# Proof of a+b whole square in Geometric method

## Formula

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

The $a+b$ whole square formula can be derived in geometrical method. In this method, the areas of square and rectangle are used to derive this algebraic identity geometrically.

We know that

1. The area of square is ${(side)}^2$
2. The area of rectangle is $length \times width$

### Proof

Consider a square. Geometrically, the length and width are equal in it. So, the area of the square is equal to square of its side.

Assume, the length of the side is divided into two unequal parts. The length of one part is $a$ and the length of the other part is $b$. Therefore, the length of the each side is $a+b$.

Therefore, the area of this square is ${(a+b)}^2$.

Now take the ruler and draw a perpendicular line to the horizontal side at the point where the side is divided into unequal parts. Similarly, repeat the same action to the vertical side.

The two perpendicular lines divide the big square into four rectilinear figures in which two figures are square and the other two are rectangle.

Now, calculate the area of each geometrical figure.

1. The first figure is a square and its side is $a$. Therefore, the area of this square is $a^2$.
2. The second and third figures are rectangle figures. The lengths of its sides are $a$ and $b$. Therefore, the area of each rectangle figure is $ab$.
3. The fourth figure is a square but length of its each side is $b$. Therefore, the area of this square is $b^2$.

Geometrically, the area of big square is equal to the sum of areas of all four rectilinear figures.

Therefore, the area of the big square is ${(a+b)}^2$. The areas of two small squares are $a^2$ and $b^2$ and the areas of two small rectangle figures are $ab$.

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

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

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

Geometrically, it is proved that square of $a+b$ can be expanded as $a$ squared plus $b$ squared plus $2ab$.

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