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Whenever I search for some kind of reasoning behind the formula for area of rectangle(length$*$width) it is introduced by counting unit squares. I have recently asked a question where my understanding seems to have fallen apart.

They were trying to convince me that the formula itself does not depend on the shape that tessellates the enclosed space of a rectangle. I looked for an answer from existing questions here and found yet another answer where it is explained in terms of unit squares.

This confuses me and I would like for someone to show why they were trying so hard to make me see that it does not depend on unit squares. What makes this formula special? What is it derived from if not from the simple fact of counting unit squares? Can anyone show how the formula would work then by counting unit triangles or unit rectangles?

Martin Sleziak
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Michael Munta
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  • I think people are trying to tell you that we do not always count unit squares to find area, even though the final result is measured in unit squares. If we derived a formula for area of rectangle as width times length, we do not really need to count unit squares every time, right? – Vasili Nov 09 '21 at 17:17
  • You need some definition of area for your question. And the area is defined for rectangles that way. They are like the building blocks of areas of more complicated sets. – Johan Aspegren Nov 13 '21 at 13:04
  • You can define the area via some other sets, say, disks. Then your formula can be proved. Anyway, those rectangle areas are what we mean by areas. If you have some set function that does not agree with that formula, it's not area. – Johan Aspegren Nov 13 '21 at 13:44

3 Answers3

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For definiteness, let's assume explicitly we're working in Euclidean plane geometry. Particularly, we have concepts of points and lines, subsidiary concepts of polygons, and numerical measures of length and angle.

The area of a rectangle (four-sided polygon bounded by two pairs of parallel segments meeting at right angles, including the boundary segments) is defined to be the product of the side lengths. Congruent rectangles (i.e., differing by a Euclidean motion) obviously have the same area.

Less obviously but pretty clearly, if a rectangle is subdivided into two rectangles sharing one side, the area of the large rectangle is the sum of the areas of the subrectangles. Inductively, if a rectangle is subdivided into finitely many non-overlapping subrectangles, the area of the large rectangle is the sum of the areas of the subrectangles.

It's reasonable to try extending this concept of area to non-rectangular subsets of the plane. Here's a naive approach: Let $P$ be a subset of the plane. To invent a name, let's say a "tile covering" of $P$ is a finite collection of rectangles such that

  • The set $P$ is contained in the union of the rectangles;
  • Distinct rectangles share no interior points. (Our rectangles must generally all be aligned on the same Cartesian axes.)

Each tile covering of $P$ has an area, the sum of the areas of the selected rectangles. The area of a tile covering is an upper bound for what we mean by the area of $P$. Now consider all tile coverings of $P$, or rather, consider the set of real numbers that are the area of some tile covering of $P$. The greatest lower bound of this set is a "candidate" for the area of $P$; we might call this greatest lower bound the outer area of $P$ and denote it $OA(P)$.

Dually, we might similarly look at the set of all "tile inclusions," finite collections of non-overlapping rectangles whose union is contained in $P$. We could define the area of a tile inclusion to be the sum of the areas of its rectangles. Finally, we could consider the least upper bound of the areas of all tile inclusions of $P$. This number $IA(P)$, the inner area of $P$, is another candidate for the area of $P$.

Finally, we might say that the set $P$ has area if $IA(P) = OA(P)$, and define the common value to be the area of $P$, acknowledging that many subsets of the plane do not have area in this sense. On the bright side, polygons do have area in this sense. Triangles, particularly, have area equal to one-half the base times the height (regardless which of three choices we make of base-and-height!). This should be believable, but it's probably also easy to see that calculating the area of a general region, such as a disk or other set whose boundary is not a finite collection of line segments, is generally a non-trivial undertaking.

Now to the question, What if we used sets other than rectangles as the basis of area? Any choice for an "area primitive" (such as squares, triangles, L-shaped hexagons, or what) assigns a numerical measure to some plane sets and not to others using the strategy outlined above. The question, as I see it, amounts to How do we know our "new" measure of area coincides with the rectangle measure of area? Unsurprisingly, the issue comes down to whether or not the "new" area of rectangles themselves coincides with length times width.

For squares as primitives (with the area of a square defined to be the square of the side length), it turns out rectangles have the same area as before. This is related to other answers that speak of counting how many squares of given size fit into a rectangle, or cover a rectangle.

For triangles as primitives (with the area defined to be one-half the base times height), the new area of rectangles again coincides: A rectangle can be diagonally subdivided into two congruent right triangles. For isosceles right triangles...you get the idea. Consequently, any polygonal primitive for which tile coverings exist leads to the same definition of area, since every polygon can be "tile covered" by triangles.

As others have commented and answered, area is not some Universal, True Concept, but a mathematical definition made in the context of a mathematical model. Even speaking mathematically, the preceding does not adequately define area of (e.g.) subsets of a sphere or other smooth surfaces in Euclidean three-space. In "reality" matters are even more complicated. Mandelbrot, at the very latest, articulated that length (e.g., the coast of Britain) is a scale-dependent quantity. Area has the same character when we consider real objects: tree leaves, lungs, the surface of the earth.

Searching for Jordan content, Lebesgue measure, and geometric measure theory will unearth as many additional details as desired.

Andrew D. Hwang
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  • Ok, this is a very good answer, but it's a bit difficult for me to visualize it. I'm not that proficient in maths so I try to think intuitively. I will try asking like this. So, we have a formula for rectangle length times width. With actual numbers for example it's $9×3=27$. Now, if, as you say, it doesn't matter which area primitive we choose then a rectangle that is $9$ by $3$ has $27$ unit squares perfectly covering it and also it has $27$ unit triangles covering it? I find that hard to believe... it should have more triangles, no? How is the formula universal then? – Michael Munta Nov 14 '21 at 17:37
  • As you note, unit squares and unit right isosceles triangles do not have the same area. My claim (related to your example, assuming I understand) is that if we use unit right-isosceles triangles to cover a region instead of squares (therefore using about twice as many fundamental units since a square can be split into two non-overlapping triangles), then the "area" of a polygonal region reckoned using squares will be the same as the area reckoned using right isosceles triangles. [...] – Andrew D. Hwang Nov 14 '21 at 18:39
  • If it matters, even when we speak of using "unit squares", we need to use squares of varying "unit" size. The shorter the side length, the more accurately we can approximate the area of a region. The same is true if we use triangles of some sort. But in general there is no "natural" way to pass from a tile covering by squares to a tile covering by triangles. It's best to think of calculating area as a "primitive-dependent process" that happens (for a large class of plane subsets) to yield the same numerical value for area regardless of the choice of primitive. – Andrew D. Hwang Nov 14 '21 at 18:43
  • So we come down to this point and this is the whole essence I am trying to understand: the length times width gives us the area in unit squares, same numbers as above would be $27$ unit squares. I am not refuting the fact that we can convert these unit squares into an area represented by isosceles triangles after it has been first calculated for unit squares. Then we have $54$ unit triangles as a result. But, $9×3$ will never give $54$, it is tuned to give a precise result in square units. So I am trying to understand the formula before it has been calculated and converted to triangles. – Michael Munta Nov 17 '21 at 08:31
  • If we wanted to get unit isosceles triangles from the start we would have to do length times width times $2$ and then we would have a formula tuned to count unit triangles. This is what I'm trying to say that when we look at the formula from that angle then the formula does not work for any shape other than unit squares. Am I making any sense now? – Michael Munta Nov 17 '21 at 08:35
  • I wouldn't say you aren't making sense, only that I may not understand what underlies your question. :) <> When people speak of "counting unit squares", that can't be taken in the sense of using a single size of square: We have to be willing to subdivide arbitrarily finely, i.e., to fix a temporary unit of length, make a tile covering by "unit squares", then consider squares of (say) half the side length and repeat, ad infinitum. For instance, the area of a disk of radius 1 cm cannot be approximated in the sense of my answer using squares with 1 cm sides. [...] – Andrew D. Hwang Nov 18 '21 at 12:47
  • If we do the same with triangles (e.g.), we can consider unit-area triangles (perhaps right isosceles with legs $\sqrt{2}$, or equilateral with sides $2/3^{1/4}$), form a tile covering, count the triangles to obtain an approximate area, then subdivide and repeat to obtain succissively better approximations. The claim is that in this situation, or in any such situation, the notion of area of a plane region will have the same numerical value ("in the limit as the tile size decreases to zero"). If this doesn't clarify, we might ought to move this to asynchronous chat.... – Andrew D. Hwang Nov 18 '21 at 12:56
  • Making the triangles so infinetisemally small would make the fact that if they are triangles or squares or circles irrelevant, is that what you are trying to say? This is a calculus concept and something I did not want to involve in this discussion because my motivation was related to the question I linked and concerns the method how ancient people might have derived this formula and I kinda doubt they thought this way at that moment in time. So, say that we can't make them infinitesimally small and suppose we have one standard triangle(half a unit square) that is our unit of area.[...] – Michael Munta Nov 18 '21 at 13:20
  • We calculate the area by counting how many triangles fit, but we also notice we can derive a formula to count these triangles. Will that formula be length times width, then? Or length times width times $2$? – Michael Munta Nov 18 '21 at 13:22
  • Scattered brief comments: (i) The area of every parallelogram (e.g. rectangle) is its base times height by geometric rearrangement, so the "one-half base times height" formula for area of a triangle can be deduced without calculus if we grant that area is additive under dissection and unchanged by Euclidean motions. (ii) If we estimate the area of a large polygon by tile covering with unit-area triangles, we count the triangles, just as we would with squares. (iii) As Aristotle and Archimedes put it, the modern definition entails a potential infinity, not an actual one. – Andrew D. Hwang Nov 18 '21 at 13:42
  • How can we move this discussion to chat? – Michael Munta Nov 18 '21 at 14:07
  • Let us continue this discussion in chat. – Michael Munta Nov 18 '21 at 16:51
  • If I might ask one more question, these set-like explanations you provided in your answer, what branch of mathematics studies these things in terms of sets? I would like to read more about this because I don't fully understand your answer. – Michael Munta Nov 27 '21 at 11:04
  • Like, you call a part of the plane a set, what is it a set of? And what means that $P$ is contained in the union of rectangles? – Michael Munta Nov 27 '21 at 11:12
  • Here are two search terms and initial references: Jordan content (which uses finite coverings), Lebesgue measure (which permits countably-infinite coverings). Note that in each, the area of a rectangle is defined to be the product of its linear dimensions. Both pages have diagrams that should help. (I probably ought to have included some in the answer.) – Andrew D. Hwang Nov 27 '21 at 15:24
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Let's say you're given two lengths. Actual lengths, not convertible to numbers without choosing a unit of measurement. Adding lengths is easy: simply attach two segments of these lengths and take the combined length as a result. Now say you want to define what it means to multiply them. When multiplying lengths, the natural result should be an area, since areas scale just as the product of two lengths would, no matter which units you choose. Now which area should it be? The obvious choice would be the area of a rectangle with the given lengths as side lengths. It is easy to see that this definition satisfies distributive laws.

With this choice, the area of a rectangle is equal to the product of its side lengths, by definition of the product. Also it should now be clear why multiplying a unit of length $u$ by itself is called the square of that unit: because $u \cdot u$ is now the area of a square with side length $u$. This is also the reason why multiplying something with itself is called squaring.

There could be other possible definitions of a product of lengths, for example the area of an ellipse with the given lengths as axes, which still satisfies the distributive law, but is not as natural or convenient. In common usage and generality, we define multiplication of multidimensional volumes in such a way that the $m$-volume of a unit $m$-dimensional hypercube times the $n$-volume of a unit $n$-dimensional hypercube is always equal to the $(m+n)$-volume of a unit $(m+n)$-dimensional hypercube (or more generally, the $m$-volume of any $m$-dimensional shape times the $n$-volume of any $n$-dimensional shape is equal to the $(m+n)$-volume of the cartesian product of both shapes). This spares mathematicians a whole bunch of unit conversion.

Magma
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Say, we agree on that a scaling of the unit does not change the number maximal number of scaled unit squares in a rectangle. Let us suppose that we have a rectangle of integer height $a = 1$ and rational width $b = q = \frac{n}{m}$. For these rectangles we have an area $1*b.$ Which is the same by scaling my $m$ if we have $a=m$ and $b=n$ for $n,m \in \mathbf{N}$. Then for height and width $c,d \in \mathbf{R}$ the area can be calculated via scaling. Say we wan't to calculate the area of the rectangle $[x,x+c]\times[y,y+d].$ We scale the unit to $\min(\frac{1}{c},\frac{1}{d})$, say to $\frac{1}{c}.$ Then we have $c_1 = 1$ and $d_2 = \frac{d}{c}.$ We have a sequence of rationals going to $d.$ So we can define $$\text{Area}([x,x+c]\times[y,y+d]) = c\lim_{i=1}q_i,$$ where you have $\lim_{i=1}q_i = d.$ For each of those approximations the definition of area is justified by the unit square method. Then we must check that this definition, that extends the notion of unit square calculations satisfies what we except from the notion of area. It is preserved by euclidean motions.

Johan Aspegren
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