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

The $a-b$ whole square formula can be derived in geometric method. In this method, a square is divided into one small square and two rectangles. The relation between areas of them is used in deriving this algebraic identity.

Take a square and assume the length of each side is $a$. So, the area of the square is $a^2$.

Draw a perpendicular line to any two opposite sides of the square and it divides them into two parts.

Assume, the length of one division of both sides is $b$ and obviously the length of the other division of same sides is $a-b$. The perpendicular line splits the square into two rectangles.

Leave the small rectangle and divide the big rectangle into two parts but remember the perpendicular line should divide opposite sides such that the parts of each divided opposite line have lengths $a-b$ and $b$.

The big rectangle is split into one square and one small rectangle.

The process has transformed the square whose area is $a^2$, into two rectangles and one small square. Now, find the areas of them.

The length of each side of small square is $a-b$. So, the area of this square is ${(a-b)}^2$.

The lengths of both sides of big rectangle are $a$ and $b$. Therefore, the area of this rectangle is $ab$.

Finally, the lengths of both sides of small rectangle are $a-b$ and $b$. Therefore, the area of this rectangle is $b(a-b)$.

It is time to obtain the expansion of $a$ minus $b$ whole square formula. The area of square whose side’s length is $a-b$ can be derived by subtracting the sum of areas of both rectangles from the actual square whose area is $a^2$. The relation can be expressed in mathematical form as follows.

${(a-b)}^2 = a^2 -[ab+b(a-b)]$

$\implies {(a-b)}^2 = a^2 -(ab+ba-b^2)$

$\implies {(a-b)}^2 = a^2 -(2ab-b^2)$

$\implies {(a-b)}^2 = a^2 -2ab + b^2$

$\therefore \,\,\,\,\, {(a-b)}^2 = a^2 + b^2 -2ab$

Save (or) Share

Copyright © 2012 - 2017 Math Doubts, All Rights Reserved