Sunday, June 11, 2006

Pythagorean Theorem

After a long break from posting, I'm back with something a bit more fundamental. I'd like to offer a graphical version of Euclid's proof to the Pythagorean Theorem (I'm just going to draw the shapes, rather than name them by their vertices).

First, though, we need to make sure that you're comfortable with the fact that shearing a shape doesn't change its area. In the following diagram, I've shown that shearing a rectangle doesn't change its area, because you can get the same shape by chopping off a triangle from one side of the rectangle and putting it on the other side.



So you can see that if shearing a parallelogram maintains its area, shearing any triangle (since a triangle is just half of a parallelogram) should maintain its area, as well.

Now, we'll move on to the Pythagorean Theorem. We'll take an arbitrary right triangle, and attach squares to each of the sides.



Now, we'll show that the c square has the area of both the a square and the b square. We'll accomplish this by showing that, in the following diagram, the two blue rectangles have the same area, and the two green rectangles have the same area as well.



First, notice that these two triangles (in the following diagram) are identical, and so they have the same area.



Now shear the first triangle so that it becomes half of the a square.



Then shear the second triangle so that it becomes half of the left portion of the c square.



And since the two brown triangles have equal area, the two blue rectangles must have the same area as well (since they each have twice the area of one of the brown triangles)



Now, we will use the same logic on another pair of identical triangles.



Shear the first triangle so that it becomes half of the b square.



Shear the second triangle so that it becomes half of the right portion of the c square.



And since these two triangles have equal area, the two green rectangles must have the same area as well.



And so the c square has an area equal to the sum of the areas of the a square and the b square. In other words c2 = a2 + b2

proof_02
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7 comments:

  1. Anonymous8/1/06, 5:49 PM

    WOW!

    I'm 33 years old, and only now do I finally 'get' Pythagorean Theorem, thanks to your mathproofs blog. :)

    I always remembered the rules and all the number-plugging I needed to do, but seeing it done graphically, I now not only know it, I understand why. Cool!

    So thanks... from a fellow RAD guy ;)

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  2. Simon Horlick10/10/06, 12:49 AM

    Thats great, I don't think anyone has ever explained it in such an easy to understand way. Thanks :)

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  3. Anonymous2/12/07, 7:20 AM

    there's a much more simple graphical proof of this theorem.
    Use two squares with sides a+b length, split on into 2 squares(one a*a and one b*b) and two rectangles(both a*b). Split the other into a big square of c*c and four triangles (l=b, h=a). Then split the two rects from the first square into four equal triangles (l=b, h=a). by subtracting the equal area of the four triangles from both squares, we're left with a c*c square that must be equal to two squares, one a*a, the other b*b.

    easier to see graphically
    -Adam from UMASS

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  4. Anonymous9/12/07, 9:49 AM

    shear brilliance

    ReplyDelete
  5. Anonymous4/24/10, 1:02 PM

    thank you

    ReplyDelete
  6. Anonymous8/29/10, 5:31 AM

    Sweet website, I hadn't come across mathproofs.blogspot.com earlier in my searches!
    Keep up the fantastic work!

    ReplyDelete
  7. How to do Limits in Calculus3/1/12, 2:54 AM

    nomis me to agree with your comment it was well defined and great explanation of pythagorean theorem and special point was its picture explanation.

    ReplyDelete