# Proof of $(a+b)(a-b)$ formula in Geometric Method

## Formula

$(a+b)(a-b) \,=\, a^2-b^2$

### Proof

$a$ and $b$ are two literals. They form two binomials $a+b$ and $a-b$ by summation and subtraction respectively. The product of them is written as $(a+b)(a-b)$ in mathematics. Its expansion is equal to $a^2-b^2$. It is used as an algebraic identity in mathematics and can be derived in geometrical approach in three simple steps on the basis of areas of square and rectangle.

#### Subtracting areas of squares

1. Take a square, whose length of each side is $a$ units. Therefore, the area of the square is $a^2$.
2. Draw a small square with the side of $b$ units at any corner of the square. So, the area of the small square is $b^2$.
3. Subtract the square whose area is $b^2$, from the square whose area is $a^2$. It forms a new geometric shape, whose area is the subtraction of the area of small square from actual square. So, the area will be $a^2-b^2$.

#### Dividing the shape as two rectangles

1. Draw a straight line to split the geometric shape. It divides the geometric shape as two rectangles.
2. Look at the upper rectangle. Geometrically, the length of this rectangle is $a$ and its width is $a-b$.
3. Similarly, consider the lower rectangle. The length of the rectangle is $a-b$ and its width is $b$.

#### Proving the Algebraic identity

The width of the upper rectangle is $a-b$ and the length of the lower rectangle is also $a-b$ geometrically. If the lower rectangle is rotated by $90^\circ$, then the lengths of both rectangles become same and it is useful to join them together as a rectangle.

1. Firstly, separate both rectangles.
2. Rotate the lower rectangle by $90^\circ$ and then join both rectangles.
3. Look at the new rectangle, whose length is $a+b$ and width is $a-b$. Therefore, the area of this rectangle is $(a+b)(a-b)$.

In first step, it is derived that the area of the subtracted shape as $a^2-b^2$ but in this case, the area of the big rectangle is determined as $(a+b)(a-b)$.

In fact, the subtracted figure is transformed as a big rectangle. Thus, the areas of both geometric shapes should be equal geometrically.

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

Geometrically, it is proved that the product of the binomials $a+b$ and $a-b$ is equal to $a^2-b^2$.