## Thursday, April 28, 2016

### PARCC Practice Test Question 7 (Day 150)

Chapter 7 of Morris Kline's Mathematics and the Physical World is called "The Dimensions of the Heavenly Spheres." In this chapter, Kline measures the size of the earth, moon, and sun.

"On he passed, far beyond the flaming walls of the world, and traversed in mind and spirit the immeasurable universe." -- Lucretius, 1st century BC Roman poet

Kline begins,

"According to history, Euclidean geometry arose from man's effort to measure areas of land, and the literal meaning of the word geometry, namely, earth-measure, supports history."

And so in this chapter, Kline truly performs geometry, or earth-measure. He introduces the three main trig ratios of sine, cosine, and tangent. Then he proceeds to measure the sizes of the three main bodies of the earth, moon, and sun.

This sounds familiar -- and it should, because we just did it earlier this month, in Lesson 13-5. The U of Chicago text provides us with the earth-moon distance, 240,000 miles, the earth-sun distance of 93,000,000 miles, and the moon's radius of 1080 miles, and then similar triangles are set up to determine the sun's radius.

Kline does likewise. But let's imagine this problem from the perspective of the ancient Greeks, who weren't given the three values (1080, 240,000, and 93,000,000) found in the U of Chicago text. How were they able to estimate the radius of the sun? The answer is -- using trig. Kline writes:

"...the trigonometric methods of obtaining such seemingly inaccessible facts as the distance to the sun and the size of the sun are far simpler than the geometric proof that the area of a circle is pi r^2."

Kline writes about Hipparchus, the second century BC Rhodes mathematician. But the story begins with Erathosthenes, the 3rd century BC Cyrenian who first proved that the earth is round. By measuring shadows at two different locations, he obtained a value of about 24,000 miles for the circumference of the earth, and thus a value of about 4000 miles for the radius of the earth.

So Hipparchus used this value to find the distance to the moon. Here E is the center of the earth, M is the center of the moon, P is a point on earth's surface where the moon is on the horizon, and Q is another point on earth's surface where the moon is directly overhead. This makes EPM a triangle, and just as in Lesson 13-5, MP is tangent to the earth and so angle P is a right angle. Since the moon is directly overhead at Q, Q is on EM. Hipparchus measured the value of angle E using the system of longitude that he invented, and obtained 89 degrees, 4 minutes, 12 seconds, according to Kline (that is, 89 + 4/60 + 12/3600 degrees, or abour 89.07 degrees). So we obtain:

cos E = PE/EM = 4000/EM
.0163 = 4000/EM
.0163EM = 4000
EM = 245,000

According to Kline, Hipparchus obtained a value 280,000 miles. This value can be used to find the radius of the moon. This time, Q is a point on the moon's surface such that PQ is tangent to the moon, so now PMQ is a right triangle with the right angle at Q. Angle MPQ (which Kline writes as angle A can be measured on the earth to be about 15 minutes (1/4 of a degree). So we write:

sin A = QM/PM
sin 15' = QM/240,000
QM = 240,000 sin 15'
QM = 1056

The distance to the sun can be estimated the same way as the distance to the moon, but the angles involved are much harder to measure. According to Kline, Hipparchus obtained 10,000,000 miles, which is off by a full order of magnitude. Once these values are obtained, now we can find the radius of the sun using the method mentioned in the U of Chicago text.

Today is Day 150 on the blog calendar. It marks the midpoint of the third trimester -- in some districts, this is known as the end of the fifth hexter.

Today's also a special day because there's a featured Google Doodle. Hertha Marks Ayrton was a 19th century British engineer -- but she was also a mathematician! The following website describes Ayrton in more detail:

https://www.agnesscott.edu/lriddle/women/ayrton.htm

From 1881 to 1883, Marks worked as a private mathematics tutor, as well as tutoring other subjects. In 1884 she invented a draftsman's device that could be used for dividing up a line into equal parts...

Dividing up a line into equal parts -- hey, doesn't that sound familiar?

CCSS.MATH.CONTENT.HSG.GPE.B.6
Find the point on a directed line segment between two given points that partitions the segment in a given ratio.

Let's get back to her biography:

as well as for enlarging and reducing figures.

That is, as well as for performing dilations on figures! So we see that two of Ayrton's mathematical accomplishments are directly related to the Common Core Geometry standards.

Now Kline tells us that mathematics matters only when it can be applied to science. So let's look to see how Ayrton applied her mathematics:

Marks began her scientific studies by attending evening classes in physics at Finsbury Technical College given by Professor William Ayrton, whom she married in 1885. She assisted her husband with his experiments in physics and electricity, becoming an acknowledged expert on the subject of the electric arc. She published several papers from her own research in electric arcs in the Proceedings of the Royal Society of London and The Electrician, and published the book The Electric Arc in 1902.

An "electric arc" occurs when electrons travel between two electrodes in an arclike direction. And an electric field propagates from an electric charge in round waves in all directions. So everything goes back to ratios, lines, circles, and spheres.

Question 7 of the PARCC Practice Test is on -- what else -- circles. (It's too bad that we couldn't have a dividing segments question or a dilation question in Ayrton's honor, but this is the next best thing.)

7. The figure illustrates an informal argument for the area of a circle. The circle is divided into congruent sectors, and the sectors are rearranged to form a shape that resembles a parallelogram, as shown. As the number of sectors increases, the rearranged shape more closely resembles a parallelogram with area A, given by the formula A = bh, where b is the base and h is the height of the parallelogram.

Select the correct value for b and h to develop the area of the circle in terms of r, the radius of the circle.

(Both b and h lead to drop-down menus. The menu for b provides choices pi, r, pi r, 2pi, and 2pi r, while the menu for r provides choices pi, r, r * r, 2r, and 2pi.)

This problem demonstrates a very well-known proof of the circle area formula -- the answer is that the base is pi r and the height is r. It appears in many Geometry texts, including the U of Chicago text in Lesson 8-9. Let's recall what Kline writes about the derivation of the circle area formula again:

"...the trigonometric methods of obtaining such seemingly inaccessible facts as the distance to the sun and the size of the sun are far simpler than the geometric proof that the area of a circle is pi r^2."

And yet the PARCC expects students to provide the geometric proof that the area of a circle is pi r^2.

Notice that this is not how other authors derive the circle formulas. Dr. Hung-Hsi Wu actually starts with the area of a circle and then uses the formulas for perimeter/area of a regular polygon (which approaches the circle in the limit) to obtain the circumference of a circle. Dr. Franklin Mason goes the other way -- he starts with the circumference and uses regular polygons to derive the area.

Unfortunately, the way I taught the pi lessons on the blog ended up as a hodgepodge. My original intention was to follow Wu's lesson, but then I gave the circumference formula first. I ended up reteaching circumference as part of a seventh grade lesson I saw as a sub earlier this month.

By the way, a week later I ended up speaking to that seventh grade teacher again. He told me that he actually taught the seventh graders the circle area derivation given by U of Chicago and PARCC. He said that it was a bit confusing for them -- but again, remember what Kline said about the formula.

Notice it's also possible to begin with the area of a circle as in Wu, then divide it into sectors to form the parallelogram. This time, the area and height of the parallelogram are known, and the students find the height and therefore the circumference.

PARCC Practice EOY Question 7
U of Chicago Correspondence: Lesson 8-9, The Area of a Circle
Key Theorem: Circle Area Formula

The area A of a circle with radius r is A = pi r^2.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.GMD.A.1
Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.

Commentary: Lesson 8-9 of the U of Chicago text gives the desired proof. In fact, the first exercise on the worksheet is the same as the first question from the text.