Area Geometry High School: Geometry Right Triangles

Using a bad base



I keep on seeing this in my Geometry classes this year. Tasked with finding the area of a right triangle, kids move toward the hypotenuse even if two of the other sides are given. Then they end up stuck looking for a height that they can’t find.

I’m pretty convinced — based on talking to kids and looking at their work — that this is all about how they see right triangles. These kids must be seeing hypotenuses as bases, and it must feel weird for them to treat the legs as bases. Or maybe instead it’s about the height? Maybe it feels strange to them to use a leg as a height?

9 replies on “Using a bad base”

Agreed. It might be a good exercise to revisit the area discussion when teaching the concept that the altitude to the hypotenuse is the geometric mean between the two segments formed on the hypotenuse. Again, after teaching Heron’s formula, a good exercise would be to demonstrate finding the area of a particular right triangle using three different metholds.

I’ve seen this too. My theory is it is a reflex response: students think, you’ve given me a right triangle, my job is to use the pythagorean theorem. And then, since you had me find this length of 10, you must want me to use it somehow.

I find, as a rule, most students ignore what you are actually asking them. They look at the givens, and they think, what processes do I know how to do from these givens? Only when they get stuck do they go back and read what the question was asking for (and often not even then).

There’s another interesting reflex thing I noticed: they see that perpendicular sign in the triangle, and go right to Pythagorean theorem. But if you label that right angle as 90 degrees instead of with a little square, you will find a significant proportion of kids suddenly don’t know how to do the problem.

But maybe, on the other hand, it is as you say, a question of which side students feel comfortable seeing as the “base”. I think some students have in their head a picture of “height” of ABC as the length of a perpendicular dropped from C to *the interior* of AB always. A perpendicular that gets dropped from C to a point outside the segment AB throws some people; but I could see that the edge case, where the perpendicular gets dropped from C to an *endpoint* of AB, might be trickier still. It doesn’t look like their canonical mental image of a “height” for triangle.

Andrew, you probably have a point. Also it might have to do with the totally reasonable idea that an object only has one hight. For example a human being, or a building. This combined with the mental image of a triangles hight leads to this.

“Listen class, this is i common mistake… ” would be my response.

Hmm, interesting! That never occurred to me, but it sounds right to me now that you point it out. Ordinary objects have one direction which is canonically identified as the “height”, and that doesn’t change if the object’s orientation changes. (As in your example: if I lie down, that doesn’t change my height.) Maybe “height” is a misleading name. I’ll have to watch for that in future!

Yes!! Because it can be approached in so many ways, depending on what level of work the students have done. And any way you approach it, it’s a complex question.

It could be a complex similar triangles question because it requires a strong understanding of what makes triangles similar, since the angles match up in a short of unexpected way and the two similar triangles don’t have the same orientation, making it challenging to see which side corresponds with which.

It could be a complex trigonometry problem because first students need to do an inverse trig function to find the angle, then apply it to another triangle to find the height. In fact, this is a good first step in having students derive the sine rule.

But I actually like it best as an area problem because it asks the students to see equivalence relationships. The student must understand that any side can be the base as long as there is a matching perpendicular height. And that any combination of 1/2 base*matching height I’d the area of this figure. That leads us from the simple A=1/2bh to 1/2bh=1/2bh to 1/2(6)(8)=1/2(10)(h). This is sort of an intro to solving systems of linear equations, but also just flexible thinking that I want to encourage.

The fun thing is that a student can approach this one problem based on their own current knowledge and still come to the answer if they’re willing to tackle a multi-step complex problem.

A possible reason for this might be the usual drawing of a triangle with a base and an altitude, with the altitude drawn being interior to the triangle. A leg doesn’t quite feel like an altitude, you know? It’s already a leg; it’s not inside the triangle, it’s on the boundary. People may just be unsure how to handle boundary cases like that (or just not able to find the altitude there) and so turn to a non-boundary case. There is of course also the case where the altitude is *exterior* to the triangle — another non-paradigmatic case I would expect people to get wrong, though that special case seems to be warned about more.

It is a bit funny because if you want to demonstrate where the rule comes from, a right triangle with its two legs is really the easiest case. But people don’t always explain that…

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