I am doing an investigation to look at shapes made up of other shapes (starting with triangles, then going on squares and hexagons

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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 possible combinations of P, D and T (or later Q and H). Then I will continue making tables of different numbers of that shape, make a graph containing all the tables and then try to devise a working formula.

As I progress, I will note down any obvious or less obvious things that I see, and any working formulas found will go on my ‘Formulas’ page. To save time, perimeter, dots enclosed, triangles etc. are written as their formulaic counterparts. My tables of recordings will include T, Q or H. This is because, whilst it will remain constant in any given table, I am quite sure that this value will need to be incorporated into any formulas.

Triangles

To find the P and D of shapes composed of different numbers of equilateral triangles, I drew them out on isometric dot paper. These tables are displayed numerically, starting with the lowest value of T. Although T is constant in the table, I have put it into each row, as it will be incorporated into the formula that I hope to find. I am predicting that there will be straightforward correlation between P, D and T. I also expect that as the value of P increases, the value of D will decrease. I say this because a circle is the shape with the largest area for its perimeter, and all the area is bunched together. When my triangles are bunched together, many of their vertexes shared 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):

15 Triangles (T=15):

16 Triangles (T=16):

20 triangles (T=20):

It is obvious with all these tables that as P increases, D decreases. The two values are inversely proportionate. As t remains constant, I suspect that some combination of P and D will give T, on account of one going up and the other going down.

            If you look at all these tables, you will see that where D=0, P is always 2 more than T. This can be written as P-2 +/- D=T. The reason I have written +/- D is because, as D is 0, it can be taken away or added without making any difference. 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 Ð 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. However if I were to change to formula to P-2-2D, then 12-2-10=0, which is incorrect.

Now I shall test this with all different values of P and D, but with a constant T of 20. I will substitute P and D for their numerical values in the T=20 table, and use them in the above formula (P-2+2D=T) If the formula works, all equations will balance to give T as 20. So;

P=12 and D=5  Р 12-2+10=20          C

P=14 and D=4  Р 14-2+8=20            C

P=16 and D=3  Р 16-2+6=20            C

P=18 and D=2  Р 18-2+4=20            C

P=20 and D=1  Р 20-2+2=20            C

(I have already tested P=22 and D=0 above). All the different shapes in the T=20 table work with this formula. Therefore I will test the formula with all the other tables below, so I can be sure that it does work with all numbers of triangles. If it works with all four of the numbers of triangles that 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    Р 8-2+4=10              C

P=10 and D=1  Р 10-2+2=10            C

P=12 and D=0  Р 12-2+0=10            C

Join now!

Where T=15…

P=11 and D=3  Р 11-2+6=15            C

P=13 and D=2  Р 13-2+3=15            C

P=15 and D=1  Р 15-2+2=15            C

P=17 and D=0  Р 17-2+0=15            C

And where T=16…

P=10 and D=4  Р 10-2+8=16            C

P=12 and D=3  Р 12-2+6=16            C

P=14 and D=2  Р 14-2+4=16            C

P=16 and D=1  Р 16-2+2=16            C

P=18 and D=0  Р 18-2+0=16            C

This proves that (for triangles at least) the formula P-2+2D=T works. This can be rearranged to give D=(T+2-P)/2 and T= P+2D-2. I do not need to test these two new formulas, as they have simply been rearranged from the existing one, ...

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