Question

DERIVE PYTHAGORAS THEOREM​

Answer 1

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

DONE!

Answer 2

Answer:

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Given:- In triangle XYZ , Angle Y = 90°

Find:- XZ²= XY²+YZ²

Solution:- From triangle XOY and XYZ

${ \sf{ x = x}} \: \: { \sf{ \to{(common \: side)}}}$

${ \sf{o = y}} \: \: { \to{ \sf{(both \: are \: equal \: to \: 90)}}}$

So, ∆ XOY ~ ∆ XYZ (A•A Similarity)

${ \implies{ \sf{ \frac{xo}{xy} = \frac{xy}{xz} }}}$

Do cross multiplication:-

XO × XZ = XY²

From, ∆YOZ and ∆XYZ

${ \sf{z = z }}\ \: \: { \to{ \sf{common \: angle}}}$

${ \sf{o = y}} \: \: \: \: { \sf{ \to{both \: are \: 90 \: degree}}}$

so, ∆ YOZ ~ ∆ XYZ ( A•A similarity)

XO × XZ + OZ × XZ = (XY² + YZ²)

⇒ (XO + OZ) × XZ = (XY²+ YZ²)

⇒ XZ × XZ = (XY² + YZ²)

⇒ XZ² = (XY² + YZ²)

Hence proved✔