The Pythagorean Theorem: A Modern Proof
Date: 04/14/97 at 05:24:34 From: Rosalyn Adams Subject: Proving the Pythagorean Theorem I am currently doing research on the Pythagorean Theorem for my Algebra II/Trig. class. I've looked at a lot of math Web sites, but none seems to answer my question. I know that the Pythagorean Theorem works and I can show how it works, but what I really need to know is why it works. Also would you have any ideas for my project, which consists of explaining the theorem and its important parts using mathematical language and symbolism?
Date: 04/14/97 at 10:39:55 From: Doctor Mitteldorf Subject: Re: Proving the Pythagorean Theorem Dear Rosalyn, Pythagoras's proof of his theorem is rather hard to follow, but there is a modern proof that is much easier. It does involve several lines of algebraic manipulation, which you must be prepared to do: Draw a horizontal line which you'll divide into two parts, labeled a and b. Now draw a vertical line at each side, and another horizontal line on the top, so you have a square of side a + b. Divide your square into four sections. There's an a^2 in one corner, a b^2 in the diagonally opposite corner, and two side-rectangles of area a*b. Now start over again and re-draw the same square, but split the sides differently. Each line going around the square clockwise should be divided long-short, long-short, long-short, long-short. Now when you connect the 4 dividing points on each line segment, you can make a diagonal square in the middle. In other words, your figure consists of a diagonal square whose side is the hypotenuse c, surrounded by four a-b-c right triangles. Now if you express the area of the big square in two different ways and clear out the algebra, you will be left with the Pythagorean theorem. Please spend a little time with it - try to make sense out of it on your own, and work with the algebra aiming for a^2 + b^2 = c^2. If you still have trouble, please write back. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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