I guess this is a good time to give my answer for the challenge of how many different trapezoids there are to draw. At the least it’ll provide an answer to people who seek on Google the answer to how many trapezoids there are to draw. In principle there’s an infinite number that can be drawn, of course, but I wanted to cut down the ways that seem to multiply cases without really being different shapes. For example, rotating a trapezoid doesn’t make it new, and just stretching it out longer in one direction or another shouldn’t. And just enlarging or shrinking the whole thing doesn’t change it. So given that, how many kinds of trapezoids do I see?
I make it out to be six.
Here’s the way I reasoned it. For simplicity, I’m assuming the two parallel bases are horizontal. And I’m assuming the lower base is the longer one; this is the way trapezoids keep getting drawn. I may as well go with the universal standard. I’m also assuming that just making both bases or both legs longer doesn’t by itself change the trapezoid. But I do think there are some differences, and this is how I come up with six.
The difference I was thinking of came about from a comment Chiaroscuro made about how to prove the area formula for a trapezoid. His idea, and a good one, was based on slicing the trapezoid up into three shapes. One of them would be the triangle on the left-hand side, one would be the rectangle in the center, and one would be the triangle on the right-hand side. This proof works just fine for the standard-model trapezoid, where the shorter base is on top, and it’s centered above the longer base. It also works fine if the center of the shorter base is directly above the right end of the lower base, but it falls apart in other cases: if you have a right trapezoid, for example, there’s only the one triangle to lop off. If the upper base isn’t anywhere above the lower base, the proof doesn’t work, but we could repair it.
That inspires the differences that I see, though: How many of the ends of the upper base are above the lower base? And put that way, there are only three possible answers. Either both ends of the upper base are above the lower base, or just one end of the upper base is above the lower, or else neither end of the upper base is above the lower. If both ends of the upper base aren’t above the lower base, they might be either to the left or to the right of the lower base; I’ll draw them going over to the right side. This implies there are three kinds of trapezoids; let me show how we get more of them.
This is the first case with both ends of the upper base above the lower base. And this suddenly shows how to make six cases out of these three possibilities: the ends of the upper base are directly above the ends of the lower base. This is the shape we might more commonly call a rectangle, and we can argue about whether rectangles are trapezoids, but let’s suppose they are or the number of possibilities drops to five.
This is the second case of having both ends of the upper base above the lower base. One end is directly above an end of the lower base; the other is between the ends of the lower base. This is the right trapezoid shape so handy in working out the rules for integration. This is the only example of a right trapezoid that we need, since it wouldn’t really be different if the vertical leg were on the right-hand side rather than the left. And we’re taking by assumption the longer base to be on the bottom, so we don’t need to consider the case of the longer base being above the shorter.
This is the third example of both ends of the shorter base being above the longer base, and it’s also the standard-issue trapezoid, the one shown in textbooks when writing out formulas for areas. This is also one where the area formula could be found by chopping the segment up into three pieces. Do it by dropping vertical lines at the points where the upper base has its ends. You get two triangles and a rectangle.
Figuring out the area of the rectangle should be easy — obvious even — although figuring out the area of each of the triangles may be a bit mysterious. After all, you need the base and the height of each triangle to find its area. The height is obvious; it’s how far apart the parallel bases are. The width, though … that seems to depend heavily on the angle the legs make. It’s worth thinking about how to find the area of these two triangles.
This is the first example of a trapezoid where the upper base has only one end point above the lower base. The right end of the top stretches out past the right end of the bottom. (It wouldn’t be different if the upper base stretched out past the left end instead.) This is probably the second-most-obvious case of a trapezoid at all. The area formula could also be worked out by chopping it up along vertical lines, slicing the trapezoid at the left end of the upper base and at the right end of the lower base.
There’s the same problem in figuring out the areas of both of these triangles, since there’s so much variability in the bases of the two triangles. But if you spotted how to work out the area of the two triangles above then this won’t give you any trouble.
Here’s the second example of just the one end of the upper base being above the lower base. It may not look very different from the previous one, but the difference shows up if you try slicing vertically at the upper-left or the lower-right ends. The left end of the upper base is directly over the right end of the lower base, and so, there’s just the one slice, and it cuts the trapezoid into two triangles, with no rectangle.
This is the example where neither end of the upper base is above the lower base. I don’t see any obvious special cases to this, unless you want to count where the two legs happen to be parallel and make the trapezoid into a parallelogram. If you do try cutting this with vertical slices at the left end of the upper base and the right end of the lower base you come out of it with two short triangles and a brand-new trapezoid, which is just a mess to deal with.
There might be another case worth considering: what about the parallelogram, where the upper base is far enough to the right of the lower base that no point of it is above the lower? And I think we don’t need to pay any special attention to this, because if we rotate that figure a quarter-turn, making what had been the two legs into the bases, then we have a parallelogram with one end of the upper base above the lower base, which I claim we already covered with the fourth kind of trapezoid.
Of course, I claim this, but does that mean I’m right? It may be fun to spend some time convincing yourself of this, or finding a counter-example. Perhaps a seventh kind needs to be added after all.