3 id="sec-1.2">The very slanted parallelogram </h3>
<div class="outline-text-3" id="text-1.2">
<p>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.
<div id="outline-container-1.2.1" class="outline-4">
<h4 id="sec-1.2.1">The long and winding road </h4>
<div class="outline-text-4" id="text-1.2.1">
<p>The hypothesis is that it should be possible to fit this
parallelogram into the <img src="ltxpng/sp-en_f46f0eb142f1e88fce09b363508e32381833ebbc.png"/> times <img src="ltxpng/sp-en_627f79684ded3e16a85d44bf8bbd232ea90c2b77.png"/> rectangle, so the area
of the parallelogram should be <img src="ltxpng/sp-en_c32770d5b8b96729ed7d6ac209b867c8c92df510.png"/>. One way to go
about it is to stretch the rectangle so that it fills the
<img src="img/paral-6.png" class="clean-center" alt="img/paral-6.png" />
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:
<img src="img/paral-7.png" class="clean-center" alt="img/paral-7.png" />
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 <img src="ltxpng/sp-en_c32770d5b8b96729ed7d6ac209b867c8c92df510.png"/> 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:
<img src="img/paral-8.png" class="clean-center" alt="img/paral-8.png" />
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
<img src="img/paral-9.png" class="clean-center" alt="img/paral-9.png" />
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.
<div id="outline-container-1.2.2" class="outline-4">
<h4 id="sec-1.2.2">Jaime and Ján's solution </h4>
<div class="outline-text-4" id="text-1.2.2">
<p>It is smarter and much simpler. Take again the original
<img src="img/paral-jaime-1.png" class="clean-center" alt="img/paral-jaime-1.png" />
and cut it like this
<img src="img/paral-jaime-2.png" class="clean-center" alt="img/paral-jaime-2.png" />
We can move the green triangle to the left, like this
<img src="img/paral-jaime-3.png" class="clean-center" alt="img/paral-jaime-3.png" />
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 href="#*The==simple==case">a simple parallelogram</a> 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 <img src="ltxpng/sp-en_c32770d5b8b96729ed7d6ac209b867c8c92df510.png"/>, so
we know that the slanted parallelogram is also <img src="ltxpng/sp-en_c32770d5b8b96729ed7d6ac209b867c8c92df510.png"/>.
<div id="outline-container-1.2.3" class="outline-4">
<h4 id="sec-1.2.3">Doetoe's solution </h4>
<div class="outline-text-4" id="text-1.2.3">
<p>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
<img src="img/paral-doetoe-1.png" class="clean-center" alt="img/paral-doetoe-1.png" />
and draw a box around it:
<img src="img/paral-doetoe-2.png" class="clean-center" alt="img/paral-doetoe-2.png" />
Now just slide the lower right triangle until it finds its
match with the other one:
<img src="img/paral-doetoe-3.png" class="clean-center" alt="img/paral-doetoe-3.png" />
There you have it, so general and beautiful and simple.
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<a class="author" href="">Juan Reyero</a>
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