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By area, I mean that the space occupied by any shape. For example, the space occupied by the rectangle - is length multiplied by breadth. If I don't consider volume - or I consider the thickness has unity then the area is nothing but a volume - and volume is the space occupied by the shape. I am just trying to understand when I multiply the length with breadth - how come it covers the entire region surrounded by the rectangle. This is the formula that I have learnt from my childhood. However, I am unable to visualize it. How come multiplication gives the area - or space occupied by the rectangle considering its thickness as unity?

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    $\begingroup$ I guess you can look at an arbitrary small square and simply say that it has an area of one, where area means the amount of space it takes up (which we all agree has some meaning...). The area of a rectangle is now "how many of these squares fit inside the rectangle?". If we fit a squares horizontally and b vertically then we have ab squares and an area of ab. After that it is sums of fractions of squares then limits of series of squares and... $\endgroup$ – Paul Sep 1 '14 at 9:22
  • $\begingroup$ Possible duplicate of this question. $\endgroup$ – Lucian Sep 1 '14 at 11:04
  • $\begingroup$ No, I think this question is different from why unit is square. $\endgroup$ – dexterous Sep 1 '14 at 11:34
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Let's have a rectangle with sides $5$m and $4$m. How many squares with a side of length $1$m (unit squares) can you put in the rectangle? The answer is $20$, which can be seen if you drew straight lines that would divide the rectangle into squares:

\begin{array}{|c|c|c|c|} \hline \hphantom{i}& \hphantom{i} & \hphantom{i} &\hphantom{i} & \hphantom{j}\\ \hline & && &\\ \hline & && &\\ \hline && & &\\ \hline \end{array}

Since there are $4$ rows with $5$ squares each, there are $5+5+5+5$ squares which (maybe coincidentally) equal $4\cdot 5$. This simply says only that the rectangle has $20$ times bigger area than the unit square. It is very appropriate to define the area of unit square as $1\text{m}^2$, since in that way the area of every rectangle can be computed as product of lengths of it's sides and consequently, knowing the area and one of the sides, you can simply get the other as $$\frac{20\text{m}^2}{4\text{m}} = 5 \frac{\text{ m}^2}{\text{m}} = 5 \text{m}\text{,}$$ i. e. you can calculate with units (meters in the case here) as they were numbers.

What about the rectangle with sides of length $2.1$m and $4.2$m? Well, just cut off appropriate part of the unit squares in the last row/column. Again, the area is $2.1\cdot 4.2$ square meters.

Now, we have rectangles covered. What about more complicated shapes? Intuitively speaking, we are dealing with strange shapes like circles for example, in the same way as we are dealing with the rectangle with non-integer lengths of sides. We only have to cut off some parts of unit squares in a more complicated way. If we want to cover a circle, you can imagine that those cuts must be finer and finer when you are approaching the edge.

That's about visualization. The truth behind the finer and finer cutting and so on, is integration. First of all, the area of rectangle with sides $a$ and $b$ is defined as $ab$ (which is well motivated and can be written as integral). Then, this must be somehow reasonably generalized to more complicated shapes. Read Wikipedia's artilce (section formal definition).

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  • $\begingroup$ So, basically it is dividing a figure into unity, and the integral multiple of unity is nothing but an area of the figure. So, I think, circle area is an approximate number that's why they have pi in it. You can never break such figures into a definite parts. I am just thinking now, when mechanical Engineers build precision instrument comprising of circles - what they do? Is it more empirical in nature? I am saying this because pi is a kind of indefinite number. Because pi value can be computed till the universe is there. $\endgroup$ – dexterous Sep 1 '14 at 10:29
  • $\begingroup$ You cannot break such figures into definite parts, true, but circle area is precisely $4\pi r^2$ and you can draw a circle with area precisely $4\pi$ by using compasses. I believe that the diameter is what concern mechanical engineers, not the area. It is worth mentioning that you cannot make a circle with the area $1$. $\endgroup$ – Antoine Sep 1 '14 at 11:41

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