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I've seen how to visualize the formula for getting the area of a parallelogram. The first picture shows 2 ways which give the same result of Area = base * height.

http://tinypic.com/view.php?pic=2na1kr7&s=5 (link to first pic)

However, the first time I thought about it before being show the actual formula, I thought it was Area = side * side.

http://tinypic.com/view.php?pic=2rhpbvk&s=5 (link to second pic)

The 2 sides of the parallelogram are 2 vectors, x and y.

If you place copies of vector y length of x times along the vector x, and then you add all lengths of vector y's (which is the same as length of vector y * length of vector x), it seems like you also get the area of the parallelogram.

I know that you'll end up with a different result since the height is always less than the length of the vector y (unless they were parallel) but the method in the second picture seems correct.

I also haven't seen the second picture's method being taught online so I'm guessing it's actually incorrect. What I want to know is what makes it incorrect because it really seems like it's correct despite giving a different answer from the usual base * height formula.

Thanks in advance for any help.

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  • $\begingroup$ "If you place copies of vector y length of x times along the vector x, and then you add all lengths of vector y's (which is the same as length of vector y * length of vector x), it seems like you also get the area of the parallelogram." is what is causing the trouble $\endgroup$ – Amr Jun 29 '13 at 20:52
  • $\begingroup$ The magnitude of the cross-product of the vectors $x$ and $y$ should give you the area of the parallelogram, not the product of the magnitudes of the vectors $x$ and $y$. $\endgroup$ – Maazul Jun 29 '13 at 20:58
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To visualise this, we must think of everything in 2 dimensions. We cannot use 1 dimensional constructions (lines) to make up the areas.

Imagine a parallelogram and we want to find its area similar to the method you described here

If you place copies of vector y length of x times along the vector x, and then you add all
lengths of vector y's (which is the same as length of vector y * length of vector x), it
seems like you also get the area of the parallelogram.

Construct a very small rectangle, height $\delta$ and width the base of the parallelogram. It will have a bit of the left (or right) edge sticking out, but this is countered by the fact the right (or left) edge doesn't quite fit in.

Now, to use this to find the area, imagine stacking all of these rectangle on top of one another each one shifted slightly to best fit inside the parallelogram. Note that the number of rectangles needed is actually related to the height of the parallelogram, not the length of the other edge and that number is $h/\delta$, where $h$ is the height of the parallelogram.

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  • $\begingroup$ Nice explanation that gets to the heart of the matter. $\endgroup$ – André Nicolas Jun 29 '13 at 21:23
  • $\begingroup$ I see what you mean and it makes a lot of sense now! One thing about the number of rectangles needed though. If the height δ of the small rectangle was equal to 1, wouldn't that make the number of rectangles needed just y/1=y? y is greater than the height so it isn't the height. $\endgroup$ – user84399 Jun 29 '13 at 21:27
  • $\begingroup$ Yes, that should say $h/ \delta$ where $h$ is the height of the paralellogram. $\endgroup$ – rwolst Jul 3 '13 at 10:28
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In the second picture, the area of the parallelogram is shown as the sum of lots of small parallelograms, which continue to follow the area = base $\times $ height rule even in the limit as their width tends to $0$.

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  • $\begingroup$ Sorry about that, there shouldn't be any spaces between those copies of y. So they're just lines stuck beside each other. $\endgroup$ – user84399 Jun 29 '13 at 21:29

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