Pythagoras Theorem is the formula for working out the hypotenuse of a right-angled triangle when the other two sides of the triangle are given. The formula for Pythagoras Theorem is:
a²+b²=c².
In this formula, a is the shortest side, b is the middle side, and c is the shortest side.
Pythagorean Triples are when the three sides of the right-angled triangle are all whole numbers. The first five Pythagorean Triples are:
3, 4, 5-because 9(3²) +16(4²) =25(5²)
5, 12, 13-because 25(5²) +144(12²) =169(13²)
7, 24, 25-because 49(7²) +576(24²) =625(25²)
9, 40, 41-because 81(9²) +1600(40²) =1681(41²)
1, 60, 61-because 121(11²) +3600(60²) =3721(61²)
As you may have noticed, the last two numbers (b and c) have a difference of 1. You may have also noticed that the first and last numbers (a and c) are odd numbers and the middle numbers (b) are even.
This investigation is about finding a formula for various different purposes including finding the nth Pythagorean Triple, the perimeter of a Pythagorean Triple and the area of a Pythagorean Triple.
Firstly, I will put the values of the first five Pythagorean Triples into table:
Term number (n)
Short side (a)
Middle side (b)
Long side (c)
3
4
5
2
5
2
3
3
7
24
25
4
9
40
41
5
1
60
61
Just by looking at the table, I can work out the formula for finding the short side of a Pythagorean Triple in terms of n. Just in case, I worked it out:
0 1 2 3 4 5
3 5 7 9 11
2 2 2 2 2
I wrote down the first five terms of the sequence, I worked out the difference between the terms, and then, using the difference, I worked out the zeroth term of the formula.
I found that this sequence was a linear sequence as it only had one difference.
The general formula for a linear sequence is bx+c.
To work out b, you have to find the second difference of the sequence.
In this case, it is 2 so b=2.
To work out c, you have to find the zeroth term.
In this case it is 1 so c=1
So the formula for finding the short side of a Pythagorean Triple in terms of n is:
2n+1
Term number (n)
Short side (a)
Middle side (b)
Long side (c)
3
4
5
2
5
2
3
3
7
24
25
4
9
40
41
5
1
60
61
Now, I had to work out the formula for finding the middle side of a Pythagorean Triple in terms of n:
0 1 2 3 4 5
0 4 12 24 40 60
4 8 12 16 20
4 4 4 4
As you can see, this sequence is different to the previous sequence. It has a second difference, so it is termed as a quadratic sequence.
The general formula for a quadratic sequence is ax²+bx+c.
To work out a, you have to divide the second difference by 2.
In this case, the second difference is 4, and 4÷2=2, so a=2.
You cannot work out b before you have worked out c.
To work out c, you have to find the zeroth term.
In this case, it is 0 so c=0.
The formula I have worked out so far is 2n²+0.
Now I have to work out b.
To work out b, you have to substitute 1 ...
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The general formula for a quadratic sequence is ax²+bx+c.
To work out a, you have to divide the second difference by 2.
In this case, the second difference is 4, and 4÷2=2, so a=2.
You cannot work out b before you have worked out c.
To work out c, you have to find the zeroth term.
In this case, it is 0 so c=0.
The formula I have worked out so far is 2n²+0.
Now I have to work out b.
To work out b, you have to substitute 1 into the formula that you have worked out so far.
2(1²) +0
=2(1) +0
=2+0
=2
So b=2.
So the formula for finding the middle side of a Pythagorean triple in terms of n is:
2n²+2n+0
Term number (n)
Short side (a)
Middle side (b)
Long side (c)
3
4
5
2
5
2
3
3
7
24
25
4
9
40
41
5
1
60
61
Now, I had to work out the formula for finding the long side of a Pythagorean triple in terms of n:
0 1 2 3 4 5
5 13 25 41 61
4 8 12 16 20
4 4 4 4
As you can see, this sequence is different to the previous sequence. It has a second difference, so it is termed as a quadratic sequence.
The general formula for a quadratic sequence is ax²+bx+c.
To work out a, you have to divide the second difference by 2.
In this case, the second difference is 4, and 4÷2=2, so a=2.
You cannot work out b before you have worked out c.
To work out c, you have to find the zeroth term.
In this case, it is 1 so c=1.
The formula I have worked out so far is 2n²+1.
Now I have to work out b.
To work out b, you have to substitute 1 into the formula that you have worked out so far.
2(1²) +0
=2(1) +0
=2+0
=2
So b=2.
So the formula for finding the long side of a Pythagorean Triple in terms of n is:
2n²+2n+1
Now, I had to work out the formula for finding the perimeter of a Pythagorean triple in terms of n:
To work out the perimeter of a shape, you have to add all the sides together.
So, to work the formula for finding the perimeter of a Pythagorean triple, you can just add together the formulas for finding the short, middle, and the large sides of a Pythagorean triple in terms of n:
2n+1+2n²+2n+2n²+2n+1.
This simplifies to equal:
4n²+6n+2
So, the formula for finding the perimeter of a Pythagorean triple is:
4n²+6n+2
Now, I had to work out the formula for finding the area of a Pythagorean triple in terms of n.
To work out the area of a triangle, you have to multiply the base by the perpendicular height, and then divide it all by 2.
So, to work out the formula for finding the area of a Pythagorean triple, you can just multiply together the formulas for finding the short and the middle sides of a Pythagorean triple in terms of n:
(2n+1)(2n²+2n)
2
This simplifies to equal:
4n³+6n²+2n
2
Then you divide the numerator by the denominator to equal:
2n³+3n²+n
So, the formula for finding the area of a Pythagorean triple in terms of n is:
2n³+3n²+n
I have now completed working out the formulas for finding the short side, middle side, long side, the perimeter and the area of Pythagorean triples in terms of n.
Now, I have to prove that the formulas for finding the short, middle and long sides work.
I can do this using one of two different methods:
You can substitute n with 2 in each of the formulas and check to see if the results match with the second Pythagorean triple which is 5, 12, 13.
Or:
You can add the squares of the formulas for finding the short, middle sides of Pythagorean triples and check to see if they equal the square of the formula for finding the long side of Pythagorean triples.
I used both methods to ensure that the formulas were correct:
Short side: 2n+1
2(2) +1
=4+1
=5
Middle side: 2n²+2n
2(2²) +2(2)
=2(4) +4
=8+4
=12
Long side: 2n²+2n+1
2(2²) +2(2) +1
=2(4) +4+1
=8+5
=13
The sums I have worked out are 5, 12, 13. This proves that the formulas were correct.
Now I will use the second method:
(2n+1)²+ (2n²+2n) ²= (2n²+2n+1)²
This is the same as:
(2n+1) (2n+1)+ (2n²+2n) (2n²+2n) = (2n²+2n+1) (2n²+2n+1)
Now I will fully multiply out the brackets:
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 also proves that the formulas were correct.
Now I will extend my investigation and investigate whether the formulas work if I double or triple any of them:
First of all, I will double the formulas and then check them.
2(2n+1)
=4n+2
So the formula for the short side doubled is:
4n+2
2(2n²+2n)
=4n²+4n
So the formula for the middle side doubled is:
4n²+4n
2(2n²+2n+1)
=4n²+4n+2
So the formula for the long side doubled is:
4n²+4n+2
Now I will investigate these formulas:
Short side (doubled): 4n+2
4(2) +2
=8+2
=10
Middle side (doubled): 4n²+4n
4(2²) +4(2)
=4(4) +8
=16+8
=24
Long side (doubled): 4n²+4n+2
4(2²) +4(2) +2
=4(4) +8+2
=16+10
=26
The results are 10, 24, 26.
Now I have to check if the small side squared plus the middle side squared equals the long side squared:
00(10²) +576(24²) =676(26²)
This proves that doubling the formulas also produce Pythagorean triples.
Now I will use the second method:
(4n+2)²+ (4n²+4n) ²= (4n²+4n+2)²
This is the same as:
(4n+2)(4n+2)(4n²+4n)(4n²+4n)= (4n²+4n+2) (4n²+4n+2)
Now I will fully multiply out the brackets:
6n²+8n+8n+4+16n4+16n²+16n³+16n³=16n4+32n³+32n²+16n+4
16n²+16n+4+16n4+32n³+16n²=16n4+32n³+32n²+16n+4
16n4+32n³+32n²+16n+4=16n4+32n³+32n²+16n+4
This also proves that doubling the formulas also produces Pythagorean triples.
Now, I will triple the formulas and then check them:
3(2n+1)
=6n+3
So the formula for the short side tripled is:
6n+3
3(2n²+2n)
=6n²+6n
So the formula for the middle side tripled is:
6n²+6n
3(2n²+2n+1)
=6n²+6n+3
So the formula for the long side tripled is:
6n²+6n+1
Now I will investigate these formulas:
Short side (tripled): 6n+3
6(2) +2
=12+2
=14
Middle side (tripled): 6n²+6n
6(2²) +6(2)
=6(4) +12
=24+12
=36
Long side (tripled): 6n²+6n+3
6(2²) +6(2) +3
=6(4) +12+3
=24+15
=39
The results are 14, 36, 39.
Now I have to check if the small side squared plus the middle side squared equals the long side squared:
96(14²) +1296(36²) =1521(39²)
This proves that tripling the formulas also produces Pythagorean triples.
Now I will use the second method:
(6n+3)²+ (6n²+6n) ²= (6n²+6n+3)²
This is the same as:
(6n+3)(6n+3)(6n²+6n)(6n²+6n)= (6n²+6n+3) (6n²+6n+3)
Now I will fully multiply out the brackets:
36n²+18n+18n+9+36n4+36n²+36n³+36n³=36n4+72n³+72n²+72n+9
36n²+36n+9+36n4+72n³+36n²=36n4+72n³+72n²+36n+9
36n4+72n³+72n²+36n+9=36n4+72n³+72n²+36n+9
This also proves that tripling the formulas also produces Pythagorean triples.
Now I will try and find the general formula for working out different families of Pythagorean triples.
The formula for the short side of a Pythagorean triple is
2n+1
For the other families, I used the formulas:
4n+2 or 2(2n+1)
And 6n+3 or 3(2n+1)
So the general formula for finding the short side in different families of Pythagorean triples is:
x(2n+1)
This means that the general formula for finding the middle side in different families of Pythagorean triples is:
x(2n²+2n)
This also means that the general formula for finding the middle side in different families of Pythagorean triples is:
x(2n²+2n+1)
Now I will prove these formulas by squaring them, multiplying out the brackets and then checking that both sides are equal.
X(2n+1) +x(2n²+2n) =x(2n²+2n+1)
2xn+x+2xn²+2xn=2xn²+2xn+x
(2xn+x)²+(2xn²+2xn)²= (2xn²+2xn+x)²
(2xn+x) (2xn+x)+ (2xn²+2xn) (2xn²+2xn) = (2xn²+2xn+x) (2xn²+2xn+x)
4x²n²+2x²n+2x²n+x²+4x²n4+4x²n³+4x²n³+4x²n²=4x²n4+4x²n³+2x²n²+4x²n³+4x²n²+2x²n
+2x²n²+2x²n+x²
8x²n²+4x²n+x²+8x²n³+4x²n4=4x²n4+8x²n³+8x²n²+4x²n+x²
This can be rearranged to form:
4x²n4+8x²n³+8x²n²+4x²n+x²=4x²n4+8x²n³+8x²n²+4x²n+x²
As you can see, both sides are equal. This proves that my general formulas were correct
Conclusion
From this investigation, I learnt about Pythagorean triples and how to find them, I learnt the formula for finding the area and perimeter of Pythagorean triples, I learned the formulas for finding other families of Pythagorean triples, I learned the general formulas for finding other families of Pythagorean triples and I also learnt about different methods for proving formulas that you have found.
To recap, all the formulas that I found can be found below:
The formula for finding the short side of a Pythagorean triple is:
2n+1
The formula for finding the middle side of a Pythagorean triple is:
2n²+2n
The formula for finding the long side of a Pythagorean triple is:
2n²+2n+1
The formula for finding the perimeter of a Pythagorean triple is:
4n²+6n+2
The formula for finding the area of Pythagorean triples is:
2n³+3n²+n
The formula for finding the short side of a Pythagorean triple, doubled is:
4n+2
The formula for finding the middle side of a Pythagorean triple, doubled is:
4n²+4n
The formula for finding the long side of a Pythagorean triple, doubled is:
4n²+4n+2
The formula for finding the short side of a Pythagorean triple, tripled is:
6n+3
The formula for finding the middle side of a Pythagorean triple, tripled is:
6n²+6n
The formula for finding the long side of a Pythagorean triple, tripled is:
6n²+6n+1
The general formula for finding the short side in different families of Pythagorean triples is:
x(2n+1)
The general formula for finding the middle side in different families of Pythagorean triples is:
x(2n²+2n)
The general formula for finding the long side in different families of Pythagorean triples is:
x(2n²+2n+1)
Overall, I think this investigation went very well and I do not think I made any mistakes in working out any of the formulas so I think it was a great success.