Question

prove Pythagoras geometrically

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Answer 1

Answer:

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Answer 2

Proof of the Pythagorean Theorem using Algebra

We can show that a2 + b2 = c2 using Algebra

Take a look at this diagram ... it has that "abc" triangle in it (four of them actually):

Squares and Triangles

Area of Whole Square

It is a big square, with each side having a length of a+b, so the total area is:

A = (a+b)(a+b)

Area of The Pieces

Now let's add up the areas of all the smaller pieces:

First, the smaller (tilted) square has an area of: c2

Each of the four triangles has an area of: ab2

So all four of them together is: 4ab2 = 2ab

Adding up the tilted square and the 4 triangles gives: A = c2 + 2ab

Both Areas Must Be Equal

The area of the large square is equal to the area of the tilted square and the 4 triangles. This can be written as:

(a+b)(a+b) = c2 + 2ab

NOW, let us rearrange this to see if we can get the pythagoras theorem:

Start with: (a+b)(a+b) = c2 + 2ab

Expand (a+b)(a+b): a2 + 2ab + b2 = c2 + 2ab

Subtract "2ab" from both sides: a2 + b2 = c2

hope it helps

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