Area:
I want to find a formula to allow me to find the area for any odd Pythagorean triple. I know that the area is: (height x base)
2
So, I know the formula for height being the middle side and the base being the shortest side so I can assume that if I substitute the formulas into this equation then I will find the formula for area.
(height x base)
2
(2n² + 2n) (2n+1)
2
Multiply this out to get 4n³ + 6n² + 2n
2
Then divide this by two to get 2n³ + 3n²+ n (this is the formula I will be using to check it is correct in my tables and formulae).
I will further prove this formula by using sequences.
3,4,5 5,12,13 7,24,25 9,40,41 11,60,61
6 30 84 180 330
+24 +54 +96 +150
+30 +42 +54
+12 +12
Here not even the second constants are the same, therefore meaning that the end number, 12, is divided by 1 X 2 X 3 (6) because of it being the third difference = 2. Below is the rule for every difference from one to 6 so you can tell what you divide it by to get that part of the sequence:
1st difference = divide by 1
2nd difference = divide by 1 x 2
3rd difference = divide by 1 x 2 x 3 (which is the one we will be using)
4th difference = divide by 1 x 2 x 3 x 4
5th difference = presumably the formula for working out differences keeps going on depending on what the difference of the number is.
I.e. 6th difference = divide by 1 x 2 x 3 x 4 x 5 x 6 etc.
Also the third constant means that the nth term contains n³.
Therefore the first part of the formula = 2n³. We shall then substitute this into the next part of the formula table.
Here I will substitute 2n³ into my table to find the rest of the formula. The difference is the amount between perimeter and 2n³:
Term number: 1 2 3 4 5
Area: 6 30 84 180 330
2n³: 2 16 54 128 250
Difference: +4 +14 +30 +52 +80
+10 +16 +22 +28
+6 +6 +6
Second difference means that the six is divided by 1 x 2 equalling three. Then the formula for the nth term contains n², which means the second part of the formula is 3n².
Now I will find the last part of our equation, which I will again put into the form of a table.
Term number: 1 2 3 4 5
Difference: +4 +14 +30 +52 +80
+10 +16 +22 +28
+6 +6 +6
Value of 3n²: 3 12 27 48 75
I have found a similarity between these two parts (difference and 3n²). I will make this easier by moving the constants out of the way.
Term number: 1 2 3 4 5
Difference: +4 +14 +30 +52 +80
+1 +2 +3 +4 +5
Value of 3n²: 3 12 27 48 75
In this table it is clear to see that you add the term number every time so therefore the last part of the formula is +n, this makes up all of the formula to give evidence that my formula was correct in my first statement and is 2n³ + 3n²+ n.
Here you can notice a similar pattern between the area formula and the perimeter formula. First I factorised the area formula and I got:
2 (4n² + 6n + 2) which means, 2 x area = perimeter, as you can see easier
N N
in the table I am going to draw below.
This proves in my formulae are similar to each other. Therefore another statement I have made is correct.
Proving:
To prove my formulas for ‘a’, ‘b’ and ‘c’ are correct. I decided to incorporate my formulas into a² + b² = c²: -
* a² + b² = c²
* (2n + 1)² + (2n² + 2n)²
= (2n² + 2n + 1)²
* (2n + 1) (2n + 1) + (2n² + 2n) (2n² + 2n)
= (2n² + 2n + 1) (2n² + 2n + 1)
* 4n² + 2n + 2n + 1 + 4n4 + 4n² + 4n³ + 4n³
= 4n4 + 8n³ + 8n² + 4n + 1
* 4n² + 4n + 1 + 4n4 + 8n³ + 4n²
= 4n4 + 8n³ + 8n² + 4n + 1
* 4n4 + 8n³ + 8n² + 4n + 1
= 4n4 + 8n³ + 8n² + 4n + 1
This proves that my ‘a’, ‘b’ and ‘c’ formulas are correct
Trigonometry:
Using trigonometry on the first five triples I will find the angles for each and draw a diagram for each:
c = cos –1 = 4
5
= 36.86
= 36.9˚ (1 dp)
b = 180 – 90 – 36.9 = 53.1˚
a = 90˚
c = cos –1 = 12
13
= 22.61
= 22.6˚ (1 dp)
b = 180 – 90 – 22.6 = 67.4˚
a = 90˚
c = cos –1 = 24
25
= 16.26
= 16.3˚ (1 dp)
b = 180 – 90 – 16.3 = 73.7˚
a = 90˚
c = cos –1 = 40
41
= 12.68
= 12.7˚ (1 dp)
b = 180 – 90 – 12.7 = 77.3˚
a = 90˚
c = cos –1 = 60
61
= 10.38
= 10.4˚ (1 dp)
b = 180 – 90 – 10.4 = 79.6˚
a = 90˚
After I looked at the graph there were no sequences or patterns apart from they all add up to 180˚ and b + c add up to 90˚. Also angle b increases when the sides are longer while c angle decreases when the sides are longer.
Conclusion:
Throughout my whole investigation I have only used odd triples and numbers, but there are also even numbers in triples and Pythagoras’ work. I would expect the same result and formulae as those of odd numbers, and maybe in the future I can investigate this.
I am also amazed at the formulae we had to find and how well it fitted together with our table and substitutions. We found the formulae for finding each side of a right-angled triangle that conforms to Pythagoras’ theorem.
I also came to a conclusion that the formula for area and the formula for perimeter are extremely similar. For some extra work I looked at trigonometry and angles and found out that angles didn’t have any correlation apart from having 180˚ between them.