(a+b)^2 answer it qiuckly
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(a+b)2=a2+2ab+b2
✔️✔️Proof :-
(a+b) ^2 is similar to (a+b)*(a+b)
So (a+b)*(a+b)=a*(a+b) + b*(a+b)
solving brackets :
a*(a+b) + b*(a+b)
=(a*a) +(a*b) +(b*a) +(b*b)
=a^2 + ab + ba + b^2
=a^2 + 2ab + b^2
(a+b) ^2 = a^2 + 2ab + b^2
✔️✔️✔️✔️For geometric proof :-
✔️Area of part 1 :
Part 1 is a square of length a.
Therefore area of part 1 = a2 ----------- (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 = b2 ---------------(iv)
So, Area of square of length
(a+b) = (a+b)2 = (i) + (ii) + (iii) + (iv)
so we get...
✔️(a+b)2 = a2 + ba + ba +b2
i.e. ➖(a+b)2 = a2 + 2ab + b2➖
✔️✔️This formula is also used in proving Pythagoras Theorem....
✔️✔️Proof :-
(a+b) ^2 is similar to (a+b)*(a+b)
So (a+b)*(a+b)=a*(a+b) + b*(a+b)
solving brackets :
a*(a+b) + b*(a+b)
=(a*a) +(a*b) +(b*a) +(b*b)
=a^2 + ab + ba + b^2
=a^2 + 2ab + b^2
(a+b) ^2 = a^2 + 2ab + b^2
✔️✔️✔️✔️For geometric proof :-
✔️Area of part 1 :
Part 1 is a square of length a.
Therefore area of part 1 = a2 ----------- (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 = b2 ---------------(iv)
So, Area of square of length
(a+b) = (a+b)2 = (i) + (ii) + (iii) + (iv)
so we get...
✔️(a+b)2 = a2 + ba + ba +b2
i.e. ➖(a+b)2 = a2 + 2ab + b2➖
✔️✔️This formula is also used in proving Pythagoras Theorem....
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Step-by-step explanation:
(a+b)^2 = a^2ab+b^2
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