# Why (a+b) 2 = a2+b2+2ab ?

Why** (a+b) ^{ 2} = a^{2}+b^{2}+2ab ?**

Ever wondered how was the above formula derived?

Probably the answer would be yes and is simple. Everybody knows it and when you multiply (a+b) with (a+b) you will get a plus b whole square.

**(a+b) * (a+b) = a ^{2} +ab + ba + b^{2} = a^{2 }+ 2ab + b^{2}**

But how did this equation** a plus b whole square** became generalized.

Let’s prove this formula geometrically.( Please refer to the pictures on the side)

- Consider a line segment.
- Consider any arbitrary point on the line segment and name the first part as ‘
**a’**and the second part as ‘**b**’. Please refer to**fig a**. - So the length of the line segment in
**fig a**is now (a+b). - Now, let’s draw a square having length (a+b). Please refer to
**fig b**. - Let’s extend the arbitrary point to other sides of the square and draw lines joining the points on the opposite side. Please refer to
**fib b**. - As we see, the square has been divided into four parts
**(1,2,3,4)**as seen in**fig b.** - The next step is to calculate the area of the square having length
**(a+b).** **As per**to calculate the area of the square : we need to calculate the area's of parts 1,2,3,4 and sum up.**fig b ,****Calculation**: Please refer to**fig c**.

**Area of part 1 :**

Part 1 is a square of length a.

Therefore area of part 1 = **a ^{2 }**---------------------------- (i)

**Area of part 2 :**

Part 2 is a rectangle of length : b and width : a

Therefore area of part 2 = length * breadth =** ba **-------------------------(ii)

**Area of part 3:**

Part 3 is a rectangle of length: b and width : a

Therefore area of part 3 = length * breadth =** ba **--------------------------(iii)

**Area of part 4:**

Part 4 is a square of length : b

Therefore area of part 4 =** b ^{2 }**----------------------------(iv)

So, Area of square of length (a+b) = (a+b)^{2 }= (i) + (ii) + (iii) + (iv)

Therefore :

(a+b)^{2} = a^{2} + ba + ba +b^{2}

i.e.** (a+b) ^{2 }= a^{2} + 2ab + b^{2}**

Hence Proved.

This simple formula is also used in proving The Pythagoras Theorem.Pythagoras Theorem is one of the first proof in Mathematics.

In my view, in mathematics when a generalized formula has been framed there will be a proof to prove and and this is my small effort to exhibit one of the proof's.

Will definitely come up with some more.

Also please find the video proof.

## Comments

good

not bad

Actually, it is proved by using Pascal's triangle, with the general formula: (a+b)^n = (n 0) a^n * b^0 + (n 1) a^n-1 * b^1 + ..... + (n n-1) a^1 b^n-1 + (n n) a^0 * b^n-1

Or shorter (a+b)^n = n Σ k=0 (n k) a^n-k * b^k

This is fantastic; I had no idea, and I have never seen it explained in this way. Great job!

in the same way, do u know (a-b)^2

nice prooof

really helpful proof

explainted very well.

very useful

very good

coooool :D

Loved it :)

(y)

Not Bad but it can be shorterand thanks for giving me answer . THANK YOU HUBPAGE.COM.THANK YOU VERY MUCH

thanx

wwhat awomderfull

tnx was helpful

thankyou

How d formula z used in daily life

it is very useful for our learnig

Thanks. Very useful

Tnx useful for me

Spcl thnx

Good

(a-b)2=a2+b2-2ab.

Good job

very useful

good job good site

Very useful and interesting

can you plz proo another formula

How many a and how many b in a+bwhole square..?

thanks for this

I really liked the proof and I never knew that it can be proved in this way tooooo......

Thank you sir

Nice explanation it is easy to learn

thakyou

Very help ful

Good, like it

hello bawa! today I really found out the mystry of the Universe.

Thanks for this but can you give some questions related to this and its answer .please

good

Thanks for the article . It was very helpful to me

How are you friend

Nice sir...........""

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