Yes, Lessons 8-1 through 8-3 must be skipped in order to maintain the digit pattern, since Days 81 through 83 were finals days. Lesson 8-1 is on perimeter, 8-2 is on tessellations, and 8-3 is an intro to area, including the area of a rectangle.
If you want, here's how you can incorporate the three missing lessons into today's lesson. For Lesson 8-1, all the students really need to know is "just add up all the sides." The U of Chicago text also provides two special cases -- a rectangle (or kite) and an equilateral polygon. But these examples can be covered quickly. In fact, a good Warm-Up activity for today could be to provide three polygons -- perhaps a scalene triangle, a rectangle, and a regular pentagon -- and then ask the students to find their perimeters by adding the lengths of all the sides.
Lesson 8-2 is on tessellations, and since this lesson isn't as important, we can just skip it. Yes, many traditionalists will be happy to hear that we're skipping tessellations! On one hand, I like the idea of teaching tessellations, since these have translation symmetry -- just as isosceles trapezoids/triangles have reflection symmetry and parallelograms have rotation symmetry. But on the other hand, if we want to reach Lesson 8-4 today then something must be left out, and this is it. (The modern Third Edition of the text teaches tessellations in Chapter 7 rather than Chapter 8.)
Indeed, the same thing happened last year in my eighth grade class. My plan was for the first activity after winter break to be the tessellation activity from the Illinois State text. But then we had the Common Planning meeting where we had to address which standards we have taught and which standards needed to be covered before the SBAC. Even though tessellations are indirectly related to the Common Core Standards, they aren't actually mentioned in the standards. Therefore I ended up skipping the tessellation activity. (There was a Square One TV song, "Tessellations," that I'd hoped to sing in class, but I never did.)
Lesson 8-3 is an intro to area. It's possible to cover this by folding it into today's Lesson 8-4. The important thing is to cover all parts of the Area Postulate:
a. Uniqueness Property: Given a unit region, every polygonal region has a unique area.
b. Rectangle Formula: The area of a rectangle with dimensions l and w is lw.
c. Congruence Property: Congruent figures have the same area.
d. Additive Property: The area of two nonoverlapping regions is the sum of the areas of the regions.
And so we hand students a grid filled with shapes whose area we want to find. We fulfill part a by pointing to the unit squares. Some of the figures are rectangles, so that gives us part b. It's better to try part d next, with figures such as a L-shape and other complex rectilinear polygons. Finally, we can demonstrate part c with the reflection or rotation image of a previously found area. Afterwards the students can attempt to find the areas of the irregular regions of Lesson 8-4.
Even with these possible additions, I didn't create a new worksheet for this year. Teachers can decide whether to follow my suggestions to incorporate Lessons 8-1 to 8-3 with this worksheet.
I like the idea of starting the second semester with an activity after a long winter break. Both Lessons 8-2 and 8-4 naturally lend themselves to activities, and I chose 8-4 for today. On the other hand, teachers who lean towards traditionalism will more likely use today to combine Lessons 8-1 and 8-3, skipping both 8-2 and 8-4. In any case, students will be ready for Lesson 8-5 tomorrow.
I believe that having finals on Days 81 to 83 makes the digit pattern work out the best. If the first semester had been only 80 days (as it is in the LAUSD), then finals would cut Chapter 7 short. And if the semester had been closer to 90 days, then the more important parts of Chapter 8 are skipped.
The only other workable possibility is for finals to be Days 91 to 93. This is because Chapter 9 (on 3D figures) isn't as important as, say, Chapter 10 (when we find their volumes), and so lessons in Chapter 9 are expendable. But few schools would start early enough in August to have 93 days in the first semester.
And besides, I like including Chapter 8 with second semester rather than first semester. I always like to think from the perspective of students having just completed Algebra I -- they want something new, not more algebra. Our first semester with Chapters 1 through 7 contain very little calculation at all, much less algebra. What little calculation does appear is most in Chapter 3, where we must add and subtract to find angles that add up to 90 or 180. The only multiplication or division that appears is by a factor of two -- bisection.
Then once we reach Chapter 8, multiplication immediately appears in the area formulas. Chapters 10 (volume), 12 (similarity), and 14 (trig) are all heavy on multiplication, division, and algebra. And so in the name of confining the hardest number crunching to semester two, we begin it with Chapter 8.
Two years ago I set up Lesson 8-4 by keeping the area of a circle in mind. And hey -- by working on the quarter-circle, part c of the Area Postulate can be used to estimate the area of the whole circle.
This is what I wrote two years ago about today's lesson:
Lesson 8-4 of the U of Chicago text is on the areas of irregular regions. And as I mentioned earlier this week, while the text focuses on the areas of triangles and the shapes of lakes, I'm going to focus on the areas of circles in preparation for pi.
The source of this is Dr. Hung-Hsi Wu. Let's see what he has to say about approximating pi:
We start by drawing a quarter unit circle on a piece of graph paper. In principle, you should get the best graph paper possible because we are going to use the grids to directly estimate π. Now, perhaps for the first time in an honest mathematics textbook, you are going to get essential information about something other than mathematics: the grids of some of the cheap graph papers are not squares but nonsquare rectangles, and such a lack of accuracy will interfere with a good estimate of π. If you are the teacher and you are going to do the following hands-on activity, be prepared to spend some money to buy good graph paper.
So far, this appears to be exactly what the U of Chicago is doing in Lesson 8-4, except with the area of a lake instead of a circle. The text writes, "you might cover the lake with a tessellation of congruent squares." (Notice that here we have tessellations again, since after all, a grid is nothing more than a square tessellation.)
Back to Wu:
So to simplify matters, suppose a quarter of a unit circle is drawn on a piece of graph paper so that the radius of length 1 is equal to 5 (sides of the) small squares, as shown. (Now as later, we shall use small squares to refer to the squares in the grid.) [boldface Wu's]
The square of area 1 then contains 5^2 small squares. We want to estimate how many small square are contained in this quarter circle. The shaded polygon consists of 15 small squares in the grid. There are 7 small squares each of which is partially inside the quarter circle. Let us estimate the best we can how many small squares altogether are inside the quarter circle. Among the three small squares in the top row, a little more than 2 small squares are inside the quarter circle; let us say 2.1 small squares. By symmetry, the three small squares in the right column also contributes 2.1 small squares. As to the remaining lonely small square near the top right-hand corner, there is about 0.5 of it inside the quarter circle.
Notice that Wu's counting technique is a little finer than the U of Chicago's. The U of Chicago would recommend that we simply count half of the squares on the boundary. Then we would estimate that there are 3.5 squares on the boundary rather than Wu's 4.7. Counting only 3.5 squares on this boundary gives a rather inaccurate value of pi, namely 2.96. We can refine this by noticing that two of the squares on the boundary are almost full -- the upper left and lower right squares. So the squares on the boundary should count more like 4.5 than 3.5. This gives us a value of 3.12 for pi -- not quite as good as Wu's 3.152, but much better than 2.96. Our relative error is 0.69%, which is about twice as much as Wu's, but still within one percent.
Because of symmetry, we only need to count the quarter circles rather than the whole circle. And dividing the unit into fifths is convenient because of the 3-4-5 Pythagorean triple -- both (3/5, 4/5) and (4/5, 3/5) are grid points on the unit circle. This makes the squares easier to count. Using tenths is also convenient since we can double 3-4-5 to 6-8-10, plus now the unit square contains 100 squares, a number it's very easy to divide by. The total number of shaded squares should be about 78.5 in the quarter circle and 314 in the whole circle.
The first question I included is about how many dots per square inch there are on the monitor of a Macintosh computer. Naturally, this U of Chicago question is based on 1990's technology. Of course, the current standard has many, many more pixels per square inch. If we stick to Macs, since these were mentioned in the original question, the new MacBook Pro laptop has a 2880*1800 resolution -- a far cry from the 512*342 resolution mentioned in the text.
I could have -- and perhaps should have -- changed this question to reflect current technology. But I definitely want to keep this type of question -- this is exactly the sort of question that appears on standardized tests, such as the PARCC and SBAC.
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