Proof of Distributive property of Multiplication over Addition

The distributive property of multiplication over addition can be proved in algebraic form by the geometrical approach. It is actually derived in mathematics by the area of a rectangle.

Introduction to Basic Geometric steps

basic geometric steps for distributive law

Take a rectangle but its dimensions are unknown. Assume, the width of this rectangle is $a$.
Divide the rectangle across its length at a point for dividing it as two different rectangles. Assume, the length of one rectangle is $b$ and the length of second rectangle is $c$.

You can observe that the widths of both rectangles are same.

Find Sum of Areas of Rectangles

The length and width of first rectangle are $b$ and $a$ respectively. Now, find the area of this rectangle.

areas of rectangles for distributive rule

$Area \,=\, b \times a$
$\implies$ $Area \,=\, a \times b$
$\,\,\, \therefore \,\,\,\,\,\,$ $Area \,=\, ab$

The length and width of second rectangle are $c$ and $a$ respectively and find the area of this rectangle.

$Area \,=\, c \times a$
$\implies$ $Area \,=\, a \times c$
$\,\,\, \therefore \,\,\,\,\,\,$ $Area \,=\, ac$

Now, find the sum of the areas of the two rectangles.

$\,\,\, \therefore \,\,\,\,\,\,$ $Area \,=\, ab+ac$

Evaluate the Area of Rectangle

Now, join both rectangles to get the actual rectangle. The lengths of both rectangles are $b$ and $c$ respectively and their sum is equal to the length of the actual rectangle.

area of rectangle for distributive property

Therefore, the length and width of the rectangle are $b+c$ and $a$ respectively and find the area of this rectangle.

$Area \,=\, (b+c) \times a$
$\implies$ $Area \,=\, a \times (b+c)$
$\,\,\, \therefore \,\,\,\,\,\,$ $Area \,=\, a(b+c)$

The area of the rectangle is $a(b+c)$ and it is divided as two rectangles. So, the area of the rectangle is equal to the sum of the areas of the two small rectangles.

$\therefore \,\,\,\,\,\,$ $a(b+c) \,=\, ab+ac$