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  • 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.

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