Triangle Number
1 = 2 x 2
2 = 3 x 4
3 = 4 x 6
4 = 5 x 8 = 2n2 + 2n
5 = 6 x 10
6 = 7 x 12
7 = 8 x 14
8 = 9 x 16
Longest side = 2n2 + 2n + 1
Perimeter = A + B + C
Area = A x B x 0.5
Box Methods
Shortest² = (2n + 1) ²
= 4n² + 4n + 1
Middle² = (2n² + 2n)²
= 4n4 + 8n³ + 4n²
Longest² = (2n² + 2n + 1)²
= 4n4 + 8n³ + 8n² +4n + 1
A + B = C
4n² + 4n + 1 + 4n4 + 8n³ + 4n² = 4n4+ 8n³ + 8n² +4n + 1
My results show that shortest² + middle² = longest². I have Pythagoras’ theorem.
For this table, I am using the term a, b, b + 2
Formulas
Shortest side = 2n + 4. This is because:
2 x 1 + 4 = 6
2 x 2 + 4 = 8
2 x 3 + 4 = 10 etc
Middle side = n² + 4n + 3
Longest side = n² + 4n + 5
Box Methods
Shortest² = (2n + 4)²
= 4n² + 16n + 16
Middle² = (n² + 4n + 3)²
= n4 + 8n³ + 22n² + 24n + 9
Longest² = (n² + 4n + 5)²
= n4 + 8n³ + 26n² + 40n + 25
A + B = C
4n² + 16n + 16 + n4 + 8n³ + 22n² + 24n + 9 = n4 + 8n³ + 26n² + 40n + 25
Once again, I have proved Pythagoras’ theorem.
For this table, I will be using the term a, b, b + 4 which I chosen to research myself.
Formulas
Shortest side = 4n + 8
Middle side = 2n² + 8n + 6
Longest side = 2n² + 8n + 10
Box Methods
Shortest² = (4n + 8)²
= 16n² + 64n + 64
Middle² = (2n² + 8n + 6)²
= 4n4 + 32n³ + 88n² + 96n + 36
Longest² = (2n² + 8n + 10)²
= 4n4 + 32n³ + 104n² + 160n + 100
A + B = C
16n² + 64n + 64 + 4n4 + 32n³ + 88n² + 96n + 36 = 4n4 + 32n³ + 104n² + 160n + 100
All of my research has proved that Pythagoras’ theorem is correct, even if you come at the theorem at different angles; you’re sure to find out that Pythagoras is correct.
