Question

derivation of pythagoras theorem

Answer 1

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 A = c2

And there are four triangles, each one has an area of A =½ab
So all four of them combined is A = 4(½ab) = 2ab

So, 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

Step-by-step explanation:

Pythagoras' theorem :-

→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Step-by-step explanation:

It's prove :-

➡ Given :-

→ A △ABC in which ∠ABC = 90° .

➡To prove :-

→ AC² = AB² + BC² .

➡ Construction :-

→ Draw BD ⊥ AC .

➡ Proof :-

In △ADB and △ABC , we have

∠A = ∠A ( common ) .

∠ADB = ∠ABC [ each equal to 90° ] .

∴ △ADB ∼ △ABC [ By AA-similarity ] .

⇒ AD/AB = AB/AC .

⇒ AB² = AD × AC ............(1) .

In △BDC and △ABC , we have

∠C = ∠C ( common ) .

∠BDC = ∠ABC [ each equal to 90° ] .

∴ △BDC ∼ △ABC [ By AA-similarity ] .

⇒ DC/BC = BC/AC .

⇒ BC² = DC × AC. ............(2) .

Add in equation (1) and (2) , we get

⇒ AB² + BC² = AD × AC + DC × AC .

⇒ AB² + BC² = AC( AD + DC ) .

⇒ AB² + BC² = AC × AC .

$\huge \green{ \boxed{ \sf \therefore AC^2 = AB^2 + BC^2 }}$

Hence, it is proved.