# Pythagorean Theorem

**Topics:**Pythagorean theorem, Triangle, Right triangle

**Pages:**9 (2616 words)

**Published:**October 7, 2012

More than 4000 years ago, the Babyloneans and the Chinese already knew that a triangle with the sides of 3, 4 and 5 must be a right triangle. They used this knowledge to construct right angles. By dividing a string into twelve equal pieces and then laying it into a triangle so that one side is three, the second side four and the last side five sections long, they could easily construct a right angle. A Greek scholar named Pythagoras, who lived around 500 BC, was also fascinated by triangles with these special side ratios. He studied them a bit closer and found that the two shorter sides of the triangles squared and then added together, equal exactly the square of the longest side. And he proved that this doesn't only work for the special triangles, but for any right triangle. Today we would write it somehow like this: a2 + b2= c2. In the time of Pythagoras they didn't use letters yet to replace variables. (They weren't introduced until the 16th century by Vieta.) Instead they wrote down everything in words, like this: if you have a right triangle, the squares of the two sides adjacent to the right angle will always be equal to the square of the longest side. We can't be sure if Pythagoras really was the first person to have found this relationship between the sides of right triangles, since no texts written by him were found. In fact, we can't even prove the guy lived. But the theorem a2 + b2= c2 got his name. Another Greek, Euclid, wrote about the theorem about 200 years later in his book called "Elements". There we also find the first known proof for the theorem. Now there are about 600 different proofs. Today the Pythagorean theorem plays an important part in many fields of mathematics. For example, it is the basis of Trigonometry, and in its arithmetic form it connects Geometry and Algebra.

PROOFS

Euclid's Proofs

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The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details are next.

Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.

For the formal proof, we require four elementary lemmata:

1. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (side-angle-side). 2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude. 3. The area of a rectangle is equal to the product of two adjacent sides. 4. The area of a square is equal to the product of two of its sides (follows from 3). Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.

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Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF The proof is as follows:

1. Let ACB be a right-angled triangle with right angle CAB. 2. On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.[10] 3. From A, draw a line parallel to BD and CE. It will...

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