Well, my student is still slightly confused with solving and setting up proportions. But he was more concerned with a question from Section 7-2 of the Glencoe text, on similar polygons. There is a group of similarity problems in the Glencoe text that are flawed. I don't remember the exact problem off the top of my head, but I found a problem in the U of Chicago text that contains the same flaw.

Here is Question 22 from Section 12-5 of the U of Chicago text:

In the figure [...], the two right cones are similar.

*QS*= 10,

*MZ*= 5.

*T*and

*L*are centers of the bases.

a. What is the ratio of similitude?

The fact that these are similar

*cones*is not the flaw -- the actual problems from the Glencoe text have the same flaw despite having similar

*triangles*instead of cones. Let's ignore the cones and only consider the triangular cross sections

*MZN*and

*SQR*, Let's also say that enough angle measures are given for us to conclude that

*MZ*and

*SQ*are corresponding sides of similar triangles, with their lengths given as above.

So we should have enough information to calculate the ratio of similitude -- also known as the "scale factor" in the Glencoe text. But now here's the flaw -- the answer could be either 1/2 or 2! Notice that if we imagine a similarity transformation mapping

*MZN*to

*SQR*, then the scale factor is 2, but if we instead imagine a similarity transformation mapping

*SQR*to

*MZN*, then the scale factor is 1/2. We obtain two different answers -- which are always reciprocals of each other -- depending on which triangle is taken to be the preimage of the similarity transformation and which triangle is taken to be its image.

My student assumed that, since English reads from left to right, the triangle closer to the left edge of the paper --

*MZN*-- must be the preimage, and so he obtained a scale factor of 2. We don't know what the intended answer in the U of Chicago text is, since this is an even numbered problem. But in the Glencoe text, where the corresponding question was an odd-numbered problem, the scale factor was given as 1/2, so that the triangle closer to the

*right*edge of the paper was the preimage. And so the teacher corrected the assignment using Glencoe's answer key -- and therefore deducted points for my student's "incorrect" answer!

So we see the flaw that commonly appears in similarity problems. Students are given two triangles with enough information to conclude that they are similar and are asked to find the scale factor, but there are two possible scale factors,

*r*and 1/

*r*, depending on which triangle is taken to be the preimage and which one is taken to be the image. Of course, if there is a similar statement somewhere in the problem, such as "tri.

*SQR*~ tri.

*MNZ*," then one can infer that

*SQR*is the preimage. But if there is nothing but the

*pictures*of the two triangles, then which triangle is the preimage is up to an arbitrary heads-or-tails

*coin flip*of the author. I don't believe that students should be penalized for making the opposite coin flip choice, choosing

*r*instead of 1/

*r*, as my student was.

In the Glencoe text, sometimes the leftmost triangle was taken to be the preimage, and other times, the rightmost triangle was taken to be the preimage. The only choice that consistently gave the correct preimage was that the triangle whose vertices were first

*alphabetically*was the preimage. So between triangles

*DEF*and

*JKL*,

*DEF*was the preimage even it was pictured to the right of

*JKL*. Of course, in our U of Chicago example,

*M*and

*N*come before

*Q, R, S,*but

*Z*comes after all of these other letters in the alphabet, but in the Glencoe text, it was always clear which vertices come first. But nowhere in the text was it stated that students should choose the alphabetically first triangle as the preimage -- yet this should be made

*crystal clear*if students are to lose points for making the wrong choice!

In the earliest sections of Chapter 12 in the U of Chicago, the focus is on dilations, and so triangles are usually labeled as

*ABC*and

*A'B'C'*. So there is no doubt that triangle

*ABC*is the preimage and the triangle with

*primed*vertices is the image. But after Section 12-5, some of the questions could be interpreted either way. For example, the first question of Section 12-6 asks students to find the ratio of the diagonals, lateral areas, and volumes of two similar boxes. But notice that U of Chicago doesn't force students to make an arbitrary choice -- the answers are given in the back of the text are given as 5/7 (or 7/5), 25/49 (or 49/25), and 125/343 (or 343/125).

And what's noteworthy is the first section of Section 12-8, the U of Chicago text asks the students to find

**a**ratio of similitude between two similar triangles -- not

**the**ratio of similitude. It's only the Glencoe text that forces students to find "the" scale factor.

I went back to double check the test that I posted here on the blog for Chapter 12. Fortunately, any question there that asks for a scale factor makes it clear which figure is the preimage or image.

But enough about the Chapter 12 Test -- let's worry about the Chapter 8 Test that I'm posting today on the blog. Here are the answers:

1. This is a triangle tessellation. The triangle is isosceles, so it shouldn't be too hard to tessellate.

2. One can find the area by drawing a square grid and estimating how many squares are taken up on the grid. Since the shape happens to be an ellipse, one can also find its area using the formula for the area of an ellipse -- pi *

*a**

*b*, where

*a*and

*b*are the major and minor axes. That's right -- I had to slip in a reference to pi this week.

3. 5/2 or 2.5.

4. 12 square units.

5. 40 square units.

6. 6 mm. This is a trick if one forgets the 1/2 -- especially with 3-4-5 right triangles featuring in the last few questions, but here the legs are 6 and 4, not 3 and 4.

7. 10 feet.

8. 4.5

*s*.

9. One can find the area by drawing a diagonal to triangulate the trapezoid. Then one adds up the area of the two triangles.

10. Answers may vary. The simplest such rectangle is long and skinny -- 1 foot by 49 feet.

11. Choice (a). The triangles have the same base and height, therefore the same area.

12. 3

*s*^2.

13. 13 feet.

14. 6 minutes.

15. 780,000 square feet.

16. 133 square feet.

17. 1/4 or .25. Probability is a tricky topic -- the U of Chicago assumes that the students already know something about probability. Then again, it should be obvious that the smaller square is 1/4 of the larger square.

18. 4,000 square units.

19. 13.5 square units.

20.

*a*(

*b*-

*c*) or

*ab*-

*ac*square units.

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