## Wednesday, April 27, 2016

### PARCC Practice Test Question 6 (Day 149)

Chapter 6 of Morris Kline's Mathematics and the Physical World is "The Laws of Space and Forms," and this chapter is all about our favorite branch of mathematics -- geometry!

"Geometry...is the science that it hath pleased God hitherto to bestow on mankind." -- Thomas Hobbes, 17th century English philosopher

Kline begins:

"The story is told that the Greek philosopher Aristippus and some friends were shipwrecked on what appeared to be a deserted island near Rhodes. The company was downcast at its ill fortune when Aristippus noticed some geometric diagrams drawn on the beach sand. 'Be of good cheer,' he told his companions, 'I see traces of civilized men.'"

In other words, wherever there is civilization, there is geometry. Kline discusses the development of geometry, beginning with the Ancient Egyptians and Babylonians, and moving to the Greeks and -- who else? -- Euclid.

Kline writes that even Euclid's contemporaries wondered why the ancient sage spent so much time with intuitively obvious theorems, such as SAS, the Triangle Inequality, and the theorem that the shortest distance between a point and a line is perpendicular to the line. Kline answers:

"Many geometrical theorems indeed seem obvious, but if they can be deduced from axioms, then the theorems are utterly beyond question."

He then gives the Triangle Midsegment Theorem as an example of a theorem that isn't obvious and requires proof. More to the point, if we relied only on intuition, rather than proof, we may end up accepting results that appear to be true, but are actually false.

Kline devotes the rest of this chapter to two main results. One of which is a continuation of the Isometric Theorem from yesterday, and the other involves reflections and light.

Yesterday, we looked at Kline's algebraic proof that among all rectangles with the same perimeter, the square has the most area. Well today, he adds a geometric proof of the same result. The following is his geometric proof:

Let ABCD be any rectangle with perimeter p. If the rectangle is not a square then one side, AB, is larger than the other. On the larger side let us construct a square, EBHG, with perimeter p. [That is, E is on AB and C is on BH, according to the diagram in the book -- dw] Now AE + EB + BC is half the perimeter of the rectangle. Hence

(1) AE + EB + BC = p/2.

But since the square has the same perimeter as the rectangle, half of its perimeter is also p/2; in symbols,

(2) EB + BC + CH = p/2.

If we subtract equation (2) from equation (1) we obtain

(3) AE - CH = 0.

Addition of CH to both sides of this equation yields

(4) AE = CH.

Now the smaller side, AD, of the rectangle must be less than any side of the square because the square has the same perimeter as the rectangle. Hence

(5) AD < GH.

The area of rectangle II [ADFE -- dw] is AE * AD and the area of rectangle III [CFGH -- dw] is CH * GH. But equation (4) tells us that AE = CH, while equation (5) tells us that AD < GH. Hence the area of rectangle II is less than the area of rectangle III.

Since the area of the entire rectangle ABCD is the sum of the areas of I [BCFE -- dw] and II and the area of the square is the sum of areas I and III, the square has greater area than the rectangle. QED

Kline ends his proof by writing, "We have proved precisely the same result that was established in the preceding chapter. It is a matter of taste whether one prefers the geometric proof to the algebraic one."

Now Kline considers the case where one of the sides of the rectangle whose area is to be maximized lies along a river. The maximum area occurs when the side along the river is twice the length of the other side -- and he proves this simply by reflecting the rectangle across the river!

Speaking of reflections, Kline moves on to the reflection of light. He writes that the first person to write about the properties of light was Euclid himself, in his work simply titled Optics. Euclid writes that the angle of incidence equals the angle of reflection.

Another theorem was proved by the mathematician Heron in the first century A.D. -- this is the same Heron whose famous formula produces the area of a triangle given its three sides. This time, Heron proves that the light actually follows the shortest path from a point to the mirror to a second point. I have seen this fact mentioned in other texts (but not the U of Chicago text), and it shows us how, for example, to get from a starting point to a river (where we can fill a bucket with water) and from there to a burning house in the shortest distance. Kline writes that this same idea shows up with billiard balls (and this does appear in the U of Chicago text, Lesson 6-4).

I'd like to give Heron's proof here, but this post is getting long enough, and we still need to get to today's PARCC question.

Before I get to the PARCC question, I want to make a comment about the blog calendar. Today is Day 149 in one of the two districts where I work. In the other district, today is Day 155. That second district has an Early Start Calendar, where the first semester ended before Christmas. As it turns out, the second district has already completed its SBAC testing, before the AP.

The blog calendar is based on my first district, where the SBAC is after the AP. It is on what I call a Middle Start Calendar -- but next year it will complete its transition to an Early Start Calendar. In fact, the first day of school will be exactly the same in both districts. I'm curious as to when the first district will give the SBAC next year, now that it's also on an Early Start Calendar.

Question 6 of the PARCC Practice Test is about rectangles and triangles:

6. In the diagram, quadrilaterals FBAG and CDEF are rectangles.

(Here is some other info from the diagram: A on CE, B on FC, BC = 3, CD = 12, G on EF, EG = 7.)

How long is DE rounded to the nearest tenth?

There are two things we need to know in order to solve this problem. First, we need to know that the opposite sides of a rectangle are congruent. We can obtain several lengths this way -- since CD = 12, its opposite side EF is also 12. This means that GF = 5, and so its opposite side AB = 5.

The other thing we need is that the opposite sides of rectangles are parallel. Since DE | | FC, the alternate interior angles CED and ECF are congruent.

This gives us two pairs of congruent angles in triangles ABC and CDE -- the right angles ABC and CDE are congruent, and the angles ACB (same as ECF) and CED. Thus by AA Similarity, triangles ABC and CDE are similar. This allows us to set up a proportion:

DE / BC = CD / AB
x/3 = 12/5
x = 36/5 = 7.2 exactly

We are told to round off the answer to the nearest tenth, but the answer DE = 7.2 is exact.

This is an interesting problem. It requires looking at similar triangles, but it's not obvious which triangles in this problem are similar. This is more complicated than anything that appears in Chapter 12 of the U of Chicago text, where it's always obvious which triangles are similar. This is the type of problem that traditionalists want to see more of, instead of dilation questions or whatnot. It requires students to think -- which is why many students will have trouble with this problem.

Today's question also brings up the idea of slope and its relationship to similar triangles -- which, as we know, is an eighth grade Common Core Standard. We wish to show that the slope along line CE is constant -- that is, the slope along CA equals the slope along AE.

But how do we find the slope along CA? Well, that equals the rise CB (which is negative) divided by the run BA. And the slope along AE equals the rise AG divided by the run GE. In other words, the triangles CBA and AGE are slope triangles.

As usual, we can prove that the slope triangles are congruent. They already have one pair of congruent angles -- the right angles CBA and AGE. And the angles ACB and EAG are congruent because they are corresponding angles, as CB | | AG. So triangles CBA and AGE are similar slope triangles, and so the slopes CB / BA and AG / GE are equal.

Here's an interesting question -- is it possible to solve this problem without similar triangles? As it turns out, it is possible. Instead of similar triangles, we use area. Let h be the value of DE that we are trying to find -- h standing for height (of the rectangle CDEF, of course).

Area(CBA) + Area(AGE) + Area (FBAG) = Area (CFE)
15/2 + 7(h - 3)/2 + 5(h - 3) = 12h/2
15/2 + (7h - 21)/2 + 5h - 15 = 12h/2
15 + 7h - 21 + 10h - 30 = 12h
17h - 36 = 12h
5h = 36
h = 7.2

This is clearly not the method intended by PARCC. After all, we look at the alignment document (provided at the same link where I'm getting the questions), and we see the standard for this question listed as G-SRT.5, where SRT stands for similarity and right triangles. But we could have sneaked this question into Chapter 8 (area) of the U of Chicago text instead of Chapter 12 (similarity).

Notice that this proof is somewhat more algebraic than the similarity proof -- that is, the students' mental burden is placed more on calculating the areas and solving the equation than on figuring out which triangles are similar. This goes back to what Kline writes about how it's a matter of taste whether an algebraic proof or a geometric proof of the same result is better.

But I've mentioned before that in many ways, similarity and area are equivalent. Many questions that are solved via similarity can be answered using area instead, and this is one of them. I wonder whether any student who sees this problem on the PARCC, but fails to notice the similar triangles, would try to find the area of the triangles and rectangles instead.

Sometimes I reflect on Common Core Geometry and how the course is organized. Many people don't like the reliance of Common Core Geometry on transformations. I like the idea that many statements that are postulates in pre-Core Geometry, such as SAS (both similarity and congruence), the slope formula, and so on, are provable in Common Core Geometry. But many, especially traditionalists, don't consider the extra level of rigor to be worth confusing students with reflections and dilations.

One idea that has been forming in my mind is replacing similarity and dilations in many Common Core proofs with area. This is already often done with the Pythagorean Theorem, but we can do it with the slope formula as well, as suggested in today's post. In some ways, this makes sense -- students have been learning about the area of a rectangle since third grade, years before they learn anything about similarity or dilations.

But there's one major problem with this idea -- beginning with the Area Postulate of Chapter 8 requires knowing that the opposite sides of a rectangle are parallel and congruent. After all, how we can say that the area of a rectangle is length times width if we don't know that the length of the top of the rectangle equals the length of the bottom of the rectangle? And I don't see how we can prove these rectangle properties without some sort of postulates about congruence. Well, it was just a thought.

PARCC Practice EOY Question 6
U of Chicago Correspondence: Lesson 12-9, The AA and SAS Similarity Theorems
Key Theorem: AA Similarity Theorem

If two triangles have two angles of one congruent to two angles of the other, then the triangles are similar.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.B.5Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.

Commentary: This question is a bit more sophisticated than any of the problems in the U of Chicago text for this lesson. Naturally, I include several more exercises similar to this PARCC problem, where students must search for similar triangles, on the worksheet.