

Proofs for the Pythagorean Theorem
Name: yendor
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1995
Question:
I recently heard that there was another proof to the pythagorean theorem. I
was wondering if someone would tell me what it is. Please do not use
trigonometry for I am severely lacking in an understanding of it. Since it
is impossible for you to know which proof I am aware of, it would be helpful
to explain both.
Replies:
Here is a derivation that is based on area. Draw a square. Starting with
the upper left corner, go right some distance, say, less than half way,
along the top edge and mark the edge at that spot. Denote this distance by
A, and denote the distance from this point to the upper right corner of the
square by B. [Thus, our square has sides of length (A + B)]. At the right
upper corner, go down a distance A and mark this point, then draw a line
from this point to the first point you marked. This will give a right
triangle with legs A and B whose right angle is the upper right corner of
the square. Let the length of its hypotenuse be denoted by C. Continue
this: move a distance A left from the square's lower right corner, and
connect this point to the second point, etc., until you have drawn an
inscribed square of side C within the original square. Now, compute areas.
On one hand,the area of the original square is
(A + B)^2 = A^2 + 2*A*B + B^2.
It is also equal to the area of the four triangles plus the area of the
inscribed square; this is 4*(A*B/2) + C^2 = 2*A*B + C^2. So we have two
expressions for the same area; therefore those expressions are equal:
A^2 + 2*A*B + B^2 = 2*A*B + C^2
Subtracting 2*A*B from both sides yields the theorem.
rcwinther
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Update: June 2012

