How Two Trapezoids Make This Simpler
[ More of Trapezoid Week! Here we make finding the area simpler by doubling the number of trapezoids on the screen. ]
Figuring out the area of a trapezoid based on making it the difference between two triangles works all right. “All right” carries with it a sense of inadequacy. The complaints against it are pretty basic. The first is that it doesn’t work for everything which might be called a trapezoid. Maybe we don’t want to consider parallelograms and rectangles to be particular kinds of trapezoids, but, why rule them out if we don’t have to? The second point is the proof is a little convoluted, requiring us to break out of thinking about trapezoids to remember details of similar triangles. It’d be nice if we had a more direct way of proving things.
Here, as suggested by Andrew Bressette several days ago, is a proof that avoids the complications of the difference-between-triangles routine. It also avoids the restriction of not working on parallelograms or rectangles. In trade, though, we need to become convinced of a little point from geometry, and we need to remember the area of a parallelogram. That’s not a complicated formula. A parallelogram has two pairs of parallel lines making it up. The length l is the length of one line of one of those pairs. The altitude a is the distance between the parallel lines of that same pair. The area of the parallelogram is that length l times that area a. If the parallelogram is a rectangle, with the pairs of parallel lines themselves perpendicular to each other, that turns into the length of the rectangle times its height, which is comfortingly familiar.
In the diagram we take the trapezoid, labelled as the figure connecting the points A, B, E, and D, to keep it consistent with the diagram used before. b1 is the length of one base, let’s say from point A to point B. b2 is the length of the other base, the distance from point D to point E.
The trick is to create a duplicate polygon, one with the same lengths and angles. This trapezoid we rotate 180 degrees and slide over to place against either the leg BE or the leg EA. I put it up against BE for convenience. This new trapezoid is the one bounded by the points F, E, B, and G.
The new trapezoid corresponds to the original one: the length from F to E is the same as the length from A to B, that is, it’s length h1. The length from B to E is the same as from E to B. The length from B to G corresponds to that from E to D, that is, that length is h2 again. The length from G to F is the same as that from D to A.
Now we have to call on a bit of geometric reasoning. I claim that the line from A to B is parallel to that from B to G, that is, if we drew the line from point A to point G, it would have point B laying on it already. Similarly, the line from D to E is parallel to the line from E to F, and the line from D to F contains point E on it already.
The easiest way to convince anyone of this is to have them cut a couple trapezoids out of two sheets of construction paper and lay them against each other. Maybe index cards if construction paper is hard to come by. Hold the two pieces one in front of the other, and make a cut from the lower-left corner to somewhere on the upper edge, then another cut from the upper-right corner to somewhere on the lower edge. Turn the back sheet around and slide it over and you’ll be sold.
This should give us cause to wonder what we mean by a proof, in mathematics. The demonstration should be convincing, although it may take a couple different attempts to be really, really sure. But as mentioned back in the start, this is just an example; we might have something that’s right for the trapezoids we happened to work with and false for ones in general. Or something false for even a couple special cases of trapezoids. We need proof.
I’m going to have to prove that by appealing to authority, since I don’t want to go into all the geometry needed to back it up. The first piece is to say that the measure of the angle connecting the line AB to BE is the same as the measure of the angle conneting the line FE to EB. (This is why in my diagram I’ve given them little wedges of the same color; I use a similar color-coding to highlight the other pairs of angles with the same measure, although one need not pay attention to the color. The real proof is in the narrative following.) The trapezoids these angles came from were identical, and these are the corresponding angles of the two trapezoids; if they’re not the same, something alarming went wrong somewhere.
Similarly, the measure of the angle between line BE and ED is the same as the measure of the angle between EB and BG. That’s from the same reasoning: they’re corresponding angles of two trapezoids which are identical in shape.
The next piece is that if we add together the measure of the angle connecting AB to BE and the measure of the angle connecting BE to ED, we get 180 degrees. There are a couple of ways to show that, all of which amount to the properties of having one line cross a pair of parallel lines. I ask that you accept that; if there’s real interest, I’ll make the case that’s so. I just want to use this fact to go on to something else interesting.
The measure of the angle between EB and BG is equal to the measure of the angle between BE and ED. And that means the measure of the angle between AB and BE plus the angle between EB and BG has to be 180 degrees. Or, another way, the angle between AB and BG is 180 degrees; it’s a straight line from AB through to BG.
By the same reasoning, just with different letters, it’s a straight line right from DE through to EF.
So this tells us that the polygon with vertices A, G, F, and D, has at least two parallel lines, one of them of length b1 plus b2 — the length from A to B plus that from B to G — and another of length b2 plus b1 — the length from D to E plus that from E to F. We’re really well-set-up to use the area of a parallelogram formula.
Although, do we know for certain this is a parallelogram? It’s hard to see how the two parallel bases could have the same lengths if the legs AD and GF weren’t parallel also, but do we know they have to be?
We can dispose of that fear pretty easily. The measure of the angle between DA and AB has to be equal to the measure of that between GF and FE, because they’re corresponding parts of equal trapezoids. And the measure of the angle between ED and DA has to be equal to the measure of that between BG and GF, for the same reasons. We have a pair of equal angles: that between AG and GF has the same measure as that between FD and DA; that between GF and FD has the same measure as that between DA and AG. We’ve got a parallelogram.
The length of the sides of this big parallelogram AGFD is b1 + b2, and the distance between the two parallel bases of that length is a. So the area of this parallelogram is the length times that distance, (b1 + b2) * a. The parallelogram was made up of two equal trapezoids, so, the area of each of those trapezoids — the trapezoid ABED and the trapezoid FEBG — has to be half that. So the trapezoid has area (1/2) * (b1 + b2) * a.
Same result, different argument. It’s nice having multiple lines of proof that get to the same point. And this is probably a better proof than the difference-of-triangles one. It works for every trapezoid even if we include parallelograms or rectangles in the group (of course, it’s a bit silly to use the area of a parallelogram formula to figure out the area of a parallelogram by a more complicated fashion), and we don’t have to resort to subtraction. We can do everything with addition and division. There’s a little bit of geometric reasoning needed, so we can be sure that the two trapezoids combine into a bigger parallelogram, but it’s a fairly straightforward bit of reasoning.
Can we do better yet?