Last year I posted student work from the Mathematics Assessment Project task Hopewell Triangles. This year I want to talk about the conversations that we had.
Students worked on the task for 15-20 minutes by themselves. For those who didn’t need that long, I had two other tasks from Illustrative Mathematics on the back of the handout: Seven Circles I and Shortest Line Segment from a Point P to a Line L.
After students worked by themselves, I collected their responses in the TNS document I had sent. They started talking about their results with their groups. Apparently I didn’t make it clear to the whole class that I would be collecting their responses, because everyone had not keyed in their responses, but I looked at the responses that I did get to monitor student progress while they were talking with their groups. I noticed that one student had answered question 4 incorrectly.
So I went to his group and listened to them as they constructed a viable argument and critiqued the reasoning of others. Alex explained how he had arrived at his wrong answer. Can anyone figure out Alex’s misconception?
In our whole class discussion, I made sure that Alex shared his thinking. We all have something to learn from other’s misconceptions and we have created a classroom where it is okay to learn from incorrect thinking. It turns out that Alex used inverse tangent to calculate the acute angle measures. He then concluded that the triangle was right because it had a right angle (since the two acute angles were complementary), not realizing his circular logic in assuming that the triangle was right to use right triangle trigonometry for his acute angle calculations.
Can we show that Alex’s thinking doesn’t work in conjunction with what we know about the other triangles?
Triangle 3 is a 45°-45°-90° triangle, and so its acute angles are 45°. At least half of the students recognized that they had already calculated one of the acute angles for Triangle 1 in problem #2. Then we can show by the Angle Addition Postulate that we have a problem when the shaded triangle is right since 53°, 90°, and 45° don’t sum to 180°.
Most students used the Pythagorean Theorem to show that the shaded triangle was not right, which was our next whole class conversation. One student told us that the hypotenuse of Triangle 3 was 5√2, which he knew because of our work with Special Right Triangles. Another student used the Pythagorean Theorem on Triangle 2 to get its hypotenuse of 7√5. And then another showed us that (5√2)2+(7√5)2≠(15)2. This would have been the end of the conversation in my class when I first started teaching. But because our focus is not only on correct answers but also on logical arguments and understanding, a student stopped us from continuing by asking a question. Why is the hypotenuse of Triangle 2 7√5 instead of 7√3? What a good question. Why is the hypotenuse of Triangle 2 7√5 instead of 7√3?
How did you get 7√3? The triangle has side lengths 7 and 14, so I assumed the triangle was a 30°-60°-90° triangle. Another student cleared up that misconception. Side lengths of x and 2x are for a 30°-60°-90° triangle only when 2x is the length of the hypotenuse.
The last whole class conversation happened because I overhead a group talking about how they knew Triangle 1 was similar to Triangle A. I had asked them to remember that part of the conversation because I was going to ask them to discuss it with the whole class later. One girl said that she knew the triangles were similar because she saw that 3-4-5 was proportional to 9-12-15. But she was having a hard time visualizing how the triangles were similar. How could we show that the two triangles are similar to each other?
How do we define similarity? Two figures are similar if there is a dilation (and if needed, a sequence of rigid motions) that will map one figure onto the other. Can we show that to be true for Triangle A and Triangle 1?
We need several rigid motions along with the dilation to show that Triangle A and Triangle 1 are similar:
Reflect Triangle A about the side that is 3 units.
Rotate Triangle A’ 270° about the vertex of the angle opposite the side that is 3 units. Let’s call that point X’.
Translate Triangle A’’ by a vector that goes from X’’ to the left vertex of the given rectangle.
Dilate Triangle A’’’ about point X’’’ by a scale factor of 3.
I was amazed at the mathematics in this lesson and the opportunity for students to make connections. What a great task! If I had written the task, it wouldn’t have occurred to me to include a triangle that students might mistake for a 30°-60°-90° triangle. Instead, the students were given the opportunity to construct a viable argument and critique the reasoning of others. It might not have occurred to me to include a calculation in an early question (#2) that students could use again in the final question (#4) – I would have likely changed the angle measures to have them perform two different calculations. Instead, the students were given the opportunity to look for regularity in repeated reasoning. If I had written the task and included a triangle similar to one of the others, I doubt it would have occurred to me to make one of them have a different orientation than the other. Instead, the students were given the opportunity to look for and make use of structure.
And so the journey continues, with great tasks like this one to make student misconceptions evident and to correct those misconceptions …