Question

Prove Pythagoras theorem

Answer 1

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It is the theorem that relates the squares of the two peprendicular sides of a right triangle with the square of its hypotenuse. This theorem, makes possible to know any one of the lenghts of the three sides of a right triangle by knowing the lengths of the other two sides. Also, it can be used to other triangles and it can give us the law of cosines, through which we can calculate the length of the third side of a triangle by knowing the lengths of the other two sides and their containing angle. In general, if the sides of a right triangle are a, b and c, with the hypotenuse being a, then we can write:

a^2 = b^2 + c^2

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Answer 2

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):



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