Tuesday, May 24, 2011
The terrible, horrible, no good, very bad math problem
Teachers typically take pains to ensure that their exam problems are clearly stated and have unambiguous answers. This is especially true in math, where precision is at a premium. I try to make it clear exactly what I want and what form it should be in (such as “exact” or “rounded to the nearest hundredth”). The last thing any exam grader wants is a problem that is easily misinterpreted, because that just leads students astray. Even worse, however, is the problem where an incorrect calculation can produce the right answer. Then you're really stuck. You have to read the solution especially carefully to make certain the student didn't get the right answer by a lucky fluke.
I have put myself in that situation. It happened in a suite of problems designed to test my students' ability to compute the perimeter, area, or volume of certain standard geometric shapes. Having quizzed them on various rectilinear figures (break it up into rectangles, kids!) and circles and boxes, I began to challenge them with combinations of the basic forms.
I wasn't surprised that the results were a mixed bag, but I felt we were making progress. Each time I quizzed them on a variant of the basic shapes, more students were picking up on the standard formulas and the tweaks required to apply them to the hybridized shapes. With an excess of confidence, I then reached a bit too far and outdid myself: I adjoined a quarter-circle and a triangle. To make matters much, much worse, I made some very bad choices for the dimensions. Can you see my mistake(s)?
As before, I asked my students to find the perimeter and the area of the shape. You can check that the perimeter is (6π + 30) ft and the area is (36π + 30) ft2. Thanks to my carelessness, I got lots of quasi-correct solutions. First of all, the perimeter of the 5-12-13 triangle is numerically equal to the area: the perimeter is 30 ft and the area is 30 ft2. That meant students were able to get the “right” answer by employing the perimeter formula when the area formula was required. And, yes, some of my students did exactly that.
It gets better. For the area computation, they needed to compute πr2 and take one-quarter of it. A full circle of radius 12 ft would have an area of 144π ft2, so a quarter-circle's area would be 36π ft2. Several students, however, apparently reasoned thus: The arrow labeled with the “12 ft” goes all the way across, so it must be a diameter. I need to use the radius, which is half of that, so I'll use 6 ft. It's a circle, so the area is πr2. Therefore I get 36π square units.
In grading this problem, it really didn't much matter if the answer happened to be right. It was much too likely that a right answer could be obtained by accident. I corrected this exercise very carefully and very slowly.
I hope I learned my lesson.