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• Level: GCSE
• Subject: Maths
• Word count: 1042

# Mathematics GCSE - Hidden Faces.

Extracts from this document...

Introduction

Shloimi Werjuka        Mathematics Coursework

Mathematics GCSE

Hidden Faces

In order to find the number of hidden faces when eight cubes are placed on a table, in a row, I counted the total amount of faces (68), which added up to 48. I then counted the amount of visible faces (26) and subtracted it off the total amount of faces (48-26). This added up to 22 hidden sides.

I then had to investigate the number of hidden faces for other rows of cubes. I started by drawing out the outcomes for the first nine rows of cubes (below):

I decided to show this information in a table (below):

I decided to show this information on a graph (below):

From this information I have noticed that the number of hidden faces are going up by three each time. In order to find the number of hidden faces for other rows of cubes, it is necessary to have a rule.

Row 2

Row 3

Row 1

Instead of trying to find the number of hidden faces I looked at the visible faces and I took that away from the total amount of faces.

Middle

I can see that 3n is 36 and then I will minus 2. So 36-2 = 16, which is correct, so I now know that the formula is correct.

Another way of working out the nth term is to use the graph. Using the formula y=m+c. The gradient is 3/1=3 and the line passes the y-axis at –2. So the formula is y=3-2. So the nth term would be 3n-2.

Moving on, I now need to work out how to calculate the number of hidden faces when there are more than one row of cubes, I therefore need to find patterns to find the formulas for different number of rows. So I drew the first three diagrams for each number of rows (below):

From these diagrams I can see a few patterns. I can see that with one row, when another cube is placed at the end another three hidden faces are added. With this information I can work out the rest of the column with one row. I noticed that with two rows when another two cubes are placed at the end then another eight hidden faces are added.

Conclusion

256, this equals 240. The number of shown faces is (42) 2 (the front and back) + (52) 2 (both sides) + (45) 1 (only top, bottom is hidden), this adds up to 56. So 240 –56 =184 hidden faces.

Now I will show the above as a formula that will work for all cuboids.

The total amount of faces is (width  height  length  number of faces per cube) WHL6 = 6LWH. The number of visible faces = (2WH) (front and back) + (2LH) (both sides) + (2 WL) (top), this equals 2WH+2LH+WL. So the formula for calculating the number of hidden faces of any cuboid is: 6LWH-(2WH+2LH+WL)

= 6LWH-2WH-2LH-WL.

In order to make sure this is the correct formula, I will check it on another cuboid and see if it works.

H = 3

W = 6

L = 4

I will now work out the number of hidden faces for this cuboid using the formula above. 6LWH-2WH-2LH-WL = 6463-(263)-(243)-(64) = 348 hidden faces. I will now check that 348 is the correct answer by working it out a different way. I counted the visible sides, and then subtracted that from the total amount of faces and I ended up with 348. So from this I can see that the formula is correct.

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