Pythagoras’ Theorem:
C
A B
Pythagoras’ Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides
i.e. BC2 = AB2 + AC2
(Refers to the diagram above)
In a triangle such as:
5
3
4
32 + 42 = 52
The sum of the lengths of the two short sides equals the length of the hypotenuse. If the three numbers are all whole numbers, this is called a Pythagoras Triple, a topic I am going to investigate in this project.
1. (smallest number)2 + (middle number)2 = (largest number)2
(a)
52 + 122 = 132
25 + 144 = 169
(b)
72 + 242 = 252
49 + 576 = 625
Both sets of numbers above satisfy the condition.
2.
(a)
Area = 12 x 5 = 30 square units
2
Perimeter = 5 + 12 + 13 = 30 units
(b)
Area = 24 x 7 = 84 square units
2
Perimeter = 7 + 24 + 25 = 56 units
Table of results
Calculations to find out the next results in the sequence:
I discovered that for the shortest side the difference was always +2 (3,5,7,9….). For the middle and longest side the difference was always the same, and followed the pattern of the 4 times tables (e.g. 8,12,16,20…).
Finding the nth term for each column in Table 1:
Column A
Trial and error was my method to find the nth term.
When n=1
2n = 2 x (1) = 2
2n + 1 = 2 x (1) + 1
2 + 1 = 3
When n=2
2n + 1= 2(2) +1
= 4 + 1
= 5
This formula works, therefore for this column:
2n + 1 is the correct formula
For column B, the second differences were the same, so I knew it was a squared formula.
Column B
I then used trial and error
When n=1
2(12) + 2(1)
2 + 2 = 4
When n=2
2(32) + 2(3)
18 + 6 = 24
This formula works, therefore for this column:
2n2 + 2n is the correct formula
I noticed that the numbers in column B were one more than in Column C. Therefore the formula would be the same as for column B, just adding 1.
2n2 + 2n + 1 is the correct formula
For the perimeter, I knew that:
P = S + M + L
Perimeter = Short Side + Middle Side + Long Side
So, this would mean adding the nth term for all three
P = S + M + L
P = (2n+1) + (2n2+2n) + (2n2+2n+1)
P = 4n2 + 6n + 2
P = 2 (2n2 + 3n + 1)
For the area, I knew that:
Area = ½ base x height
= ½ mid length x short length
=ms/2
Algebraically this equals the following:
= (2n2 + 2) (2n + 1)
2
= n (n+1) (2n+1)
To prove that these are correct, I did the following:
S2 = (2n+1) = (2n+1) (2n+1)
= 4n2 + 2n + 1 + 2n
= 4n2 + 4n + 1
M2 = (2n2 + 2n) (2n2 + 2n)
= 4n4 + (4n3 + 4n3) + 4n2
= 4n4 + 8n3 + 4n2
L2 = (2n2 + 2n + 1) (2n2 + 2n + 1)
= 4n2 + 4n + 1 + 4n2 + 8n3 + 4n2
= 4n4 + 8n3 + 8n2 + 4n + 1
Shows that S2 + M2 = L2 is correct.
Enlarging Triangles
Using column S, where all numbers are even.
Table 2 (Multiply by 2 from Table 1)
Scale factor = 2
Multiply by 3 from Table 1
Table 3
Scale factor = 3
Multiply by 4 from Table 1
Table 4
Scale factor = 4
Having noticed this tend in enlarging the triangles, I thought there may be a way to create a formula, which would apply to ALL enlargements, however high or low the scale factor.
So if I used the symbol E to stand for ‘Enlargements’ and then placed this in front of all the possible nth terms calculated, it would mean that they could be enlarged by a scale factor of ‘E’.
S = E (2n+1) M = (2n2+2n) L = E (2n2+2n+1)
For the area, I knew that the units are UNITS SQUARED, so….
If the ratio of lengths is 1:E
1:E2
Then Ratio of areas is 1:E2
Therefore, this means that for the area column:
A = E2 n (n+1) (2n+1)
Perimeter is a length so he ratio of perimeter is 1:E
P = E (4n2 + 6n + 2)
Conclusion:
In this project I have investigated several possible Pythagorean triples, and by trial and error found formulas to express this.
I have found that for any whole number ‘n’ a Pythagorean triple is given by:
S = 2n + 1
M = 2n2 + 2n
L = 2n2 + 2n + 1
Further triples can be found by enlarging these by ‘E’.
S = E (2n + 1)
M = E (2n2 + 2n)
L = E (2n2 + 2n + 1)
