Proving the Pythagorean Theorem using Congruent Squares
Date: 12/5/95 at 15:14:58 From: Anonymous Subject: Proof of Pythagorean Theorem Dear Dr. Math, A friend of mine is irked because of constant use of the Pythagorean Theorem, which he has not seen proven. I have seen it proven before, but could not quite recount the proof. I have heard the theorem called the "most proved theorem," so this ought to be an easy and a very prolific search, but could you find a proof of the theorem. Thanks, Johnny Vogler
Date: 3/6/96 at 15:15:45 From: Doctor Dusty Subject: Re: Proof of Pythagorean Theorem This is a very neat little proof of the Pythagorean Theorem, but it might look a little funky because of the graphics. Here it goes: b a b a _____.__ _____.__ a | | a a | | . . . | b | | | | b | | b b | . |_____.__| |__._____| a b a a b First of all, construct two congruent squares, both with sides a + b. You know that the area of each square should be (a + b)^2 Now connect the points in the first square to create two squares and two rectangles. When you add the areas of those four up, you get a^2 + ab + ab + b^2 which is also equal to the area of the entire square, (a + b)^2. Connect the points of the second square to form another square and call the length of each side of the inner square c. You now have four right triangles and one square. Add the areas of the 5 figures together and you have .5ab + .5ab + .5ab + .5ab + c^2 The areas of both squares are equal; we agreed to that in the beginning. Set the area equation of the first equal to that of the second. a^2 + 2ab + b^2 = 2ab + c^2 Subtract 2ab from both sides and you have a^2 + b^2 = c^2 where a is one leg of the right triangle, b is the other leg, and c is the hypotenuse. (I got that from looking at the second square.) I just think that this proof is really neat. I hope it calms your friend down. -Doctor Dusty, The Math Forum
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