1. Key Stone Problems… Key Stone Problems… next Set 9 © 2007 Herbert I. Gross

2. You will soon be assigned problems to test whether you have internalized the material in Lesson 9 of our algebra course. The Keystone Illustrations below are prototypes of the problems you'll be doing. Work out the problems on your own. Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve each problem. Instructions for the Keystone Problems next © 2007 Herbert I. Gross

3. As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

4. next #1a. Some geometry textbooks define two regions to be equal if they have the same area. In this context, is it true or is it false that the relation “is equal to” is an equivalence relation? Keystone Illustrations for Lesson 9 next Answer: True © 2007 Herbert I. Gross

5. Answer: True Solution for #1a: For “is equal to” to be an equivalence relation, three things have to be true… next © 2007 Herbert I. Gross next (1) It must be reflexive. That is, every geometric region must have the same area as itself. This is obviously true.

6. Solution for #1a: next © 2007 Herbert I. Gross next (2) It must be symmetric. That means that if the first region has the same area as the second region, it is also true that the second region has the same area as the first region. This is also a truism. For example: if the first region has an area of 36 square inches, then the second region also has an area of 36 square inches.

7. Solution for #1a: next © 2007 Herbert I. Gross next (3) It must be transitive. That means if the first region has the same area as the second region, and the second region has the same area as the third region, then the first region must have the same area as the third region.

8. Solution for #1a: next © 2007 Herbert I. Gross For example, suppose the first region has an area of 36 in 2. The fact that it is equal to the second region means that the area of the second region is also 36 in 2. The fact that the second region has the same area as the third region means that the area of the third region is also 36 in 2. Therefore, the first region is equal to the third region.

9. next #1b. Some geometry textbooks define two regions to be equal if they have the same area. Is it also true if two regions are equal (as defined above), they also have the same perimeter? Explain. Keystone Illustrations for Lesson 9 next Answer: False © 2007 Herbert I. Gross

10. There are different ways to define what we mean by equal. For example, some geometry books define two regions to be congruent if they have identical size and shape. next © 2007 Herbert I. Gross Note next Even then however, the two regions can be different with respect to their position in space.

11. We can show that if two regions (for example, two rectangles) are congruent, they have to have the same perimeter. next © 2007 Herbert I. Gross Discussion next 7 1017 3 1013 20

12. Answer: False Solution for #1b: Getting back to the question at hand, all we have to do to show that the statement is false is to find two regions that have the same area but not the same perimeter. next © 2007 Herbert I. Gross next So, for example, consider the rectangle R 1 whose dimensions are 4 in. by 9 in. and the rectangle R 2 whose dimensions are 3 in. by 12 in.

13. Solution for #1b: The area of R 1 is 4 in. × 9 in., or 36 square inches and the area of R 2 is 3 in. × 12 in. which is also 36 square inches. next © 2007 Herbert I. Gross next On the other hand, the perimeter of R 1 is 2 × (4 in. + 9 in.), or 26 inches while the perimeter of R 2 is 2 × (3 in. + 12 in.) or 30 inches.

14. Solution for #1b: R 1 has an area of 36 square in. next © 2007 Herbert I. Gross next A = 4 × 9= 36 square in. = 26 inches while the perimeter of R 1 is… 567 91011 131415 171819 212223 252627 293031 8 12 16 20 24 28 32 33343536 1234 1 2 3 4 5 6 7 8 9 10111213 2324 25 26 14 15 16 17 18 19 20 21 22 P = 2 × (4 + 9)

15. Solution for #1b: The area of R 2 is 36 square inches next © 2007 Herbert I. Gross next A = 3 × 12= 36 square inches = 30 inches while the perimeter is… 56 79 1011 131415 1718 1921 2223 252627 2930 31 8 12 16 20 24 28 3233 343536 123 4 1 2 3 4 5 6 7 8 9 10 11 12 13 23 24 25 26 1415 16 17 18 19 20 21 22 27 282930 P = 2 × (3 + 12)

16. As an aside, a square is a special case of a rectangle. A square whose dimensions are 6 in. by 6 in. also has, like R 1 and R 2, an area of 36 in 2. However since each of the square’s sides is 6 in., its perimeter is 4 × 6 in. or 24 in. (unlike R 1 ’s perimeter which is 26 inches or R 2 ’s perimeter, which is 30 in.). The interesting point is that, of all the rectangles that have the same area, the square has the least (smallest) perimeter. next © 2007 Herbert I. Gross Digression

17. However, to get back to the main point, we want to emphasize that when we talk about equal or equivalent, we always mean with respect to a specific relation. next © 2007 Herbert I. Gross Equivalence For example, when we write 3 + 2 = 9 – 4, we do not mean that there is no difference between the symbols 3 + 2 on the left and 9 – 4 on the right side; but rather that they name the same number ― five. next

18. The main point, then, that we must always keep in mind is that in mathematics when we use the term equal, we mean equal with respect to a certain property. © 2007 Herbert I. Gross Final Note