## Friday, May 15, 2015

### PARCC Practice Test Question 24 (Day 169)

Question 24 of the PARCC Practice Test is on inscribed angles of a circle. As I've mentioned earlier, inscribed angles, in Section 15-3 of the U of Chicago text, is the only material from the last chapter of the text that appears on the PARCC:

The figure shows a circle with center P, a diameter BD, and inscribed Triangle BCD. The length of PC is 10. Let Angle CBD = (x) degrees and Angle BCD equal (x + 54) degrees.

Part A

What is the value of x?

Part B

Identify of the true statements about the figure.

(A) The length of CD is less than 10.
(B) The length of CD is equal to 10.
(C) The length of CD is greater than 10.
(D) Triangle CPD is equilateral.
(E) The measure of Angle CPD is less than 60 degrees.
(F) The measure of Angle CPD is greater than 60 degrees.

For this question, the key theorem is not the Inscribed Angle Theorem itself, but a corollary:

An angle inscribed in a semicircle is a right angle.

Since BD is a diameter, Arc BCD is a semicircle, and so Angle BCD is inscribed in a semicircle. So Angle BCD is a right angle. Since its measure is x + 54 = 90, we easily conclude that x = 36.

To find out which statements in Part B are true, we need to figure out all the other angle measures, so now this problem becomes similar to Question 8 of the Practice Test. First, we can use the Inscribed Angle Theorem itself to conclude that the measure of arc CD, and therefore the measure of angle CPD, must be twice 36, or 72 degrees, which is clearly greater than 60 degrees. Finally, we can find that the measure of angle CDP is 54 degrees several ways -- by noting that BCD is a right triangle, or that CDP is an isosceles triangle, or even that arc BC measures 108 degrees. (Yes, the same 54 that shows up in the measure of BCD, x + 54, is also the measure of BCD.) Since CPD has a greater measure than CDP, we can use the Unequal Angles Theorem of Section 13-7 to conclude that the side opposite the larger angle, CD, must be longer than the side opposite the shorter angle, PC, which is known to be 10. Therefore the correct answers are (C) and (F).

For me, this was a good problem because it alerted to me which sections of the last three chapters of the U of Chicago text appear on the PARCC. Recall that this year, I basically tore up Chapter 15 since I thought that only Section 15-3 would appear on the PARCC. I covered most of Chapter 13 on the blog this year, but then I wrote that I would tear this chapter up next year in order to save some more time for Chapters 8, 10, and 14 -- it appeared that outside of the sections on logic, the only geometric section of Chapter 13 that would appear on the PARCC was Section 13-5, on tangents to circles and spheres.

Well, I'm correct about Chapter 15 so far, as we still haven't seen any content from the last chapter outside of Section 15-3. But I erred on Chapter 13. The featured question for today mentions the Unequal Angles Theorem, which appears in Section 13-7, not 13-5. Fortunately, I've already covered Section 13-7 on the blog, but it does mean that next year, I can't throw out everything in Chapter 13 after 13-5, as I had planned.

Let's recall what Section 13-7 of the U of Chicago text is about. The chapter is called Exterior Angles, and it contains four theorems:

-- Exterior Angle Theorem
-- Exterior Angle Inequality
-- Unequal Sides Theorem
-- Unequal Angles Theorem

We've discussed the content of this chapter several times. It is a highly logical chapter as each theorem in the section is used to prove the next. But Dr. Franklin Mason had once used a different logical sequence of these theorems. Recall that Dr. M used to prove the Triangle Exterior Angle Inequality using only congruent triangles -- similar to the way Euclid once proved it -- without using his Fifth Postulate. Then the TEAI was combined with a Parallel Postulate to prove the Exterior Angle (Equality) Theorem. Dr. M -- presumably because this path was too tricky for students -- first made the TEAI a postulate, then dropped it from the chapter on parallel lines altogether. But he still mentions the TEAI in his chapter on inequalities in triangles -- in which his proof sequence is now identical to that in the U of Chicago.

But why does the U of Chicago wait all the way until Chapter 13 to give these theorems -- especially when it appears in Chapter 5 of Dr. M? The reason it that the proof of the Unequal Angles Theorem uses an indirect proof, and the text waits until Section 13-4 to teach indirect proof. In particular, we prove the Unequal Angles Theorem indirectly -- suppose the side opposite the larger angle were not the longer side. Well, it can't be shorter, since then the longer side would be opposite the smaller angle, contradicting the previously proved Unequal Sides Theorem. And it can't be equal, since then the angles opposite congruent sides would not be equal, contradicting the Isosceles Triangle Theorem, so therefore the longer side must be opposite the larger angle. QED

So what does this mean as far as including Section 13-7 in next year's schedule is concerned? I'm already including Sections 11-3, 13-5, and 15-3 as a short circle unit -- the final unit of my curriculum next year -- so we could squeeze in Section 13-7 here. Then again, the inequality theorems aren't necessarily connected to circles, even though today's PARCC problem includes it as Part B of a question where Part A is about circles.

I could also follow Dr. M's path and include 13-7 between Chapters 4 and 5 of the U of Chicago. We would need to cover Section 5-1, on the Isosceles Triangle Theorem, before we could do 13-7, but it would still be before the rest of U of Chicago's Chapter 5, which is on quadrilaterals and corresponds to Dr. M's Chapter 6. In fact, we see how many of the inequalities of Dr. M's Chapter 5 is spread out among several U of Chicago chapters. Section 1-9 is on the Triangle Inequality, which is a postulate in the U of Chicago but can be proved after the theorems of 13-7. And 7-8 is the SAS Inequality or Hinge Theorem, which the U of Chicago proves using the Triangle Inequality.

Well, I don't have to decide which path I wish to follow yet. But this does give me something to think about as I plan for next year.

Still, I find today's featured question to be an okay PARCC problem. There are only two parts -- not four like yesterday's. The only tricky part for students would be to make sure that both correct answers are selected for Part B. We notice that the first three choices (A)-(C) go together, as CD must be either less than, equal to, or greater than 10. Choices (D)-(F) also go together as the measure of Angle CPD must be either less than, equal to, or greater than 60 degrees. Observe that CPD is an isosceles triangle as two of its sides are radii, so if Angle CPD were 60, this would immediately make the triangle equilateral. So Choice (D) does ultimately boil down to Angle CPD = 60. So it should be clear that exactly two answers are to be chosen -- one from (A)-(C) and the other from (D)-(F).

Today is an activity day. I've decided to focus my attention for this activity on Section 13-7, since even though I covered it, I may have de-emphasized its importance. In this activity, we consider the following three inequalities for triangles:

-- Unequal Sides Theorem
-- Unequal Angles Theorem
-- Triangle Inequality

And we ask, which of these are true for pentagons? This is inspired by a questions in the Exploration Section of the U of Chicago text, asking about what inequalities apply to quadrilaterals or pentagons.

As the students will find out, the Unequal Sides and Unequal Angles Theorems certainly do not apply to pentagons. One way to see this is to consider what I call the "home" pentagon -- where "home" can refer to either a house or to home plate in baseball. This pentagon has line symmetry, three right angles, and two 135-degree angles. The two base right angles are opposite the sides of the "roof." We can make this roof arbitrarily steeper, so that the two 135-degree angles, as well as the lengths opposite the two right angles, have increased in measure. So this is a counterexample to both the Unequal Sides and Unequal Angles Theorems because we have longer sides opposite right angles and the shorter sides opposite obtuse angles.

But the Triangle Inequality -- now the Pentagon Inequality -- remains true. No side of any polygon can be longer than the lengths of the other sides combined. This is merely because the shortest distance between two points is a straight line, not a polygonal path.

PARCC Practice EOY Exam Question 24
U of Chicago Correspondence: Section 15-3, The Inscribed Angle Theorem
Key Theorem: Inscribed Angle Theorem

In a circle, the measure of an inscribed angle is one-half the measure of the intercepted arc.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.C.A.2
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

Commentary: I repeat my comments from Question 8. Most of the inscribed angle problems in the U of Chicago text give the intercepted arc measure, but a few do require students to calculate the arc measure first before finding the inscribed angle measure. In the SPUR section, Objective B, including Questions 7 through 11, are good questions to consider. We must also consider Section 13-7 and its questions on unequal sides and angles.