# Shapes Investigation

#### Essay Preview

More ↓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.

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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 Ã 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 Ã 12-2+10=20 C

P=14 and D=4 Ã 14-2+8=20 C

P=16 and D=3 Ã 16-2+6=20 C

P=18 and D=2 Ã 18-2+4=20 C

P=20 and D=1 Ã 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=10â€¦

P=8 and D=2 Ã 8-2+4=10 C

P=10 and D=1 Ã 10-2+2=10 C

P=12 and D=0 Ã 12-2+0=10 C

Where T=15â€¦

P=11 and D=3 Ã 11-2+6=15 C

P=13 and D=2 Ã 13-2+3=15 C

P=15 and D=1 Ã 15-2+2=15 C

P=17 and D=0 Ã 17-2+0=15 C

And where T=16â€¦

P=10 and D=4 Ã 10-2+8=16 C

P=12 and D=3 Ã 12-2+6=16 C

P=14 and D=2 Ã 14-2+4=16 C

P=16 and D=1 Ã 16-2+2=16 C

P=18 and D=0 Ã 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 Ã D=2

T=8+4-2 Ã T=10

And where T=15, P=11 and D=3â€¦

D=(15+2-11)/2Ã D=3

T=11+6-2 Ã T=15

And where T=16, P=10 and D=4â€¦

D=(16+2-10)/2Ã D=4

T=10+8-2 Ã T=16

And there T=20, P=12 and D=5â€¦

D=(20+2-12)/2Ã D=5

T=12+10-2 Ã 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=10â€¦

P=10+2-8 Ã P=4 DP is out by -10, or -Q

D=(10+2-14)/2Ã D=-1 DD is out by -5

Q=14+8-2 Ã Q=20 DQ is out by +10, or +Q

And where P=16, D=6 and Q=13â€¦

P=13+2-12 Ã P=3 DP is out by -13, or -Q

D=(13+2-16)/2Ã D=-0.5 DD is out -6.5

Q=16+12-2 Ã Q=26 D Q is out by +13, or +Q

And where P=16, D=9 and Q=16â€¦

P=16+2-18 Ã P=0 D P is out by -16, or -Q

D=(16+2-16)/2Ã D=1 D D is out by -8

Q=16+18-2 Ã 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 Ã Q=7+4-1 Ã Q=10 C

And where P=16, D=6 and Q=13, Q=P/2+D-1 Ã Q=8+6-1 Ã Q=13 C

And where P=16, D=9 and Q=16, Q=P/2+D-1 Ã Q=8+9-1 Ã 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=10â€¦

P=2(10-4+1) Ã P=14 C

D=10-14/2+1 Ã D=4 C

And where P=16, D=6 and Q=13â€¦

P=2(13-6+1) Ã P=16 C

D=13-16/2+1 Ã D=6 C

And where P=16, D=9 and Q=16â€¦

P=2(16-9+1) Ã P=16 C

D=16-16/2+1 Ã 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) Ã P=4 C

D=1-4/2+1 Ã D=0 C

Q=4/2+0-1 Ã 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=5â€¦

P=5+2-8 Ã P=-1 DP is out by -15, or P+1, or 3H

D=(5+2-14)/2 Ã D=-3.5 DD is out by -7.5

H=14+8-2 Ã H=20 DH is out by 20, or +3H, or x4H

And where P=18, D=4 and H=6â€¦

P=5+2-8 Ã P=-1 DP is out by -19, or P+1

D=(6+2-18)/2 Ã D=-5 DD is out by -9

H=18+8-2 Ã H=24 DH is out by +18, or +3H, or x4H

And where P=24, D= 9 and H=10â€¦

P=10+2-18 Ã P=-6 DP is out by -30, or 3H

D=(10+2-24)/2Ã D=-6 DD is out by -15

H=24+18-2 Ã 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 Ã H=5 C

And where P=18, D=4 and H=6, H=(18+8-2)/4 Ã H=6 C

And where P=24, D=9 and H=10, H=(24+18-2)/4 Ã 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)â€¦

X=(14+4-2)/(3-2) Ã X=16/1 Ã X=16 C

And where P=24, D=5 and the shape is Sâ€¦

X=(24+10-2)/(4-2) Ã X=32/2 Ã X=16 C

And where P=34, D=4 and the shape is Hâ€¦

X=(34+8-2)/(6-2) Ã X=40/4 Ã 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: Ã‚

Max. Perimeter Ã‚

No. of sides Ã‚

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.