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• Level: GCSE
• Subject: Maths
• Word count: 1673

# Investigating T-shapes.

Extracts from this document...

Introduction

Sheena Robinson        Maths Coursework        30/04/2007

Investigative Maths Coursework

We looked at a T-shape drawn on a nine width grid like this:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81

The total of the numbers inside the T-shape is called the T-total.

 1 2 3 11 20

(1+2+3+11+20=37)

The number at the end of the stem of the T-shape is the T-number. This remains the same even if you rotate the T-shape.

Our first task was to translate the T-shape into different positions on the same sized grid and investigate the relationship between the T-total and T-number.

n = T-number                                        t = T-total

(By ‘difference between’ I mean the amount added or subtracted to get to the next number in the second column)

 n t Difference between 5n 5n - t Therefore5n - 63 20 37 100 63 t 21 42 5 105 63 t 22 47 5 110 63 t 23 52 5 115 63 t 24 57 5 120 63 t

So the equation for finding the T-total anywhere on a nine width grid if you only know the T-number is 5n – 63 = t.

You can prove his by using algebra:

 n-19 n-18 n-17 n-9 n

n + (n - 9) + (n – 18) + (n – 17) + (n – 19)

=         5n – (9 + 18 +17 + 19)

=        5n – 63

This is so because each time you move up a row on a grid you take away one grid width and down you add it.

Middle

10 width:

 n-21 n-20 n-19 n-10 n

n + (n – 10) + (n – 20) + (n – 19) + (n – 21)

=        5n – (10 + 20 + 19 + 21)

=        5n – 70

11 width:

 n-23 n-22 n-21 n-11 n

n + (n – 11) + (n – 22) + (n – 21) + (n – 23)

=        5n – (11 + 22 + 21 + 23)

=        5n – 77

12 width:

 n-25 n-24 n-23 n-12 n

n + (n – 12) + (n – 24) + (n – 23) + (n – 25)

=        5n – (12 + 24 + 23 + 25)

=        5n – 84

This can be put into a table to solve the equation for a T-shape for any width grid.

w = width

 w t Difference between Therefore5n – 7w 8 5n – 56 t 9 5n – 63 7 t 10 5n –70 7 t 11 5n – 77 7 t 12 5n -84 7 t

Again we can prove this using algebra:

 n-2w-1 n-2w n-2w+1 n - w n

n + (n – w) + (n - 2w) + (n – 2w + 1) + (n – 2w – 1)

=        5n – (7w + 1 – 1)

=         5n – 7w

Now I have the equation for finding the T-total of this T shape on any grid I can move on to the next and final task. This was to investigate the relationship between the T-total when using a transformation or combination of transformations.

There would be no point in studying both rotation AND reflection because a diagonal reflection is the same as a 90º rotation and horizontal the same as 180º.

For this same reason there is no point in studying reflection AND translation as vertical reflection is a translation right or left.

Conclusion

In full the equation for finding the T-total for a rotation of 90º and THEN a translation of any vector is:

5(n + x – yw) + 7 = t

To prove this I will take a 90º rotation ad then a random translation.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81

The translation has a vector of         (4)

(-2)

This means that x = 4, y = -2 and n = 10. The grid width (w) is 9.

5(n + x – yw) + 7

=         5(10 + 4 – (-2 x 9) + 7

=        5(10 + 4 + 18) + 7

=        (5 x 32) + 7

=        160 + 7

=        167

This can be proved by doing the long calculation for the T-total of the new rotation and translation of the shape.

32 + 33 + 34 + 25 + 43 = 167

In conclusion you have to know what to add on to the 5n or 5(n + x – yw) to get the T-total. This is different for every rotation and there is no way of knowing. You can remember it quite easily though. 7w for 0º and 180º. 7 for 90º and 270º. Whether to add or subtract his is simple to deduce. If you times the T-number by 5 and all the other numbers are less than the T-number you will get an answer too high s you will have to minus something. And visa versa.

Whilst doing this investigation  I tried to work systematically and clearly to prove my points thoroughly. I am satisfied with my answers and would like to use the extra knowledge I have gained in years to come.

This student written piece of work is one of many that can be found in our GCSE T-Total section.

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