May 20, 2003, 05:57
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#1
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Emperor
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Geeeeometry (Help!)
Most of you are probably familiar with the Pythagorean Theorem. In a right triangle, a² + b² = c²
I need the most simple way to prove it. Does anybody know?
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May 20, 2003, 06:11
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#2
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I know but i'm too lazy to write it in English.
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May 20, 2003, 06:12
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#3
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"Beware of he who would deny you access to information, for in his heart he dreams himself your master" - Commissioner Pravin Lal.
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May 20, 2003, 06:20
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#4
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Thanks Eli, I think the second link will be most helpful, I'll go have a look at it.
I just have to figure out what "perpendicular" means.
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May 20, 2003, 06:30
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#5
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At right-angle to.
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I'm building a wagon! On some other part of the internets, obviously (but not that other site).
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May 20, 2003, 06:35
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#6
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Deity
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I think you can start bt the Cosine Rule and work it as a special case.
Cosine Rule: a 2 = b 2 + c 2 - 2bc*cos A
A = 90°
cos(A) = 0
Q.E.D.
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May 20, 2003, 06:38
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#7
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Thanks guys. Learning, learning.
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May 20, 2003, 06:42
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#8
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Prince
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if you can get your hands on the book 'Fermats Last Theorem' theres a (proper) proof in the appendics of that (my copy is at work)
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May 20, 2003, 07:19
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#9
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Warlord
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I remember that once an Applet proving the Pythagorean Theorem won a Sun Contest, i was unable to find it, anyway the following seems to me well done too.
http://www.utc.edu/~cpmawata/geom/geom7.htm
Obviously you need a Java-enabled browser, just try the link, if it starts you have it
P.S. No, i found it, it's here:
http://www.math.ubc.ca/~morey/java/pyth/
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May 20, 2003, 07:36
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#10
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Prince
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Quote:
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Originally posted by Urban Ranger
I think you can start bt the Cosine Rule and work it as a special case.
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Yeah, but then you have to prove the Cosine Rule.
Although there is a trigonometrical proof somewhere...
I like Angelo's first one the best. I think that is the one Pythag himself used.
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Concrete, Abstract, or Squoingy?
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May 20, 2003, 12:08
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#11
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Emperor
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Stephen Hawkings did a good proof of it in the Brief History of Time.
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May 20, 2003, 12:15
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#12
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Deity
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Quote:
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Originally posted by Immortal Wombat
Yeah, but then you have to prove the Cosine Rule.
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Well, you could, but you're not asked to do it.
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(\__/) 07/07/1937 - Never forget
(='.'=) "Claims demand evidence; extraordinary claims demand extraordinary evidence." -- Carl Sagan
(")_(") "Starting the fire from within."
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May 20, 2003, 12:42
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#13
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Prince
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The area of the large square can be found in two ways:
1) Area of large square = (x+y)^2
2) Measure each element in the large square. The area of each triangle is 0.5xy. The area of the tilted square is z^2
Area of large square = 4*(area of each triangle)+ area of tilted square
= 4 (0.5xy)+z^2
1) and 2) give different expressions but must be equivilent because the represent the same area
so (x+y)^2 = 4(0.5xy)+z^2
= x^2+y^2+2xy = 2xy+z^2
= x^2+y^2+z^2 which is the theorem.
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May 20, 2003, 13:22
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#14
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Emperor
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or you could just make a right triangle that's 3x4x5
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May 20, 2003, 13:27
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#15
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something a humanities major would say.
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May 20, 2003, 14:52
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#16
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Guest
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Thats called an example orange, not a proof.
With that triangle, you could claim all right triangle have sides a,b,c such that a+b=c+2... which of course is not true
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