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  • Level: GCSE
  • Subject: Maths
  • Word count: 2332

Maths Coursework on Diagonal Differences

Extracts from this document...

Introduction

Naila Parveen         Maths Coursework

Maths Coursework on Diagonal Differences

Introduction:

I am given a 10 by 10 grid. I am going to find the diagonal difference of the different size boxes (e.g. 3 by 3, 4 by 4) within the 10 by 10 grid, by multiplying the opposite corners which result in the two answers, we then deduct these two to get the final answer for that size.

This is the grid that I will use to help me investigate the diagonal difference.

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Aim:

I am going to investigate the diagonal difference of a 2 by 2 box inside a 10 by 10 grid. I will then try to find a formula which should relate to the diagonal difference of each square, I will then further this investigation by trying to find the diagonal difference of a 5 by 5 grid and an 11 by 11 grid, and find the formula to see if they are the same for a 10 by 10 grid. I will also do an extension by doing a rectangle as well as a square and then find the diagonal difference and the formula for this.

I am going to find the formula by finding the diagonal difference of all the sizes within the 10 by 10 grid, and then try to find any pattern, which would help me in finding the formula by drawing a grid. This method would be useful because it will show me

...read more.

Middle

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51 x 95 = 4845

55 x 91 = 5005

Diagonal difference: 160

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15 x 59 = 885

19 x 55 = 1045

Diagonal difference: 160

The diagonal difference for any 5 by 5 box is 160. I predict that to get the 6 by 6 I will have to square 7 and then minus one and then times this by 10 (as shown below)

52 x 10

For a 6 by 6 box it should be

(6-1)2 x 10

52 x 10

25 x 10

250

The general formula for this is (n-1)2 x 10

So the grid for the general formula would look like the box shown below

image00.png

image01.png

Now I am going to try out the rectangular boxes for this I am going to use algebraic formula for each box.

I am going to start of by a 2 by 3 box.

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35 x 47 = 1645

45 x 37 = 1665

Diagonal difference: 20

Now in algebra:

X

X+1

X+2

X+10

X+11

X+12

    (x+2) (x+10) – X (x+12)

= x2 + 2x + 10x + 20 -(x2 + 12)

= x2 + 12x + 20 – x2 -12x

= 20

Now I am going to try another 2 by 3 box but this time I am going to try out my algebraic box first and then the number box.

X

X+1

X+2

X+10

X+11

X+12

    (x+2) (x+10) – X (x+12)

= x2 + 2x + 10x + 20 -(x2 + 12)

= x2 + 12x + 20 – x2 -12x

= 20

Now I am going to do a number box for this formula

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51

52

53

The difference for this should be 20; this is because it should be the same as the other 2 by 3 boxes that I have done above.

41 x 53 = 2173

43 x 51 = 2193

Diagonal difference: 20

Now I am going to try a 2 by 4 rectangular box. I predict that the difference would be 30.

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55 x 68 = 3740

58 x 65 = 3770

Diagonal difference: 30

I was correct but to make sure I will do at least 2 more 2 by 4 boxes

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14 x 27 = 378

24 x 17 = 408

Diagonal difference: 30

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82 x 95 = 7790

85 x 92 = 7820

Diagonal difference: 30

Now I am going to do an algebraic formula for a 2 by 4 rectangular box

X

X+1

X+2

X+3

X+10

X+11

X+ 12

X+13

    (x+3) (x+10) – X (x+13)

= x2 + 3x + 10x + 30 -(x2 + 13)

= x2 + 13x + 40 – x2 -13x

= 40

The answer matches my boxes above. This shows that my algebra is accurate!

Now I am going to try out a 2 by 5 rectangular box.

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55 x 69 = 3795

59 x 65 = 3835

Diagonal difference: 40

I am going to try one more to verify that my answer is correct, but I think that my results should be accurate but just to make sure I am going still going to try this out on an algebraic box.

X

X+1

X+2

X+3

X+4

X+10

X+11

X+12

X+13

X+14

(x+3) (x+10) – X (x+13)

= x2 + 3x + 10x + 30 -(x2 + 13)

= x2 + 13x + 40 – x2 -13x

= 40

Now I am going to try out another algebraic and number box just to verify my working out and to make sure that everything is accurate.

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32 x 46 = 1472

36 x 42 = 1512

Diagonal difference: 40

X

X+1

X+2

X+3

X+4

X+10

X+11

X+12

X+13

X+14

...read more.

Conclusion

Width

(W)

Length

(L)

Differences

3

4

60

6 x 10

(2x3)x10

3

5

80

8 x 10

(2x4)x10

3

6

100

10 x 10

(2x5)x10

3image03.png

7image04.png

120image05.png

12 x 10image06.png

(2x6)x10

image07.png

I have noticed that this formula works for squares as well as rectangles. I have also noticed that the differences are always one less than the box size, so this made it easier for me to obtain a formula.

The formula for this is (W-1) (L-1) x 10

Now I am going to draw a table using the formula above to predict the boxes with the width of 4.

Sizes of rectangles

Width

(W)

Length

(L)

Differences

4

5

120

1 x 10

4

6

150

2 x 10

4

7

180

3 x 10

Now I am going to try and work out the algebraic formula for working out the diagonal differences for all squares.

So for a 6 by 6 box I predict that the diagonal difference would be 250

To show this I will do a number grid and also in algebra. The general differences formula that I predict is (n-1)2 x 10

Now to see if it works!

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45 x 100 = 4500

50 x 95 = 4750

Diagonal difference: 250

Now to show a 6 by 6 box in algebra.        

X

X+1

X+2

X+3

X+4

X+5

X+10

X+11

X+12

X+13

X+14

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X+52

X+53

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X+55

  (x+50) (x+5)-x(x+55)

= x2+50x+5x+250-(x2+55x)

= x2+55x+250-x2-55ximage02.pngimage02.pngimage02.pngimage02.png

= 250

I have noticed that my formula works for a 10 by 10 grid size. So now I am going to change the grid size to see if it has the same results or different. I am also trying to find out if this formula will work for any grid size!

...read more.

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