• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1433

Investigation in to How many tiles and borders is needed for each pattern

Extracts from this document...

Introduction

GCSE Maths Coursework                                                                Tanbir Hussain 10R

Borders

Introduction: In this I am going to find out how many tiles and borders is needed for each pattern. I am also going to make predictions. I am gong to work out a formula for the tiles and borders needed for the nth pattern. I will also work out a formula for the total tiles for each pattern.  

Diagrams:

1.

4 Borders                                               5.

5 Tiles                                                    

9 Altogether                                                                                                                                                        

 2.

12 Borders

13 Tiles

25 Altogether

24 Borders

61 Tiles

85 Altogether

3.

                                          6.

16 Borders

25 Tiles

41 Altogether

4.

20 Borders

41 Tiles

61 Altogether

                                        28 Borders

                                        85 Tiles

                                        113 Altogether

7.

...read more.

Middle

Test Prediction:

I have tested the prediction by showing anther diagram of the 8th pattern. This pattern has 145 tiles, 36 borders and a total of 191 tiles together.

36 Borders

145 Ties

191 Altogether

Patterns:

I have noticed a pattern where the total tile of the 1st pattern is the number of tiles for the 2nd pattern. This applies for each of the patterns.

Pattern No

No Of Tiles

No Of Borders

Total Tiles

1

5

8

13

2

13

12

25

I also have noticed that if you add pattern No.1s border with the 1st difference you can get the second patterns number of borders.

For example:

n   Tiles  Borders   1st Difference

1     5       8 +4    8+4=12

2    13     12                      

3    25     16              +4    16+4=20

4    41     20

Formula:

I found out the formulas was a quadratic because it has a 2nd difference. For example in No of tiles:

5    13    25    41

    8    12    16

       4      4

I worked out the formula for the tiles, borders and the total tiles altogether. I used the general equation that is an +bn+c to work out the tiles and borders for the nth

  1. From the table I have found a formula to calculate the border.

For the nth term the formula is 4n+4

  1. Anther formula I found was to work out the number of tiles for the nth term.

For the nth term the formula is 2n+2n+1

  1. Anther formula I found was to work out the total tiles for each pattern

For the nth term the formula is 2n +6n+5

Test The Formula:

1. Formula to calculate the number of borders is 4n+4

For example: when n=2

(4 2)+4

8+4=12

So then the 2nd pattern will have 12 borders.  

2. Formula to calculate the number of tiles is 2n+2n+1  

For example: When n=3

(3 ) 2=18

2  3=6

+1

=18+6+1

=25

So then for the 3rd pattern you will get 25 tiles.

3. Formula to calculate the total tiles for each pattern is 2n +6n+5

For example: When n=4

(4 ) 2=32

6  4=24

+5

=32+24+5

=61

Conclusion:

I have finished this part of my investigation and here is my result:

Here is the list of patterns I found:

n   Tiles  Borders   1st Difference

1     5       8 +4             8+4=12

2    13     12                      

3    25     16              +4           16+4=20

4    41     20

Looking at my patterns I found the following formulas:

1. To work out the number of tiles for the nth term the formula is

2n +2n+1

2. To work out the number of borders of the nth term the formula is

4n+4.

3. To work out the total borders for the nth term the formula is

2n +6n+5

In this in part of my investigation I have learnt that how many number of tiles and the number of borders is needed for any pattern that I have investigated.

I found this investigation quite easy but I had a little bit of trouble on finding the formula but eventually I solved by using the formula

an +bn+c.

a= ½ the 2second difference          b= the total of                        c is equal to when                                                              

                                                        a+c=1st difference                  n=0          

Extension On Borders

Introduction:

In this part of my investigation I am going to extend the patterns. I am going to repeat the procedure with the extended patterns.

Diagrams:

1.                                                                 5.

10 Borders                         

8 Tiles

18 Altogether

2.

14 Borders

18 Tiles

32 Altogether

26 Borders

3.                                                                        72 Tiles

...read more.

Conclusion

+4n+2

For example: When n=5

(5 ) 2=50

4  5=20

+2

=50+20+2

=72

So then for the 5th pattern you will get 72 tiles.

3. Formula to calculate the total tiles for each pattern is 2n +8n+8

For example: When n=6

(6 ) 2=72

8  6=48

+8

=72+48+8

=128

Conclusion:

I have finished this part of my investigation and here is my result:

Here is the list of patterns I found:

n   Tiles  Borders   1st Difference

1     8       10 +4    10+4=14

2    18      14                      

3    32      18              +4    18+4=22

4    50      22

Looking at my patterns I found the following formulas:

1. To work out the number of borders of the nth term the formula is

4n+6.

2. To work out the number of tiles for the nth term the formula is

2n +4n+12

3. To work out the total borders for the nth term the formula is

2n +8n+8

In this in part of my investigation I have learnt that how many number of tiles and the number of borders is needed for any pattern that I have investigated.

I found this investigation quite easy but I had a little bit of trouble on finding the formula but eventually I solved by using the formula

an +bn+c.

a= ½ the 2second difference          b= the total of                        c is equal to when                                                              

                                                        a+c=1st difference                 n=0          

Page  of

...read more.

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

Related GCSE T-Total essays

  1. T-Total Maths

    T=5N+7 I tested that when: T-number=62 T-total=277 N = 9 10 11 12 T = 52 57 63 67 T-number T = (5x9) +7 = 45+7 T = 52 T-total This equation has produced its first correct answer. I will carry on and test T-shape I know the T-total for N = 10 T = (5 x 10)

  2. Math Coursework-T-Total Investigation

    (Example is shown on the grid) If the T number is 19, there will be no grid left on the left hand side of the T-shape and so that part will be stuck out. The same happen to all the numbers in the first row of the grid.

  1. T-Totals Investigation.

    With this I can see that 40�8= 5. From looking at the previous vertical translations on the 9 by 9 grid and the ones done here on the 8 by 8 grid, I can see that the number that is added to get to the next total, when divided by the width of the grid (the grid number)

  2. T-total Investigation

    that the pattern between the T-no was 1 and the pattern between the T-total is 5 and so therefore 5t is the first part of my formula. I then multiplied the T-no by 5 and then took away the T-total from the answer.

  1. Maths Coursework- Borders

    This would give the next cross-shape a total of 113 squares Working out Formulae Next I must work out the rule for finding the number of black, white and total number of squares to make any cross-shape built up in this way.

  2. T-Total Investigation

    Now that I have done the upside down rule, I will do it with the T on its side. I will do it on a 9x9 grid first. I have got the T-Number as12 and it is facing right. I will do the same as before to find out what I need to minus to get the T-Total.

  1. Borders - Investigation into how many squares in total, grey and white inclusive, would ...

    d = 63 U4 = 64a + 16b + 4c + d = 129 U2 - U1 = 7a + 3b + c = 18 = U5 U5 x2 = 14a + 6b + 2c = 36 = U8 U3 - U1 = 26a + 8b + 2c = 56

  2. Connect 4 - Maths Investigation.

    For Height4: [H(4L-9)] 4(4L-9) = 16L - 36 ~ to get the rule for height 4 you need to take away 9L and add 18. The Final rule for Connect 4 is H(4L-9)-9L+18 Test Final Rule To make sure that the final rule is correct I will have to test it on a grid that I haven't tried.

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