Question

$\boxed {\begin {array}{c}\\ \sf Prove\: that \\ \sf {a^2+b^2= (a+b)^2-2ab \\ \end {array}}$​

Answer 1

Step-by-step explanation:

Given Question:-

Prove that a^2+b^2 = (a+b)^2-2ab

Proof:-

Consider a square of side (a+b) units

Then area of a square = side×side sq.units

Area of the square = (a+b)(a+b) sq.units

Area of the square = (a+b)^2 sq.units-------------(1)

Now

Draw two lines as shown in the figure

It is divided into 4 parts

Area of the 1st part = Area of the square of side

'a' units = a×a = a^2 sq.units ---------(2)

Area of the 2nd part = Area of the rectangle of

length a and breadth b units

=>ab sq.units ----------(3)

Area of the 3rd part = Area of the square of the

side b units = b×b = b^2 sq.units ----------(4)

Area of the 4th part = Area of the rectangle of

length a units and breadth b units

= ab sq.units --------------(5)

Total area of the square =

Area of (1st+2nd+3rd+4th )parts

=>(a+b)^2 = a^2+ab+b^2+ab

=>(a+b)^2 = a^2+2ab+b^2

=>(a+b)^2 - 2ab = a^2+b^2

Answer:-

a^2+b^2 = (a+b)^2 - 2ab

Hence proved