Monday, March 30, 2015

Section 10-2: Surface Areas of Pyramids and Cones (Day 137)

Spring break is over -- for me, at least. The geometry student I tutor actually attends a religious school that takes two weeks off for Easter, so he's still off another week. Yeah, he's lucky!

Last weekend, the county library held its biannual book sale. Usually, book sales are held the first Saturday in April and October, but the spring sale moves up to the last Saturday in March if necessary to avoid Easter weekend.

As I mentioned after the last book sale in October, I often purchase used math textbooks at these book sales, especially if they are texts for classes that I may consider teaching some day. It's been a while since I've last found some geometry texts, but I did see some pre-algebra texts which may be intended for seventh graders. Let me discuss these texts, especially the geometry units.

One of the texts was published by Merrill, the other by McDougal Littell. I ended up purchasing the latter, which is dated 2001. I actually recognize this text from when I spent one month in an advanced seventh grade math classroom back in 2012. Geometry is covered in Chapters 8 through 10. Chapter 8 covers points, lines, polygons, transformations, and similarity. The transformation section covers reflections and translations (but not dilations in the similarity section), but of course, this is an old pre-Common Core text, so transformations aren't used to define congruence. Chapter 9 is officially called "Real Numbers and Solving Inequalities," but the real numbers portion of the chapter segues from square roots to the Pythagorean Theorem and to the Distance Formula.

That takes us to Chapter 10. As it turns out, much of Chapter 10 of this seventh-grade text matches up with the same numbered chapter of the U of Chicago geometry text. Here are the sections:

Section 10.1: Circumference and Area of a Circle
Section 10.2: Three-Dimensional Figures
Section 10.3: Surface Areas of Prisms and Cylinders
Section 10.4: Volume of a Prism
Section 10.5: Volume of a Cylinder
Section 10.6: Volumes of Pyramids and Cones
Section 10.7: Volume of a Sphere
Section 10.8: Similar Solids

It's often interesting to see how much surface area and volume appears in pre-algebra texts. Wee see that this text gives all of the volume formulas, while only the cylindric solids have their surface areas included in the text.

But let's keep in mind that this text was specifically written for the old California state standards that we had before the Common Core -- and these standards are still relevant only in that they appear in the CAHSEE exit exam that students must still take to graduate. Surface area and volume appear on the CAHSEE only because they are technically seventh grade standards. In practice, only boxes and their unions have their surface areas and volumes appear on the test. As we can see, this corresponds to Sections 10.3 and 10.4 on the test. The second half of Chapter 10 doesn't appear on the CAHSEE.

The final chapter, Chapter 12, of this text is on polynomials. This chapter actually goes a bit beyond the seventh grade standards -- most notably, Section 12.5 is "Multiplying Polynomials" and actually teaches the FOIL method of multiplying two binomials. I was only in the classroom that taught using this text for a month, but I was told that the honors class would cover Chapter 12 around the start of the second semester, with the rest of the chapters taught in numerical order. (Non-honors classes would not cover Chapter 12 at all.) The next section, Section 12.6, may also seem a bit advanced for a pre-algebra class -- "Graphing y = ax^2 and y = ax^3" -- but not only does it appear in the seventh grade standards, but also on the CAHSEE. Algebra I teachers may cover the graph of y = ax^2 slightly after the date of the CAHSEE exam, but it nonetheless appears on the test because it's not an Algebra I standard, but a seventh grade standard! Even when we look at the CAHSEE Released Test Questions, we see that y = ax^2 and y = ax^3 appear not in the Algebra I section, but actually the Algebra and Functions (i.e., seventh grade standards) section.

If we compare this to the Common Core Standards, we see that much of Chapter 10 of the McDougal Littell text corresponds to an eighth grade standard in Common Core:

Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.

This is to be expected. The Common Core Standards are based on Algebra I in ninth grade, while the California Standards were based on Algebra I in eighth grade. So many eighth grade Common Core Standards must have been seventh grade standards in California.

Before we leave the McDougal Littell text, let me note that Section 4.3 is on "Solving Equations Involving Negative Coefficients." I point this out only because, if you recall, last month I subbed for a sixth grade teacher who unwittingly assigned a worksheet with a few negative coefficients, and the sixth graders were confused -- as they should have been, since this was a seventh grade standard.

I didn't purchase the Merrill Pre-Algebra text, so I don't recall how old the text is. But I glanced at it and noticed that all of the equations that appear in Chapter 10 of the U of Chicago Geometry text also appear in this text, with the exception of the equations involving a sphere. That is, the surface area formulas of all cylindric and conic solids appear in this text. This is unusual since, as we've seen, neither the CAHSEE nor the Common Core Standard expect students to learn the more complex surface area formulas before high school Geometry. Since today's lesson is Section 10-2 of the U of Chicago text, which is on surface areas of pyramids and cones, I want to discuss what I remember about the Merrill lesson on these surface areas.

Both Merrill and the U of Chicago give the lateral area of the pyramid as the sum of the areas of its triangle lateral faces. But only the U of Chicago gives the formula for a regular pyramid, which it defines in Section 9-3 as a pyramid whose base is a regular polygon and the segment connecting the vertex to the center of this polygon is perpendicular to the plane of the base. The formula for the lateral area of a regular pyramid is LA = 1/2 * l * p.

But now we must consider the surface area of a cone. The Merrill text does something interesting here, as it considers the area of the net of the cone. We cut out the circular base and a slit in the lateral region, and then flatten this lateral region. What remains is a sector of a circle. Then the Merrill text simply gives the area of this sector as pi * r * s (where s, rather than l, is the slant height) without any further explanation.

The U of Chicago text, meanwhile, gives a limiting argument for the surface area of the cone, as its circular base is the limit of regular polygons as the number of sides approaches infinity. But there is Exploration Question 25, where the Merrill demonstration is done in reverse -- we begin with a sector of a disk and fold it into a cone.

But neither tells us why the area of the sector (and thus the lateral area of the cone) is pi * r * l. Let me give a demonstration of why the area of the sector is pi * r * l.

We begin with the area of a circle, pi * R^2. The reason why I used a capital R is to emphasize that the radius of the circle that appears in Question 25 is not the radius r of the base -- indeed, it's easy to see that the radius of the circle becomes the slant height l. So the area of the circle is pi * l^2 -- that is, before we cut out the sector. We want to fit the area after we cut it.

Let's recall another formula for the area of a circle given by Dr. Hung-Hsi Wu: A = 1/2 * C * R -- and once again, R = l, so we have A = 1/2 * C * l. But neither one of these gives us the circumference or area of a sector. If we let theta be the central angle of a sector, we obtain:

x = theta / 360 * C
L.A. = theta / 360 * A
        = theta / 360 * 1/2 * C * l

For lack of a better variable, I just let x be the arclength of our sector. But here I let L.A. be the area of the sector, since these equals the lateral area of the cone we seek. The big problem, of course, is that we don't know what angle theta is for the cone to have a particular shape. But we notice that we can simply substitute the first equation into the second:

L.A. = 1/2 * theta / 360 * C * l
        = 1/2 * x * l

And what exactly is the arclength x of our sector? Notice that once we fold the sector into a cone, the arclength of the sector becomes the circumference of the circular base of the cone! And this we know exactly what it is -- since the radius of the base is r, its circumference must be 2 * pi * r:

L.A. = 1/2 * (2 * pi * r) * l
        = pi * r * l

as desired. QED

I incorporate this demonstration into my lesson. As this is the first lesson after a vacation, I wanted to give the students something fun to do -- so let's have them cut out sectors and make cones!

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