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• Level: GCSE
• Subject: Maths
• Word count: 4940

# Investigating Structures - In this investigation I will be looking at different joints used in building different sized cube shaped structures.

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Introduction

Investigating Structures

In this investigation I will be looking at different joints used in building different sized cube shaped structures. I have found that there are four different types of joint:

3-joint                4-joint                5-joint                6-joint

The aim of my investigation is to be able to use the length of the cube’s edge (n) to find out the number of each different joint in that cube. I have taken n to be the number of rods on a cube’s edge.

I will start my investigation by drawing simple cube structures on isometric paper. I will draw four structures. A 1 x 1 x 1 cube, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube and a 4 x 4 x 4 cube. These are shown on the isometric paper included.

After looking at each I will count the number of each type of joint in each cube, and put my results in a table to make it easier to analyse them. Here is the table:

 Length of n 3-Joint (c) 4-Joint (e) 5-Joint (y) 6-Joint (X) Total Joints (t) 1 8 0 0 0 8 2 8 12 6 1 27 3 8 24 24 8 64 4 8 36 54 27 125

I now have my results, and shall set about finding out their formulae. Firstly I will do the 3-Joint’s formula. After looking at the table, I notice that the pattern is that there always 8 3-Joints in a cube. So the formula must be:

c = 8

This formula is as above because 3-Joints only occur on the corners of a cube. As there are only eight corners on any sized cube, the answer must be eight.

I will now look at the formula for the 4-Joint. The first immediate pattern I notice is that the results for the 4-Joint go up in steps of 12. So there must be a 12 in the formula. If I multiply n by 12, I will always get an answer, which is too large by 12.

Middle

8 + 48 + 96 + 64 = 216

These are the results I found. A picture of a 5 x 5 x 5 cube has been drawn, and is on isometric paper. I manually counted the number of each joint in this cube; they came to the same totals as shown above.

Extension (1)

I have completed my original problem, and will now extend it. I will find the same type of joints, except I will vary the size of one of the lengths. This will create a square-faced cuboid.

As in the previous investigation, I will be looking for the 3-Joints, 4-Joints, 5-Joints and 6-Joints. To start off I will draw the first four square-faced cuboids. These are a 1 x 1 x 2 cuboid, a 2 x 2 x 3 cuboid, a 3 x 3 x 4 cuboid and a 4 x 4 x 5 cuboid. These are drawn on isometric paper.

I manually counted the number of each joint type in each cuboid and have put them in a similar table as to the one in the previous investigation:

 Length of n Length of m 3-Joints (c) 4-Joints (e) 5-Joints (y) 6-Joints (X) Total Joints (t) 1 2 8 4 0 0 12 2 3 8 16 10 2 36 3 4 8 28 32 12 80 4 5 8 40 66 36 150

I now have my results, so I will now find and explain the formulae for each joint type. Firstly I will do the 3-Joint. After looking at my results I notice that, as before, all the answers are 8. Therefore the formula is again:

c = 8

Like a cube, a cuboid can only have eight corners, and as we know 3-Joints only occur on corners, then there will ALWAYS be eight 3-Joints on any sized cuboid. This is the reason for the above formula.

I will now find and explain the formula for the 4-Joints. The formula for the 4-Joints in the cube was 12(n – 1). This was because there are 12 edges on a cube.

Conclusion

(m(p+1)+p(m+1))(n+1)

This formula totals the number of vertical and horizontal formulae in any cuboid or cube. Now we must find all the rods, which are in-line with n. To the side of the largest face, the number of rods going backwards is always one more than the number of vertical rods in-line with m. Therefore in our formula somewhere is (m+1). At the top of the largest face, there is also 1 more rod, than the number of horizontal rods in-line with p. This proves that in the end of our formula there will be (p+1). If I multiply (m+1) with (p+1), I get the number of rods in-line with n, between two of these large ‘faces’. So, if I multiply my result by n, this will give me the total number of rods in-line with n (n(m+1)(p+1)). So final formula therefore is:

TR = (m(p+1)+p(m+1))(n+1)+n(m+1)(p+1)

As I stated before, this formula will work on a square-faced cuboid, and a cube. If it needs to be used on a square-faced cuboid, all you need to do is substitute p with n. If you need to use it on a cube, you just need to replace both p and m with n. The reason that 1 is added onto the first p, is because you always have one more vertical rod going across than p. This reason applies to all the values which need to have 1 added onto them. I have explained the formula in the above paragraphs.

Edward Maxwell-Lyte 11N

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

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