Question

Prove the theorem Pythagoras theorem​

Answer 1

Step-by-step explanation:

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

DONE!

Now we can see why the Pythagorean Theorem works ... and it is actually a proof of the Pythagorean Theorem.

This proof came from China over 2000 years ago!

There are many more proofs of the Pythagorean theorem, but this one works nicely.

• Hope it helps you.
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Answer 2

Answer:

Pythagoras stated relationship among 3 sides in a right triangle,& also in an obtuse triangle & an acute triangle too… But Pythagoras Theorem is the relationship in a right triangle… & the other two are extensions of Pythagoras theoram…. as these 2 are proved by the right triangle theorem.

The theorem is stated as follows:

(1) In any right triangle the square of the hypotenuse is equal to the sum of the squares of the other 2 sides.

ie, if we construct a square on the hypotenuse another squares on the other two sides of the triangle.

Then area( square on the hypotenuse) = area( square on the base) + area( square on the perpendicular)

Like in right triangle ABC right angled at ‘B'

AC²= AB² + BC²……………(1)

(2) Now the extension of Pythagoras theorem in an obtuse triangle ABC , obtuse angled at B is→

AC²= AB²+ BC² + 2BC.BX ( where BX is the projection of AB on BC

(3) Now the extension of Pythagoras theorem in an acute triangle ABC, considering AC side opposite to acute angle B is→

AC²= AB² + BC² - 2BC.BX ( where BX is the projection of AB on BC.