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  • Level: GCSE
  • Subject: Maths
  • Word count: 7605

T-Shapes Coursework

Extracts from this document...

Introduction

Maths Investigation 2

Higher Tier Task - “T Shapes

I was given a number grid, like Fig 1.1. On it, a “T Shape” was to be placed in any possible position on the grid, such that all of the shape remained within the boundaries. The task was to investigate the patterns connecting the numbers in the “T” themselves, and also patterns in combinations of these numbers.

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Fig 1.1

Any number of investigations was possible, but I chose an investigation in which I would add up all the numbers in the “T”.

Section 1: 3x1 “T” on Width 8 Grid

1) Introduction

In this section, a 3x1 “T” will be used. This means that there are 3 squares along the top of the “T” (the “wing”, in yellow), and 1 on the bottom (the “tail”, in green). This is as shown originally in the task brief.

2) Method

To discover any patterns, I will calculate the sum of the numbers in the “T” for 5 sequential locations of the box within the grid. As it would be impractical to do all possible calculations, 5 should be enough to display any patterns that may lie therein. The “Middle Number” will be the number in the wing that is adjacent to the tail. It is highlighted in red on Fig 1.1.

3) Data Collection

Here are the results of the 5 calculations for 3x1 “T” on Width 8 Grid:

Middle Number

Sum of Wing

Sum of Tail

Total Sum

(Wing + Tail)

2

6

10

16

3

9

11

20

4

12

12

24

5

15

13

28

6

18

14

32

4) Data Analysis

From the table, it is possible to see a couple of useful patterns:

  1. The Sum of the Wing is always 3 times the Middle Number;
  2. The Sum of the Tail is always 8 more than the Middle Number;

5) Generalisation

It can be assumed that for all possible locations of the 3x1 “T” on the width 8 grid, these patterns will be true. Therefore, the following logic can be used to create a formula where:

n

...read more.

Middle

n will continue to be used for the Middle Number in the “T” i.e. the location of the “T” upon the grid.

2) Method

Varying values of w will be tested to give different widths of wings for the “T”. The widths will be 5, 7 and 9. Only odd numbers can be used for the wing width, because the “T” always has an equal number of boxes either side of the Middle Number. With these "T”s, in 5 different locations on the width 20 grid, the Total Sum will be calculated.

Width 20 must be used for this section because when varying the widths of “T”, the grid must be wide enough to give a good range of values for data collection, and later testing.

3) Data Collection

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Fig 3.1

Fig 3.1 is the width 20 grid, with 3x1, 5x1, 7x1 and 9x1 example “T”s.

a)Here are the results of the 5 calculations for a 5x1 “T” on Width 20 Grid:

Middle Number

Sum of Wing

Sum of Tail

Total Sum

(Wing + Tail)

  30

150

  50

200

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250

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320

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350

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440

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450

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560

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680


b)Here are the results of the 5 calculations for a 7x1 “T” on Width 20 Grid:

Middle Number

Sum of Wing

Sum of Tail

Total Sum

(Wing + Tail)

  30

210

  50

260

  50

350

  70

420

  70

490

  90

580

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630

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740

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770

130

900

c) Here are the results of the 5 calculations for a 9x1 “T” on Width 20 Grid:

Middle Number

Sum of Wing

Sum of Tail

Total Sum

(Wing + Tail)

  30

270

  50

  320

  50

450

  70

  520

  70

630

  90

  720

  90

810

110

  920

110

990

130

1120

4) Data Analysis

From the tables (a)-(c), it is possible to see that when only the wing width is varied, only the Sum of the Wing changes. In fact, if we take one constant Middle Number, 25, from each of the above tables, we get the following:

Middle Number

Wing Width

(w)

Sum of Wing

Sum of Tail

Total Sum

(Wing + Tail)

25

5

125

35

160

25

7

175

35

210

25

9

225

35

260

From these tables, it is possible to see a useful pattern:

  1. The Sum of the Wing equals the Middle Number multiplied by the Wing Width.
...read more.

Conclusion

 + n + gl + g}                                        There are “l” number of “{n + n + 10l + 10}”simage07.png

lx {2n + g(l + 1)}                                          Factorisedimage05.png

½ xlx {2n + g(l + 1)}

Now, using the already justified “wn”, the Total Sum formula can be created:

Total Sum=

=

=

[Sum of Wing] + [Sum of Tail]

[wn] + [½ xlx {2n + g(l + 1)}]

wn + ½ l {2n + g(l + 1)}

8) Conclusion

After this justification, it can now be said that for every possible wxl “T” on a Width g Grid, the Total Sum of all of the squares contained within it is wn + ½ l {2n + g(l + 1)}.

Investigation Conclusion and Evaluation

From the simple study of a 3x1 “T” on a width 8 grid, I have been able to progress all the way to the formula to find the sum of the “T” on any width grid. Patterns were easy to spot early on, however when advancing into more complex areas of the project, like the last section where three different variables were altered, spotting patterns became more difficult and knowledge from the prior sections was required to find the formula.

Overall, the best way of presenting the results was in tables because there was an obvious pattern of values in the “Difference” column, because they were all in one line. A graph or diagram would not have been as suitable because the patterns would not have been as apparent, except in a linear progression but this is easily seen in a pattern as the numbers incrementing by a constant amount each time.

There are limitations on the formula though. The width of the wing must be never create a situation where any boxes are outside the grid. This relies on the middle number being chosen carefully to ensure, even if the wing width appears it will fit on the grid (like a Wing Width of 8 on a Width 50 Grid), that it is not so close to the edge that the “T” is not wholly on the grid.

The final justified formula is that the Total Sum  =  

wn + ½ l {2n + g(l + 1)}

...read more.

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