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

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In a lovely series of videos James

Starting from some simple postulates, namely:

- area is additive over finitely many parts that do not overlap;
- congruent figures have equal area;
- a rectangle of length and width has area

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 and height has area

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 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?”

Gary,

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.

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

1 | Alexander Bogomolny

February 24, 2011 at 11:31 pm

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?

Gary Ernest Davis

February 25, 2011 at 7:28 am

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.