So the rule for the Middle Side is……. 2n2 + 2n
To check if this formula is correct I will apply it to the nth term. I should end up with the length of the middle side.
Nth Term 2n2 2n 2n2 + 2n Middle Side
The table proves that 2n2 + 2n is the rule for middle side.
I do not see any reason to work out the rule for the Hypotenuse this way. This is because it can clearly be seen that the length of the hypotenuse for any of these triangles is just the middle side plus 1.
So the rule for the Hypotenuse is……. 2n2 + 2n + 1
Shortest Side
2n + 1
With these equations I can work out the length of any
Middle Side side on a Pythagorean triangle just by knowing what
2n2 + 2n number the triangle is in the sequence.
Hypotenuse
2n2 + 2n + 1
Now that I have these three rules I can calculate the lengths of the sides of the next triangles in the sequence.
Triangle 4
Shortest side = ( 2 x 4 ) + 1 = 9
Middle side = 2( 4 x 4 ) + 2 x 4 = 32 + 8 = 40
Hypotenuse = 2( 4 x 4 ) + ( 2 x 4 ) + 1 = 41
Perimeter = 9 + 40 + 41 = 90
Area = ½( 9 x 40 ) = 180
Triangle 5
Shortest side = ( 2 x 5 ) + 1 = 11
Middle side = 2( 5 x 5 ) + ( 2 x 5 ) = 50 + 10 = 60
Hypotenuse = 2( 5 x 5 ) + ( 2 x 5 ) + 1 = 61
Perimeter = 11 + 60 + 61 = 132
Area = ½ ( 11 x 60 ) = 330
Triangle 6
Shortest side = ( 2 x 6 ) + 1 = 13
Middle side = 2( 6 x 6 ) + ( 2 x 6 ) = 72 + 12 = 84
Hypotenuse = 2( 6 x 6 ) + ( 2 x 6 ) + 1 = 85
Perimeter = 13 + 84 + 85 = 182
Area = ½ ( 13 x 84 ) = 546
Triangle 7
Shortest side = ( 2 x 7 ) + 1 = 15
Middle side = 2( 7 x 7 ) + ( 2 x 7 ) = 98 + 14 = 112
Hypotenuse = 2( 7 x 7 ) + ( 2 x 7 ) + 1 = 113
Perimeter = 15 + 112 + 113 = 240
Area = ½ ( 15 x 112 ) = 840
Now I will collect this new information into a table.
Now that I have the rules for the sides of the triangle I will work out the rules for the perimeter and area using them.
Perimeter
Perimeter = a + b + c
So if I let a, b, and c be the rules for each of the sides it should give me the rule for the perimeter. This way I will be able to work out the perimeter of any triangle just by using the nth term of the triangle.
Shortest Side + Middle Side + Hypotenuse = Perimeter
a + b + c = Perimeter
( 2n + 1 ) + ( 2n2 + 2n ) + ( 2n2 + 2n + 1 ) = Perimeter
2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter
If I collect all the like terms I will end up with the rule for the perimeter.
2n2 + 2n2 + 2n + 2n + 2n + 1 + 1 = 4n2 + 6n + 2
4n2 + 6n + 2 = Perimeter
To see if this is the correct formula I will test it with the nth term. If it is correct I should get the perimeter of the triangles.
Nth Term 4n 6n 2 4n2 + 6n + 2 Perimeter
This proves that this is the correct formula for working out the perimeter for any triangle in the sequence.
The next thing I am going to do is to work out the rule of the area for these triangles.
Area
½ ( a x b )
Above you can see the equation that is used to work out the area of any triangle. The 2 values needed to do this are the lengths of the shortest and middle sides.
In a similar way to the method I used to work out the perimeter, I can substitute the rule for the shortest and middle side into the equation. This should give me the rule to work out the area of any triangle in the sequence.
The formula : ½ ( a x b ) becomes……. ½ x ( 2n + 1 ) x ( 2n2 + 2n )
I can simplify this formula to get the rule for the area.
When I multiply out the brackets: ( 2n + 1 ) ( 2n2 + 2n )
They become: 4n3 + 4n2 + 2n2 + 2n
: 4n3 + 6n2 + 2n
I still have to multiply it by ½ so….
½ ( 4n3 + 6n2 + 2n ) = 2n3 + 3n2 + n
Area = 2n3 + 3n2 + n
I am quite certain that this is the rule for working out area. Just to make it is correct I will test it with the nth term, if it is correct I should end up with the area of the triangles.
Nth Term 2n 3n n 2n3 + 3n2 + n Area
The table proves that this is the correct formula and it will work for any triangle in the sequence. You will notice that I did not use the sequences to work out the rules for the perimeter and area, like I did for the sides. I thought that this was a more useful way of working out the rules and a more efficient way because it took less time. I would have ended up with the same answer either way. I feel that this method also gives a better representation of how the sides, perimeter and area of pythagorean triangles are linked
Now I will begin to investigate further into the relationships between the sides, perimeters and areas of pythagorean triangles.
There are different methods to find a missing length on a Pythagorean triangle. Without having to use the rules which I have worked out. Some methods are easier than others; however they all lead to the same answers.
Pythagoras’ theorem a2 + b2 = c2
With this method 2 of the sides must already be known.
There are three formulas. Each one works out one side of a right angled triangle.
a2 + b2 = c2 ( Hypotenuse )
c2 – a2 = b2 ( Middle Side )
c2 – b2 = a2 ( Shortest Side )
Below is an example of how these work.
With this triangle the shortest side is
Missing. So if I wanted to find the
length of this side I would use the formula for the
shortest side.
c2 – b2 = a2
When I substitute the 2 sides into this formula I can work out the shortest side.
132 – 122 = a2
169 – 144 = a2
25 = a2
√25 = a
a = 5
This method would work very well with any right angled triangle.
Another way to work out missing sides would be to use trigonometry. For this to work you still need to know 2 sides of the triangle. This can be used as an alternative to the Pythagoras method, but I don’t recommend it because it is more complicated than Pythagoras, but still gets you the right answer.
Sin = O/H Cos = A/H Tan = O/A
These are the trig ratios used in trigonometry to work out the length of a side I need an angle and 1 side. To work out the angle I need 2 sides.
Below you can see how I would use trigonometry to find a missing side.
If I wanted to find the missing side in the above triangle. I would first find the angle marked as X on the triangle. The two sides involved are the Adjacent and Hypotenuse sides so I would use the Cosine ratio.
Cos = A/H
Cos = 60/61
Cos = 0.984
Cos-1 0.984 = 10.3 o
Now I can use one of two ways to work out the missing side opposite to the angle.
1.) Sin 10.3 = O/H 2.) Tan 10.3 = O/A
Sin 10.3 = O/61 Tan 10.3 = O/60
Sin 10.3 x 61 = O Tan 10.3 x 60 = O
10.9 = O 10.9 = O
11 = O 11 = O
The answer should actually be 11. I got 10.9 because I rounded the angle off to 1 decimal place. The length of the sides should be rounded off to the neares unit each time to get the correct answer.
I have also noticed that the angle shown as
X on the triangle becomes smaller and smaller
as the sequence of triangles progresses.
Although the angle will continue to decrease in size with each new triangle. It will never fall below the value of zero. f the angles were plotted on a graph it would produce a curve such as Y = 1/X.
The ends of such a curve never touch the
X or Y axis although they move closer and
closer to them all the time. There is no end
to these curves the go on forever. If I could work out the gradient of the angles in pythagorean triangles, I would be able to calculate the angle in a triangle without the need for 2 sides. I would only need the nth term of the triangle and 1 side. The nth term to work out the angle and the side to work out any of the other two missing sides.
However, I am not going to work out the gradient of the curve that would be made by the angles. This is because this is just an alternative method to using the rules for the sides and Pythagoras. Using the rules and Pythagoras would be a much simpler way of working out missing sides.
At the beginning of my coursework I began initially by looking at odd numbers as the shortest sides of the triangles. Now I will look at triangles where the shortest sides are even numbers. For these to by pythagorean triples they must also satisfy the condition…..
a2 + b2 = c2
Below is the table of the triangles for which the shortest side was an odd number.
I can see immediately from the table that to get the shortest side as an even number I will need to change the terms of the sequence.
If……. 0 1 2 3 4 5
Give…….. 1 3 5 7 9 11
Then……. 0.5 1.5 2.5 3.5 4.5
Must give…… 2 4 6 8 10
I can see that if I go in between the current terms I will get even numbers.
If I use the nth terms :: 0.5, 1.5, 2.5, 3.5, 4.5,5.5.
Instead of the nth terms :: 1, 2, 3, 4, 5, 6
I can apply the new terms to the rules, which I used to for the triangle that had the shortest side as an odd number. This will give me the properties of the triangles that have the shortest side as an even number.
I do not actually need the rules to work out the hypotenuse, perimeter and area of these triangles. I can work out everything with just the shortest and middle sides. Because these triangles are pythagorean triples the rules I worked out at the beginning of my coursework still apply to these triangles.
Triangle 1 = 0.5
Shortest side = ( 2 x 0.5 ) + 1 = 2
Middle Side = 2( 0.5 x 0.5 ) + ( 2 x 0.5 ) = 1.5
Hypotenuse = 1.5 + 1 = 2.5
Perimeter = 2 + 1.5 + 2.5 = 6
Area = ½ ( 2 x 1.5 ) = 1.25
Triangle 2 = 1.5
Shortest side = ( 2 x 1.5 ) + 1 = 4
Middle Side = 2( 1.5 x 1.5 ) + ( 2 x 1.5 ) = 7.5
Hypotenuse = 7.5 + 1 = 8.5
Perimeter = 4 + 7.5 + 8.5 = 20
Area = ½ ( 4 x 7.5 )
Triangle 3 = 2.5
Shortest side = ( 2 x 2.5 ) + 1 = 6
Middle Side = 2( 2.5 x 2.5 ) + ( 2 x 2.5 ) = 17.5
Hypotenuse = 17.5 + 1 = 18.5
Perimeter = 6 + 17.5 + 18.5 = 42
Area = ½ ( 6 x 17.5 ) = 52.5
Triangle 4 = 3.5
Shortest side = ( 2 x 3.5 ) + 1 = 8
Middle Side = 2( 3.5 x 3.5 ) + ( 2 x 3.5 ) = 31.5
Hypotenuse = 31.5 + 1 = 32.5
Perimeter = 8 + 31.5 + 32.5 = 72
Area = ½ ( 8 x 31.5 ) = 126
Triangle 5 = 4.5
Shortest side = ( 2 x 4.5 ) + 1 = 10
Middle Side = 2( 4.5 x 4.5 ) + ( 2 x 4.5 ) = 49.5
Hypotenuse = 49.5 + 1 = 50.5
Perimeter = 10 + 49.5 + 50.5 = 110
Area = ½ ( 10 x 49.5 ) = 247.5
I will now collect all of this information into a graph to make it easier to see.
You will notice that I have changed the nth terms back into whole numbers and they now begin with 1 and not 0.5.
I expect to find now that the equations to find the properties have also changed. This is because they are a completely new set of numbers.
However, because I used the formulas which I found at the beginning of the coursework to get these numbers. The rules would not have changed that much. All I have changed in this table which effects the equations is the nth term.
So to continue using the existing nth term. I would just have to replace n in each equation by ( n – 0.5 ).
So the rules for triangles with the shortest side as an even number would be …..
Shortest Side = 2( n – 0.5 ) + 1
Middle Side = 2( n – 0.5 )2 + 2( n – 0.5 )
Hypotenuse = 2( n – 0.5 )2 + 2( n – 0.5 ) + 1
Perimeter = 4( n- 0.5 )2 + 6( n – 0.5 ) + 2
Area = 2( n – 0.5 )3 + 3( n – 0.5 )2 + ( n-0.5 )
Suppose I did not know where in the sequence a triangle is but I know at least 1 property of the triangle. For example the shortest side and with this knowledge I need to know the area of the triangle.
This problem gives birth to a whole new range of formulas and rules which link the properties of the triangles. I will now attempt to decipher all of these formulas.
To begin with I will work out how to calculate each property of a pythagorean triangle with just the knowledge of what the shortest side is.
The fact that I already know how to get each property from just the nth term will be extremely useful. This means that all I need to do is reverse the formula that is used to get the shortest side from the nth term. This will give me the location of the triangle in the sequence. Then I just apply it to each of the other formulae and I will be able to work out every other property from just the shortest side.
First I must reverse the formula for the shortest side.
2n + 1 will become…… ( n – 1 )
2
This should give me the nth term from any length on the shortest side.
This means I can apply this to the other formulas.
Formula to give middle side from the shortest.
Formula to give the hypotenuse from the shortest side.
Formula to give the perimeter from the shortest side.
Formula to give the area from the shortest side.
The equations have turned out quite well and they will work for any triangle in the sequence.
For the next step I intended to work out the formula to get the nth term from the length of the middle side. I experienced great problems trying to get this formula and my attempts have been unsuccessful. I do not think that it is possible with the level of mathematics I am currently capable of applying.
I think that the level of mathematics required is too advanced and I do not yet have the ability to handle such tasks. However because I was able to do it for the shortest side there must be a way of doing it for the rest of the sides.
I will try to explain in more detail what I was trying to do.
2n2 + 2n = middle side
This is the equation which gives the length of the middle side from the nth term.
I needed to work out the equation which gives the nth term from the length of the middle side.
Once I had this equation I could just have substituted it into the rules for the sides, perimeter and area, like I did with the shortest side.
For example……
4n2 + 6n + 2
In this formula, n would have become the equation which gives n from the middle side. Then this formula would have given me the perimeter with only the knowledge of the middle side of the triangle.
I will not try to go any further with this because the equations for the perimeter and area are even more complex than the equation for the middle side.
Conclusion
In my coursework I have found out that there is a strong link between pythagorean triples. The sides, perimeter and area of the triangles which are made from these pythagorean triples showed a distinct pattern. Because of this I was able to find a rule for each property of the right-angled triangles. These rules allowed me to be able to find the properties of any pythagorean triple in the sequence. I could easily find many pythagorean triples beyond those give at the beginning of the coursework. I have also found some formulas that link the sides, perimeter and area of the triangles together. If I was not limited by my abilities I might have found all the equations and this would have allowed me to work out all the properties of the pythagorean triangle with either one of the sides, the perimeter or the area. I was unable to reverse the rule for the middle side. I needed to do this to find the links between the middle side and the other sides, perimeter and area. Either I do not have the knowledge to do this or it is impossible. I was able to do it for the shortest side so there must be a way for the middle side and I don’t know how to do it..
If I did this piece of coursework again I would definitely spend a greater amount of time trying to figure out the formulas that I have given up on. By doing this I might have found them and I would have been able to make more links between the sides, perimeter and the area. I would also have gone into more depth with trigonometry and worked out the gradient of the curve created by the angles in the triangles.
