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Pyhtagorean Theorem
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Introduction
Jaymin Patel 10Q Page of
Pythagoras’ theorem states that a²+b²=c² 
As you can see, a is the longest side, b is the middle side and c is the longest side (hypotenuse). The point of this coursework is to find b when a is an odd number and all of the sides are positive integers. Then after that I will go looking when a is a positive number.
The numbers 3, 4 and 5 work in Pythagoras’ theorem,
3²+4²=5²
because 3²=3×3=9
4²=4×4=16
5²=5×5=25
and so 3²+4²=9+16=25=5²
The numbers 5, 12 and 13 also work,
5²+12²=13²
because 5²=5×5=25
12²=12×12=144
13²=13×13=169
and so 5²+12²=25+144=169=13²
The numbers 7, 24 and 25 also work,
7²+24²=25²
because 7²=7×7=49
24²=24×24=576
25²=25×25=625
and so 7²+24²=49+576=625=25²
3, 4 and 5
Perimeter= 3+4+5=12
Area= ½ ×3×4=6
5, 12 and 13
Perimeter= 5+12+13=30
Area= ½ ×5×12=30
7, 24 and 25
Perimeter= 7+24+25=56
Area= ½ ×7×24=84
From the first three terms I have realised that: -
- a increases by 2 each time
- a is equal to the formula ×2+1 from n
- b is always even
- c is always odd
- c is always +1 of b
- b=(a×n) + n
I have also added 2 more terms using what I think are my formulae.
‘n’ | ‘a’ | ‘b’ | ‘c’ | Perimeter | Area |
1 | 3 | 4 | 5 | 12 | 6 |
2 | 5 | 12 | 13 | 30 | 30 |
3 | 7 | 24 | 25 | 56 | 84 |
4 | 9 | 40 | 41 | 90 | 180 |
5 | 11 | 60 | 61 | 132 | 330 |
Here are my formulas.
Middle
‘n’ | ‘a’ | ‘b’ | ‘c’ | area | perimeter |
1 | 3 | 4 | 5 | 6 | 12 |
2 | 5 | 12 | 13 | 30 | 30 |
3 | 7 | 24 | 25 | 84 | 56 |
4 | 9 | 40 | 41 | 180 | 90 |
5 | 11 | 60 | 61 | 330 | 132 |
6 | 13 | 84 | 85 | 546 | 182 |
7 | 15 | 112 | 113 | 840 | 240 |
8 | 17 | 144 | 145 | 1224 | 306 |
9 | 19 | 180 | 181 | 1710 | 360 |
10 | 21 | 220 | 221 | 2310 | 462 |
Now I have to prove that my formulae for a, b and c work. To do this I must incorporate 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
4n⁴+8n³+8n²+4n+1 = 4n⁴+8n³+8n²+4n+1
This proves that my formulae for a, b and c are all correct!
Now for part 2, I will try to solve the problem of Pythagorean triples with even numbers for a.
The numbers 6, 8 and 10 work for Pythagoras’ theorem,
6²+8²=10²
because 6²=6×6=36
8²=8×8=64
10²=10×10=100
and so 6²+8²=36+64=100=10²
The numbers 10, 24 and 26 also work,
10²+24²=26²
because 10²=10×10=100
24²=24×24=576
26²=26×26=676
and so 10²+24²=100+576=676=26²
Conclusion
So 2n+n²-1+n²+1
= 2n+2n²
= 2n²+2n
Here is how I got to my b and c formulae.
Getting b
a²+b²=c²
=c²= (b+2) ²
=(2n) ²+b²=(b+2) ²
=4n²+b²=b²+4b+4
=4n²-4=4b
=n²-1=b
=b=n²-1
Getting c
By looking at b you can see that c is always +2 than b.
So c is n²-1+2
=n²+1
Here is a table with all of the results for even numbers.
‘n’ | ‘a’ | ‘b’ | ‘c’ | Perimeter | Area |
1 | 2 | 0 | 2 | 4 | 0 |
2 | 4 | 3 | 5 | 12 | 6 |
3 | 6 | 8 | 10 | 24 | 24 |
4 | 8 | 15 | 17 | 40 | 60 |
5 | 10 | 24 | 26 | 60 | 120 |
6 | 12 | 35 | 37 | 84 | 210 |
7 | 14 | 48 | 50 | 112 | 336 |
8 | 16 | 63 | 65 | 144 | 504 |
9 | 18 | 80 | 82 | 180 | 720 |
10 | 20 | 99 | 101 | 220 | 990 |
Now I have to prove my formulae for a, b and c. To do this I have to incorporate a²+b²=c²
a²+b²=c²
= (2n)²+(n²-1) ²=(n²+1) ²
= (2n)(2n)=4n²
(n²-1)(n²-1)=n⁴-n²-n²+1
(n²+1)(n²+1)=n⁴+n²+n²+1
= 4n²+n⁴-n²-n²+1=n⁴+n²+n²+1
= n⁴+2n³+1=n⁴+2n³+1
This proves that my formulas for a, b and c are all correct.
This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.
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