Lesson 8-4: Areas of Irregular Regions (Day 117)

This is what I wrote last year 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 Day.

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. And maybe I'll do so at a later time. 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.