• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  33. 33
    33
  34. 34
    34
  • Level: GCSE
  • Subject: Maths
  • Word count: 10854

I am going to investigate how changing the number of tiles at the centre of a pattern, will affect the number of border tiles I

Extracts from this document...

Introduction

Contents Page 1 ~ Introduction Page 2 ~ Patterns for 2 centre tiles Page 5 ~ Patterns for 3 centre tiles Page 8 ~ Patterns for 4 centre tiles Page 11 ~ Patterns for 5 centre tiles Page 14 ~ Summary for patterns with a single row of centre tiles Page 16 ~ Patterns for 4 centre tiles Page 19 ~ Patterns for 6 centre tiles Page 22 ~ Patterns for 8 centre tiles Page 25 ~ Patterns for 10 centre tiles Page 28 ~ Summary for pattern with a double row of centre tiles Page 30 ~ Summary for single and double rows of tiles Page 31 ~ Patterns for 6 centre tiles Page 34 ~ Patterns for 9 centre tiles Page 37 ~ Patterns for 12 centre tiles Page 40 ~ Patterns for 15 centre tiles Page 43 ~ Summary for patterns with a triple row of centre tiles Page 44 ~ Conclusion Borders Coursework Introduction For my experiment I am going to investigate how changing the number of tiles at the centre of a pattern, will affect the number of border tiles I will need. I will do this to find patterns and a formula, to link back to each set of patterns. Each formula will be tested by using a larger border, but with the same number of centre tiles, this will ensure my formula is correct. I will then try to find a general formula, that will enable me to predict the border for any size centre tiles. I will also do the same for the total tiles in the pattern. Key N ~ Pattern B ~ Outer border tiles T ~ Total Tiles C ~ Centre Tiles D ~ First difference H ~ Height W ~ Width / ~ x(y ~ Prediction For two centre tiles Table 1: Pattern: N 1 2 3 4 5 Border Tiles: B 6 10 14 18 22 +4 +4 +4 +4 B= 4N + 2 To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). ...read more.

Middle

Total Tiles I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number. Table 2 Pattern: N 1 2 3 4 5 Total Tiles: T 20 36 56 80 108 +16 +20 +24 +28 +4 +4 +4 Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N�. The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N�, so you should now have 2N�. You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with eight centre tiles is: T= 2N� + 10N + 8 Testing my formula. I will test my formula to see if it is produces the same values as in Table 2. N=1, T = 2(1)� + 10(1) + 8 = 20. N=2, T = 2(2)� + 10(2) + 8 = 36. N=3, T = 2(3)� + 10(3) + 8 = 56. I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams. Prediction I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 8 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then: T= 2N� + 10N + 8 = 2(7)� + 10(7) ...read more.

Conclusion

+ 4)(3) + 6 Then T = 2(5)( + (2(15) + 4)(5) +15 3 3 T = 48 T = 135 If C = 12 and N = 4 Then T = 2(4)( + (2(12) +4)(4) + 12 3 T = 92 These three checks are correct so I know my formula is working. I will now use it to make predictions for diagram number, N = 7. Prediction If C = 15 and N = 7. Then T = 2(7)( + (2(15) + 4)(7) + 15 3 T = 211 This once again proves my formula is correct. Conclusion I have proved that my formulas work and that there is a pattern between them. This proves that my prediction for border tiles B = 2(2N + C + H - 2) and my prediction for total tiles T = 2N( + 2(W + H - 1)N + C was correct, therefore I could predict the total tiles and border tiles for any pattern with any number of centre tiles. E.g. Border Tiles Border Tiles in the third pattern of a diagram with nine centre tiles. If the pattern number was N=3, then: B = 2(2N + C + H - 2) B = 2(2(3) + 9 + 3 - 2) B = 32 I have checked this against my previous data and have concluded that the data matches. I can therefore conclude that I can predict any border tile pattern for any number of centre tiles. Total Tiles Total tiles in the fifth pattern of a diagram with 9 centre tiles. If N=5, C=9, W=3, H=3 T = 2N( + 2(W + H - 1)N + C T = 2(3)( + 2(5 + 3 - 1)5 + 9 T = 50 + 50 + 9 T = 109 I have checked this against my previous data and have concluded that the data matches. I can therefore conclude that I can predict any amount of total tiles pattern for any number of centre tiles. ?? ?? ?? ?? ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE T-Total section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE T-Total essays

  1. Maths Coursework- Borders

    2 a = 2 Substitute a = 2 in 1 2 + b = - 4 b = - 6 nth term = 2n2 - 6n + 5 White Squares; 0 , 4 , 8 , 12 , 16 , 20 , 24 4 , 4 , 4 , 4

  2. Maths Coursework T-Totals

    generalizations bar; When a T-Shape is rotated by 90, 180 or 270 degrees, its T-Total is larger. If we try the same rotations on a different grid width and plot the results in a table, patterns might become easier to see.

  1. Black and white squares

    and through this formula I will be able to find out how many squares would be needed to make any cross shape made in the patterns above. Formula n(n-1) x 4 + 1 2 Proving the formula n = 1 1x(1-1)= 0 0/2= 0 0x4= 0 0+1= 1 n =

  2. Border sequences

    = 1 3rd term ~ 2x3� + (2x3) + ? = 25 18 + 6 + ? = 25 ? = 1 This number is a constant therefore the number 1 is the last part in the formula. ? The formula for the nth term is: 2n� + 2n + 1 When I thought about the total number

  1. T-Total. I will take steps to find formulae for changing the position of the ...

    5 6 5 6 7 11 12 13 14 15 20 21 22 23 24 When the T is moved across one square, 5 is added to the T total because there are 5 squares in the T shape meaning 1 has been added to each number in the T adding up to a total of 5.

  2. Investigate how to calculate the total number of Winning Line

    However, the number of diagonal winning lines is not the same. It is proportional to the difference between the grid side length and the winning line length From using my results table and testing formulae I calculated a formula to discover the number of horizontal or vertical winning lines using only the grid side length and winning line length.

  1. ICT Coursework: Data Management Systems

    The price the retailer pays for the cards is important, as the total expenditure needs to also be calculated. These two sources are then used together to calculate the profit/loss. The code number of the cards is important, as when it is used in the spreadsheet, it makes looking up the prices of individual cards much quicker and easier.

  2. For this task we were required to create a model that can be used ...

    When the day and night costs have been worked out, a total needs to be worked out by adding the two values together. This gives the total cost for all the units - day and night. Standing charge, if there is one, is added on to the total.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work