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.

- 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.

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

$Area \,=\, b \times a$

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

$\,\,\, \therefore \,\,\,\,\,\,$ $Area \,=\, ab$

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.

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

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.

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.

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

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