Am doing an investigation to look at shapes made up of other shapes (starting with triangles, then going on squares and hexagons. I will try to find the relationship between the perimeter (in cm), dots
Summary I am doing an investigation to look at shapes made up of other shapes (starting with triangles, then going on squares and hexagons. I will try to find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape. From this, I will try to find a formula linking P (perimeter), D (dots enclosed) and T (number of triangles used to make a shape). Later on in this investigation T will be substituted for Q (squares) and H (hexagons) used to make a shape. Other letters used in my formulas and equations are X (T, Q or H), and Y (the number of sides a shape has). I have decided not to use S for squares, as it is possible it could be mistaken for 5, when put into a formula. After this, I will try to find a formula that links the number of shapes, P and D that will work with any tessellating shape - my 'universal' formula. I anticipate that for this to work I will have to include that number of sides of the shapes I use in my formula. Method I will first draw out all possible shapes using, for example, 16 triangles, avoiding drawing those shapes with the same properties of T, P and D, as this is pointless (i.e. those arranged in the same way but say, on their side. I will attach these drawings to the front of each section. From this, I will make a list of all possible combinations of P, D and T (or later Q
volumes of open ended prisms
Part 1 For part 1 of this piece of math coursework I will be investigating volumes of prisms, which can be made from a 24cm, by 32cm piece of card. I will be trying to determine which shape will make the prism with the largest volume. To do this, I will be exploring the volumes of triangular prism, cylinders, quadrilateral prisms, pentagonal prism, hexagonal prism, heptagonal prism and octagonal prism. I will then try to work out a formula for working out the volume of an "n" sided shape. Triangular prisms First, I will be investigating the volume of triangular prisms. We know that in a triangle, the lengths of the left and right sides must add up to more than the length of the base and we also know that the volume of any prism is the area of cross section multiplied by the length. To find the volume, we must first find the area of the 7cm cross-section. To find the area of a triangle we must h use the formula Area(a) = base (b) x height (h) 2 10cm 32cm To work out the height we must use Pythagoras's theorem Using the rules of Pythagoras, we know
Shapes Investigation - Find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape.
GCSE Maths Coursework - Shapes Investigation Summary I am doing an investigation to look at shapes made up of other shapes (starting with triangles, then going on squares and hexagons. I will try to find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape. From this, I will try to find a formula linking P (perimeter), D (dots enclosed) and T (number of triangles used to make a shape). Later on in this investigation T will be substituted for Q (squares) and H (hexagons) used to make a shape. Other letters used in my formulas and equations are X (T, Q or H), and Y (the number of sides a shape has). I have decided not to use S for squares, as it is possible it could be mistaken for 5, when put into a formula. After this, I will try to find a formula that links the number of shapes, P and D that will work with any tessellating shape - my 'universal' formula. I anticipate that for this to work I will have to include that number of sides of the shapes I use in my formula. Method I will first draw out all possible shapes using, for example, 16 triangles, avoiding drawing those shapes with the same properties of T, P and D, as this is pointless (i.e. those arranged in the same way but say, on their side. I will attach these drawings to the front of each section. From this, I will make a list of all
Investigation into the progression of patterns in 3d shapes.
Borders Coursework I am continuing my investigation into the progression of patterns further, in that, I am no longer working with 2D shapes, but instead 3D ones. I have drawn the first four patterns on separate isometric paper and the pattern number relates to its nth term, i.e. pattern 2 is the same as n = 2. In order to draw the shapes I have drawn them in separate layers, to make it easier to count the total number of squares in each pattern. When drawing each pattern, I added 1 square to each free side of the previous pattern. For instance, pattern 1 is a cube which has 6 sides; therefore I added 6 cubes to it. On the following pages you will see the diagrams that show how the 3D patterns are built up. As shown by my drawings, each pattern's number of layers increases. If we look at each individual layer we can see a significant link between the 2D and 3D patterns. First of all I looked at the centre layer, which is built up in exactly the same way as the 2D patterns. I then looked at layer A for all the patterns and found it followed the same pattern of progression, however it does not have the same formula. In pattern 1, layer A = 0 therefore it is a step behind. The progression of layer B is again the same however the formula is different. In pattern 2 layer B = 0, therefore it is two steps behind. The results of my investigation can be shown as follows:
The Painted Cube - Maths Investigations
The Painted Cube. D.K A cube is painted on all its faces. It is then cut into 27 identical cubes. How many cubes have paint on. (a) 3 faces. (b) 2 faces. (c) 1 face. (d) 0 faces. A similar cube is painted on all six faces it is then cut into 64 identical cubes. How many cubes have paint on. (a) 3 faces. (b) 2 faces. (c) 1 face. (d) 0 faces. A cube made of 27 smaller ones has a length, width and height of 3. I know this because 3*3*3= 27. 3*3*3 3 Faces. = 8 2 Faces. = 2 Faces. = 6 0 Faces. = Total. = 27 Already just by looking at the first cube I have realised that all the cubes will have 8 small cubes with three faces covered, noticed this as all cubes no matter what size always have 8 corners. A cube split into 64 identical cubes has lengths widths and heights of 4*4*4. 3 Faces. = 8 2 Faces. = 24 Faces. = 24 0 Faces. = 8 Total. = 64 My 4*4*4 cube has proven that all the faces with three sides covered will be 8. 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
Investigate the number of hidden faces when cubes are joined indifferent ways.
Hidden Faces Investigation: Investigate the number of hidden faces when cubes are joined in different ways Andrew Bambridge 10FA3 Hidden Faces Coursework I am going to investigate the number of hidden faces when cubes are joined in different ways. The aim of this task is to find a formula which is common with every cube and hidden face. In order for me to find the overall formula which ca be found by finding out a formula which can determine the outcome of number of hidden faces on 'n' cubes. I will have to start by spotting patterns and their differences then find out how each set of hidden faces are common with each other. Then try different formulas and see if it fits with the set of results. Part one: Number Of Rows 1 When a single cube is placed on a flat surface only 1 face is hidden out of 6 When two cubes are joined together and placed on a flat surface there are 4 hidden Faces out of a possible 12 When three cubes are joined there are 7 hidden faces There are 10 hidden faces There are 13 hidden faces There are 16 hidden faces Results Differences: There is a difference of +3 every time There are 4 hidden faces and 8 can be seen There are 12 hidden faces and 12 can be seen There are 20 hidden faces and 16 can be seen There are 28 hidden faces and 20 can be seen There are 36 hidden faces and 24 can be seen Results Difference: There is a
Shapes (starting with triangles, then going on squares and hexagons. I will try to find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape.
Shapes Investigation Summary I am doing an investigation to look at shapes made up of other shapes (starting with triangles, then going on squares and hexagons. I will try to find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape. From this, I will try to find a formula linking P (perimeter), D (dots enclosed) and T (number of triangles used to make a shape). Later on in this investigation T will be substituted for Q (squares) and H (hexagons) used to make a shape. Other letters used in my formulas and equations are X (T, Q or H), and Y (the number of sides a shape has). I have decided not to use S for squares, as it is possible it could be mistaken for 5, when put into a formula. After this, I will try to find a formula that links the number of shapes, P and D that will work with any tessellating shape - my 'universal' formula. I anticipate that for this to work I will have to include that number of sides of the shapes I use in my formula. Method I will first draw out all possible shapes using, for example, 16 triangles, avoiding drawing those shapes with the same properties of T, P and D, as this is pointless (i.e. those arranged in the same way but say, on their side. I will attach these drawings to the front of each section. From this, I will make a list of all possible combinations of P,
Compare the two poems 'Porphyria's lover' and 'My Last Duchess' by Robert Browning. In which way do they form part of a literary tradition?
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Cubes and Cuboids Investigation.
Cubes and Cuboids Investigation I am going to investigate the different patterns that occur with different cubes when all the faces are painted of a large cube and then that is separated into smaller cubes and then how many faces of each small cube are still painted. Here are my cubes. They are 2*2*2, 3*3*3 and 4*4*4. I am going to establish the patterns that recur as the cube gets larger. For example the number of cubes with one face painted, with two faces painted, with three faces painted and the number of cubes with no faces painted when the larger cube is split up. Here is a table: Length of cube No. of small cubes No. of small cubes with X painted faces X=3 X=2 X=1 X=0 2 8 8 0 0 0 3 27 8 2 6 4 64 8 24 24 8 Immediately I noticed that all of the cubes have 8 cubes with 3 different faces painted when they are separated. All of these 8 are the vertices of the cube and so every cube except that which has a length of one will have 8 cubes with three faces painted. This can be shown in the table: cube length (X) No. of cubes with 3 painted faces (Y) 2 8 3 8 4 8 Y=8 'font-size:14.0pt; '>The above tells us how many cubes will have three painted faces to find out how many will have two, here is a table: cube length (X) No. of cubes with 2 painted faces (Y) 2 0 3 2 4 24 Y=12(X-2) 'font-size:14.0pt; '>I noticed this formula because
Patterns With Fractions Investigations
Mathematics Coursework. Patterns With Fractions. Consider the sequence of fractions and the differences between the fractions: Term (n) 1 2 3 4 5 st Difference 2nd Difference (For rest of differences, see page11) Finding the starting fraction for the nth term: , , , , = (The general formula) Check if correct formula: Term (n) Numerator (n) Denominator (n + 1) Final Fraction 2 ? 2 2 3 ? 3 3 4 ? 4 4 5 ? 5 5 6 ? (Check On Page 11) Finding the nth term for the 1st difference: In order to find out the nth term for the 1st differences, the requirement is to subtract the 2nd fraction from the 1st fraction (the smaller fraction from the bigger fraction). - = = = (The general formula for 1st difference) Check if correct formula: Term (n) Numerator (1) Denominator (n + 1)(n+2) Final Fraction (1+1)(1+2) = 6 ? 2 (2+1)(2+2) = 2 ? 3 (3+1)(3+2) = 20 ? 4 (4+1)(4+2) = 30 ? 5 (5+1)(5+2) = 42 ? (Check On Page 11) Finding the nth term for the 2nd difference: In order to find out the nth term for the 2nd differences, the requirement is to subtract the 1st fraction from the 2nd fraction (the