# Hidden faces are those hidden after the cubes have been viewed from all angles.

Extracts from this document...

Introduction

HIDDEN FACES

Hidden Faces: Hidden faces are those hidden after the cubes have been viewed from all angles.

Introduction: I am investigating the number of hidden faces for other cuboids made from cubes. I will use visual representation to display my results in the form of graphs. I will collect my results in a table. I will start to collect my information in my table starting with one cube and building them up into rows and different sized cuboids. At the end of my investigation I hope to have a formula worked out, and also I hope to be able to find the number of hidden faces on a cuboids made up from 30 cubes.

Collecting Data: I have drawn a table to record my results. In the first column I have the number of cubes and in the second I have the number of hidden faces. In my table I have found the hidden faces for every one cube put down there is one hidden face on the bottom.

Middle

The Differences: In this sequence (using the number of hidden faces from my table) the difference between them is +3. So what ever my formula turns out to be it must have a 3 in it. If I use ‘N’ as representing the number of cubes. My formula looks like this ‘3N’ or 3xN. So If a number of cubes are put into my formula it will go through like this.

X =

Prediction: I predict that if I put a 9 and then a 7 into my formula. 9 cubes will have 27 hidden faces. And if I put 7 cubes into my formula there will be 21 hidden faces.

The Final Piece: If I put a 9 into my formula it gives me 3x9=27. And if I put a 7 in it will come out like this 3x7 =21. So my prediction was right but if you count the number of hidden faces without using the formula 9 cubes has 25 hidden faces and 7 cubes has 19 hidden faces. So each time the number of hidden faces is -2 more than I predicted.

Conclusion

18 cubes stacked = 3.5 x 18 = 63

So my formula so far is wrong. Because each time I put in a number of cubes the number of hidden faces is four more than I predicted. Each time I am having to – 4 each time to get the right number of hidden faces. My formula is now complete and looks like this 3.5n – 4.To prove this I am going to input different amounts of cubes.

26 stacked cubes = 3.5 x 26 = 91

91 – 4 = 86 hidden faces

30 stacked cubes = 3.5 x 30 = 105

105 – 4 = 101 hidden faces

36 stacked cubes = 3.5 x 36 = 126

126 – 4 = 122 hidden faces

So my second formula also works. My second way of working it out like I did for cubes in a row which is 6n – S this also works using stacked cubes

Conclusion: In conclusion my investigation shows how my formulas help to find the number of hidden faces. My first formula works only for cubes in a row. My second formula only works for cubes stacked two high in a line. But the amazing thing is that the formula that works for both is the other way of working out the number of hidden faces 6n- S. This formula involves a lot more working out but can be effective if used.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month