# 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

minded

very nice.........and helpful

Solution of (6×+8)^2=6×^2+2ab+8^2

Answer in hindi

Actually good but as well as now its common.. way to prove but it will b more better if u find out some more way 2 prove ,becoz that may give u more greetings from everyone

Then also u tried ur best to help us so ..good job keep it up

9771090194

Diploma Electrical Engnnering

Mathematics =3 semester Question and Answer send me please VVI Question

How a2B 2 +aB2

A+b =a2*bxv

Tens

Vvvvv good

Could u give me this kind of geometrical explanation for (a+b)³.?

Very nice and helpful to me thanx for it..........

Very Good.

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

How are you friend

Thanks for the article . It was very helpful to me

good

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

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

Good, like it

Very help ful

thakyou

Nice explanation it is easy to learn

Thank you sir

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

thanks for this

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

can you plz proo another formula

Very useful and interesting

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

Good

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Tnx useful for me

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How d formula z used in daily life

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(y)

very good

very useful

explainted very well.

really helpful proof

nice prooof

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

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

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

not bad

good

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