• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # painted sides of a cube

Extracts from this document...

Introduction

Painted Sides of a Cube  By Connor McInnes

Here is a 3 x 3 x 3 cube: These 3 cubes all represent 3 x 3 x 3 cubes, the first one has the blocks shaded (pink) that will be painted 3 sides, the second cube has the blocks shaded (blue) that will have 2 sides painted, and the third cube has blocks shaded (green)

Middle

There is a definite pattern for the cube and the sides painted.
After looking at the first 4 cubes, the sides painted look as such:

In a 2 x 2 x 2 cube there are:
0 blocks with 0 sides painted.
0 blocks with 1 side painted.
0 blocks with 2 sides painted.
8 blocks with 3 sides painted.

In a 3 x 3 x 3 cube there are:
1 blocks with 0 sides painted.
6 blocks with 1 side painted.
12 blocks with 2 sides painted.
8 blocks with 3 sides painted.

In a 4 x 4 x 4 cube there are:
8 blocks with 0 sides painted.

Conclusion

For the cubes with 3 sides painted, it will always be 8. The eight painted cubes are the 8 corners of any and all cubes.

The spreadsheet below takes you through the first nine cubes with n x n x n sides. Now, we have found these four values for our cubes with n sides:

(n-2)3, 6(n-2)2, 12(n-2), 8

Looking at the four values, one notices that the values are the products of a binomial expansion.

n3= ((n-2) +2)3

The expansion of the binomial looks as such:

n3= (n-2)3 + 6(n-2)2 + 12(n-2) + 8

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

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Hidden Faces and Cubes essays

1. ## I am doing an investigation to look at shapes made up of other shapes ...

If you look at the ones above, the first is straightforward, the second has a 'divide by' at the beginning of the formula, and the last one has it at the end, as well as a set of brackets. I will try to get them all looking the same, with

2. ## Shapes Investigation I will try to find the relationship between the perimeter (in cm), ...

Q or H for X, which will represent all or any of the shapes. So the formulas I have are: T=P+2D-2, Q=P/2+D-1 and H=(P+2D-2)/4. I think that to get my universal formula, I need to have all these three in the same 'format'.

1. ## gcse maths shapes investigation

to the existing formulas to get a new, working formula for squares. So where P=14, D=4 and Q=10... P=10+2-8 � P=4 DP is out by -10, or -Q D=(10+2-14)/2� D=-1 DD is out by -5 Q=14+8-2 � Q=20 DQ is out by +10, or +Q And where P=16, D=6 and Q=13...

2. ## mathsI will try to find the correlations between the perimeter (in cm), dots enclosed ...

4 16 28 3 16 30 2 16 32 1 16 34 0 16 Firstly I will test my previous formulas, P=T+2-2D, D=(T+2-P)/2 and T= P+2D-2, to see if they hold true - of course, substituting T with Q. If the formulas still hold true, I will be able to

1. ## I am doing an investigation to look at shapes made up of other shapes.

However, as this is in effect a formula triangle (of sorts), all indices (D, P and T) must be incorporated. With T=20 and P=12, P-2 +/-D=T � 12-2 +/-5=T. So T=15 or 5. If I were to make it P-2+2D=T, then that would mean that 12-2+10=20, therefore T=20, which is correct.

2. ## Am doing an investigation to look at shapes made up of other shapes (starting ...

I have looked at, I think this will be sufficient evidence that it will continue with all other numbers of triangles. So where T=10... P=8 and D=2 � 8-2+4=10 C P=10 and D=1 � 10-2+2=10 C P=12 and D=0 � 12-2+0=10 C Where T=15...

1. ## Shapes (starting with triangles, then going on squares and hexagons. I will try to ...

If the formula works, all equations will balance to give T as 20. So; P=12 and D=5 � 12-2+10=20 C P=14 and D=4 � 14-2+8=20 C P=16 and D=3 � 16-2+6=20 C P=18 and D=2 � 18-2+4=20 C P=20 and D=1 � 20-2+2=20 C (I have already tested P=22 and D=0 above).

2. ## I am doing an investigation to look at shapes made up of other shapes ...

dots with many other triangles, therefore there are much more dots enclosed than if the triangles were laid in a line 10 Triangles (T=10): P= D= T= 8cm 2 10 10cm 1 10 12cm 0 10 15 Triangles (T=15): P= D= T= 11 3 15 13 2 15 15 1 • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 