PBL for Engagement and Empowerment in the Mathematics Classroom

A square pyramid is a pyramid with a square base and four triangular lateral faces. The slant height is the distance from the vertex of the pyramid along a lateral face to the midpoint of a base edge. If the slant height is 10 and an edge of the square is 12, what is the altitude of this pyramid?

This is an example of a problem that probably is closest to what a traditional textbook reading would do for a student on a typical night’s homework. We do assume that students have had past experience with three-dimensional solids in middle school curriculum and have heard the term ’square pyramid’ before, although this is their first encounter with it in our book (it’s also defined in the reference section of our book too). The problem comes complete with a few other definitions just in case they aren’t familiar with slant height as well. However, although the definitions are in the problem, it can often seem that students don’t use their logic or problem solving skills to really understand what these new terms mean in the problem.

When I gave this problem last week, the student that put it on the board did it incorrectly because she assume that the slant height was actually the lateral edge (the length of the congruent sides of the isosceles triangle that is the lateral face of the pyramid). So even though the defintion of the slant height was right there in the problem, why was this student unable to transfer this definition to the problem? Well, there could be many reasons. When students read a problem for the first time, I theorize that new material takes time to become a part of their toolkit, or set of skills that they would use. So in this case, even though the definition is inthe problem, and possibly she even understood the definition, transfering that to the problem situation was difficult since the new idea of slant height was “too new”. She actually new to use the pythagorean theorem, just used it with the wrong triangle.

I also theorize that it is easier for students to visualize the right triangle that is formed by the lateral edge, the altitude and the square base because the lateral edge is physically drawn in (in a diagram that students generally draw) already. We are going to change that for next year and actually draw in the slant height in a diagram in the text.

There are a number of problems in our curriculum where a great deal of assumptions are made about students reading the definitions and applying them to the problem at hand. This result is more positive on specific problems when students have the motivation to complete the problems on their own. The kids who are finding the curriculum instrisically challenging and interesting don’t have a problem realizing that applying the definition is an expectation. I’ve also found that the more you expect from them on a regular basis, the more they begin to fulfill that expectation too.

OK, well maybe that title isn’t really what this problem is about, but many teachers might think this problem is too hard for a lot of geometry students. Initially, we were hesitant to include it in our discussion of circles, since we don’t discuss the all-to-familiar unit circle really at all in our geometry course discusison of trigonometry. But this problem is really more of a discussion of the special right triangle (45-45-90) or if students choose the prior method of the intersection of the line and circle (see previous blog entry). This is a really interesting problem with different solutions, so it’s fun to see how different students approach it.

The vertices of a square with sides parallel to the coordinate axes, lie on a circle centered at the origin with radius 5. Find the vertices of the square.

What have we discussed up until now? The students have formalized the idea of the equation of a circle centered at the origin as x^2+y^2=r^2 (from the Pythagorean Theorem) so they should be able to come up with the equation of the circle. They also know the terms of the radius, diameter, coordinate axes and show be able to set up the picture.

When I did this problem in class, many students started by drawing rectangles inscribed in the circle with dimensions of 6 x 8, realizing that the diagonal would be 10. This was a great start. They realized that somewhere in between the 6 x8 and 8 x 6 rectangle was the square they were looking for. They were then able to actually draw the diagram with that square in the circle. They could then approximate the coordinates which helped a lot in the class discussion. I felt very proud of students who were able to come to class with even that done. However, other students got even farther.

This is where I wish I were able to draw some diagrams on my blog. I need to do a little more research with Edublogs to see if this is possible. Most students then drew the diameter as the hypotenuse of an isosceles right triangle and either used the Pythagorean theorem to solve for the side, or remembered the ratio of the 45-45-90 triangle, and simply divided 10 by sqrt(2). Once they realized the side of the triangle would be 10/sqrt(2), they just needed to transfer that information to the coordinate plane. This is where a good diagram was important.

Since the origin was at the center of the circle, and the radius was half the diameter, most students could easily extrapolate the fact that you would simply divide the side of the isosceles triangle in half to obtain the coordinates. In other words, you’d just go over 5/sqrt(2) and up 5/sqrt(2) to get to the vertex in the first quadrant and do the same for the other quadrants (taking into account the negatives for left, right, up and down).

This was an amazing problem for discussion since another student did the problem by finding the intersection of the line y=x and the circle x^2+y^2=25. This student had been able to independently transfer the skill of substituting the line equation into the circle equation from a previous problem. This was the minority of the class however, but still someone was else to do it. I was impressed. If you’d like to try this problem it was #8 on p.50 at my curriculum

Here’s a great problem relating parallelograms and area -

A parallelogram has 10-inch and 18-inch sides and an area of 144 square inches.
a. How far apart are the 18-inch sides?
b. How far apart are the 10-inch sides?
c. What are the angles of the parallelogram?
d. How long are the diagonals?

OK, what’s the pre-existing knowledge or discussion that has happened before the problem is presented to the students? So far, the students have discussed the fact that the area of a parallelogram is simply the base times the height (or altitude, perpendicular distance between the parallel sides), but have not done a huge number of repetitive area problems, I’d say not even two or three of them. They’ve done a dissection discussion problem a few pages back about how to cut off a right triangle and move it over and the parallelogram basically becomes a rectangle, and that’s why it’s the same area formula.

So, in my class, most students got part a, pretty quickly, just divinding 144 by 18 and getting 8. The other type of previous problem the students have done is being given the area of a triangle and all the sides, using each side in turn to find each of the three altitudes, so they area used to turning the triangle around to find each of the heights. So now we are seeing if they can transfer that method to a parallelogram in this problem. This took getting all students to the board in my class.

Let me back up for a second. We also had covered problems that had discussed both the tangent and sine functions for right triangle trigonometery (which are introduced before cosine in our curriuculum). Hence, the students were desperately trying to find the height to the side of 10 using a right triangle, which could not work, of course, without being able to break up the side of 18. Some of them guessed lucky making a 6-8-10 triangle, which did get them the right answer, but they could not justify it – no beans in my classroom.

So, we finally had to get them to draw the parallelogram with the 10 as the base, then students finally realized they could use the area once again. This was the a-ha moment and it spread like wildfire. Thank God. Once we got through this point, they were able to find the angles with the trig no problem.

However, finding the diagonals took some manipulating of right triangles with the heights. The shorter one was easier with the right triangle within the parallelogram. This created a right triangle with legs of 8 and 12 and the hypotenuse was the diagonal. Putting a right triangle outisde the parallelogram made a bigger right triangle with legs of 18 and 8 and that hypotenuse was the longer diagonal. Since I expected so many students to not have totally completed this problem, but only have attempted it, I made time for this to be done at the board in pairs during classtime. It was time well spent, and many students wrote this problem up in their journals – very well, I might add.

If you’d like to try this problem in class it was p. 47 #2.

OK, so there’s a problem that goes like this:

Sketch the circle whose equation is x^2+y^2=100. Using the same system of coordinate axes, graph the line x+3y=10, which should intersect the circle twice – at A =(10,0) and at another point B in the second quadrant. Estimate the coordinates of B. Now use algebra to find them exactly. Segment AB is called a chord of a circle.

OK, what have the kids seen before this? They have been introduced the equations of circles that are centered at the origin. Simply, we did this by asking them to plot points that were say 5 units away from the origin and we discussed from the standpoint of the pythagorean theorem. This happened about 3 pages prior to this question. We’ve also done a few problems about circumcenters and the smallest circle that fit around a right triangle. However, this is the first circle of a radius 10, I believe. This is also the first time, they are asked to graph a line and a circle on the same coordinate axes.

This is also the first time they are being asked to approximate the point of intersection. The point does end up being a lattice point (what we refer to as a lattice point is a point that has integer valued coordinates), so the students probably will be able to come to class with the answer, which happens to be(-8,6) if they did a decent job of graphing it well.

What they have not done before is algebraically substituting in an expression for y. So this is the interesting part – to see if anyone thinks to do this. Most student probably won’t. In the past, many student remember solving systems of equations in algebra class, but they get confused when they see the quadratic terms. Once prodded in the right direction to ’substitute’ however, they realize that the linear equation is easy to solve for x, and see what they can do. This quadratic is actually not that hard to solve, pretty simple factoring involved and no quadratic formula, so it’s’ a nice first attempt at this type of problem. I will report back with what happens in my class tomorrow.

This problem also introduces the terminology of a chord of a circle. I generally like to have the students come up with a definition at this point so that we can formally have one to move forward in the problems as well. Overall, lots of good stuff in this problem, mostly generated by the students, even the substitution idea sometimes.

If you’d like to try this problem it is # 4 on p.49 at my website. You can get pdf files for the entire curriculum there. Have a look.

So I have this new idea for a direction for this blog and I’m hoping that it might take off. I’ve archived all the old “transition” entries hoping that those can still hang out and be categorized in case I need to refer to them in some future entry. But what I’m hoping to turn this blog into now is something of an organizational place for the categorization of our curriculum at my school. Since our geometry curriculum went PBL, I’ve been looking for a way to categorize the problems. Every summer though my colleagues and I rewrite and edit the problems so categorizing them by problem number and page actually doesn’t really work.

but then I thought, a blog would really be awesome because what it does it create this hypertext categorization that will let us connect the problems by categorizes and tags. I’m going to try it and see what happens. We can always go in and edit and and later on see if it makes a lot of sense or not. The only thing that will be problematic is that there seems to be no equation editor to speak of. I need to write to someone at edublogs and see if there is one somewhere that can be accessed. I’ll see what I can do about that.

Wish me luck.