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

# Maths Coursework - Grid Size

Extracts from this document...

Introduction

## 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.

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.

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.

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

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