# Rotation Of "T"- Shapes Clockwise

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Introduction

## Rotation Of “T”-Shapes Clockwise

I shaded “T”-shapes on all the girds from 5 by 5 to 10 by 10, so that I could make a sequence and then find the formulae for all the grids. I rotated the “T”-shapes 90° clockwise until I had made a full rotation, (refer to diagram 1A) but did not change the “T”-number as I rotated the “T”-shapes.

I then drew tables to show the “T”-numbers and the “T”-totals for the grids.

Once I had drawn the tables and calculated the “T”-totals for all the grids I realised that there is a difference of 7 between the differences of “T”-totals as the grid size increases by one. E.g. In the 5 by 5 grid there is a difference of 42 between the “T”-totals of the “T”-totals of the first rotation and in the 6 by 6 grid there is a difference of 49 between the “T”-totals of the first rotation. 49-42 = 7 (refer to diagram 1B.) After I had

Middle

= 33 + ( 42 + 7 )

= 33 + 49

= 82

“T”-total = pt + ( 7g + 7 )

= 36 + ( 7g + 7 )

= 36 + ( 49 + 7 )

= 36 + 56

=92.

This formula worked for both grids so I concluded that this was the correct formula for the first rotation.

Now I needed to find the formula for the second rotation. From the information I had from the tables I had drawn I realised that the difference between the “T”-totals of the second rotation was less than the difference between the “T”-totals of the first rotation. So I subtracted the difference between the “T”-totals of the second rotation in the 6 by 6 grid (35)from the difference between the “T”-totals of the first rotation of the 6 by 6 grid (49) (refer to diagram 1B). 49 – 35 = 14. After I found that there was a difference of 14 I knew that there was there was a difference of 7 between 7g which for the 7 by 7 grid is 49 and 35.This was because I had added 7 to 42 to find the formula for the first rotation. So if you imagine 35, 42 and 49 on a number line. Then there is a difference of 14 between 35 and 49 and 42 being the mid-point between the two numbers.

Conclusion

=134 – ( 49 + 7 )

= 134 –56

=78 this is correct so I tried to use the same formula on the 8 by 8.

“T”-total = pt – ( 7g + 7 )

= 151 – ( 56 + 7 )

= 151 – 63

= 88 this was correct so I concluded that this was the correct formula for the third rotation.

When I was looking for the formula for the fourth rotation I realised that just like the first and third rotation. The difference between the “T”-totals of the second and fourth rotations where the same except that instead of adding 42 you subtract 42 in the 7 by 7 grid ( refer to diagram 3). So as I had done to find the formula for the third rotation I subtracted ( 7g – 7) from the previous “T”-total to get the next “T”-total in the sequence. So I tried the formula

“T”-total = pt - ( 7g – 7 )

= 78 – ( 49 – 7 )

= 78 – 42

=36 this was correct so I tried this formula on the 8 by 8 grid.

“T”-total = pt – ( 7g – 7 )

=88 – ( 56 – 7 )

=88 - 49

= 39

Now I have found the formulae for all the rotations where pt = previous “T”-total in the and g = grid size

“T”-total = pt + ( 7g + 7 ) for the first rotation.

“T”-total = pt + ( 7g – 7 ) for the second rotation.

“T”-total = pt – ( 7g + 7 ) for the third rotation rotation.

“T”- total = pt – ( 7g – 7 ) for the fourth rotation.

( refer to diagram 4 )

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

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