The terrible, horrible, no good, very bad math problem

A boon for Lucky Larry

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.

You may be familiar with the classic Norman window, which is a semicircle mounted atop a rectangle, the semicircle's flat side coinciding with one side of the rectangle. In finding the perimeter, the student must make a judicious use of the circumference formula for a circle and the perimeter formula for the rectangle. The answer is the sum of half the circumference of the full circle and the three exterior sides of the rectangle (omitting the one adjoined to the semicircle). Not too hard. The computation of the area is even easier: half the area of the full circle plus the area of the rectangle (that length times width business beloved by all).

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.

¡Caramba!

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.