Republic of Mathematics blog

Why does a triangle have an area?

Posted by: Gary Ernest Davis on: February 23, 2011

In a lovely series of videos James Tanton clarifies the formula for the area of a plane triangle.

Starting from some simple postulates, namely:

  1. area is additive over finitely many parts that do not overlap;
  2. congruent figures have equal area;
  3. a rectangle of length L and width W has area L\times W

James shows that, if a triangle has an area then the area of the triangle must be half the length of the base of the triangle multiplied by its height.

Unfortunately there is nothing in the assumptions, or the demonstration, that allows us to show that a triangle HAS an area.

We have postulated rectangles have areas, and we have postulated that we can add up areas provided we do not overlap figures, and that we can move figures with areas around by congruences so as to have figures with the same area.

All we can get from (1)-(3) is figures that are non-overlapping unions of a finite number of congruent transformations of rectangles.

Triangles are not such beasts.

So we have a problem in asserting that a triangle even HAS an area.

Of course we can add it in as a postulate:

4. Every triangle has an area.

Then James’s argument in the videos is perfectly valid, and establishes a formula for the area of a triangle.

Well what’s the fuss you may ask?

Once we have triangles having area, and knowing a formula for their area, we can surely split all polygons into non-overlapping triangles and show that polygons have areas, and obtain a formula for that area.

Indeed we can.

But now, from the perspective of general polygons we see that triangles are somehow more fundamental than squares or rectangles: we can build polygons out of non-overlapping triangles, but we cannot generally build them out of non-overlapping congruent transformations of rectangles.

So the assumptions (1)-(4) are a little redundant, and maybe we should have assumed:

(A) area is additive over finitely many parts that do not overlap;

(B) congruent figures have equal area;

(C) a triangle of base length B and height H has area \frac{B\times H}{2}

Now it’s straightforward to see that a rectangle has an area and that area is its length times its width.

The point is, as James says several times in the videos, area, even of plane figures, is not such a simple notion as we might at first think.

Area as “amount of stuff” just does not cut it.

Once we get to odd regions, such as circular regions we are forced to abandon the assumption of only a finite number of parts in assumption (1).

Once we abandon that assumption and allow infinitely many non-overlapping parts then we CAN prove from (1)-(3) that every triangle HAS an area, just as we can prove that every circular region has an area.

A notion of area based on finite additivity – as in assumption (1) – is very restricted, and one has to make ad hoc assumptions as to what plane figures actually HAVE an area.

Infinite additivity (strictly speaking, countably-infinite additivity) allows us to do much more.

So let me say again, as James Tanton says in the videos: area is NOT straightforward, it is NOT simply “amount of stuff”, it has more to do with decomposition into parts known to have area than anything else, and passing to plane figures such as circular regions forces us into thinking about decompositions into infinitely-many parts.

And, just for good measure (if you will pardon the pun) triangles are probably more fundamental plane figures than are rectangles.


As Alexander Bogomolny (@CutTheKNotMath) points out, if one takes \frac{B\times H}{2} as the area of a triangle as a definition, one is obliged to show it does not matter which base, and which height.

Alex has a very nice post on area at CTK Insights: “Does Triangle Have Area?”

7 Responses to "Why does a triangle have an area?"

The situation in which a rectangle has been assigned area while a triangle has not leads to a question: Is there a non-measurable set such that its union with a congruent copy of itself is measurable?

Yes, that’s jut is, Alex. If triangles were non-measurable still a union of two congruent right triangles could produce a square which is measurable.

This is not too different from Lebesgue measurability. If we allow the axiom of choice we get non-measurable sets and, in 3D, the Banach-Tarski paradox.

The latter, seems to depend critically on what notion of “congruence” one has – that is, upon what is the allowed group of motions.


I’ve been thinking of it and, to my surprise, I now believe this is possible. It might work this way.

Let A be non-measurable on [0, 1], F the characteristic function of A.

G = F, on [0, 1]
G(x) = -F(-x), on [-1, 0].

Draw a tent over [-1, 1] with y = x + 1 and y = -x + 1 and cut off the protruding pieces of G’s graph. The result is a deformed right triangle which, when rotated 180 degrees fills up the deformities. The union of the two is a square.


I think it goes to illustrate James Tanton’s point that area (= measurability) is not so simple and “intuitively obvious”.

Starting with James Tanton’s 3 postulates, can’t we show that shearing a rectangle produces a parallelogram that conserves the rectangle’s area? Though we haven’t yet defined the area of a triangle or even whether a triangle HAS area, “We have postulated rectangles have areas, and we have postulated that we can add up areas provided we do not overlap figures, and that we can move figures with areas around by congruences so as to have figures with the same area.” Thus we can translate triangular parts of the parallelogram to restore the original rectangle. We can also divide the parallelogram into two congruent triangles, and according to postulate 2, congruent figures have the same area. Therefore, if the area of a rectangle is bh, then the area of a triangle derived from shearing and splitting is 1/2 bh.

– Michel Paul


there are a couple of issues here. The first has to do with allowable transformations. It is true, AFTER (but not before) we have defined area, that a shear preserves area. James Tanton talks about congruences but does not say explicitly what those congruences are – these, by definition are area-preserving transformations (since “area” is defined in terms of them!). We could, of course, include shear transformations among those preserving area as part of the definition of area.
The second issue has to with whether when we have a plane figure with area, a “part” of that plane figure also has area. This is a deep question related to the foundations of set theory, even in the case of Lebesgue measure. So it is by no means a consequence of the definition of area that James Tanton gives, using rectangles, that even if we admit shear transformations as allowable “area-preserving” transformations, as a component of the definition of area , that we could take a part of a sheared rectangle and assert it had an area. Indeed we could just cut a rectangle in two if that were the case and assert the area of the triangle is half that of the rectangle.

That, of course, is what is commonly done in school mathematics because there “area” is treated as a perceptual given. Unfortunately, taking area as a perceptual given leads to the unfortunate empirically observed situation where many people believe the area of a region does not change as its perimeter changes because “the amount of stuff” does not change. Of course there is no “stuff” and area is not about “amount of stuff”. It is about decomposition into pieces that are already declared to have a known area (“tiling” if you wish).

As Alexander Bogomolny points out, Hilbert’s approach was to take triangles as the more fundamental object and DEFINE the area of a triangle as half the base length times the height, with the proviso that now we must show this definition is independent of which of the 3 sides of the triangle we choose as as the base.

[…] Towards the end of February, James Tanton posted a series of videos discussion the notion of area and Paul Curry’s paradox. As a result a discussion sprang up on Twitter, which raised the question as to why a triangle has area. The issue has been discussed by both Alexander Bogomolny in a post entitled Does Triangle Have Area? , posted at CTK Insights; and Gary Davis at Why does a triangle have an area?. […]

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