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  • Level: GCSE
  • Subject: Maths
  • Word count: 4680

"Multiply the figures in opposite corners of the square and find the difference between the two products. Try this for more 2 by 2 squares what do you notice?" Investigate!

Extracts from this document...

Introduction

Gurprit Singh Khela

Opposite Corners-coursework

“Multiply the figures in opposite corners of the square and find the difference between the two products.

Try this for more 2 by 2 squares what do you notice?”

Investigate!

In this investigation I will research various squares and rectangles within selected grids such as 10 by 10, 11 by 11 and so forth. I will find patterns between the differences and the squares and rectangles within the grids. By the end of this investigation my aim is to achieve a formula that will connect and link the shape within the grid whether it is a square, rectangle etc. to the size of the grid itself. If I have time I may possibly go into 3 Dimensions and investigate the addition of the next dimension and see how this will affect the 2D work so far and how it can be linked.

To start off with I will take a 2 by 2 square out of the top left hand corner of my grid which for now is 10 by 10 size.

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2 x 11 =22

1 x 12 =12

I have multiplied the opposite corners and now will subtract the smaller number from the larger one as explained in the question above.

22 – 12 = 10

By subtracting the 2 totals I have found the difference which is 10.I wonder if this difference of 10 will remain constant if I change the position of the 2 by 2 square on the grid. To test this I will move the square along a place and see what the difference is.

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3

12

13

3 x 12 = 36

2 x 13 = 26

This is interesting because the difference seems to stay constant. This is possibly because I have only moved it across the grid. I will now randomly take a square from the bottom right hand corner of the grid and see if the difference stays at 10.

...read more.

Middle

2 + 44x + 160

(x2 + 44x + 160) – (x2 + 44x) = 160

My theory is correct and the difference has been proven to be 160.I have now found a pattern and have enough results to come up with a formula that will give me the results for any size square on a 10 by 10 size grid. I should be able to use this formula easily and at ease by cutting out all the working as I have done above for each square. Here is my table of results:

Square Size

Difference

First Difference

Second Difference

2 x2

10

3 x 3

40

30

20

4 x 4

90

50

20

5 x 5

160

70

20

6 x 6

250

90

20

7 x 7

360

110

20

8 x 8

490

130

20

9 x 9

640

150

20

10 x 10

810

170

20

I have now predicted the rest of the results as I did with the 5 by 5 square. I can do this because I know the second difference will remain the same i.e. 20.

image12.png

A graph can be shown to represent the table and how the different totals change. Looking at the difference it is increasing rapidly which is true as it is growing with an ever increasing first difference. As you can see the first difference is at a slant. It is increasing but what’s different is that it has a stable gradient that does not change therefore it is linear. Looking at the main difference it has a curved shape suggesting the equation of the line may be quadratic or squared equation as it resembles y = x2 .

This is where we can make formulae from the results of the table. The formula should connect the size of the square (n) which will be the length of one side of the square to the difference and save the hassle of the working out. Looking at the

...read more.

Conclusion

2 x y rectangle 10(y – 1)

3 x y rectangle 20(y – 1)

Notice that there is no squaring of the bracket this is because there is only one term being used whereas with the squares there was one term representing both the width and the length. Again I notice when I take away 1 from the width and multiply by 10 the answer comes to the coefficient of (y – 1) this now proves useful as I have used the width to create another variable inside the equation. I will label the width of the rectangle as z. Now when I add this information I have into the formula I will have the following:

(y – 1){10(z – 1)}

y = the length of shape

z = width of the shape

Now I need to test this formula with a result which I have not done. I will find the difference for a rectangle with the width 4 and length 2.Technically speaking this should have the same answer as 2 x 4.Using the formula I have got the answer:

(2 – 1){10(4 – 1)}

Difference is 30

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32

2 x 31 = 62

1 x 32 = 32

Difference is 30

There is no need to re do the rectangle with algebra because the difference matches the one from the 2 x 4 rectangle. Therefore I have achieved the formula that I set out to do and conveniently enough this formula will also work for any size square as well.

(2 – 1){10(4 – 1)}image03.png

Also by looking at the formula to make it work with any size grid I can change the grid number which is here. From my previous work this will change to 11 when I investigate an 11 by 11 grid this is linked to the second difference which was double the grid size as I changed them. For a 10 by 10 grid the second difference was 20 and for an 11 by 11 it changed to 22.Both of the second differences had to be divided by 2 which gave the coefficient of n. There for I believe that this can be changed to g which will represent the grid size. In conclusion I have amassed the formula:

(y – 1){g(z – 1)}

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...read more.

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