Why does a triangle have an area?

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

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

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 $L$ 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;

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.

Postscript

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?”