Lesson 8-6: Areas of Trapezoids (Day 86)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

This tin can's volume is 64pi cubic units. Its lids were removed and then the remaining can was cut vertically so the can was made into a rectangle. The rectangle's area is 16 sq. units. The radius of the can's base is ____pi.

Well, the can in this problem is clearly a cylinder, and so we may use the cylinder formulas from Chapter 10 of the U of Chicago text. Notice that "the rectangle's area" is just another way of referring to the lateral area of the cylinder. This produces the following equations:

V = pi r^2 h = 64pi
L.A. = 2pi rh = 16

This is a system of nonlinear equations in two variables, r and h. We could use substitution to solve this system, but notice that we can also eliminate h by dividing the first equation by the second. This is convenient, since Pappas is asking us for r anyway:

pi r^2 h = 64pi
2pi rh = 16

r/2 = 4pi
r = 8pi

Pappas already provides us with the pi. So all we need to full in the blank is the 8 -- and of course, today's date is the eighth. The math is correct, and Pappas makes no error in stating this problem. But there's something that I don't like about this problem.

If we substitute in r = 8pi, we find that h = 1/pi^2. Now Pappas only refers to square and cubic units and never specifies the actual units, but let's imagine that these are inches. So the radius of our tin can is 8pi inches, or about two feet. The diameter is twice this, or about four feet. But the height is only 1/pi^2 inch, or about 0.1 inch. Now try to imagine a tin can that's four feet wide and 0.1 inch tall.

Of course, you might counter that this shows that "one unit" isn't one inch. But no matter what "one unit" is, we've found the width of the can to be nearly 500 times its height, which is unreasonable for any size can. In other words, Pappas chooses dimensions for this problem that, while mathematically correct, don't accurately describe any object in the real world.

Let's try rewriting this problem so that it makes sense. We want the answer to the new problem to remain 8, to match the date. Real cans that you buy in the supermarket don't have radii of 8 inches or even diameters of 8 inches, but a can of diameter 8 cm is reasonable. So let's try:

This tin can's volume is 160pi cubic cm. Its lids were removed and then the remaining can was cut vertically so the can was made into a rectangle. The rectangle's area is 80pi sq. units. The diameter of the can's base is ____ cm.

V = pi r^2 h = 160pi
L.A. = 2pi rh = 80pi

r/2 = 2
r = 4
2r = 8

Therefore the diameter is 8 cm -- and of course, today's date is the eighth.

Lesson 8-6 of the U of Chicago text is called "Areas of Trapezoids." In the modern Third Edition, areas of trapezoids appear in Lesson 8-5.

Today is the first lesson of the second semester, and this is the lesson I've been bringing up for a while now. The digit pattern tells us that we should begin with Lesson 8-6. But the first semester finals took place on Days 83-85, thus blocking Lessons 8-3 through 8-5. Our challenge is to cover the missing material from Lessons 8-3 to 8-5, especially triangle area in Lesson 8-5. So let's dive in.

In Lesson 8-3, the Area Postulate tells us that area satisfies four properties. Three of these are Uniqueness, Congruence, and Additivity. The fourth is the rectangle formula, A = lw. But by starting with Lesson 8-6 today, we're essentially postulating the formula not of a rectangle, but of a trapezoid:

A = (1/2)h(b_1 + b_2)

So now all we need to do is derive formulas for the parallelogram, rectangle, and triangle. The area of the parallelogram is easy, since today's Lesson 8-6 already provide for this. Last year I wrote:

• The text uses inclusive definitions, so a parallelogram is a trapezoid. If you're wondering why there's a section for areas of trapezoids but not of parallelograms, this is why. Recall that the most useful fact about a trapezoid that isn't isosceles is its area formula.

Thus we find the area of a parallelogram by using the trapezoid formula. For the parallelogram, both bases are equal, so b_1 = b_2 implies:

A = hb

And now you might notice that a rectangle is a parallelogram, so now we have the rectangle formula as well. Notice that the variables here are A = hb instead of A = lw, but there's nothing stopping us from calling the dimensions of a rectangle "base" and "height" instead of "length" and "width."

This leaves us only with the triangle area -- the subject of Lesson 8-5. I think that there are three different ways to proceed at this point:

• Follow the missing Lesson 8-5 precisely -- branch into three cases depending on whether the triangle is right (half a rectangle), acute, or obtuse.
• Since we have a parallelogram area formula, just take half a parallelogram. This is how some other Geometry texts present the area of a triangle, specifically those that teach a parallelogram formula before a triangle formula.
• Consider a trapezoid with a base of length 0. It's easy to imagine how as the base of a trapezoid approaches zero, the trapezoid's area approaches the triangle's.

The third method is the most interesting to me, since it means that a single formula works for all the shapes that we're teaching today. But I suspect that the second method will be the easiest for the students to understand. The first method, which requires three cases, is a bit much on a day when we're giving the areas of so many other shapes. In any case we obtain the formula:

A = (1/2)hb

For today's lesson, I'll take last year's Lesson 8-6 worksheet. I won't change it, so it means that the vocab term "triangulated polygon" is meaningless, and so the answer to #2 must now change from last year's (c) to this year's (b).

It also includes a "review" question on triangle area, but it's not "review" at all. We might consider this question to be sufficient, but to be safe, let's add an extra worksheet from last year's Lesson 8-5 to today's post. Polygon area is one of the more important lessons of the course.

And that's all I have to say today. No, I can't even find a contrived reason to start writing about computer music again. (Even though I have more to say about 16-bit music, it's all useless unless I have a 16-bit player.)

So I guess that's all, folks!