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• Level: GCSE
• Subject: Maths
• Word count: 1501

# 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

+18               +26             +34             +42

+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.

Middle

+10     +16    +22    +28

+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        +80

+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.

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 PerimeterA + b + c Area(h x b)2 Formula2 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:

Conclusion

## = 12.7˚ (1 dp)

b = 180 – 90 – 12.7 = 77.3˚

a = 90˚

## = 10.4˚ (1 dp)

b = 180 – 90 – 10.4 = 79.6˚

a = 90˚

 n Shortest side2n + 1 Middle side2n² + 2n Longest side2n² + 2n +1 a˚ b˚ c˚ 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.

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