Therefore the formulae for three faces is 8.
I will now try and find a formulae for a 1 side painted cube, to do this I will use a 5*5*5 cube.
I am only using one side of the cube because all the sides are the same. If n equals the length of the cube, and you then take away the surrounding cubes (which have two and three faces covered) you then get (n-2) and because it is an area you square it, it becomes (n-2)2. A cube has 6 faces so I also have to times (n-2)2 which is one face by 6. My formulae for 1 side of the cube being covered in paint is 6(n-2)2.
I will now work out the formulae for edges. As with the cubes with painted faces I can work out one edge then multiply by 12.
Again it is the number of cubes take away 2. This gives me (n-2) and since there are also 12 edges I have to include this giving me a formulae of 12(n-2).
To find the formulae for 0 painted cubes it is the total number of small cubes minus the outer layer of the large cube.
If I take a length of small cubes, as long as they are not all made up from the outer layer they will look like this.
Because there are two small cubes on the ends of the strand it will make the formulae n-2 and because the 0 sided faces will make up an area I will have to cube it, making the formulae for a 0 sided cube (n-2)3.
Using a 6*6*6 cube and my formulae’s I will check to see if my work for cubes is correct.
A) 12(n-2)
12(6-2) = 48
B) 6(n-2)2
6(6-2)2 = 96
C) (n-2)3
(6-2)3 = 64
I know that my formulae’s have worked because 6*6*6 equals 216 and so do all of my answers 8, A, B and C when added together.
Cuboids.
Now that I have my formulas for cubes I will attempt cuboids. I think that the formulae for 3 faces covered will be 8 because the same as a cube the cuboid will have 8 corners.
3*3*4
From this first cuboid I have noticed that the formulae for 3 faces covered will be 8, the same as for cubes. My predictions were correct.
3*4*5.
A cuboid is different to a cube in the way that all the lengths are not necessarily the same. So because of this reason I need to label the sides L, W and H, which stands for Length, Width and Height.
To find out the formulae for 1 face covered in paint I first worked out the total surface area of the cubes in the large cube, which gave me 2(LH+LW+WH). A LH gives me the blue, a LW gives me the red and WH gives me the green. I then have to times it by 2 because there are two of each. I then had to include 2(-4L –4W –4H +12) because after I have the area for a whole side I only need the inner part.
So I take away the length, width and height cubes, which are the edges of the side to get the area for 1 painted cube. So that I don’t take off too much I also have to add 12 to this because some of the length, width and heights overlap each other.
To get the formulae for 0 faces I noticed that it is the cuboid takeaway the outer layer.
L*W*H gives me the whole large cube, so if I take away 2 from each of them then it takes away the outer layer for me.
The formulae for 0 faces covered in paint is (L-2)(W-2)(H-2).
The formulae for 2 faces covered in paint is 4(L+W+H-6). I found this out because a length, width and height added together gives me one of four combinations to get all the cubes covered in paint. I then take away the 6 because you take away 2 away from each of the lengths, widths and heights.
To make sure all of my formulae’s work I will test them on 3*3*5 cuboid.
A) 4(L+W+H -6)
= 4(3+3+5 -6)
= 4(5)
= 20
B) 2(LH+LW+WH) –2(-4L–4W–4H +12)
= 2(15+9+15) –2(-12-12-20 +12)
= 2(39) –2(-32)
= 78 – 64
= 14
C) (L-2)(W-2)(H-2)
= (3-2)(3-2)(5-2)
= (1)(1)(3)
= 3
To check my answers I will add up the totals which I got using my formulae’s and check to see if they mach the total of 3*3*5= 45. 8 + 20 + 14 + 3 = 45
I know that all of my formulas are correct because the total of my answers I got from using my formulas match the answer I got from multiplying 3, 3 and 5 the lengths of my cuboid.
