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
I have also added 2 more terms using what I think are my formulae.
Here are my formulas. They are formulas on how to get from n to all of the others: -
-
n to a- 2n+1
-
n to b- 2n²+2n
-
n to c- 2n²+2n+1
-
n to perimeter- 4n²+6n+2
-
n to area- 2n³+3n²+n
Here is how I got to the area and perimeter formulae.
We all know that the area of a triangle has a formula of area= ½ ×a×b. so with that knowledge, you substitute a and b with their formulae to get the formula.
So (2n+1)(2n²+2n)
2
= 4n³+2n²+4n²+2n
2
=2n³+n²+2n²+n
=2n³+3n²+n
=n(2n²+3n+1)
We also know that to get the perimeter of any shape, not just a triangle, you add up the lengths of all of the sides. So with this knowledge, once again, you substitute a and b with their formulae.
So (2n+1) ²+(2n+1)
= 4n²+6n+2
= 2(2n²+3n+1)
= 2n(2n+3)+2
Here is how I got my formulae for sides a, b and c.
- Take the first five terms, 3, 5, 7, 9, 11. You can see that all of these numbers are just the odd numbers. You can see that the formula 2n+1 works because they are all the consecutive odd numbers.
- From looking at my table I could see that n×a+n=b. so then I substituted a for its formula and I got 2n²+2n.
- Side c is +1 of b so you just put the formula of b and then +1. So 2n²+2n+1.
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
The numbers 20, 99 and 101 also work,
20²+99²=101²
because 20²=20×20=400
99²=99×99=9801
101²=101×101=10201
and so 20²+99²=400+9801=10201=101²
6, 8 and 10
Perimeter=6+8+10=24
Area= ½(6×8)=24
10, 24 and 26
Perimeter=10+24+26=60
Area= ½(10×24)=240
20, 99 and 101
Perimeter=20+99+101=220
Area= ½(20×99)=990
Here is a table with some terms on it.
Here are my formulae on how to get from n to all of the others:-
-
n to a- 2n
-
n to b- n²-1
-
n to c- n²+1
-
n to perimeter- 2n²+2n
-
n to are- n³-n
Here is how I got to my area and perimeter formulae.
We all know that the area of a triangle has a formula of area= ½(a×b). So with that knowledge, you substitute a and b with their formulae to get the formula.
So (2n)(n²-1)
2
= n(n²-1)
= n³-n
We also know that to get the perimeter of any shape, not just a triangle, you add up the lengths of all of the sides. So with this knowledge, once again, you substitute a and b with their formulae.
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.
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.