I was recently asked "Why is the area of a circle irrational?", to which I replied that it was not necessarily irrational—there are of course certain values for $r$ that would make $\pi r^2$ rational. She proceeded to clarify, "But the area of a square of side length $1$ is rational, yet the area of a circle of radius $1$ isn't. What's so special about the square?"

My answer to this was of course "We measure area in square units, hence the area of a unit square is one square unit." Perfectly contented, I headed home. On the way back, however, it dawned on me that this was unsatisfactory. Why must we measure area in square units? Area is one of those quantities that one could scale by any constant, such as $1 \over \pi$, and have almost every property preserved. Is there a fundamental reason why we don't say the area of a unit square is $1 \over \pi$ circular units?

This further confused me when I noticed that the phrase $x^2$ being spoken as "x squared" is a consequence of using the unit square, not a justification for it. If the Ancient Greeks used circular units, we would no doubt pronounce $x^2$ as "x circled"...

I looked for a justification for using the unit square as the basis unit for area, and of course the obvious one is calculus. The Fundamental Theorem of Calculus provides an easy definition of area, thanks to integrals. The area of a unit square is simply:

$$\int_0^1 dx$$

And since the integral is the antiderivative, it is convenient to say the area of a unit square is $1$.

But I am still not sure this is unsatisfactory, as the intuitive connection between integrals and area relies on the concept that the area of a rectangle is $lw$. Had this been off by a factor of $1 \over \pi$, the connection between antiderivatives and areas would certainly be more complicated... but many mathematical formulae have $\pi$ or $1 \over \pi$ in them, and it would be a stretch to proclaim them inelegant.

In the end, is there something fundamental about the unit square? Why not a unit triangle or unit circle? Or is it so merely because the Ancient Greeks did it that way?

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    $\begingroup$ We use lines because a line is the shortest distance between two points. It would only make sense to use circular units if we measured distance with semicircular arcs. $\endgroup$
    – Brad
    Commented May 18, 2014 at 19:33
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    $\begingroup$ Your integral for the area of a unit square is dimensionally incorrect. It should be a double integral: $$\int_0^1\int_0^1\:dx\:dy$$ $\endgroup$
    – 小太郎
    Commented May 19, 2014 at 7:13
  • $\begingroup$ @小太郎 I'm not sure how it's dimensionally incorrect. From x = 0 meters to x = 1 meter, for f(x) = 1 meter, the integral of f(x) should be 1 square meter. I've vaguely seen area being done with double integrals, but all the examples of measurement of area under a curve in my textbooks was done with single integrals. $\endgroup$
    – prosfilaes
    Commented May 19, 2014 at 9:12
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    $\begingroup$ Why the area of a circle with radius 1 is irrational while a square of side 1 is not, is purely due to how we defined the value 1. We could have said 1 = the number of times the diameter of a circle goes into its circumference and had the area of a circle being r<sup>2</sup>. $\endgroup$
    – Holloway
    Commented May 19, 2014 at 14:54
  • $\begingroup$ Why square units? - Why not ? :-) $\endgroup$
    – Lucian
    Commented May 19, 2014 at 17:36

8 Answers 8


Filling areas with circles is a futile endeavour, since there's always a free space left over in between neighbouring discs. Furthermore, there's a direct connection between the geometric shapes called rectangles, and multiplication, inasmuch as the latter's commutative property becomes self-evident when one pictures a $m\times n$ grid, which obviously holds as many elements, then rotates it at a $90^\circ$ angle, thus transforming it into an $n\times m$ grid, which obviously holds just as many elements as before being rotated, since it is still the same grid. So there are both practical as well as theoretical advantages to using rectangular shapes.

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    $\begingroup$ Although rectangles certainly stack better, it is just as impossible to partition a disk into finitely many squares than the partition a square into finitely many circles (even if "partition" is relaxed to allow parts to overlap along their boundaries). With infinitely many parts, both are possible. $\endgroup$ Commented May 19, 2014 at 0:57
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    $\begingroup$ @MarcvanLeeuwen: I did not have an already-shaped area in mind: That would have been self-serving. Imagine that you are free to choose any shape you want, as long as the resulting surface is contiguous. Even with these liberties in place, circles are not adequate for the task. $\endgroup$
    – Lucian
    Commented May 19, 2014 at 1:47
  • $\begingroup$ I expect that another important factor is the ease with which rectangular dimensions can be calculated -- as Lucian said, there is a direct connection between rectangles and multiplication. Lucian's answer addresses both the elegance and efficiency of using "square" units. $\endgroup$
    – adam.r
    Commented May 19, 2014 at 3:36
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    $\begingroup$ It's worth considering tesselation in higher dimensions, too. The square generalises to the (hyper)cube, which can also tesselate. The other two tesselating polygons - the triangle and the hexagon - don't work as well. The triangle is analogous to the tetrahedron, which doesn't tesselate, and the hexagon doesn't have a natural analogue. $\endgroup$
    – mudri
    Commented May 19, 2014 at 7:34
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    $\begingroup$ You can't fill a circle with unit circles because they don't tesselate perfectly. $\endgroup$
    – Marcin
    Commented May 19, 2014 at 16:04

If you want to know the area of something in terms of how many copies of some standard shape cover that area, the standard shape needs to be something that tiles the plane in a tessellation. Otherwise -- for example, if you use a unit circle or a unit regular pentagon -- when you try to cover a given area with a shape that cannot tile the plane, for most areas you will inevitably have gaps between the unit shapes or places where the unit shapes overlap or both.

The only regular polygons that tile the plane are the regular hexagon, the square, and the equilateral triangle.

(I suppose you could specify that areas are covered by close-packed circles, but each of those circles corresponds to a regular hexagon, which together completely cover the same area without overlaps or gaps in a hexagonal tiling, so you might as well use a hexagon as your unit of area).

Some people measure the area of geometric shapes in units of "unit tetras" or "unit triangles" instead of the classic "unit cubes" or "unit squares".

Likewise, if you want to know the volume of something in terms of how many copies of some standard shape fill that volume, the standard shape needs to be a space-filling polyhedra that fills space to form a honeycomb.

(I suppose you could specify that volumes are filled by close-packed spheres, but each of those spheres corresponds to a rhombic dodecahedra, which together completely cover the same area without overlaps or gaps in a rhombic dodecahedral honeycomb, so you might as well use a rhombic dodecahedron as your unit of area, like honeybees do).

The only Platonic solid that is space-filling by itself is the cube. However, Buckminster Fuller shows that sometimes it is convenient to use the regular tetrahedron with edge length 1 as a unit of volume. The tetrahedral-octahedral honeycomb fill space, as in a octet truss, composed of regular octahedrons of edge length 1 (which has a volume of exactly four unit tetras) and unit tetras.

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    $\begingroup$ It is also interesting to see how things behave for large dimensions. The $n$-dimensional hypercube is "large" compared to some other natural "standard measure shapes". For example for dimension $1000$, i.e. inside $\mathbb{R}^{1000}$, the volume of a solid $1000$-ball (or $1000$-dimensional disk) of radius one is approx. $3\cdot 10^{-886}$, and the volume of a $1000$-dimensional regular simplex of edge length one is approx. $2\cdot 10^{-2717}$. $\endgroup$ Commented May 19, 2014 at 12:17

Geometry comes from geo-metry, earth measuring. The original motivation behind geometry was to define "this is my land, that is yours, here is the border, and mine is bigger". For this to work, two requirements must be met:

  • Two adjecent fields must be connected without "loss", like the star-shaped space between circles.
  • A border must be easy to define, like a straight line between two landmarks.
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    $\begingroup$ This still does not explain why squares an not for instance equilateral triangles are used as basic unit for surface measurement. $\endgroup$ Commented May 19, 2014 at 1:00
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    $\begingroup$ @MarcvanLeeuwen It's easier to tile something with squares than with triangles. $\endgroup$
    – Jack M
    Commented May 19, 2014 at 13:18

area measured in circular units

Is there a fundamental reason why we don't say the area of a unit square is $1 \over \pi$ circular units?

We do, in fact, measure area in circular mil far more often than area in square mil.(*)

The area of a circle with a diameter of 1 mil is 1 circular mil.

A solid wire -- assuming a circular cross section, which is true for the vast majority of wire -- with a diameter of $d$ mil, has an area of $d^2$ circular mil.

The area of a square of width $w$ mil is ${4 \over \pi}w^2$ circular mils.

(*) Does anyone actually use square mils, other than as a teaching aid to help explain circular mils? I honestly hardly ever use circular mils, but I hear they are apparently the standard unit for wires connected to lightning rods and certain other safety wire size mandates.


I find your reception of squared as being related to a square is odd. So basically questioning the whole question. As others mentioned, there is nothing wrong with having irrational numbers, they can just not be displayed properly. Therefore we have constants like $\pi$ to pull out. But to the essence: The reason why there is a $2$ in the squred result, in the "area" is because you have got two integrals. You have to integrate in two dimensions. Be it ($x$ and $y$) or ($r$ and $\phi$). But it is two dimensions - so two integrals - so a two on the unit. And there is in my opinion no relation to a figure that is rectangular.


'Irrational' doesn't mean there's anything wrong or weird about a number - it just means it can't be expressed as a ratio. Literally ratio-nal. Just like $4$ and $3$ are integers, but $\frac{3}{4} = 0.75$ which is not an integer. They're all still numbers.

You got an irrational result because the formula for circle-area has a factor of $\pi$, which is irrational. Actually, if your radius was a rational multiple of $\sqrt{\frac{1}{\pi}}$ (which is irrational) then you WOULD end up with a rational result but this is kind of a pretty special case.

You measure area in square units because its convenient for the majority of maths you use (if drawing circles was 'cheaper' than drawing two perpindicular lines or if it made the maths simpler then we'd probably do that instead).


Area is in square units simply because the units are part of the equation. Since X.X=X2, inch x inch = inch2.


The question posed to you was why are the areas of the circle and square incommensurate? It doesn't matter which one you use as a reference. More precisely: the areas of a circle and square, where either figure is inscribed in the other, are not only irrational multiples of one another, but are transcendental multiples of one another. That is: the ratio of their areas can't be expressed as a whole number ratio, nor as any other algebraic number; i.e. as the root of any polynomial with whole number coefficients.


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