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I understand the formula that the two sides must be multiplied, but what is the reason for this? I think of area as the space an enclosed figure occupies in a two-dimensional plane.

It seems to me that if you multiply one side by the other you are actually increasing one side by the length of the other. Sort of like stacking the bottom line until it reaches the top edge of the left line. The problem is, this stacking depends on the thickness of the line and if I make this line infinitely thin, we can still calculate the area... So, what gives?!

line stacking

What was the logic used to come up with this incredible formula? I do not know much about advanced calculations so keep it simple. Thank you all in advance.

user3704920
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  • Consider it the definition of area. It is foundational for other notions of area. – Simon S Oct 27 '15 at 14:27
  • @SimonS not sure what you mean. Isn't area the space 2 dimensional enclosed figures occupy? – user3704920 Oct 27 '15 at 14:28
  • The thickness of the lines is just an artifact of the way we have to draw things so they are visible to our eyes. Mathematically, a line has no thickness whatsoever, no even an atom's width. We define a rectangular area typically using Cartesian coordinates, e.g., the rectangle prescribed by the vertices $(0,0), (a,0), (a,b), (0,b)$ with $a, b \geq 0$. And then we define the area of that rectangle to be $ab$. – Simon S Oct 27 '15 at 14:31
  • It kind of started with finding area of square of side with unit length. It has been defined as the unit area. Then later on finding area of rectangle with integral length. In the end it is generalised for all real length as the multiplication of two adjacent sides' length. – Bhaskar Oct 27 '15 at 14:35
  • Draw a $3$ by $3$ grid. This grid is made up of nine small squares. Those one-by-one squares have an area of $1$ by definition, so the whole thing has an area of $9$. This argument works for any $m$ by $n$ rectangle, as long as $m$ and $n$ are whole numbers. – Akiva Weinberger Oct 27 '15 at 14:40

3 Answers3

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What is the area of a rectangle of $w$ width and unit height? There are $w$ unit squares so the area is $A=w \times (1 \mbox{ unit square}) = (w\mbox{ unit squares})$

Now stack these rectangle to get $h$ height. You need $h$ of them and hence the total area is $A=h \times (w\mbox{ unit squares}) = (h \cdot w\mbox{ unit squares})$

John Alexiou
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Here is a heuristic argument.

It all started with counting how many objects of the same kind (say, barrels of wine, sacks of grains, etc), are there when they are packed in rectangular arrays. Quite early in human history - I am willing to conjecture this -- people understood that if you pile $7$ rows of barrels, and each row has, say, $6$ barrels, then you don't have to count all the $42$ barrels separately - it is sufficient to count one row, and just take the result $7$ times. Hence the word "times" we use for multiplication.

That multiplication of the lengths of the sides of a rectangle give it's area is just the realisation that if you pack your objects in an ideal manner -- you fill up the area completely, as well as the lengths of the sides.

uniquesolution
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  • Here you mention the actual objects that occupy a space, but in terms of a square, what is it that is occupying this space? – user3704920 Oct 27 '15 at 14:38
  • Use your imagination :) we can fill up the square with lots of things. – uniquesolution Oct 27 '15 at 14:39
  • that does not answer my question. You define the area by the things inside of it. But just 4 lines composing a square does not have anything inside to define its area. – user3704920 Oct 27 '15 at 14:40
  • Of course it answers your question. A square has inside it anything i choose to put inside it. Also, I insisted that my argument is heuristic, so that it would futile on your part to insist that I will develop a theory of measure here based on the notion of content. This is not the intention of my argument, although it can be done completely rigorously. – uniquesolution Oct 27 '15 at 14:44
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Ok, finally got it thanks to all answers above. My logic was wrong because we do not define an area just using plain lines and measure the space they occupy. That would not make any sense. Instead we have a unit of measure as mentioned by @akivaweinberger and @ja72, for example feet. A square that masures 5 feet in each side would mean that it is filled with smaller squares of 1 feet each, filling the circle with 25 squares. Thanks all.

user3704920
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