**Additional Materials**

**Materials Required**

**Estimated Time Needed**

*(Times are approximate and will depend on the needs of the students.)*

Homework

- Improve Individual Solutions to the Assessment Task (10 minutes)
**Make sure students have their original individual work on the***Table Tiles*task to hand.**Give them a fresh, blank copy of the***Table Tiles*task sheet and of the grid paper.

*Read through your original responses, and think about what you have learned from this lesson.**Using what you have learned, try to improve your work.***If a student is satisfied with his or her solution, ask the student to try a different approach to the problem and to compare the approach already used.**

- Solutions
Size of tabletop (cm) 10 x 10 20 x 20 30 x 30 40 x 40 50 x 50 60 x 60 Number of quarter tiles 4

4

4

4

4

4

Number of half tiles 0

4

8

12

16

20

Number of whole tiles 1

5

13

25

41

61

For a tabletop of side length x, n =

^{x}⁄_{10}is the number of tile diagonal widths in the side length.The number of quarter tiles is always 4. The number of half tiles is 4(n − 1) = 4 (

^{x}⁄_{10}− 1).The number of whole tiles is n

^{2}+ (n − 1)^{2}= 2n^{2}− 2n + 1 = 2(^{x}⁄_{10})^{2}− 2(^{x}⁄_{10}−1)^{2}= (^{x}⁄_{10})^{2}+ (^{x}⁄_{10}− 1)^{2}.

- Analysis of Student Responses to Discuss
**Leon's method**Leon drew three diagrams showing tabletops with systematically increasing side lengths. He colored the diagrams to pick out rows of whole squares parallel to the diagonal of the tabletop.

Width of Tabletop 10 cm 20 cm 30 cm Left column 1 content goes here 1 1 + 3 +1 1 + 3 + 5 + 3 + 1 Leon wrote the sum of the numbers to show the total number of whole tiles. He did this in an organized way. He predicted the number of whole tiles in the next diagram accurately.

Leon could check the next diagram to see if his conjecture is correct, but this would not be a proof. He could use his method to predict the number of whole tiles in the next diagram for any diagram he has drawn, but that method would not enable him to calculate the number of whole tiles for tables of any size. To do that, he might attempt to articulate the relationship between the number of whole tiles across the diagonal and the number of odd numbers to sum (assuming that students do not recognize that 1 + 3 + 5 + … + 2n −1 = n

^{2}).**Gianna's method**Gianna shaded alternate horizontal rows of squares. In the first two diagrams she wrote numbers in the whole squares. These record the number of whole squares in each horizontal row of that tabletop.

For the diagram of the 30 cm tabletop this gives:

In the first row:

3 3 3 In the second row:

2 2 In the third row:

3 3 3 In the fourth row:

2 2 In the fifth row:

3 3 3 Gianna picked out the pattern as 3 rows of 3 whole tiles, and 2 rows of 2 whole tiles, to find the total number of whole tiles is 3 × 3 + 2 × 2.

Gianna could then generalize to show that in the nth tabletop there would be n rows of n whole tiles, and n-1 rows of n-1 whole tiles, in total n

^{2}+(n − 1)^{2}whole tiles.**Ava's method**Ava drew one diagram showing how the length of a 40 cm tableside is made of two 5 cm edges of quarter tiles and three 10cm half tiles, and another similar diagram for the 50 cm table. Ava systematically organized data in a table, although it is not clear where the data came from. She found differences and second differences between numbers of tiles in tables of increasing side length. She did not show a way of calculating the number of tiles of various types given an arbitrary table size, and she did not use algebra.

Ava could next use her table of data to derive formulas for the number of tiles in any table. She might do this by using the first and second differences with the standard quadratic sequence algorithm.