Why is the area of a $3$-by-$3$ square $3\times 3$ and not $3+3$?

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When we need to find the area of a square, we multiply the sides. For example, the area of a square with one side as $3$ cm will be $3\times 3 \text{ cm}^2$.

My question is: What is the logic behind this? Why can't we just add the sides and get the area? For example, $3+3=6\text{ cm}^2$.

Thanks!

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It is good to ask why! That is how all the science was discovered. For simplicity, one could think of the "area" in (centimeters, for example) as:

the number of squares of size 1 cm required to occupy a given region.

To be more accurate the region has to have no depth, it has to be flat, i.e. 2-dimensional only.

In the example below we have an Orange square of area $1\ \mathrm{cm}^2$ ($1\ \mathrm{cm}$ squared) and we want to fill the big square with such Orange squares. The area represents the number of Orange squares required.

You can count the number manually one or notice that:

The required number is $3+3+3=9$

The number $9$ here tells us that we need $9$ such Orange tiles (squares).

There is no magic, its all about the definition you set for the area. People agree on one definition and then device means to calculate the value corresponding to the definition.

or use multiplication to speed things up to get the correct answer $3 \times 3=9$

I hope this helps a bit.

enter image description here

Edit: In response to the comment, here is what a $25m$ by $0.5m$ would look like after filling it with square tiles each having side of $1m$. Each tile can be split in half. In the drawing, the rectangle uses 2 halves except at the last tile, it uses only 1 half a tile. You don't need further adjustments.

enter image description here

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You can fit $9$ one-by-one squares into one three-by-three square.

What we imagine as "area" has the property that if something has area $x$, and some other thing has area $y$, then the two things together have area $x+y$. This propety can only hold for squares if their area is the product, not sum of their side lengths.