Today, I finally had the opportunity to sub in a geometry class. The teacher for whom I am subbing has three sections of geometry (periods 1, 2, and 6) and two sections of statistics (3 and 5). I never took stats in high school -- only in college -- so I'm weaker in stats than I am in geometry.
Oh, and by the way, you may be wondering how I could be teaching a geometry class when I wrote earlier that my school district has switched to integrated math for Common Core. Well, the colleges who look at student transcripts will accept either the traditional pathway of Algebra I, Geometry, and Algebra II, or an integrated pathway, but not a mixture of both. So all current sophomores and above have been grandfathered in to the traditional pathway in order to avoid sending confusing transcripts to the state colleges.
And speaking of stats, notice that statistics and probability are integrated with Common Core mathematics as well. Here in California, the traditional pathway, for schools and districts that choose it, includes statistics with Algebra I and probability with Geometry.
The textbook for today's geometry class is published by McDougall Littell, dated 2007, and is aligned to the old pre-Common Core California State Standards. Here is the table of contents:
1. Essentials of Geometry
2. Reasoning and Proof
3. Parallel and Perpendicular Lines
4. Congruent Triangles
5. Relationships within Triangles
7. Right Triangles and Trigonometry
9. Properties of Transformations
10. Properties of Circles
11. Measuring Length and Area
12. Surface Area and Volume of Solids
As it turns out, the class is currently in Chapter 4: Congruent Triangles -- which corresponds exactly to Chapter 7 of the U of Chicago text, the current chapter for this blog. And so I definitely want to compare what I'm posting on this blog to what's happening in an actual classroom.
Here are the sections of Chapter 4 of the McDougal Littell text:
4.1 Apply Triangle Sum Properties
4.2 Apply Congruence and Triangles
4.3 Prove Triangles Congruent by SSS
4.4 Prove Triangles Congruent by SAS and HL
4.5 Prove Triangles Congruent by ASA and AAS
4.6 Use Congruent Triangles
4.7 Use Isosceles and Equilateral Triangles
4.8 Perform Congruence Transformations
Notice that the "congruence transformations" referred to in Section 4.8 of the text are isometries -- that is, translations, reflections, and rotations. But these are not used to prove SSS, SAS, or ASA as they would be in a Common Core class. Indeed, instead SAS and coordinate geometry are used to prove that a translation is an isometry ("congruence transformation"). Had this been a Common Core class, not only would Section 4.8 be the first section of this chapter, but all of Chapter 9, which covers transformations in general, would precede Chapter 4.
Indeed, many pre-Common Core Geometry teachers omit sections like 4.8 altogether. I notice that the regular teacher assigned questions 1-25 from the review section of this chapter, but the questions in that section go up to 29. So the teacher skipped the questions from Section 4.8 -- which implies that he probably skipped 4.8 completely. This can never happen in a Common Core class, but officially this class is a pre-Common Core class.
As for the congruence statements, SSS, SAS, and ASA are all postulates -- and this is typical for most pre-Common Core texts. What is unusual is that SSS, SAS, and ASA are in three different sections. I also notice that the HL Theorem is stated before it can be proved. There's a proof of HL based on AAS in the back of the book, but notice that AAS itself doesn't appear until the next section. I assume that the text places HL here in order to keep the number of sides straight. (In 4.3 there are three pairs of congruent sides, in 4.4 there are two pairs, in 4.5 there is only one pair.)
Today's post will be based on the math that I actually covered in class today. I was originally going to cap off the chapter with a lesson on the Angle Bisector and Median Concurrency Theorems, but to me, it's better for me to post something that I covered in an actual class with real, live students, so that I can describe to you readers how the students are actually learning the material. This has priority over trying to catch every single Common Core Standard. But I do want to post those Concurrency Theorems at some point since they will prepare us for similarity in the second semester. (Actually, Chapter 5 of this text gives coordinate proofs of the Concurrency Theorems, but these will be unacceptable in a Common Core class.)
Let's discuss the exercises that appear here in the review section of Chapter 4. Since this is a review section, I decided to play the game that I mentioned back in October -- the one where I begin with the students guessing my age and weight before asking geometry questions. But the first three questions in the review section are vocabulary questions, and I think these questions took too much time based on how my game works -- I want them solving problems, not writing sentences. By sixth period, I refined the game so that I just gave them the answers to 1-3 and began with question #4.
Now #4 is critical, because it refers to exterior angles of a triangle. Now the Triangle Exterior Angle Equality states that the exterior angle is the sum of the two remote interior angles -- as opposed to the Triangle Exterior Angle Inequality, which states merely that the exterior angle is greater than (not necessarily the sum of) the remote interiors. The Equality requires Euclidean geometry, whereas the Inequality is true in both Euclidean and hyperbolic geometry.
Now even though this is a review of Section 4.1 of this text, the students were very confused when I taught this to them first and second periods. Part of the confusion was that in first period, I gave a point to the group that gave me the value of x when the text was asking for the actual measure of the exterior angle. Of course, the other students knew I was wrong only because it was an odd-numbered problem to which the answer is in the back of the text. By sixth period, I decided to include the Triangle Exterior Angle Equality when discussing question #4, even though the question asks only for a drawing and not for a formula, in order to prepare the students for Questions 6-8. Also, I switched so that every fourth, rather than every third, question is a question where I allow every group to earn a point, so that only even-numbered questions are allowed for this opportunity.
In sixth period the students had a little trouble with HL -- the often forgotten Congruence Theorem. I must point out that they did seem to figure out the exterior angle problems a little better. This is why I'm posting the Exterior Angle Equality today -- if students are going to have trouble with exterior angles, why try throwing the Concurrency Theorems at them?
I decided to include questions on exterior angles, congruent triangles, as well as explanations for why two triangles are or aren't congruent. I also decided to include a plan for a proof question on this review worksheet, even though this question doesn't really fit well into my game.
Meanwhile, here's the congruence question that I promised you from yesterday. As I mentioned, this is based on a question asked on a test for one of the students I tutor:
What type of quadrilateral has diagonals dividing it into four congruent triangles?
The correct answer is a rhombus, but how do we prove this? Again, we may refer to the quadrilateral as ABCD and the point where AC and BD intersect as O. But the problem is that we only know that AOB is congruent to the triangle with vertices B, O, and C -- but is it triangle BOC, or is it COB, or possibly even BCO? Nowhere in the question is it stated which vertices correspond -- which nearly every congruence problem states!
We ask, to which angle in triangle BOC does angle AOB correspond? We notice that AOB is actually an exterior angle of triangle BOC -- which means that it's greater than the measure of either of its remote interior angles, BCO and CBO. (Hey, there's that Triangle Exterior Angle Inequality again, but notice that only the Inequality is needed for this one, not the full Equality.) So AOB can't be congruent to either of them, hence it can't correspond to either of them. So AOB must correspond to angle COB, and since these congruent angles form a linear pair, they must be right angles.
Now we know that AOB, BOC, COD, and DOA are congruent right triangles, but notice that we still don't know which sides correspond. But notice that in triangles AOB, we don't know whether leg AO corresponds to BO or CO of the next triangle. Yet we do know that the sides opposite the known right angle -- that is, the hypotenuses -- must correspond! So AB is congruent to BC, and likewise to CD and DA, the other hypotenuses. These four hypotenuses are the four sides of the quadrilateral ABCD, and they are all congruent. Therefore, the figure is a rhombus. QED
Technically speaking, we should prove the converse as well. If a quadrilateral is a rhombus, then its diagonals divide it into four congruent triangles. But this follows easily from well-known properties of the rhombus.
This whole proof is somewhat advanced for a geometry student -- so I don't know exactly what the teacher was expecting the students to do. (Just guess it's a rhombus and prove the converse?) The student ended up writing "square" and didn't write much of a proof -- yet the teacher must have been feeling super-nice and awarded him 3 out of 4 points! (He's clearly not a Putnam grader!)