How is $(a + b)^{2} = a^{2} + 2 a b + b^{2}$

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Solution

REF.Image
Consider square PQRS with side = a+b
Divide PQRS in 4 rectangles, area af square = area of rectangled
area of square = $(a + b)^{2} = S^{2}$
$= (a + b) (a + b)$
$= a^{2} + a b + b a + b^{2}$
$= a^{2} + 2 a b + b^{2}$
Hence proved

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Similar questions

If $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ then what is the value of $a^{2} + b^{2}$

(a + b)^{2}

expands as

a^{2} + b^{2} + 2 a b

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. Question 3

Which of the following is correct?
a)

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

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

(a - b)^{2} = a^{2} - b^{2}

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

Q. Evaluate:

\sqrt{\frac{a^{2} + 2 a b + b^{2}}{a^{2} - 2 a b + b^{2}}}

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How is (a+b)2=a2+2ab+b2

REF.ImageConsider square PQRS with side = a+bDivide PQRS in 4 rectangles, area af square = area of rectangledarea of square = (a+b)2=S2=(a+b)(a+b)=a2+ab+ba+b2=a2+2ab+b2Hence proved

How is $(a + b)^{2} = a^{2} + 2 a b + b^{2}$

REF.Image
Consider square PQRS with side = a+b
Divide PQRS in 4 rectangles, area af square = area of rectangled
area of square = $(a + b)^{2} = S^{2}$
$= (a + b) (a + b)$
$= a^{2} + a b + b a + b^{2}$
$= a^{2} + 2 a b + b^{2}$
Hence proved