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

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

split shapes for 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.

joining two rectangles
  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 $a^2-b^2$.



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