### Pythagoras’ theorem states that aï¿½+bï¿½=cï¿½

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,3ï¿½+4ï¿½=5ï¿½because 3ï¿½=3ï¿½3=94ï¿½=4ï¿½4=165ï¿½=5ï¿½5=25and so 3ï¿½+4ï¿½=9+16=25=5ï¿½The numbers 5, 12 and 13 also work,5ï¿½+12ï¿½=13ï¿½because 5ï¿½=5ï¿½5=2512ï¿½=12ï¿½12=14413ï¿½=13ï¿½13=169and so 5ï¿½+12ï¿½=25+144=169=13ï¿½The numbers 7, 24 and 25 also work,7ï¿½+24ï¿½=25ï¿½because 7ï¿½=7ï¿½7=4924ï¿½=24ï¿½24=57625ï¿½=25ï¿½25=625and so 7ï¿½+24ï¿½=49+576=625=25ï¿½3, 4 and 5Perimeter= 3+4+5=12Area= 1/2 ï¿½3ï¿½4=65, 12 and 13Perimeter= 5+12+13=30Area= 1/2 ï¿½5ï¿½12=307, 24 and 25Perimeter= 7+24+25=56Area= 1/2 ï¿½7ï¿½24=84From the first three terms I have realised that: -* a increases by 2 each time* a is equal to the formula ï¿½2+1 from n* b is always even* c is always odd* c is always +1 of b* b=(aï¿½n) + nI have also added 2 more terms using what I think are my formulae.’n”a”b”c’PerimeterArea134512625121330303724255684494041901805116061132330Here are my formulas.

They are formulas on how to get from n to all of the others: -1. n to a- 2n+12. n to b- 2nï¿½+2n3. n to c- 2nï¿½+2n+14.

n to perimeter- 4nï¿½+6n+25. n to area- 2nï¿½+3nï¿½+nHere is how I got to the area and perimeter formulae.We all know that the area of a triangle has a formula of area= 1/2 ï¿½aï¿½b. so with that knowledge, you substitute a and b with their formulae to get the formula.So (2n+1)(2nï¿½+2n)2= 4nï¿½+2nï¿½+4nï¿½+2n2=2nï¿½+nï¿½+2nï¿½+n=2nï¿½+3nï¿½+n=n(2nï¿½+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) ï¿½+(2n+1)= 4nï¿½+6n+2= 2(2nï¿½+3n+1)= 2n(2n+3)+2Here is how I got my formulae for sides a, b and c.a) 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.b) From looking at my table I could see that nï¿½a+n=b.

so then I substituted a for its formula and I got 2nï¿½+2n.c) Side c is +1 of b so you just put the formula of b and then +1. So 2nï¿½+2n+1.’n”a”b”c’areaperimeter13456122512133030372425845649404118090511606133013261384855461827151121138402408171441451224306919180181171036010212202212310462Now I have to prove that my formulae for a, b and c work. To do this I must incorporate aï¿½+bï¿½+cï¿½: -aï¿½+bï¿½=cï¿½(2n+1)ï¿½+(2nï¿½+2n) ï¿½=(2nï¿½+2n+1) ï¿½(2n+1)(2n+1) + (2nï¿½+2n)(2nï¿½+2n) = (2nï¿½+2n+1)(2nï¿½+2n+1)4nï¿½+2n+2n+1+4n4+4nï¿½+4nï¿½+4nï¿½ = 4n4+8nï¿½+8nï¿½+4n+14n4+8nï¿½+8nï¿½+4n+1 = 4n4+8nï¿½+8nï¿½+4n+1This 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,6ï¿½+8ï¿½=10ï¿½because 6ï¿½=6ï¿½6=368ï¿½=8ï¿½8=6410ï¿½=10ï¿½10=100and so 6ï¿½+8ï¿½=36+64=100=10ï¿½The numbers 10, 24 and 26 also work,10ï¿½+24ï¿½=26ï¿½because 10ï¿½=10ï¿½10=10024ï¿½=24ï¿½24=57626ï¿½=26ï¿½26=676and so 10ï¿½+24ï¿½=100+576=676=26ï¿½The numbers 20, 99 and 101 also work,20ï¿½+99ï¿½=101ï¿½because 20ï¿½=20ï¿½20=40099ï¿½=99ï¿½99=9801101ï¿½=101ï¿½101=10201and so 20ï¿½+99ï¿½=400+9801=10201=101ï¿½6, 8 and 10Perimeter=6+8+10=24Area= 1/2(6ï¿½8)=2410, 24 and 26Perimeter=10+24+26=60Area= 1/2(10ï¿½24)=24020, 99 and 101Perimeter=20+99+101=220Area= 1/2(20ï¿½99)=990Here is a table with some terms on it.

‘n”a”b”c’PerimeterArea12024024351263681024244815174060510242660120Here are my formulae on how to get from n to all of the others:-1. n to a- 2n2. n to b- nï¿½-13. n to c- nï¿½+14. n to perimeter- 2nï¿½+2n5.

n to are- nï¿½-nHere is how I got to my area and perimeter formulae.We all know that the area of a triangle has a formula of area= 1/2(aï¿½b). So with that knowledge, you substitute a and b with their formulae to get the formula.

So (2n)(nï¿½-1)2= n(nï¿½-1)= nï¿½-nWe 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+nï¿½-1+nï¿½+1= 2n+2nï¿½= 2nï¿½+2nHere is how I got to my b and c formulae.Getting baï¿½+bï¿½=cï¿½=cï¿½= (b+2) ï¿½=(2n) ï¿½+bï¿½=(b+2) ï¿½=4nï¿½+bï¿½=bï¿½+4b+4=4nï¿½-4=4b=nï¿½-1=b=b=nï¿½-1Getting cBy looking at b you can see that c is always +2 than b.So c is nï¿½-1+2=nï¿½+1Here is a table with all of the results for even numbers.

‘n”a”b”c’PerimeterArea12024024351263681024244815174060510242660120612353784210714485011233681663651445049188082180720102099101220990Now I have to prove my formulae for a, b and c. To do this I have to incorporate aï¿½+bï¿½=cï¿½aï¿½+bï¿½=cï¿½= (2n)ï¿½+(nï¿½-1) ï¿½=(nï¿½+1) ï¿½= (2n)(2n)=4nï¿½(nï¿½-1)(nï¿½-1)=n4-nï¿½-nï¿½+1(nï¿½+1)(nï¿½+1)=n4+nï¿½+nï¿½+1= 4nï¿½+n4-nï¿½-nï¿½+1=n4+nï¿½+nï¿½+1= n4+2nï¿½+1=n4+2nï¿½+1This proves that my formulas for a, b and c are all correct.