how do I prove the pythagorean thereom?

Answer 1

An easy way is to draw a square, mark off two sections of each side, call one a and one b, so that the 4 points between the a and b on each side would connect to make another square inside the big square. Call the sides of this little square c. The area of the big square is (a+b)^2 = a^2 + 2ab + b^2

It's also the sum of the areas of the 4 right triangles (each ab/2; 4ab/2 = 2ab) and the little square inside (c^2)

so a^2 + 2ab + b^2 = 2ab + c^2

subtract 2ab from both and you have it

Answer 2

Proving the Pythagorean Theorem has been the pass time of ameteur and professional mathematicians for centuries. One president of the United States, James Garfield, has even given an original proof. But Garfield was an unusually gifted man, unlike many of our most recent residents of the Oval Office.

Here is one proof often found in math books and other sources.

First, draw a square and divide each side into line segments with ratio a/b. One segment is "a" units long and the other is "b" units long. Although one could divide each side evenly, we want the proof to be of a general case of a triangle, not a special case, so divide the sides unevenly. Just make sure you divide all sides the same way.

Now, from each point of division on each side, draw another line segment connecting it to the points of division on the two adjacent sides. When you have done this, you will be left with a skewed square inside the larger square (i.e. the inner square will be tilted so that its sides don't parallel the sides of the outer square). Each side of the inner square now has a length which we shall call "c" units long.

Next we need to find the area of the larger square. Since the length of each edge is (a+b) units long, the outer square is (a+b) x (a+b) square units in area.

Note this: (a+b) x (a+b) = (a+b)² = a² + 2ab + b², is the area of the outer square.

Now, we calculate the area of the inner square. Since its edge is "c" units long, its area is c x c = c² units.

Notice from your construction that you have created 4 triangles which frame the edges of the inner square. These triangles are all right triangles with equal bases and equal heights, whose hypotenuses are all "c" and whose areas we can calculate easily, because we have a formula for doing that. If you have made "a" longer than "b," use "a" as the height of each triangle and "b" as the base of each triangle. If you have made "b" longer, use it as the height, and "a" as the base.

Recall that the area formula for a right triangle, or any other triangle for that matter, is: A = ½ bh, where b and h are the length of the base and height respectively. Depending on which segment length represents the base or height of the triangle, we plug those values into this formula to obtain this:

A = ½ ab.

So the area of each of these triangles is given by ½ ab. But we have 4 triangles, so we need to multiply the preceding formula by 4 to find the total area of all the triangles. When we do this, we get:

A(4 triangles) = 2 ab.

Now we get to the heart of the matter. We note that the area of the larger square is equal to the sum of the area of the smaller square plus the area of the 4 triangles. Mathematically then, this is the case:

(a+b) x (a+b) = (a+b)² = a² + 2ab + b² = c² + 2 ab --->

a² + 2ab + b² - 2ab = c² + 2 ab - 2ab ---> a² + b² = c².

Thus, we have proven the Pythagorean Theorem: The sum of the squares of each leg of a right triangle equals the square of the hypotenuse of the triangle.

Answer 3

'Stone' and 'namenot' clearly do not know a thereom from a hole in the ground ☺

'hayharbr' has the right of it. And, BTW, the proof he presents was originally published by John Calvin Coolidge, a pretty fair amateur mathematician and the 30'th President of the United States. All in all there are, IIRC, some 200+ proofs of the Pythagorean Theorem. They range from the fairly simple (such as Coolidge's proof) to some that are quite sophisticated (and difficult to follow ☺)


Doug

Answer 4

a^2 + b^2 = c^2

Answer 5

even though this is not really a prove but the phythagorean theroem can be derived from the cosine law

c^2=a^2+b^2-2bc cosC

c^2=a^2+b^2-2bc cos90
c^2=a^2+b^2-2bc(0)
c^2=a^2+b^2

Answer 6

It is unprovable. That is why it is called a thereom instead of Pythagorean Law.

Answer 7

prove it? or use it?

I can still use it but its been 15 years since geometry class so I'm not sure how to prove it anymore. Sorry.

Answer 8

This page
http://www.cut-the-knot.org/pythagoras/index.shtml
has a lot of different proofs. Enjoy.

Answer 9

create a triangle to scale and find the sides using trigonomic fuctions (sine, cosine tangent ) then plug them into the equation and show that it works possibly.

Answer 10

You can use an actual model, like making a triange out of straws or toothpicks and measuring it.