# Investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers.

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Introduction

## Beyond Pythagoras

## The aim of this piece of my coursework is to investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers. I will collate the relevant data and formulae for the nth terms by using a grid and from this data I should be able to make predictions on the nth terms of Pythagorean Triples. I will keep a narrative of what I am doing and discovering, referring within it to each process of my investigation.

The numbers 3, 4 and 5 satisfy the condition 32+42=52,

Because 32= 3x3 =9

42= 4x4 =16

52= 5x5 =25

And so… 32+42=9+16=25=52

I now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) 2+ (middle number) 2= (largest number) 2.

- 5, 12, 13

52+122 = 25+144 = 169 = 132.

- 7, 24, 25

72+242 = 49+576 = 625 +252

Here is a table containing the results:

a | b | c | area | perimeter |

3 5 7 9 11 | 4 12 24 40 60 | 5 13 25 41 61 | 6 3 84 180 330 | 12 30 56 90 132 |

2n +1 | 2n² + 2n | 2n³ + 2n + n² +n | 2n³ + 2n + n² + n | (2n + 1) + (2n (n+1)) + (2n² + (2n + 1) |

I looked at the table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.

I already know that the (smallest number)² + (middle number)² = (largest number)² So, therefore, I know that there will be a connection between the numbers written above. The problem is that it is obviously not:

(Middle number)² + (largest number)² = (smallest number)²

Because: 122 + 132 = 144+169 = 313

52 = 25

Middle

121 = 60.5

2

Lower bound = 60, Upper bound = 61.

Middle Side = 60, Largest Side =61.

132 = Middle number + Largest number

169 = Middle number + Largest number

169 = 84.5

2

Lower bound = 84, Upper bound = 85.

Middle Side = 84, Largest Side =85.

152 = Middle number + Largest number

225 = Middle number + Largest number

225 = 112.5

2

Lower bound = 112, Upper bound = 113.

Middle Side = 112, Largest Side =113.

172 = Middle number + Largest number

289 = Middle number + Largest number

289 = 144.5

2

Lower bound = 144, Upper bound = 145

Middle Side = 144, Largest Side =145.

192 = Middle number + Largest number

361 = Middle number + Largest number

361 = 180.5

2

Lower bound = 180, Upper bound = 181.

Middle Side = 180, Largest Side =181.

212 = Middle number + Largest number

441 = Middle number + Largest number

441 = 220.5

2

Lower bound = 220, Upper bound = 221.

Middle Side = 220, Largest Side =221.

I now have 10 different triangles, which I think is easily enough to find a relationship between each side.

Shortest side²= Middle side + Longest side

The way I mentioned above and on the previous pages, is quite a good way of finding the middle and longest sides. An easier and faster way to work out the sides would be by using the nth term. I will now try to work out the nth term for each side (shortest middle and longest).

The formula I will work out first is for the shortest side.

1 | 3 5 7 9 11

2 2 2 2

The difference between the lengths of the shortest side is 2. This means the equation must be something to do with 2n. So, I will write down all the answers for 2n.

There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1.

Conclusion

Area = ½ b h

b = Base

h = Height

I should think that the ‘base’ side means the adjsacent side, and the ‘height’ side means the opposite side. However, I will continue to refer to them as ‘base’ and ‘height’ from now on. So:

Area = ½ (Shortest Side) X (Middle Side)

= ½ (2n +1) x (2n² 2n)

= (2n +1) (2n² + 2n)

2

I will now check this using the first 3 terms.

(2n + 1) (2n² + 2n) = ½ b h

2

(2 x 1 + 1) (2 x 12 + 2 x 1) = ½ x 3 x 4

2

3 x 4 = ½ x 12

2

12 = 6

2

6 = 6

My formula works for the first term.

(2n + 1) (2n² + 2n) = ½ b h

2

(2 x 2 + 1) (2 x 22 + 2 x 2) = ½ x 5 x 12

2

5 x 12 = ½ x 60

2

60 = 30

2

30 = 30

My formula also works for the 2nd term.

(2n + 1) (2n2 + 2n) = ½ b h

2

(2 x 3 + 1) (2 x 32 + 2 x 3) = ½ x 7 x 24

2

7 x 24 = ½ x 168

2

168 = 168

2

168 = 168

My formula works for all 3 terms. So…

Area = (2n +1) (2n² 2n)

2

If I am given a right-angled triangle I can always apply an enlargement to it (for example, I can double all lengths) and get another right-angled triangle. This means that from a given triple a, b, c I can produce many more Pythagorean triples na, nb, nc for any whole number n.

For example, starting with the triple 3, 4, 5 and taking n = 5 we get the new triple 15, 20, 25. Sometimes we can reverse this process by starting with a triple and then reducing the lengths of the sides to get another triple. This does not always work; if we start with 3, 4, 5, for example, and halve the lengths of the sides we do not get a triple of whole numbers. However, sometimes we do; for example, by halving lengths the triple 10, 24, 26 converts into the triple 5, 12, 13.

This student written piece of work is one of many that can be found in our GCSE Pythagorean Triples section.

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