# Maths Coursework- Borders

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Introduction

Joanna Burton 10s

23rd June 2004

## Maths Coursework- Borders

##### QUESTION

## Figure below shows a dark cross-shape that has been surrounded by white squares to create a bigger cross-shape;

## The bigger cross-shape consists of 25 small squares in total.

The next cross-shape is always made by surrounding the previous cross-shape with small squares.

### Part 1- Investigate to see how many squares would be needed to make any cross-shape built in this way.

### Part 2- Extend your investigation to 3 dimensions.

##### Introduction –

##### I am doing an investigation to see how many squares would be needed to make any cross-shape built up in this way. Each cross-shape is made by using the previous cross-shape and adding another layer of white squares, making all the inner squares black. The first cross-shape in the sequence is a single black square.

###### To start my investigation I must draw the first 7 cross-shapes. This will enable me to see a pattern in the shapes so I can make a table and record how many black and white squares there are in each cross-shape I have drawn. From my table I must use the results to work out formulae for black, white and total number of squares.

After this I will test the formulae on a pattern I have already drawn and on one I have not already drawn.

I will be working systematically in my investigation because if I work in a

particular order it will be easier for to see a pattern and links in the sequences.

Middle

Total no. of squares;

2n2 – 2n + 1

2 × 32 – (2 × 3) + 1 = 18 – 6 + 1

= 13

White squares;

4n – 4

(4 × 3) – 4 = 12 – 4

= 8

Black squares;

2n2 – 6n + 5

2 × 32 – (6 × 3) + 5 = 18 – 18 + 5

= 5

This shows that the formulae I have used are correct as the equations give the correct results for cross-shape pattern 3.

Now I will test the formulae on pattern 9 - a cross-shape I don’t know the answer to.

Total no. of squares;

2n2 – 2n + 1

2 × 92 – (2 × 9) + 1 = 162 – 18 + 1

= 145

White squares;

4n – 4

(4 × 9) – 4 = 36 – 4

= 32

Black squares;

2n2 – 6n + 5

2 × 92 – (6 × 9) + 5 = 162 – 54 + 5

= 113

I must draw pattern 9 to check that the formulae have worked and I have reached the correct results.

Justification of the 2D cross-shapes and formulae

I must look at the cross-shape in a different way, by tilting the shape and using lines I can clearly see and work out how many white squares there are.

###### Using pattern 3 – to find white squares

###### there are 4 sides with 3 squares on each side so..

White squares; 3 × 4 = 12 – 4 ( - 4 as corners count twice )

4 × 3 (n) = 12

4 × 3 (n) – 4 = 8

same as 4n – 4 and 4 × ( n –1 )

##### Using pattern 4 – to find white squares

Using the rule 4n – 4 so…= 4 × 4 – 4 ( - 4 as corners are counted twice )

= 16 – 4 = 12

also using the rule 4 × ( n – 1 ) so… = 4 × ( 4 – 1 )

= 4 × 3

= 12

This shows how and why the nth term for the justification works.

##### Using pattern 3 – to find total number of squares

We need to use a quadratic rule because when the sequence is written out and the difference is worked out, the second differences are equal, which gives you n2 used in a quadratic equation.

White = 32

Black = 22 total no. of squares = 32 + 22 = 13

So… for size n, total = n2 + ( n - 1 ) 2

Conclusion

Using 3D cross-shape pattern 2

nth term = 4 n3 - 2n2 + 8 n - 1

3 3

= ( 4 × 23 ) - ( 2 × 22 ) + ( 8 × 2 ) - 1

- 3

= 7

The formula tells us that 3D cross-shape pattern 2 would have 7 squares, this is the same answer as in the table, showing us that the formula being used is correct.

Next I will try…

### Using 3D cross-shape pattern 6

nth term = 4 n3 - 2n2 + 8 n - 1

3 3

= ( 4 × 63 ) - ( 2 × 62 ) + ( 8 × 6 ) - 1

- 3

= 231

This answer also corresponds correctly with the answer in the table which indicates that the formula is successful in finding any 3D cross-shape pattern built up in this way.

### Justification of the 3D cross-shapes

I will now explain why my formula works and also different ways of coming to the same answer using a different method.

3D pattern 2 3D pattern 3

If you look at the 3D shapes from a side view they’d look like this.

Each layer corresponds with the 2D cross-shape patterns,

for example layer 1 in the 3D cross-shape corresponds with the 2D cross-shape pattern 1 as they both have the same total number of squares, same goes for layer 2 in the 3D cross-shape as this corresponds with 2D cross-shape pattern 2 also having the same number of squares.

This is how you can work out the total number of squares more easily as you can use the answers to the 2D shapes to help you as the 3D shapes are hard to draw.

Another formula possible to find the total number of squares for any 3D shape built up in this way is;

2 ( layer 1 + layer 2 + layer 3…) + layer n

or

- ( 2n2 – 2n +1 ) + 2 ( 2 ∑ r2 – 2 ∑ r + n )

By using the appropriate algebra this equation should lead to the appropriate answer.

.

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

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