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 September 17th, 2010, 06:48 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Pythagorean theorem Prove that Pythagorean theorem works. Well I know that I need to have three squares on each side of the triangle.
 September 17th, 2010, 07:19 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 The triangle should have a right angle, and have just one square on each side! I think you already know how to prove the theorem.
 September 17th, 2010, 11:10 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Pythagorean theorem "Cut the knot." There are infinite algebraic and geometric proofs.
 September 18th, 2010, 12:47 PM #4 Global Moderator   Joined: May 2007 Posts: 6,784 Thanks: 707 Re: Pythagorean theorem Trivial corollary - Place any geometric figure on the hypotenuse, then place geometrically similar figures on each leg. The sum of the areas of the figures on the legs equals the area of the figure on the hypotenuse.
 September 18th, 2010, 01:16 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 That can be achieved by constructing just one line segment, a perpendicular drawn from the right-angle vertex to the hypotenuse, since the resulting smaller triangles have the same angles as the original triangle. Their areas obviously sum to the area of the original triangle, so the Pythagorean theorem follows immediately from the theorem that the areas of similar figures are proportional to the squares on corresponding sides.
September 18th, 2010, 09:16 PM   #6
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Re: Pythagorean theorem

Quote:
 Originally Posted by mathman Trivial corollary - Place any geometric figure on the hypotenuse, then place geometrically similar figures on each leg. The sum of the areas of the figures on the legs equals the area of the figure on the hypotenuse.
Yes, that reminds me of solving this pythagorean theorem place three semicircles on each side of the right triangle. 1/2 pi (1/2 a)^2 + 1/2 pi (1/2 b)^2 = 1/2 pi (1/2 c)^2, therefore a^2 + b^2 = c^2.

 September 19th, 2010, 10:10 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 How did you obtain the result for the semicircles without already knowing the Pythagorean theorem?
 September 19th, 2010, 06:50 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: Pythagorean theorem
September 20th, 2010, 07:13 AM   #9
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Re: Pythagorean theorem

Hello, johnny!

Quote:
 Prove that Pythagorean theorem works.

This is my favorite proof . . .

Code:
a        b
*-----*-----------*
|    *    *       | a
|   *       c *   |
b |  * c            *
| *              *|
|*            c * |
*              *  | b
a |   *  c      *   |
|       *    *    |
*-----------*-----*
b        a

$\text{We have a square of side }c\text{ inscribed in a square of side }a+b$.

$\text{The total area of the square is: }\;\begin{Bmatrix}\text{4 right triangles with area }\frac{1}{2}ab \\ \\ \\ \\ \text{and a square with area }c^2 \end{Bmatrix}$

[color=beige]. . [/color]$\text{Hence: }\;A \;=\;4\left(\frac{1}{2}ab\right)\,+\,c^2 \;=\;2ab\,+\,c^2\;\;[1]$

$\text{But we know the total area is: }\;(a\,+\,b)^2 \;=\;a^2\,+\,2ab\,+\,b^2\;\;[2]$

$\text{Equate [2] and [1]: }\;a^2\,+\,2ab\,+\,b^2\;=\;2ab\,+\,c^2$

[color=beige]. . [/color]$\text{Therefore: }\;a^2\,+\,b^2\;=\;c^2$

 September 20th, 2010, 03:32 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Pythagorean theorem Hello soroban! That is my favorite as well. It is the one I have used over the years that I have had the best success with easily demonstrating the truth of this theorem to students of algebra/geometry.

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