Show that formula for area of rectangle does not depend on unit shape

Question

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?

Answer 1

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.

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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.

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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.

Answer 2

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.

Answer 3

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.