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GCSE: Hidden Faces and Cubes
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- Level: GCSE
- Questions: 75
I will 8cm 8cm use Pythagoras again and say that the height 2 = 82 - 42 h so, the height is 6.928cm, to find the area we must Multiply the height by the base: 8cm 6.928 x 8 = 55.426cm 2 = 27.713cm2 Now to find the volume: 27.713 x 32 = 886.816cm3 32cm H2 = 92 - 32 h= 8.49cm. Area = base (6) x height (8.49) 9cm 9cm 2 H = 25.455cm2 6cm Volume = area of cross section (25.455)
- Word count: 2729
and the total number of squares. Then I will test the formula on my pattern that I have drawn and the ones I haven't drawn. Finally I will extend my investigation to 3D and hopefully find a formula for it, test it and find total number of squares etc in pattern without drawing it. Patterns Pattern 1 = White 1 Black 0 Pattern 2 = White 1 Black 5 Pattern 3 = White 5 Black 8 Pattern 4 = White 13 Black 12 Patterns Pattern 5 = White 25 Black 16 Pattern 6 = White 41 Black 20 Pattern no No of white squares No of
- Word count: 2991
An experiment to find out if seeing the eyes of a well known persons face is a factor of face recognition
The results indicate that the eyes of a person's face are an important factor of face recognition. Introduction / Background: Being able to recognise other people is of great significance in our lives. You can imagine how difficult life would be if you didn't recognise your mother, brother, best friend or partner. Human beings seem to have an amazing ability to recognise faces. Standing (1973) showed participants 10,000 faces over five days. When they were shown pairs of faces, one of which they had previously been shown, together with a new one, they were able to identify the face they had seen 98% of the time.
- Word count: 2040
Investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes.
1046 The problem is to find formulae that represent the number of rods, 3 joints, 4 joints, 5 joints and 6 joints in an n x n x n cube. And then repeat for a cubiod Cubes (Sheet 1) I started the Investigation by drawing a cube shape. I thought 5 different sized cubes would be enough to work out formula and trends that may come up. I increased the cubes 1cm2 each time, starting with 1 x 1 x 1.
- Word count: 2445
The aim of my investigation is based on the number of hidden faces and faces in view of cubes that are placed on a table.
Thirteen of these faces are hidden; the number of faces hidden confirms my first formula: 30-17=13. Through confirming my first formula I will now be able to predict further sets of cubes such as a row of eight cubes placed on a table. So... 6(number of faces per cube) x8 (number of cubes)=48(number of faces all together) 27 faces in view Therefore: 48(number of faces all together)-27(number of faces in view)=21(number of hidden faces) I will now conduct an experiment with Lego cubes that connect together, thus simulating the cubes on the table. I will simulate the set of eight cubes on a row and see if the results match that of my formula.
- Word count: 2942
I am going to investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes.
Without using diagonals, this is the most amount of rods to join together. The problem is to find formulae that represent the number of rods, 3 joints, 4 joints, 5 joints and 6 joints in an nxnxn cube. And then repeat the investigation but for an xxyxz cuboid. Stradegy To carry out this investigation, I will need to spot patterns that may emerge as the cube/cuboid gets bigger. Using visual images will aid me greatly in this respect and so I will present the cubes that I am investigating on geometric or spotty paper.
- Word count: 2359
An investigation for working out hidden faces as different number of cubes are joined by making different shapes.
No of total faces 1 1 1 5 6 2 2 4 8 12 3 3 7 11 18 4 4 10 14 24 5 5 13 17 30 6 6 16 20 36 7 7 19 23 42 From the above table a simple sequence can be formed and an nth term of the sequence can be worked out Sequence for hidden faces 1 4 7 10 13 16 19 3 3 3 3 3 3 By taking the 1st difference between the numbers of above sequences a constant of 3 is observed so the nth term is 3n+a
- Word count: 2099