COURSEWORKBANK
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):
|P= |D= |T= |
|8cm |2 |10 |
|10cm |1 |10 |
|12cm |0 |10 |
15 Triangles (T=15):
|P= |D= |T= |
|11 |3 |15 |
|13 |2 |15 |
|15 |1 |15 |
|17 |0 |15 |
16 Triangles (T=16):
|P= |D= |T= |
|10cm |4 |16 |
|12cm |3 |16 |
|14cm |2 |16 |
|16cm |1 |16 |
|18cm |0 |16 |
20 triangles (T=20):
|P= |D= |T= |
|12 |5 |20 |
|14 |4 |20 |
|16 |3 |20 |
|18 |2 |20 |
|20 |1 |20 |
|22 |0 |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 Ð 12-2+10=20 C
P=14 and D=4 Ð 14-2+8=20 C
P=16 and D=3 Ð 16-2+6=20 C
P=18 and D=2 Ð 18-2+4=20 C
P=20 and D=1 Ð 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 Ð 8-2+4=10 C
P=10 and D=1 Ð 10-2+2=10 C
P=12 and D=0 Ð 12-2+0=10 C
Where T=15…
P=11 and D=3 Ð 11-2+6=15 C
P=13 and D=2 Ð 13-2+3=15 C
P=15 and D=1 Ð 15-2+2=15 C
P=17 and D=0 Ð 17-2+0=15 C
And where T=16…
P=10 and D=4 Ð 10-2+8=16 C
P=12 and D=3 Ð 12-2+6=16 C
P=14 and D=2 Ð 14-2+4=16 C
P=16 and D=1 Ð 16-2+2=16 C
P=18 and D=0 Ð 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, P-2+2D=T. However, just to make sure, I will test the new formulas once for each number of triangles. I will make sure I do not test them with a D of zero, as this would give less margin for error (I have not tested any shapes where T or P are zero)
Where T=10, P=8 and D=2,
D=(10+2-8)/2 Ð D=2
T=8+4-2 Ð T=10
And where T=15, P=11 and D=3…
D=(15+2-11)/2Ð D=3
T=11+6-2 Ð T=15
And where T=16, P=10 and D=4…
D=(16+2-10)/2Ð D=4
T=10+8-2 Ð T=16
And there T=20, P=12 and D=5…
D=(20+2-12)/2Ð D=5
T=12+10-2 Ð T=20
Without going any further, I would say this is sufficient evidence to prove that my three formulas work for triangles. It also shows that I have rearranged my first formula correctly – so if one formula works for a certain number of triangles, they all will. Also, I do not have to worry about the value of (T+2-P)/2 being anything other than a whole number (i.e. when T-P gives an odd number, then /2 to give x.5). This is because when T is an even number, P is an even number, so T+P is therefore an even number. When T is an odd number, so is P, and again T+P is an even number, which can be halved to give a whole number.
I will now move on to looking at a different shape, as I have found the formulas for triangles:
P=T+2-2D D=(T+2-P)/2 T=P+2D-2
Squares
I will now move on to finding a formula linking the number of squares (Q), the perimeter and the number of dots enclosed. I will do the same as I did for triangles – draw out shapes with different numbers of squares, record the number of hexagons in the shape, the perimeter in cm and the dots. You cannot draw squares well on isometric dot paper. The nearest you can get is a rhombus or a rectangle – however a rhombus shares all the same features of a square with regards to number of sides and how it tessellates. So instead I used squared paper. Below I have laid out tables of P and D for all the values of Q that I drew out. I have only looked at 3 different values of Q, as I do not think there is any need for more. All different values of T previously followed the same pattern, and I am quite confident this will be the same case with squares, as they are both regular tessellating shapes.
10 Squares (Q=10):
|P= |D= |Q= |
|14 |4 |10 |
|16 |3 |10 |
|18 |2 |10 |
|20 |1 |10 |
|22 |0 |10 |
13 Squares (Q=10):
|P= |D= |T= |
|16 |6 |13 |
|18 |5 |13 |
|20 |4 |13 |
|22 |3 |13 |
|24 |2 |13 |
|26 |1 |13 |
|28 |0 |13 |
16 Squares (Q=16):
|P= |D= |T= |
|16 |9 |16 |
|18 |8 |16 |
|20 |7 |16 |
|22 |6 |16 |
|24 |5 |16 |
|26 |4 |16 |
|28 |3 |16 |
|30 |2 |16 |
|32 |1 |16 |
|34 |0 |16 |
Firstly I will test my previous formulas, P=T+2-2D, D=(T+2-P)/2 and T= P+2D-2, to see if they hold true – of course, substituting T with Q. If the formulas still hold true, I will be able to save lots of time trying to find a formula linking P, D and Q. Even if they don’t, all will not be lost – all the answers may be incorrect by, say, 4. Therefore I could make slight modifications (i.e. +4) to the existing formulas to get a new, working formula for squares.
So where P=14, D=4 and Q=10…
P=10+2-8 Ð P=4 DP is out by –10, or -Q
D=(10+2-14)/2Ð D=-1 DD is out by –5
Q=14+8-2 Ð Q=20 DQ is out by +10, or +Q
And where P=16, D=6 and Q=13…
P=13+2-12 Ð P=3 DP is out by –13, or -Q
D=(13+2-16)/2Ð D=-0.5 DD is out –6.5
Q=16+12-2 Ð Q=26 D Q is out by +13, or +Q
And where P=16, D=9 and Q=16…
P=16+2-18 Ð P=0 D P is out by –16, or -Q
D=(16+2-16)/2Ð D=1 D D is out by –8
Q=16+18-2 Ð Q=32 DQ is out by +16, or +Q
From these trials it is clear that the formulas for triangles doesn’t work properly, but there is some correlation which could help me to find a formula for squares. For example, the formulas always give a value of Q to be twice its real value, and the value of P is always less than its real value by whatever Q is.
So, in theory, all I need to do to get a working formula for Q= is to change the formula from Q=P+2D-2 to Q=(P+2D-2)/2, or more simply Q=P/2+D-1. Now I shall test it, just to make sure it works.
So where P=14, D=4 and Q=10, Q=P/2+D-1 Ð Q=7+4-1 Ð Q=10 C
And where P=16, D=6 and Q=13, Q=P/2+D-1 Ð Q=8+6-1 Ð Q=13 C
And where P=16, D=9 and Q=16, Q=P/2+D-1 Ð Q=8+9-1 Ð Q=16 C
As I expected, this new formula works for finding Q – I presume that if I rearrange this to give P= and D=, these two new formulas will work. Q=P/2+D-1 can be rearranged to give P=2(Q-D+1) and D=Q-P/2+1. Although I am pretty sure that these formulas work – as the formula they are derived from is fine – I will test them just to make sure. There is no reason mathematically why the formulas should not work, unless there is no or irregular correlation between P, D and Q, but testing will iron out any mistakes I may have made.
So where P=14, D=4 and Q=10…
P=2(10-4+1) Ð P=14 C
D=10-14/2+1 Ð D=4 C
And where P=16, D=6 and Q=13…
P=2(13-6+1) Ð P=16 C
D=13-16/2+1 Ð D=6 C
And where P=16, D=9 and Q=16…
P=2(16-9+1) Ð P=16 C
D=16-16/2+1 Ð D=9 C
As expected, these formulas work fine, and I am pretty sure that they will work on all other corresponding values of P, D and Q – there is no reason for them not to. The only possible situation where my formulas might fail me is where the value of D is 0, as when D is multiplied or subtracted in the formulas, a value of 0 will not alter the result without D included. So, assuming that P=4, D=0 and Q=1, my formulas show that:
P=2(1-0+1) Ð P=4 C
D=1-4/2+1 Ð D=0 C
Q=4/2+0-1 Ð Q=1 C
These all work, so this goes to prove that my formula will work when the value of D is 0.
By modifying my existing formula for triangles I have been able to save myself lots of time trying to find a formula from scratch, and can now move on to another tessellating shape, having found the formulas:
P=2(Q-D+1) D=Q-P/2+1 Q=P/2+D-1
Hexagons
Now that I have found working formulas for both triangles and squares, I have decided to move on to hexagons, and record their relating P, D and H (number of hexagons), and then go on to find a formula linking them. I am quite confident that I will find 3 formulas for P=, D= and H=, as a hexagon is a regular tessellating shape, as are squares and equilateral triangles. When I say regular, I mean that the shape in question’s order of rotational symmetry is equal to its number of sides, and all internal/external angles of the shape are the same. Nevertheless, I will draw out tables of the P, D and H of shapes made up of different numbers of Hexagons as I have done previously with triangles and squares.
5 Hexagons (H=5)
|P= |D= |H= |
|14 |4 |5 |
|16 |3 |5 |
|18 |2 |5 |
|20 |1 |5 |
|22 |0 |5 |
6 Hexagons (H=6)
|P= |D= |H= |
|18 |4 |6 |
|20 |3 |6 |
|22 |2 |6 |
|24 |1 |6 |
|26 |0 |6 |
10 Hexagons (H=10)
|P= |D= |H= |
|24 |9 |10 |
|26 |8 |10 |
|28 |7 |10 |
|30 |6 |10 |
|32 |5 |10 |
|34 |4 |10 |
|36 |3 |10 |
|38 |2 |10 |
|40 |1 |10 |
|42 |0 |10 |
As before, I will test out the first line of each table above with the previous formula. I will test out the formulas for triangles and for hexagons, on the grounds that I do not know which formula will need less modification (and therefore less time) to find a working formula linking P, D and H. Firstly I will test out one line from each table with the formula that I found for triangles, P=T+2-2D, D=(T+2-P)/2 and T=P+2D-2. I will substitute T for H in all 3 formulas.
So where P=14, D=4 and H=5…
P=5+2-8 Ð P=-1 DP is out by -15, or P+1, or 3H
D=(5+2-14)/2 Ð D=-3.5 DD is out by -7.5
H=14+8-2 Ð H=20 DH is out by 20, or +3H, or x4H
And where P=18, D=4 and H=6…
P=5+2-8 Ð P=-1 DP is out by -19, or P+1
D=(6+2-18)/2 Ð D=-5 DD is out by -9
H=18+8-2 Ð H=24 DH is out by +18, or +3H, or x4H
And where P=24, D= 9 and H=10…
P=10+2-18 Ð P=-6 DP is out by -30, or 3H
D=(10+2-24)/2Ð D=-6 DD is out by -15
H=24+18-2 Ð H=40 DH is out by +30, or +3H, or x4H
From these trials I can see that there is obviously a pattern with the H=P+2D-2 formula, in which the answer I get is always four times the answer I need. If I were to change the formula to H=(P+2D-2)/4, I have a pretty good idea that this would work. If it did, it could then be rearranged to give P= and D=. I will now test the formula H=(P+2D-2)/4, in exactly the same way as above, but omitting P= and D= to save time.
So where P=14, D=4 and H=5, H=(14+8-2)/4 Ð H=5 C
And where P=18, D=4 and H=6, H=(18+8-2)/4 Ð H=6 C
And where P=24, D=9 and H=10, H=(24+18-2)/4 Ð H=10 C
By simply inserting a /4 onto the end of the previous formula which did not work, I have created a formula which will give you the correct value of H, if you know P and D. H=(P+2D-2)/4 can be rearranged to give P=4H+2-2D and D=2H-P/2+1. Previously, when I rearranged a working formula I then tested out the two new formulas all over again, just to make sure. However, as this proved flawless both for triangles and for squares, I am sure that there is no need – I have checked my workings thoroughly, and I can see no errors. So the formulas linking P, D and H are:
P=4H+2-2D D=2H-P/2+1 H=(P+2D-2)/4
‘Universal’ Formulas
After finding 3 successful formulas for shapes made of hexagons, I cannot think of any other regular shapes to move on to. When I say regular shapes, I am referring to those shapes for which their order of rotational symmetry is equal to their number of sides, and all their sides are the same length. Therefore, I have decided to move on to trying to find a formula that will work for Triangles, Squares (and Rhombuses) and Hexagons. I know that pentagons tessellate, and I think that heptagons do as well, but I cannot be certain without testing first. Unfortunately, as I do not have any dotted paper that will accommodate 5- and 7-sided shapes, this will not be possible - though I would have liked to do try them out.
I have predicted that somewhere in these universal formulas I will have to incorporate the number of sides the shape has.
In both squares and hexagons, the Q= or H= formula was the one with which I found a pattern first (and in the case of hexagons, the only one). In triangles I did not have an existing formula to work with, so this does not apply to them. Assuming I do find a formula in terms of T=, Q= or H=, I will be able to simply rearrange them to give D= and P=. As I am trying to get a ‘universal’ formula that will work for all 4 shapes (rhombuses included), I will later on substitute T, Q or H for X, which will represent all or any of the shapes.
So the formulas I have are:
T=P+2D-2, Q=P/2+D-1 and H=(P+2D-2)/4. I think that to get my universal formula, I need to have all these three in the same ‘format’. If you look at the ones above, the first is straightforward, the second has a ‘divide by’ at the beginning of the formula, and the last one has it at the end, as well as a set of brackets. I will try to get them all looking the same, with the ‘divide by’ signs in the same place, even if they are only /1, as it may help me when trying to find a connection between the three. I have therefore rearranged the three formulas above to give:
T=(P+2D-2)/1, Q=(P+2D-2)/2 and H=(P+2D-2)/4
Now that I have made the formulas look the same (although underneath they still do exactly the same as before), it is obvious that the only difference between them is the amount that they are divided by (i.e. /1, /2 and /4). But why are these ‘divided by’ amounts different? I suspect, as predicted, it has something to do with the amount of sides each shape has. Formulas aside, we are saying that triangles, with 3 sides, are being divided by a factor of 1. 2 are dividing squares, with 4 sides,, and hexagons with 6 sides are being divided by 4. B simplifying this whole division thing out it shows quite clearly that the formula for the shapes are being divided by the amount of sides a shape has, then taking 2 away. Therefore, I should be able to swap the /1, /2 and /4 in the triangle, square and hexagon formulas respectively with (Y-2), to give the formula X=(P+2D-2)/(Y-2).
As this is quite a complex formula in comparison to previous ones I have found, I feel it would be wise to test this formula once for triangles, squares and hexagons, just to make sure.
So where P=14, D=2 and the shape is T (i.e. three-sided)…
X=(14+4-2)/(3-2) Ð X=16/1 Ð X=16 C
And where P=24, D=5 and the shape is S…
X=(24+10-2)/(4-2) Ð X=32/2 Ð X=16 C
And where P=34, D=4 and the shape is H…
X=(34+8-2)/(6-2) Ð X=40/4 Ð X=10 C
This shows that my universal formula works correctly, and also that my predictions about the universal formula needing to take into account the number of sides of a shape. Again, this can be rearranged to give three more (an extra indices, Y, has been added to allow for an extra formula) formulas, which are P=X(Y-2)+2-2D, D=(X(Y-2)+2P)/2 and Y=P+2D-X. So the universal formulas for triangles, squares and hexagons are:
X=(P+2D-2)/(Y-2) P=X(Y-2)+2-2D D=(X(Y-2)+2P)/2 Y=P+2D-X
Further Investigation
Ideally I would like to further my investigation by branching into 3D shapes, and doing what I have been doing with 2D shapes all over again. However I feel this would be very time-consuming, and it would be much better to take 2D shapes as far as I can.
For this reason I am not going to attempt to see if I can find a formula whereby you can find out the maximum perimeter/number of dots enclosed of any shape, so long as you know how many of what shapes it is composed of. As this is not mentioned at all in the investigation booklet, I am not certain that I will be able to get a formula, but at least I can say I have tried. I will first write out the maximum perimeter for shapes made of 10 triangles, 10 squares and 10 hexagons below. Although the number of shapes is a constant value, I am putting this into the table as I am sure that the number of shapes will have to be taken into account in the formula. The number of sides of a shape may well be incorporated as well, so I have also put these into the table.
|Shape composed of:  |Max. Perimeter  |No. of sides  |
|10 Triangles |12 |3 |
|10 Squares |22 |4 |
|10 Hexagons |42 |6 |
One thing that catches my eye straight away is the maximum perimeter. For all three shapes, the values end in 2. This is -2 exists in the (P+2D-2) part of my universal formula. Once you have removed the extra 2 form each shape, you are left with a number which, when divided by the number of shapes (in this case 10) will give you a number to which you add 2 to get the number of sides of that shape. With this in mind, you can see that when you subtract 2 from the number of sides, you get a number which, when multiplied by the number of shapes, gives a figure to which you add 2 to give the maximum perimeter. In formulaic terms, this equates to the Maximum P=(Y-2)X+2, which is correct if you apply it to the table above.
I assume that this will continue to work with different numbers of shapes, so I won’t test it. Is it simply a derivation of the universal P= formula, but without the -2D part.
Conclusion
In the time available to me, I believe that I have researched the links between the P, D and X of a shape to the full extent of my ability. I found a formula which link the P, D and T or shapes made of triangles. I then extended this to squares and hexagons, and was able to construct my universal formula, which will tell you the P, D and X of any shape constructed of triangles, squares or hexagons. I found that the amount of sides of a shape must be taken into account in the formula – something not necessary when deriving a formula to work with just one shape, as this value is constant. I also found that many of my predictions I made along the way turned out to be correct.
I am certain that the reason my universal formula ends (Y-2) and not just Y is because when taking the number of sides of a shape into account, you only want to know the ones that are not touching any others. However, if all your triangles, squares etc. are touching each other in a line (the arrangement with the largest P and smallest D), two sides of any shape are always touching the edges of another shape unless they are at either end of a linear shape, and I am pretty sure this is there it -2 in T+2D-2 comes in. If the shape is not linear, but a cross, say, it will have four edges that do not touch any others. However, the two extra branches (that turn a ‘line’ shape into a cross shape) come to the centre where they join with the sides of shapes that would otherwise be free. So when the two extra branches seem to be making two new edges that do not share with any other shapes, the balance in the equation as far as the –2 is concerned is not disrupted in the slightest. This is because it merely ‘takes’ free edges form somewhere else (i.e. wherever it joins the linear part of the shape).
I also successfully found a formula that will give you the maximum perimeter of a shape, so long as you know what shape it is and how many there are.
Evaluation
I would say that this investigation has been a success. I managed to find a link between P, D and T, which progressed into a link between P, D and Q or H, then on to my universal formulas. After that I went a stage further and developed a formula that would tell you the maximum possible perimeter of a shape.
Unfortunately, my investigation was hindered by 2 things – foremost was my lack of time to carry out the investigation as far as possible (i.e. researching 3D shapes, and irregular tessellating shapes, such as ‘L’ and ‘T’ shapes), and second I did not have any dotted paper capable of drawing regular pentagons or heptagons, as I could have looked at these along with my triangles, squares and hexagons.
If I were to redo this investigation, I would make sure that I set aside enough time to do a proper job of it, and looking at my own made-up shapes to see if my formulas still applied. I also would have very much liked to have been able to move on to looking at 3D shapes, but doing this means wither physically making paper or card squares / pyramids etc, which is vastly time-consuming, or using multi-cubes, of which I have none.
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