This is what I wrote two years ago about today's lesson. Notice that I spent much of that post comparing the U of Chicago text to three other math texts (and I decided to preserve this discussion):

One of the texts was published by Merrill, the others 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.

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 it appears in the 7th grade standards.
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:

CCSS.MATH.CONTENT.8.G.C.9

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

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." For comparison purposes, let's look at the McDougal Littell Algebra Readiness text in more detail:

1. Expressions, Unit Analysis, and Problem Solving

2. Fractions

3. Decimals and Percents

4. Integers

5. Rational Numbers and Their Properties

6. Exponents

7. Square Roots and the Pythagorean Theorem

8. Equations in One Variable

9. Inequalities in One Variable

10. Linear Equations in Two Variables

The purpose of Algebra Readiness was to prepare students for Algebra I. Therefore, as we can see, there is very little geometry in this text compared to the McDougal Littell Math 7 text. The only geometry that appears is in Chapter 7, with even less geometry content than Chapter 9 of the Math 7 text (as the Distance Formula doesn't appear in the Algebra Readiness text). Area and volume are nowhere to be seen in the Algebra Readiness text (except the appendix, "Skills Review Handbook").

In many ways, Algebra Readiness was more like a Common Core 7 text than Common Core 8, as Common Core 8 contains more geometry (and even a little more algebra) than the Readiness text.

Now leave MacDougal Littell and continue with the Merrill text:

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 Lesson 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 Lesson 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.

**2018 Update:**Today is an activity day (as if we haven't had enough activities this week). The worksheet from two years ago already incorporates this section's Exploration question. Students are directed to cut out a circle sector, fold this sector into a cone, then calculate its lateral area.

Two years ago, I described how to solve a similar problem:

(The diagram shows a circle centered at

Area of sector

Notice that Chapter 8 of the U of Chicago text -- although it covers both arc length and circle area -- doesn't discuss the area of a sector. Many other Geometry texts do discuss the areas of sectors -- indeed the lesson where the formula

But still, we can use the pattern established in the U of Chicago text for arclength -- the measure of the arc is 45 degrees, so the area of the sector is 45/360 times that of the entire circle:

Area(

12.5pi = 45/360 (pi

12.5pi = (pi

100pi = pi

100 =

and of course, we only consider the positive solution. So the radius is 10 and the diameter is 20.

*P*, with*A*and*B*on the circle. Arc*AB*measures 45 degrees.)Area of sector

*PAB*= 12.5pi. The length of the diameter = ?Notice that Chapter 8 of the U of Chicago text -- although it covers both arc length and circle area -- doesn't discuss the area of a sector. Many other Geometry texts do discuss the areas of sectors -- indeed the lesson where the formula

*A*= pi*r*^2 appears is often called "Areas of Circles*and Sectors*."But still, we can use the pattern established in the U of Chicago text for arclength -- the measure of the arc is 45 degrees, so the area of the sector is 45/360 times that of the entire circle:

Area(

*PAB*) = 45/360 **A*12.5pi = 45/360 (pi

*r*^2)12.5pi = (pi

*r*^2)/8100pi = pi

*r*^2100 =

*r*^2*r*= +/-10and of course, we only consider the positive solution. So the radius is 10 and the diameter is 20.

By the way, suppose we think of the sector mentioned in this question as one that we must cut out of the circle to form a cone. The circumference of the original circle is 20pi cm, and we've cut out one-eighth of the circumference, or 2.5pi cm, leaving 17.5pi cm, So the base of the cone must have diameter 17.5 cm -- that it, it has radius 8.75 cm. The slant height is the radius of the original circle, which is 10 cm. We may use the Pythagorean Theorem to find the vertical height of the cone -- which turns out to be approximately 5 cm. Yes, it is almost exactly half of the slant height because the numbers 4-7-8 are very close to a Pythagorean triple.

Let's continue with the next proposition in Euclid:

**Proposition 15.**

This proof should be easy to modernize:

Given:

*P*,

*Q*,

~~EF~~

Prove: Plane

*P*| | Plane

*Q*

*Proof:*

Statements Reasons

1. bla, bla, bla 1. Given

2. Point

*G*in plane

*Q*so that 2. Proposition 11 from Monday (construction)

*Q*

3.

*H*in

*Q*so that

*K*in

*Q*so that

4.

5.

*BC*5. Perpendicular to Parallels (planar)

6.

*P*6. Proposition 4 from two weeks ago

7. Plane

*P*| | Plane

*Q*7. Proposition 14 from yesterday (a form of Two Perpendiculars)

Euclid's original proof should be simple enough for high school students to understand without the need to convert it to two columns. All that's need is to replace phrases such as "therefore the sum of the two angles...is two right angles" with "Perpendicular to Parallels," for example.

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