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Introduction

GCSE Maths Coursework – Growing Shapes

In this piece of coursework, I will investigate how shapes grow. Firstly, I will look at how an arrangement of shaded and unshaded squares grows; I have been given 3 diagrams. Each diagram shows how the shape grows. They do this by adding a new layer of squares to each exposed surface of the previous shape. When I finish investigating squares I may look at other shapes. I am now going to draw 6 different stages of this growth to establish if any patterns arise. Finding the Nth Term

I am now going to put the information I found from each of the shapes into tabular form so that I can understand it better and perhaps spot a pattern emerging. If I find a pattern I will use it to determine an Nth term, meaning a term that will be applicable to all shapes of this nature.

Perimeter

 Stage (N) 1 2 3 4 5 6 Sequence 12 20 28 36 44 52 1st difference +8 +8 +8 +8 +8 2nd difference

I can see a clear pattern here, the sequence for perimeter is going up regularly in 8’s, and therefore this sequence is a linear sequence and has an Nth term.

Nth term = 8N+?

From my table I can see that I need to add on 4 each time to get the number in the sequence so; Nth term for Perimeter = 8N+4

 Stage (N) 1 2 3 4 5 6 Sequence 4 8 12 16 20 24 1st difference +4 +4 +4 +4 +4 2nd difference

I can see a clear pattern here, the sequence for shaded is going up regularly in 4’s, and therefore this sequence is a linear sequence and has an Nth term.

Nth term = 4N+?

From my table I can see that 4 times N gives me the number in the sequence so; Nth term for Shaded = 4N

 Stage (N) 1 2 3 4 5 6 Sequence 1 5 13 25 41 61 1st difference +4 +8 +12 +16 +20 2nd difference +4 +4 +4 +4

I can see here that there is not much of a pattern in the 1st

Middle

I can see here that there is not must of a pattern in the 1st difference, but when I calculate the 2nd difference I can see that it goes up in 6’s, therefore this is a quadratic sequence and has an Nth term. The second difference is 6 therefore the coefficient of N² must be half of 6 i.e. 3.

Nth term = 3N²+bN+c

From my table I will solve the Nth term using simultaneous equations, I will use stages 2&3 to solve the Nth term for the unshaded squares.

1) 12+2b+c=7

2) 27+3b+c=19

2-1 is 15+b=12

b=-3

To find c: 12-6+c=7

c=1

Nth term for Unshaded = 3n²-3n+1

Total

6N

+        3n²-3n+1

3n²+3n+1

Nth term for Total = 3n²+3n+1

Predictions

The formulae I have found are:

Perimeter                                                        12N+6

Total Squares                                                3n²+3n+1

I will now do a check to see if my formulae are correct. I will test this by predicting, using my formulae, the perimeter, number of shaded squares, number of unshaded squares and the total number of squares, for the next two shapes in this sequence, stages 5 and 6. I will then calculate each of these manually and compare the answers.

Stage 5

I predict:

Perimeter - 12(5)+6=66

Total - 30+61=91 Stage 6

I predict:

Perimeter - 12(6)+6=78

Unshaded - 3(6)²-3(6)+1=91 Total - 36+91=127

I found that my predictions for stages 5 and 6 were correct, as I have counted the hexagons and found the answers to be the same as what I had predicted.

Spatial Justification

Now that I have found that my formulae are correct by making predictions then proving them, I will now use diagrams to spatially justify my formula and explain whythese formulae work.

Perimeter

Conclusion

Next I did the same thing for hexagons and found another set of formulae, I then compared the formulae for squares and the formulae for hexagons to see if there were any connections between them. I found a connection and used this to determine the general formulae for any 2D shape that tessellates.

Next I decided to see if my general formulae worked, I decided to use triangles, as, like squares and hexagons they tessellate. I checked my formulae by predicting the shaded, unshaded and total amount of triangles for the first three stages of the sequence. My predictions were correct and my general formula worked. I now believed that I had investigated 2D shapes as much as I could, and could now use my general formula for any 2D shape, I then decided to move on to 3D shapes. I chose the cube as it is very similar to the square and would be the simplest 3D shape to investigate. Now I investigated the volume of the cubes, I discovered that the cubes were made up of the square sequences built on top of each other. I used this information to create a formula for the volume of the cubes. I then used my formula to predict the next stage in the cube sequence. And found my prediction to be correct. Here is a summary of what I found:

Squares

Perimeter                                                        8N+4

Total Squares                                                2N²+2N+1

Hexagons

Perimeter                                                        12N+6

Total                                                                3n²+3n+1

Triangles

Total                                                                1.5n²+1.5n+1

General formulae for 2D

Perimeter                                                (2x)N+x

Total                                                (½x)N²+(½x)N+1

Cubes

Volume                                                4/3N+2N²+8/3N+1

-  -

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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