Why is $\text{meter} \times \text{meter}$ a legitimate unit for measuring area? How did we transition from meters, as a unit of distance, to $\text{m}^2$, as the unit of area, I do understand that for example if we define a square that is 1meter long and 1meter wide and use it as our basic unit for measuring areas, then the area of a rectangle (expressed as the number of my meter-squares)would simply be its length times its wide.
But that aside, what justifies $\text{m}\times \text{m}$ as a unit for measuring the area, in other words, why a distance "times" another distance yields an area?  
 A: All the theorems are just mere consequences of axioms which have no proof and must be assumed(they are not silly). Now the area of a square of $1m×1m$ is defined to be $1m^2$. So now in any figure, you may fit all these squares to get the actual "area". So you actually  multiply with numbers $>1$, which signifies you are fitting that many unit squares in your figure.
A: I was also thinking about the reason why we can do this. Then I thought about a world in which we have discrete lengths. The following image is constructed of such sticks. In the $x$-direction we have three sticks and in the $y$-direction we have 2 sticks. We can see that the enclosed area is consisting of $2 \times 3$ squares. Hence, if we define the area of one square as $\text{m} \times \text{m}$ the total area is equal to $6 \text{ m}^2$.

A: You can divide any surface by infinitesimally small rectangles.now in these small rectangles for infinitesimally small length there is their respective width. So, if you sum up all the areas of those rectangles then you get the total area of that surface no matter how it looks like. 
A: The concept of measuring area in the way we do goes back to the very early days of mathematics, when lengths were measured relative to the length of a reference segment and areas were (usually) measured relative to the area of a square built on the same segment.
Why a square and not some other geometric figure, such as a circle?
Possibly the reason is that squares provide a relatively easy way to measure areas:
they're highly symmetric,
they can be measured by the length of just one side (unlike general rectangles),
and the plane can be tessellated by congruent copies of a single square (unlike circles).
Moreover, humans thousands of years ago had a tendency to lay out their settlements in patterns that were either rectangular or round, so choosing equilateral triangles probably did not seem as natural a choice for a "unit area" figure.
Perhaps if our mathematics had been invented by bees, area might have been measured by regular hexagons (in the pattern of a honeycomb) rather than in the pattern of plowed fields! But we are humans, so we use human mathematics.

If we had chosen a different way to measure area, most likely it would just have introduced a constant conversion factor into our distance-to-area formulas.
After all, you can take a region such as the inside of a triangle or a parallelogram, double it in size, and then cut it into four regions all congruent to the original reason. Triple it in size, then you can cut it into nine regions congruent to the original region.
If the rule for the area of a triangle or parallelogram were something other than "multiply a length times a length times a constant," 
you would quickly run into situations where the area of the whole was clearly not the sum of areas of its parts. Back in the real world, that sort of thinking would lead to all kinds of complications with land deals.
A: A justification for why areas are understood as products of lengths requires a treatment based on measure theory.
In measure theory, we work with measurable spaces $(X, \Sigma)$, where $\Sigma$ satisfies:

*

*$X\in\Sigma$

*$S\in\Sigma\implies{X\setminus{S}}\in\Sigma$

*$\forall{n\in\mathbb{N}},S_n\in\Sigma\implies\bigcup_{n\in\mathbb{N}}S_n\in\Sigma$.

These spaces can be equipped with functions $\mu:\Sigma\rightarrow[0,\infty]$, which satisfy:

*

*$\mu(\emptyset)=0$

*$\forall{i,j\in\mathbb{N}},E_i\cap{E_j}=\emptyset\implies\mu\left(\bigcup_{n\in\mathbb{N}}E_n\right)=\sum_{n\in\mathbb{N}}E_n$.

These functions are called measures, and these are the functions that generalize the concepts of length, area, and volume, to arbitrary measurable spaces.
Imagine that you have measure spaces $(X_0,\Sigma_0,\mu_0)$ and $(X_1,\Sigma_1,\mu_1)$, where we imagine that $X_0$ and $X_1$ are spaces that have dimension $1$. For $S_0\in\Sigma_0$ and $S_1\in\Sigma_1$, $\mu_0(S_0)$ and $\mu_1(S_1)$ can be interpreted to be the "lengths" of the sets $S_0$ and $S_1$ respectively, in the context of these spaces. A very natural idea is to consider the Cartesian product of these spaces, since Cartesian products are just universally important in mathematics. The product space of these two measure spaces is given by $(X_0\times{X_1},\sigma(\Sigma_0\times{\Sigma_1}),\mu)$, where $X_0\times{X_1}$ is the Cartesian product of $X_0$ and $X_1$, $\sigma(\Sigma_0\times{\Sigma_1})$ is the $\sigma$-algebra generated by $\Sigma_0\times{\Sigma_1}$, and $\mu$ is the measure defined by $\mu(S_0\times{S_1})=\mu_0(S_0)\mu_1(S_1)$. Since $X_0$ and $X_1$ are spaces of dimension 1, $X_0\times{X_1}$ is necessarily a space of dimension 2, and since $\mu_0$ and $\mu_1$ are lengths, it must be the case that $\mu$ is a product of lengths. Since $\mu$ is also an area, it follows that area must be a product of lengths.
