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Maths hidden faces

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Maths hidden faces Introduction

In this coursework I would be investigating the number of hidden faces in different cubes and cuboids. I would provide predictions to make sure I get the right results. After that I would provide diagrams of the cubes and I would explain how I found it. In each section of a set of cubes I would provide formula’s that I would find. I would also give information that the pattern carries on.

The reason why I am doing this investigation about cubes is because to find the hidden faces and total faces which would be added near the diagrams. I am going to increase one dimension at a time while holding the other one the same. I will start by holding the width and the height by one while increasing the depth by one.

Moving on to deeper investigation I would investigate cubes that have more cubes and rows of cubes which would lead to more accurate results. I would again provide total faces and hidden faces. The reason why I would go into such deeper investigation is because to get more information about the cubes with more of them and check if the pattern carries on.

Overall of my coursework I would provide a lot of information about what I have done and how I have done it.

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3n – 2Holding width and depth by one and increasing the height = 2n-1Holding height and depth by one and increasing the width = 3n – 2Holding width and height by two and increasing the depth = 4n – 2Holding width and depth by two and increasing the height = 4n – 4Holding height and depth by two and increasing the width = 4n + 2Holding the width and height by three and increasing the depth = 5n-18Holding width and depth by three and increasing the height = the formulas vary considerably as the size of cubes increases.Holding height and depth by three and increasing the width = 5n – 18Increasing all dimensions of a cube by and going up to six = the formulas vary considerably.

    I found the formulas for particular dimensions of the cuboids. However some formulas are inconsistent as they vary with the size of the cubes. Each formula only works for a certain dimension numbers and it can’t be used to find the number of hidden faces in any cube. These formulas do not help me to find the total numbers of hidden faces in any cubes as they all keep hanging and varying.  I cannot see a prospect of finding the general formula for the total number of hidden faces in any cube so I am going to stop drawing any more cubes.

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width = 5, height = 1 and depth = 1. 6 (5, 1, 1) = 30 and (2, 5,1+2,1,1+5,5) = 17. Therefore 30-17=13 which is correct. Width=5, height = 5, and depth = 5. 6 (5,5,5)= 750 and (2,5,5,+2,5,5,+5,5) = 125 so 750- 125 = 625, which again show I am on the right way. Now my formula is consistent as it works on any dimension of a cube and finds the exact number of hidden faces. This could not be done by a formula containing only the total numbers of cubes as it limits the formula so it works only on a particular dimension. However some formulas that use only the total numbers of cubes do not work on particular dimensions at all keep varying as the size of cube increases because the formula has an insufficient amount of values. The main reason why I found the formula was because I split up my cube into different values such as sides, front and back and top, which were crucial in helping me to find the total numbers of seen faces and the rest seemed to have followed as it was rather easy to find the total number of cubes and faces. Now that I have found the general formula I have worked out further hidden faces in different dimensions and put my results on the table.  

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