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Introduction

Iain Downer

Growing Shapes

Aim

My aim is to investigate and explain the growth of congruent shapes by using different methods. I am going to try and find out the formulae for the following in this order:

• Length
• Perimeter
• Area
• Number of shapes
• Number of Lines (inner, outer)
• Number of shapes added
• Number of vertices

At the end of my investigation I am going to look at my results for both squares and triangles and compare them to see if they have anything in common or are similar in any way.

I also hope to extend my coursework and look at 3D shapes or other shapes such as pentagons, hexagons etc…

I am going to work systematically and look to spot any patterns, and in the end, establish an algebraic formula for all things listed above.

The patterns for squares are as follows

1234

The patterns for triangles are as follows

1234

5

The pattern for squares grows by adding another square to each face every time:

The pattern for triangles grows by adding another triangle to each face every time:

Growing Shapes: Squares

Perimeter

 Pattern no. (n) Perimeter 1 4 2 12 3 20 4 28 5 36

D1

As there are all 8’s in the D1 column, the formula contains 8n

 Pattern no. (n) Perimeter Perimeter – 8n 1 4 -4 2 12 -4 3 20 -4 4 28 -4 5 36 -4

Formula for perimeter of squares – Formula = 8n-4

Check

When n = 5

Perimeter = 8n - 4

= 8 × 5 – 4

= 36 Length and Width

 Pattern no. (n) Length 1 1 2 3 3 5 4 7 5 9

D1

As there are all 2’s in the D1 column, the formula contains 2n

 Pattern no. (n) Length Length – 2n 1 1 -1 2 3 -1 3 5 -1 4 7 -1 5 9 -1

Formula for length of squares – formula = 2n – 1

Check

When n = 3

Length = 2n – 1

= 2 × 3 – 1

= 5 Number of Squares/Area

 Pattern no. (n) No. of Squares 1 1 2 5 3 13 4 25 5 41

D1        D2

As there are all 4’s in the D2 column,the formula contains 2n2.

 Pattern no. (n) No. of Squares No. of Squares – 2n2 1 1 -1 2 5 -3 3 13 -5 4 25 -7 5 41 -9

D1

As there are all -2’s in the D1 column, the formula contains 2n2 – 2n.

I need to substitute in the pattern number to find out what to do next.

I am going to use 3 to start with.

(2×32) – (2×3) = 12

To get from 12 to 13, I need to add 1.

I am now going to substitute in 4 to see if it works as well.

(2×42) – (2× 4) = 24

To get from 24 to 25, I need to add 1 so the formula = 2n2 – 2n + 1 and works for everything.

Check

When n = 2

Area = 2n2 -2n + 1

= 2×22 – 4 +1

= 5 I have to use another method to find the formula so I am going to use the odd numbers method.

I notice that with pattern 4 in columns when n = 4:

2 lots of (1 + 3 + 5) = 2 x 9

1 lot of 7 = 1 x 7

2 x 9 = 2(n-1)2

7 = 2n – 1

Ts = 2(n-1)2 + 2n – 1

Number of Lines

 Pattern no. (n) No. of lines 1 4 2 16 3 36 4 64 5 100

D1        D2
As there are all 8’s in the D
2 column, the formula contains 4n2.

 Pattern no. (n) No. of lines No. of lines - 4n2 1 4 0 2 16 0 3 36 0 4 64 0 5 100 0

The formula for number of lines = 4n2

Check

Number of lines = 4n2

= 4 × 32

= 36 Number of Shapes Added

I started with pattern number 2 because there is no previous shape to add the squares onto

 Pattern no. (n) No. shapes added 2 4 3 8 4 12 5 16 6 20

D1

As there are all 4’s in the D1 column, the formula contains 4n.

 Pattern no. (n) No. shapes added No. shapes added - 4n 2 4 -4 3 16 -4 4 36 -4 5 64 -4 6 100 -4

The formula for number of shapes added = 4n – 4

Check

Number of shapes added = 4n – 4

= 4 × 3 – 4

= 8 Number of outer vertices

 Pattern no. (n) No. of outer vertices 1 4 2 8 3 12 4 16 5 20

D1

As there are all 4’s in the D1 column, the formula contains 4n.

 Pattern no. (n) No. of outer vertices No. of outer vertices - 4n 1 4 0 2 8 0 3 12 0 4 16 0 5 20 0

Formula for number of outer vertices = 4n

Check

Number of outer vertices = 4n

= 4 × 3

= 12 Growing Shapes: Triangles

Perimeter

 Pattern no. (n) Perimeter 1 3 2 6 3 12 4 15 5 21 6 24 7 30

D1        D2        D3

I can tell that this method for working out the perimeter does not work because the numbers in the difference columns alternate between positive and negative.

Instead of trying to find a formula for the all of the triangular patterns, I can try and find 2 separate formulas for the patterns where the added triangles are pointing up and the patterns where the triangles are pointing down.

 Pattern no. Pattern no. pointing up(n) Perimeter 1 1 3 3 2 12 5 3 21 7 4 30 9 5 39

D1

As there are all 9’s in the D1 column, the formula contains 9n.

 Pattern no. pointing up(n) Perimeter Perimeter – 9n 1 3 -6 2 12 -6 3 21 -6 4 30 -6 5 39 -6

Middle

24

8

4

33

10

5

42

D1

As there are all 9’s in the D1 column, the formula contains 9n.

 Pattern no. pointing down(n) Perimeter Perimeter – 9n 1 6 -3 2 15 -3 3 24 -3 4 33 -3 5 42 -3

Formula for the perimeter of a triangular pattern when the added squares are pointing down = 9n-3.

Check

Formula = 9n-3

= 9×3 – 3

= 24 Number of Triangles/Area

The yellow triangles represent the area

 Pattern no. (n) No. of Triangles 1 1 2 4 3 10 4 19 5 31

D1        D2

As there are all 3’s in the D2 column, the formula contains 1 n2

 Pattern no.(n) No. of Triangles No. of Triangles - 1 n2 1 1 -0.5 2 4 -2 3 10 -3.5 4 19 -5 5 31 -6.5

D1

As there are all -1.5’s in the D1 column, the formula contains 1 n2 -1 n.

 Pattern no.(n) No. of Triangles No. of Triangles - 1 n2 - 1 n 1 1 1 2 4 1 3 10 1 4 19 1 5 31 1

Formula for working out the Number of Triangles = 1 n2 -1 n + 1

Check

Number of Triangles = 1 n2 -1 n + 1

= (32×1.5) - (1.5×3) +1

= 24 Number of lines

The black lines represent the number of lines.

 Pattern no. (n) No. of Lines 1 3 2 9 3 21 4 36 5 57

D1        D2        D3

This method doesn’t work because you never have all the same number in the difference column.

This means that I have to work out the formula for the odd and even pattern numbers separately. This also means that I must extend the table using the pattern I have already found.

 Pattern no. (n) No. of Lines 1 3 2 9 3 21 4 36 5 57 6 81 7 111 8 144 9 183 10 225

D1        D2        D3

 Even Pattern no. (n) No. of lines 2 1 9 4 2 36 6 3 81 8 4 144 10 5 225

D1        D2

Formula for working out the number of lines in an even pattern number = 9n2

Check

No. of lines = 9n2

= 32 × 9

= 81 Odd Pattern no. (n) No. of lines 1 1 3 3 2 21 5 3 57 7 4 111 9 5 183

D1        D2

As there are all 18’s in the D2 column, the formula contains 9n2.

 Odd Pattern no. (n) No. of lines No. of lines – 9n2 1 1 3 -6 3 2 21 -15 5 3 57 -24 7 4 111 -33 9 5 183 -42

D1

As there are all 9’s in the D1 column, the formula contains9n2 – 9n.

 Odd Pattern no. (n) No. of lines No. of lines – 9n2- 9n 1 1 3 +3 3 2 21 +3 5 3 57 +3 7 4 111 +3 9 5 183 +3

Formula for working out the number of lines in an even pattern number = 9n2 -9n+3.

Check

No. of lines = 9n2 -9n + 3

= (32 × 9) -9n +3

= 57 Number of Triangles Added

The red squares represent the added triangles.

 Pattern no. (n) Triangles Added 1 0 2 3 3 6 4 9 5 12

D1

As there are all 3’s in the D1 column, the formula contains 3n.

 Pattern no. (n) Triangles Added Triangles Added – 3n 1 0 -3 2 3 -3 3 6 -3 4 9 -3 5 12 -3

Formula for working out the number of triangles added = 3n-3

Check

Triangles added = 3n-3

= 3 × 3 – 3

= 6 Width

For width I am going to measure the widest part of the shape.

 Pattern no. (n) Width 1 1 2 3 3 5 4 7 5 9

D1

As there are all 2’s in the D1 column, the formula contains 2n.

 Pattern no. (n) Width Width – 2n 1 1 -1 2 3 -1 3 5 -1 4 7 -1 5 9 -1

Conclusion

3n2 – 3n + 1

Check

No. of hexagons = 3n2 – 3n + 1

= (3 x 32) – (3 x 3) +1

= 19 Growing Shapes Extension: 3D Shapes

For 3D shapes, I am only going to work out the number of cubes.

The pattern for 3D shapes is as follows.

I drew the shapes using individual cubes which I found on AutoShapes in Word.

Area/Number of Cubes

 Pattern no. (n) No. of Cubes 1 1 2 7 3 25 4 63 5 129

D1        D2        D3

As there are all 8‘s in the D3 column, the formula contains 1 n3

 Pattern no. (n) No. of cubes No. of cubes – 1 n3 1 1 - 2 7 -3 3 25 -11 4 63 -22 5 129 -37 D1                D2

As there are all -4’s in the D2 column, the formula contains 1 n3 – 2n2

 Pattern no. (n) No. of cubes No. of cubes – 1 n3 – 2n2 1 1 1 2 7 4 3 25 7 4 63 9 5 129 12 D1

As there are all -2 in the D1 column, the formula contains 1 n3 – 2n2 + 2 n

 Pattern no. (n) No. of cubes No. of cubes – 1 n3 – 2n2 +2 n 1 1 2 2 7 8 3 25 16 4 63 64 5 129 130

I can see from the table that I need to -1 so the formula = 1 n3 – 2n2 +2 n - 1

Conclusion

Here is a table showing all the formulas I have managed to work out during my investigation.

 Squares – Perimeter 8n-4 Squares – Area 2n2-2n+1 Squares – No. of lines 4n2 Squares – Length 2n-1 Squares – Shapes added 4n-4 Squares – No. of outer vertices 4n Triangles – Perimeter Pointing up 9n-6, Pointing down 9n-3 Triangles – Area 1 n2 -1 n + 1 Triangles – No. of lines 9n2 Triangles – Width 2n-1 Triangles – Shapes added 3n-3 Pentagons – Area 2.5n2-2.5n+1 Pentagons – Shapes added 5n-5 Hexagons – Area 3n2-3n+1 3D shapes – Area 1 n3 -2n2+2 n-1

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