• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 2506

Maths Coursework - Grid Size

Extracts from this document...

Introduction

Maths Coursework – Grid Size


Maths Coursework

Aim: This piece of coursework will allow me to work out and understand a formula, which will be used to apply to any polygon put onto any size grid and find out what the product of the numbers in the corners of the polygon. This coursework will involve a lot of algebraic formulas and working-outs. It will use a letter representing the top left number of each square and adding numbers within the polygon to find my final value.

Method: I will calculate the DPD (diagonal product difference) of a square on a 10x10 grid. Then I will investigate further and change different variables as stated below. I will create formulas, which will allow me to find the DPD for any square size and any grid size. Eventually at the end of my coursework I will have a general formula that I can use for any grid size and any square size/position.

Firstly I will calculate the DPD for a square on the 10x10 square, then I will change the position so I can find a pattern between the two. I will then use algebra instead of numbers. ‘N’ will represent the top left number in the square/rectangle. I will calculate the DPD of that formula. If the answer to both the numerical and the algebraic formulas are the same then I will prove my theory by changing the position of the square/rectangle.

...read more.

Middle

+N+20N+20 – N2 – 21N

= N2-N2+N+20N– 21N+20

= 20

Replaced by numbers:

(61+20)(61+1) – 61(61+21)

= 81x62 – 61x82

= 5022 – 5002

 = 20

Second test with another position on the grid:

(63+20)(63+1) – 63(63 +21)

= 83x64 – 63x84

= 5312 – 5292

= 20

63

64

73

74

83

84

N

N+1

N+G

N+G+1

N+2G

N+2G+1

 On a 10x10 grid, G equals 10.

Including ‘g’:

(N+2G)(N+1) – N(N+2G+1)

 = (N2+N+2GN+2G) – (N2+2GN+N)

= N2-N2+2GN-2GN+2G+N– N

= 2G

I will now put G into the equation for a 10x10 grid algebraic formula:

G means grid size. For now it equals to 10.

N

N+1

N+G

N+G+1

DPD

= (N+G)(N+1) – N(N+G+1)

= (N2+N+GN+G) – (N2+GN+N)

= G

Through using my results and algebraic formulas I have tested my theory and it is correct. For an 8x8 grid the letter ‘G’ would equal to 8, no matter where the position of the square was.

I will now add G to a 2x3 rectangle.

N

N+1

N+2

N+G

N+G+1

N+G+2


I will now add H to my formula. This will involve me looking at how the height affects the DPD. First I will test it on a 10x10 2x3 rectangle. H = Height.

 (N+G(H-1))(N+1) – N(N+G(H-1)+1)        

= (N² + GHN – GN + N + GH –G)– (N² + GHN – GN + N)        

= GH-G

N

N+1

N+G(H-1)

N+G(H-1)+1

N+G(H-1)

N+G(H-1)+1

Through using my algebra I have worked out that the DPD will be whatever the grid size is.

...read more.

Conclusion

The width affects the DPD because changing the width of the grid means that the numbers being cross-multiplied will have to change. Using the following example I will explain what limitations apply to the width of the grid.

1        2        3        4        5        6        7        8        9        10


Base = 10

To work out the limitations on width you will need to find out which is the last column is will work in. The width of the rectangle in the diagram above is 4. The base of the grid is 10.

10 – 4 = 6

Putting the square in column 6 will mean that there will still be a column with nothing in it in front of the rectangle on the grid. So I add one. This gives me the last column that the rectangle will work in.

10 – 4 + 1 =7

The seventh column is the last column that the rectangle of width 4 can be placed in order to work out the DPD.

B – W + 1

The following diagram will explain the limitations, which apply to height (length).

1

2

3

4

5  

     Length = 10

6

7

8

9

10

The limitations on length of the grid are shown in a formula similar to that of the width. The height of the rectangle in the above diagram is 2. The length of the grid is 10. So to find out the last possible row that the rectangle can be placed in, we do the calculation:

10 – 2 = 8

If we place the rectangle in row 8 than there is still a free row at the bottom, row 10. So we add one.

10 – 2 + 1 = 9

Putting the rectangle in row 9 is the furthest row down that the rectangle can go.

With ‘L’ being length and ‘H’ being height:

L – H + 1

These are the limitations for the grids.

...read more.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences 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 Number Stairs, Grids and Sequences essays

  1. Marked by a teacher

    Opposite Corners. In this coursework, to find a formula from a set of numbers ...

    4 star(s)

    29999 - 29264 = difference Therefore difference = 735 Solution Check: y (n-1) � = difference 15 (8 - 1)� = difference 15 � 7� = difference Therefore difference = 735 1. Solution: 3240 = 10 (n - 1) � 3240�10 = (n - 1)

  2. Marked by a teacher

    I am going to investigate by taking a square shape of numbers from a ...

    4 star(s)

    number Left corner x right corner Right corner x left corner Products difference 5 3x21=63 5x19=95 32 6 11x29=319 13x27=351 32 7 4x22=88 6x20=120 32 I have noticed that the products difference of 3x3 squares in an 8x8 grid equal to 32.

  1. Marked by a teacher

    Number Grid Aim: The aim of this investigation is to formulate an algebraic equation ...

    3 star(s)

    = n2 + 11n + n + 11 ? [n2 + 11n + n +11] - [n2 + 11n + n] = 11 I will use no. 5 (above) and apply this to the formula 92(92 + 11 + 1)

  2. GCSE Maths Sequences Coursework

    In the next shape in the sequence there are 24 sides plus 4 from the 4 squares at the end that have 3 sides. I have discovered that if you look at the sequence number you find that it is 8 times the number you get if you add the sides not including the four end ones.

  1. Staircase Coursework

    + (n + 2) + (n + 10) + (n + 11) + (n + 20) I will now simplify this result, by adding up all the n `s and 1�s which leaves me with the following result: St = 6n + 44 I will now test the formula to see if it works, by trying it out at another example.

  2. Maths coursework. For my extension piece I decided to investigate stairs that ascend along ...

    52 53 54 55 56 41 42 43 44 45 46 47 48 33 34 35 36 37 38 39 40 25 26 27 28 29 30 31 32 17 18 19 20 21 22 23 24 9 10 11 12 13 14 15 16 1 2 3 4 5

  1. Mathematics Layers Coursework

    So on the next layer there is going to be a space missing with each combination for five combinations, so it will just be six multiplied by five because the first layer has six combinations and the second layer has five combinations.

  2. Number Grid Coursework

    brackets on the left of the minus sign, and not from the more simply factorised "a(a + 11)" term. The a2 term is present on both sides of the minus sign, as is the 11a term therefore, they cancel each other out to leave the number 10.

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