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Find a formula to enable the perimeter to be found for any odd Pythagorean triple.

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Introduction

Paul Fudio

Perimeter:

I want to find a formula to enable me to find the perimeter for any odd Pythagorean triple. I know that perimeter is a + b + c so I know the formula for the shortest, middle and longest side so I can assume that if I substitute the formulas I know, I can presume I will find the formulas for the perimeter.

The perimeter = a + b + c. Therefore I took my formula for ‘a’ (2n + 1), my formula for ‘b’ (2n² + 2n) and my formula for ‘c’ (2n² + 2n + 1). I then did the following:

      a      +        b       +           c

(2n + 1) + (2n² + 2n) + (2n² + 2n + 1) =

2n + 1 + 2n + 2n + 1 + 4n² =

4n² + 6n + 1 + 1 =

4n² + 6n + 2 = formula for perimeter

I will further prove this formula by using sequences.

3,4,5                5,12,13        7,24,25        9,40,41        11,60,61

   12                 30               56                90                132     image00.pngimage00.pngimage00.pngimage00.png

            +18               +26             +34             +42                  image00.pngimage00.pngimage00.png

                       +8                 +8              +8      

Here the first differences aren’t the same, but the second differences are. If the second difference is a constant, then the formula for the nth term contains n². The number in front of n² is half the constant difference. Therefore the first part of the formula is 4n².

...read more.

Middle

image01.pngimage01.pngimage01.png

                                 +10     +16    +22    +28

image01.pngimage01.pngimage01.png

                                       +6     +6      +6

Value of 3n²:                3        12        27        48        75

I have found a similarity between these two parts (difference and 3n²). I will make this easier by moving the constants out of the way.

Term number:        1        2        3        4        5        

Difference:                +4        +14        +30        +52        +80image02.pngimage02.pngimage02.pngimage02.pngimage02.png

                              +1      +2     +3      +4     +5                                                  

Value of 3n²:                3        12        27        48        75

In this table it is clear to see that you add the term number every time so therefore the last part of the formula is +n, this makes up all of the formula to give evidence that my formula was correct in my first statement and is 2n³ + 3n²+ n.

image34.pngimage35.pngimage06.pngimage05.pngimage04.pngimage03.png

Here you can notice a similar pattern between the area formula and the perimeter formula. First I factorised the area formula and I got:

2 (4n² + 6n + 2) which means, 2 x area = perimeter, as you can see easier

N                                            N    

in the table I am going to draw below.

N

Perimeter

A + b + c

Area

(h x b)

2

Formula

2 x area = perimeter

 N                

Check

1

12

6

2/1 x 6 =  

12

2

30

30

2/2 x 30 =

30

3

56

84

2/3 x 84 =

56

4

90

180

2/4 x 180 =

90

5

132

330

2/5 x 330 =

132

6

182

546

2/6 x 546 =

182

7

240

840

2/7 x 840 =

240

8

306

1224

2/8 x 1224 =

306

9

380

1710

2/9 x 1710 =

380

10

462

4620

2/10 x 4620 =

462

This proves in my formulae are similar to each other. Therefore another statement I have made is correct.

Proving:

...read more.

Conclusion

image13.png

image26.png

c = cos –1 = 40image08.png

                                                41

                                               = 12.68image27.pngimage28.png

                                               = 12.7˚ (1 dp)

                                b = 180 – 90 – 12.7 = 77.3˚

                                a = 90˚image24.pngimage23.png

image29.pngimage13.png

c = cos –1 = 60image07.pngimage08.png

                                                61image31.png

                                               = 10.38

                                               = 10.4˚ (1 dp)image32.png

                                b = 180 – 90 – 10.4 = 79.6˚

                                a = 90˚image23.pngimage24.png

image33.pngimage13.png

n

Shortest side

2n + 1

Middle side

2n² + 2n

Longest side

2n² + 2n +1

1

3

4

5

90

53.1

36.9

2

5

12

13

90

67.4

22.6

3

7

24

25

90

73.7

16.3

4

9

40

41

90

77.3

12.7

5

11

60

61

90

79.6

10.4

After I looked at the graph there were no sequences or patterns apart from they all add up to 180˚ and b + c add up to 90˚. Also angle b increases when the sides are longer while c angle decreases when the sides are longer.

Conclusion:

Throughout my whole investigation I have only used odd triples and numbers, but there are also even numbers in triples and Pythagoras’ work. I would expect the same result and formulae as those of odd numbers, and maybe in the future I can investigate this.

I am also amazed at the formulae we had to find and how well it fitted together with our table and substitutions. We found the formulae for finding each side of a right-angled triangle that conforms to Pythagoras’ theorem.

I also came to a conclusion that the formula for area and the formula for perimeter are extremely similar. For some extra work I looked at trigonometry and angles and found out that angles didn’t have any correlation apart from having 180˚ between them.

...read more.

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