# Pythagoras' theorem states

Categories: PythagorasState

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¿½

because 3ï¿½=3ï¿½3=9

4ï¿½=4ï¿½4=16

5ï¿½=5ï¿½5=25

and so 3ï¿½+4ï¿½=9+16=25=5ï¿½

The numbers 5, 12 and 13 also work,

5ï¿½+12ï¿½=13ï¿½

because 5ï¿½=5ï¿½5=25

12ï¿½=12ï¿½12=144

13ï¿½=13ï¿½13=169

and so 5ï¿½+12ï¿½=25+144=169=13ï¿½

The numbers 7, 24 and 25 also work,

7ï¿½+24ï¿½=25ï¿½

because 7ï¿½=7ï¿½7=49

24ï¿½=24ï¿½24=576

25ï¿½=25ï¿½25=625

and so 7ï¿½+24ï¿½=49+576=625=25ï¿½

3, 4 and 5

Perimeter= 3+4+5=12

Area= 1/2 ï¿½3ï¿½4=6

5, 12 and 13

Perimeter= 5+12+13=30

Area= 1/2 ï¿½5ï¿½12=30

7, 24 and 25

Perimeter= 7+24+25=56

Area= 1/2 ï¿½7ï¿½24=84

From 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) + 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. They are formulas on how to get from n to all of the others: –

1. n to a- 2n+1

2. n to b- 2nï¿½+2n

3. n to c- 2nï¿½+2n+1

4. n to perimeter- 4nï¿½+6n+2

5. n to area- 2nï¿½+3nï¿½+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= 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+2n

2

=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.

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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)+2

Here 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’

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 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+1

4n4+8nï¿½+8nï¿½+4n+1 = 4n4+8nï¿½+8nï¿½+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,

6ï¿½+8ï¿½=10ï¿½

because 6ï¿½=6ï¿½6=36

8ï¿½=8ï¿½8=64

10ï¿½=10ï¿½10=100

and so 6ï¿½+8ï¿½=36+64=100=10ï¿½

The numbers 10, 24 and 26 also work,

10ï¿½+24ï¿½=26ï¿½

because 10ï¿½=10ï¿½10=100

24ï¿½=24ï¿½24=576

26ï¿½=26ï¿½26=676

and so 10ï¿½+24ï¿½=100+576=676=26ï¿½

The numbers 20, 99 and 101 also work,

20ï¿½+99ï¿½=101ï¿½

because 20ï¿½=20ï¿½20=400

99ï¿½=99ï¿½99=9801

101ï¿½=101ï¿½101=10201

and so 20ï¿½+99ï¿½=400+9801=10201=101ï¿½

6, 8 and 10

Perimeter=6+8+10=24

Area= 1/2(6ï¿½8)=24

10, 24 and 26

Perimeter=10+24+26=60

Area= 1/2(10ï¿½24)=240

20, 99 and 101

Perimeter=20+99+101=220

Area= 1/2(20ï¿½99)=990

Here is a table with some terms on it.

‘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

Here are my formulae on how to get from n to all of the others:-

1. n to a- 2n

2. n to b- nï¿½-1

3. n to c- nï¿½+1

4. n to perimeter- 2nï¿½+2n

5. n to are- nï¿½-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= 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-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+nï¿½-1+nï¿½+1

= 2n+2nï¿½

= 2nï¿½+2n

Here is how I got to my b and c formulae.

Getting b

aï¿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¿½-1

Getting c

By looking at b you can see that c is always +2 than b.

So c is nï¿½-1+2

=nï¿½+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 aï¿½+bï¿½=cï¿½

aï¿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+1

This proves that my formulas for a, b and c are all correct.

## Cite this page

Pythagoras' theorem states. (2020, Jun 02). Retrieved from http://studymoose.com/pythagoras-theorem-states-new-essay

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