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  • Level: GCSE
  • Subject: Maths
  • Word count: 1457

Cube Towers Investigation

Extracts from this document...

Introduction

Aim

The aim of this investigation was to find how many hidden faces there were when various sized cuboids were built, and also to find the relationship between the number of hidden faces in a cube and how many cubes there were in the shape.

Square Based Towers

I started off the investigation using square based cuboids, eg. 1 x 1, 2 x 2, 3 x 3, etc.

1.1 Diagrams

n = 1

n = 2

n = 3

x = 1

x = 3

x = 5

n = 4

n = 5

n = 6

n = 7

x = 7

x = 9

x = 11

x = 13

1.2 Table of Results

1 x 1 CUBES

No. of Cubes (n)

Height (h)

Hidden Faces (x)

1

1

1

2

2

3

3

3

5

4

4

7

5

5

9

6

6

11

7

7

13

1.31 Pattern (WORDS)

The number of hidden faces goes up by two each time another cube is added on.

1.32 Pattern (ALGEBRA)

The formula to find the number of hidden faces in a 1 x 1 square-based cube tower is 2n – 1. The 2n comes from the pattern that each time another cube is added to the previous one(s), the number of hidden faces go up by two; but that can’t be the only part of the formula, because then the linear sequence would be 2, 4, 6, 8, 10… etc.

...read more.

Middle

12

8

2

28

12

3

44

16

4

60

20

5

76

24

6

92

28

7

108


2.31 Pattern (WORDS)

The number of hidden faces goes up by sixteen each time another row consisting of four cubes is added on.

2.32 Pattern (ALGEBRA)

The formula to find the number of hidden faces in a 2 x 2 square-based cube tower is 4n – 4. The 4n comes from the pattern that each time another four cubes are added to the previous row, the number of hidden faces goes up by sixteen.

The [ – 4 ] is taken from how sixteen relates to the number of hidden faces. When there are four cubes, the number of hidden faces is twelve, but we know the hidden faces increase by sixteen each time, so to get from sixteen to twelve, we have to subtract four. When there are eight cubes, the number of hidden faces is twenty-eight, which is 4(8) – 4, which is how I figured out the formula to be 4n – 4.

2.4 Checking My Formula

EXAMPLE #1 = 4(12) – 4

EXAMPLE #1 = 48 – 4

EXAMPLE #1 = 44

EXAMPLE #2 = 4(20) – 4

EXAMPLE #2 = 80 – 4

EXAMPLE #2 = 76

2.5 Prediction

EIGHT CUBES = 4(32) – 4

EIGHT CUBES = 128 – 4

EIGHT CUBES = 124

2.6 Checking My Prediction

3.1 Diagrams

n = 9

n = 18

n = 27

x = 33

x = 75

x = 117

3.2 Table of Results

3 x 3 CUBES

No.

...read more.

Conclusion

  1. To get the second part of the formula, use the number by which the hidden faces go up by each time again, and subtract the number of hidden faces if the height is one from it, eg, 16 – 12 = 4. So – 4 would be the second part of the formula.

The General Formula

After finding the formula for finding the number of hidden faces in each square-based tower, I tried to find a general formula that not only would be able to solve the hidden faces in all square-based towers, but also to be able to find hidden faces in rectangular-based towers as well.

I realized that what I was really looking for was just a formula that showed:

total faces – seen faces = hidden faces

The total faces could be seen as 6n.

The seen faces can also be referred to as the area.image00.pngimage01.png

image02.png

General Formula for Finding Hidden Faces in All Cuboids

6n – (2lh + 2wh + lw)

Conclusion

In conclusion, the formula for finding the number of hidden faces in all cuboids is 6n – ( 2lh + 2wh + lw ).

...read more.

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