In this lesson, I briefly describe my methodology when using challenging exit tickets as a tool for review. Then, I cover some of the talking points to consider in each problem. Each problem brings out wonderful misconceptions and highlights different strategies your students might deploy.
The exit tickets were tough in the previous lesson. Many students asked for more time so I wil give them another ten minutes to work before starting today's lesson. After I collect the Exit Tickets from yesterday, I will hand out another blank ticket to each table and post the problems and prompt on the board.
With their work fresh in mind, I will ask the students to talk about their strategies for solving the problems. Then, we'll vote as a class on the problems we want to discuss in detail. I lead this conversation. I will share the answers to each question chosen and point out the lessons I learned about their misconceptions.
Then, I plan to hand yesterday's Exit Tickets back. I'll ask the students to grade their partners work. I often give out special "grading" markers and collect the tickets at the end of class for my own review. But, I will not do this today. I think the students need a chance to review their own work as we are talking about the problems and comparing solutions. Timely feedback is key.
I designed the Line Star Problem to simultaneously look at students' understanding of supplementary, complementary, adjacent and vertical angles, while avoiding any work with transversals. Each angle is rounded to the nearest whole number, in hopes of avoiding arithmetic mistakes.
Students tend to find angles 1, 2 and 3 by using the black line to fill in the missing gap on the left. Then they find angle 2, which is vertical to the blank angle they already found. Then, they find angle 3 by either using the purple line as a reference and writing out 180 - (25 + 25 + 52) = 78 or realizing that angle three is vertical to both the 31 and 47 degree angles. A key observation to make is that the blue line is a line segment and does not cross through the center.
An alternative strategy that students is is to find angle 1 by seeing that it is vertical to the 52 degree angle or by using the grey line as a 180 degree reference.
I find that this problem is fun to review with students. Because they are seeing vertical angles out of the usual context of two lines crossing each other, they need to think carefully about the location of each line in order to find the vertical pairs. Students are also given a variety of lines to work with. With so many supplemental angles, they need to think critically about which lines to use as they solve the problem.
This problem overwhelms many students. They see the angle diagram and are desperate to have a set of steps to reach angle 1, 2, etc. However, the conversation I have with many successful students is to just start "finding what you can" and then building off of that information. The idea is to discover a solution as you work through the problem. This is a big shift in mindset that many students need to make as they transition into tougher problems. They need to get away from automatic problems that require more reflex than thought.
The wonderful part about this problem is that students will often share 5 or 6 different strategies. The enjoy seeing different approaches and celebrate algorithms that seem simplify the process.
For example, students are excited to see that angle 2 is an alternate interior angle to the 76 degree angle. The love to see the large triangle that encloses angle 1 with the same 76 degree angle and the alternate interior angle to the 53 degree angle. You can quickly discover many different approaches by starting the problem over and simply using different angles as you go.
It is important to stress the value of a discovered solution. Students need to understand that complex problems are not meant to solved automatically. They are meant to take trial and error. They are like puzzles that we unravel. That is the joy of problem solving (or at least one of many joys).
In this problem students are confronted with two variables involving two different angle pairs.
The problem reads like this:
Lines ER and CG cross at point A. Angle CAB = 10x + 100 and Angle GAR = 50x + 20. Angle DBA = 50y – 210 and angle EBD = 28y
What is the value of angles EBD + GAR?
The challenge is for students to realize that they can solve for one variable without using the other. This is because the value of the angles that involve the x value have no impact of the angles that involve the y-values.
Once they make this connection they can use their understanding of vertical and supplemental angles and algebra to quickly solve the problem.
The conversations around this problem become very interesting because students have wonderful ways to describing why the supplemental and vertical angle values are not dependent upon each other.
This is by far the toughest problem. It is one that is typically asked in high school, but it reflects the complex ideas in transversals in so many nice ways. It also gets an interesting misconception about alternate interior angles.
Here is an image of the problem: Problem Definition
Here is the doc file: topic 28 exit ticket
First of all, students need use the converse of the statement "if two parallel lines are cut by a transferal, then the alternate interior angles are equal" to "if the alternate interior angles are equal, then the lines cut by a transversal are parallel." The really tricky part here is, "which alternate interior angles actually relate lines AB and CD, the top and bottom lines of this polygon." Furthermore, students need to avoid circular logic. They often assume the lines are parallel and then set one angle in triangle ABD equal to an algebraic expression in BDC to find x. Of course, this assumes the very thing that they are trying to prove.
Here are some wonderful and interesting things that came up.
1. Many students realized that the missing angle in triangle ABD much be 44 degrees, since we know that triangles have to have 180 degrees. They also realized that the algebraic expressions in triangle BDC must also add up to 180 degrees. I love this because they use something that is guaranteed, that triangles have 180 degrees to find something that might or might not be true.
2. Many students used proof by contradiction. They assumed that the 43 degree angle does equal angle DBC and then substituted that x value into each expression of triangle BDC to show that this leads to a triangle having less than 180 degrees (something that can't happen). They then stated that lines AD and BC are not parallel. Then they went on to finish by assuming that the 44 degree angle and angle BDC are equal and used that x value to show that they get a triangle with 180 degrees.
3. Many students struggled at understanding why angles ADB and DBC show that the two lines on the side of the polygon are not parallel. This led to a great conversation around which angles relate the lines on the top and bottom of the polygon. It took this problem for them to really think about how the angles relate to the lines that form them, not just the lines around them.