When I lecture I like to improvise. I prepare notes, of course, the more detailed the more precise I need to be, but my performing instincts are most satisfied when I just go in front of the class with some key points to hit and maybe a few key lines worked out ahead of time. But I did recently make an iconic mistake, repeating the mathematics instructor’s equivalent of the lawyer asking in court a question without already knowing what the answer will be. Improvisation has to be carefully prepared.

The lecture was going over areas for basic geometric figures — rectangles, parallelograms, triangles, circles, and so on — and we’d got to the funkiest-looking formula of the shapes in the book. What’s the area of a trapezoid? The trapezoid is a four-sided figure with two parallel lines. If the length of one of those parallel lines is *b*_{1} — and since we’re choosing how to describe these things, it can be, by definition — and the length of the other of those parallel lines is *b*_{2} — once again, we can — and the distance between the two parallel lines is *a*, then we have a simple formula for the area of this trapezoid. It’s *a * (b*_{1} + b_{2})/2.

(The letters might seem a bit arbitrary. *b* for the lengths of bases is probably almost mnemonic; at least, it’s got the use of b-for-a-base-length in common with the area formula for triangles. *a* seems to be a harder match. I like thinking of it as the “altitude” of the higher base above the lower; altitude is used to describe how far other polygons reach, too.)

And I mentioned to my students that they could just memorize this, but if they liked, they could work it out from knowing the area formula for triangles. My students seemed, rather credibly, skeptical. So I thought I’d give it a try, and I did, and got stuck. Happily I’m not too uncomfortable with dead air, standing in front of a group with nothing to say, but even better someone asked a marginally related question and I was able to go to that, and leave the unfinished demonstration behind, with nobody asking aloud how we were to get from where we left off to where we wanted to go.

Afterwards I did realize the way out of where I got stuck, and after one false start — at home — figured out the derivation. This I wrote up and sent them as a shared file on the class web space, hopefully saving my honor without taking up more class time.

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## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
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I teach geometry at the high school level and just recently taught this same topic! While I definitely require students to memorize the formula for the area of parallelograms (and therefore rectangle/rhombus and square) I have a different method for trapezoids. A trapezoid is simply half of a parallelogram. If you took a copy of your trapezoid, rotated it 180 degrees and slide it over a bit then you get a parallelogram. So all my students have to do is find the area of that parallelogram and cut it in half to get there original trapezoid. Now of course students are simply performing the formula since the base of the parallelogram is the sum of the two bases but it does stick in the memories of students for perhaps a bit longer.

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Showing the formula from the pair of trapezoids together would certainly have been the wiser course. But I had started out thinking triangles and didn’t vary from that soon enough.

When I next teach a course like this I must make some construction paper figures for trapezoids and for that matter parallelograms. It may be the same things I could draw on the board, but any kind of real object seems to help disproportionately much.

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After seeing this title, I thought about this for a while. Hopefully I’m not spoiling your next column, but I’m going for the thought process from me, someone who was very good at math but hasn’t done this sort of thing in 20 years:

“So a trapezoid is a square plus two triangles. The square’s area is clearly

a*b1, so now I have to add the area of the two triangles. [thinking a while here] Or one triangle, since I can smoosh them together- I know they’re the same height! Aha! I get a triangleatall andb2 – b1wide. That’sa*(b2-b1)/2. So I’ve gota*b1 + a* (b2-b1)/2.. dang, I should simplify this.So I can factor out the a, and get everything over the divide by 2. .. I’ve got

a * ((2*b1) + (b2-b1)) / 2. Condense the b1s there!a* (b1 + b2) / 2Hmm! That looks simpler than I thought it would be for a formula.”LikeLike

Nice spot of reasoning. And not to fear; you didn’t preempt any future entries I had planned.

Also I appreciate the way you did describe going through it, since that does nicely match the sort of ‘what do I see? What can I make of this?’ approach that most mathematics really demands, too. There are also the long, disappointing false trails out there, but on a problem like this it’s hard to reach those except when it’s your homework problem.

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