One major difference that I noticed between the Glencoe and the U of Chicago texts is that Glencoe actually allows the scale factor to be negative. The U of Chicago text makes it clear that its focus is on positive scale factors:

"Any number but zero can be the magnitude, but in this book

*k*> 0 unless otherwise stated."

The only problem that I could find with a negative scale factor is in the Exploration part of Section 12-1 in the text. Question 18 reads:

Use Triangle

*JKL*of Question 5.

a. Multiply all the coordinates of the vertices by -3.

b. Describe the figure that results.

c. Generalize parts a and b.

I've briefly mentioned what a dilation of scale factor -1 would look like. Such a transformation would actually be a rotation of 180 degrees centered at the same point as the dilation. I've stated how rotations of 180 degrees and dilations of positive scale factors have several properties in common -- most notably, lines and their images under both types of transformations are parallel.

Dilations with a negative scale factor other than -1 can be thought of as the composite of a dilation with the corresponding positive scale factor followed by the 180-degree rotation. So the dilation in Question 18 above is the composite of a dilation of scale factor 3 (the dilation mentioned in Question 5) and a 180-degree rotation. The dilation and rotation commute, so it doesn't matter which one of them we perform first. The set of all dilations with nonzero scale factor, along with the translations, form a group, with the set of all dilations with positive scale factor and translations a subgroup.

Dilations with negative scale factor, like all rotations, preserve orientation. In all dimensions, the dilation of scale factor -1 is a point inversion (or point "reflection"). It preserves orientation if there are an even number of dimensions, but if there are an odd number of dimensions, then the dilation of scale factor -1 reverses orientation. This includes the 1D inversion (which is merely a reflection) and the 3D inversion (a roto-reflection).

Today I also subbed in a middle school math class. As it turned out, it was a class of seventh graders who were getting ready to take a test on angles. Printed on each test was the Common Core Standard being tested, which is:

CCSS.MATH.CONTENT.7.G.B.5

Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.

During the seventh grade class, an eighth grader was sent to my room to make up her own test. I glanced at her test and saw that it was also about angles -- except on parallel lines. In some ways, both of these tests fit better with yesterday's featured PARCC question that today's -- even though I'm comparing middle school tests to the high school PARCC question. Of course, the basic angle relationships of Chapter 3 of the U of Chicago text (the Linear Pair and Vertical Angle Theorems) come up over and over again, first in middle school and again in high school.

That's enough about yesterday's question -- now for today's. Question 31 of the PARCC Practice Test has four parts. It involves the coordinate plane and the perimeter and area of various quadrilaterals.

Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet.

In the figure, here are the vertices:

Plot of land: (0, 0), (50, 0), (70, 60), (0, 60)

Warehouse: (4, 20), (39, 20), (39, 50), (4, 50)

Parking area: (4, 0), (50, 0), (39, 20), (4, 20)

Part A:

What is the perimeter of the plot of land?

Express your answer to the nearest tenth of a foot.

Part B:

What is the area, in square feet, of the plot of land that does

**not**include the warehouse and the parking area?

Part C:

Luke is planning to put a fence along two interior sides of the parking area. The sides are represented in the plan by the legs of the trapezoid. What is the total length of fence needed?

Express your answer to the nearest tenth of a foot.

Part D:

In the future, Luke has plans to construct a circular storage bin centered at coordinates (50, 40) on the plan. Which of the listed measurements could be the diameter of a bin that will fit on the plot and be

**at least**2 feet away from the warehouse?

Select

**all**that apply.

(A) 10 feet

(B) 15 feet

(C) 18 feet

(D) 22 feet

(E) 25 feet

For Part A, three of the sides clearly have length 50, 60, and 70 feet. The remaining side requires the Distance Formula: sqrt((70 - 50)^2 + (60 - 0)^2) = sqrt(4,000) = 63.246 feet. So the total perimeter is 50 + 60 + 70 + 63.246 = 243.246 feet, or 243.2 feet to the nearest tenth.

For Part B, it's easier to find the area of the total plot of land and then subtract. We use the Trapezoid Area Formula as all three quadrilaterals are trapezoids (including the rectangle, inclusive definition):

Plot of land: 1/2 (60)(50 + 70) = 3,600 square feet

Warehouse: 1/2 (30)(25 + 25) = 750 square feet

Parking area: 1/2 (20)(46 + 25) = 710 square feet

So the net area is 3,600 - 750 - 710 = 2,140 square feet.

For Part C, one leg is clearly 20 feet, and the other requires the Distance Formula. We find the other leg as follows: sqrt((39 - 50)^2 + (20 - 0)^2) = sqrt(521) = 22.83 feet. So the total fence has length 20 + 22.83 = 42.83 feet, or 42.8 feet to the nearest tenth.

Finally, Part D is very tricky. The closest point from (50, 40) to the warehouse is clearly (39, 40), which is clearly 11 feet away. But we need a 2-foot gap, so the maximum radius is 9 feet. But now we must make sure that the other side will fit on the plot of land. The shortest distance between a point and a line is the perpendicular distance. The right side of the plot has slope 3, and so if we drop down a perpendicular it would have slope -1/3. The equation of the line passing through (50, 0) with slope 3 is found to be:

*y*= 3(

*x*- 50)

and the equation of the line passing through (50, 40) with slope -1/3 is:

(

*y*- 40) = -1/3 (

*x*- 50)

To solve this system, we substituting, we obtain:

3

*x*- 150 - 40 = -1/3

*x*+ 50/3

9

*x*- 570 = -

*x*+ 50

10

*x*= 620

*x*= 62

*y*= 3(62 - 50)

*y*= 36

The distance formula gives the distance from (50, 40) to (62, 36) as sqrt((62- 50)^2 + (36 - 40)^2), which equals sqrt(148) or 12.2 feet, which is clearly greater than 9 feet. So only the distance to the warehouse matters the most. The radius must be at most 9 feet, and the diameter at most 18 feet, so the correct choices are (A), (B), and (C).

I admit that if I were a student taking this test, I wouldn't want to do all of that work for Part D just to find out that the bin is closer to the warehouse than the edge of the plot anyway -- especially after having already worked on 30 other PARCC questions. I'm more likely just to ignore the edge and choose (A), (B), and (C) anyway -- either that, or guess that the edge is slightly closer and choose only (A) and (B).

In addition to today's worksheet -- especially since neither yesterday nor today did I have much room to include extra practice problems -- I present the worksheet on linear pairs and vertical angles on which the seventh graders practiced to prepare for today's test.

**PARCC Practice EOY Exam Question 31**

**U of Chicago Correspondence:**

**Section 11-2, The Distance Formula**

**Key Theorem: Distance Formula:**

**The distance between two points (**

*x*_1,*y*_1) and (*x*_2,*y*_2) in the coordinate plane is:**sqrt((**

*x*_2 -*x*_1)^2) + (*y*_2 -*y*_1)^2).

**Common Core Standard:**

CCSS.MATH.CONTENT.HSG.GPE.B.7

Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula.

**Commentary: The Distance Formula is in this section, although there is not much on area in this chapter at all. Section 8-6 of the text gives some good area problems involving trapezoids on the coordinate plane.**

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