# Mathematics Coursework - Beyond Pythagoras

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Mathematics Coursework - Beyond Pythagoras

1)

The numbers 3, 4, and 5 satisfy the condition

3Â² + 4Â² = 5Â²

because 3Â² = 3x3 =9

4Â² = 4x4 = 16

5Â² = 5x5 = 25

and so

3Â² + 4Â² = 9 + 16 = 25 = 5Â²

I will now have to find out if the following sets of numbers satisfy a
similar condition of (smallest number) Â² + (middle number) Â² =
(largest number) Â².

a) 5, 12, 13

5Â² + 12Â² = 25 + 144 = 169 = 13Â²

b) 7, 24, 25

7Â² + 24Â² = 49 + 576 = 625 = 25Â²

2) Perimeter
============

[IMAGE]

5

5 + 12 + 13 = 20

12

[IMAGE]

7

7 + 24 + 25 = 56

24

b)

Nth term

Length of shortest side

Length of middle side

Length of longest side

Perimeter
---------

Area
----

[IMAGE][IMAGE]

+2

+1

1

3

4

5

12

6

[IMAGE]2

5

12

13

30

30

[IMAGE]

+2

3

7

24

25

56

84

4

9

40

41

90

180

[IMAGE]

+2

5

11

60

61

132

330

+2

6

13

84

85

182

546

[IMAGE]7

15

112

113

240

840

+2

8

17

144

145

306

1224

9

19

180

181

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380

1710

10

21

220

221

462

2310

I looked at the table and noticed that there was only 1 difference
between the length of the middle side and the length of the longest
side. And also if you can see in the shortest side column, it goes up
by 2. I have also noticed that the area is

Â½ (shortest side) x (middle side).

3)

In this section I will be working out and finding out the formulas
for:

* Shortest side

* Middle side

* Longest side

In finding out the formula for the shortest side I predict that the
formula will be something to do with the differences between the
lengths (which is 2). But I don't know the formula so I will have to
work that out.

Firstly I will be finding out the formula for the shortest side.

3 5 7 9 11

2 2 2 2

The differences between the lengths of the shortest side are 2. This
means the equation must be something to do with 2n.

Let's seeâ€¦

Nth term

Length of shortest side

1

3

2n

[IMAGE]2 x 1 = 2 (wrong)

There is only a difference of +1 between 2n and the shortest side, so
this means the formula should be 2n+1. To see if I'm correct I will
now test this formula.

2n+1

Nth term

Length of shortest side

1

3

[IMAGE]2x1+1=3 (correct)

Just in case I will test this formula in the next term:

2n+1

Nth term

Length of shortest side

2

5

2x2=4

[IMAGE]4+1=5 (correct)

I now have to work out the formula for the middle side. I predict that
the formula has got to do with something about the differences in
lengths. (For instance in this case is 4), most likely 4n.However,
because 4 is the difference, the formula must be nÂ². I now believe
that the answer will have something to do with 4nÂ². So, I will now

4nÂ² work for the first term, but, it then collapses after this, as the
difference between 4nÂ² gets larger, the thing you notice is that the
difference in the 2nd term between 4nÂ² and the middle side is the
middle side for the term before. This goes for all the other terms
from the 2nd.

This means that if I subtract the previous term, then I should in

16-4 = 12

36-12 = 24

64-24 = 40

Etc.

So, the equation I have so far is:

4nÂ²- (previous middle side) = middle side

All the previous terms is, (n - 1), so if I put this into the above
formula, then it should give me my middle side.

4nÂ² - 4(1-1) Â² = middle side

This should in theory give me my middle side. I will test my theory
with the first term.

4 x 12 - 4(1-1) = 4

4 x 1 - 4 x 0Â² = 4

4 - 4 x 0 = 4

4 - 0 = 4

4 = 4

My formula works for the first term. I will now check if this formula
works for the next term.

4 x 2Â² - 4(2 - 1) Â² = 12

4 x 4 - 4 x 1Â² = 12

16 - 4 x 1 = 12

16 - 4 = 12

12 = 12

My formula also works for the 2nd term. You may be thinking that this
is the correct formula but just to check, I will check this formula
for the 3rd term.

4 x 3Â² - 4(3-1) Â² = 24

4 x 4 - 4 x 2Â² = 24

36 - 4 x 4 = 24

36 - 16 = 24

20 = 24

My formula did not work for the 3rd term. It now looks as if "4nÂ² -
4(n - 1) Â² " is not the correct formula after all. To check, I will
look to see if the formula works using the 4th term.

4 x 4Â² - 4(4 - 1) Â² = 40

4 x 16 - 4 x 3Â² = 40

64 - 36 = 40

28 = 40

My formula doesn't work for the 4th term either. I can now safely say
that

4nÂ² - 4(n - 1) Â² is definitely not the correct formula for the middle
side.

I believe the problem with 4nÂ² - 4(n - 1) was that 4nÂ², once you start
using the larger numbers, it becomes far too high to bring it back
down to the number I want for the middle side. Also, 4(n - 1) Â² is not
as small when it gets larger so it doesn't bring the 4nÂ² down enough,
to equal the middle side.

I will now look at the difference to see if I can find a pattern
there.

1 4 11 20 31

3 7 9 11

2 2 2

The difference here is 2, which means that the answer will involve 2
and nÂ². I will try 2nÂ². I can see that the difference between 2nÂ² and
the middle number is the 2 times table. The 2 times table in the nth
term is 2n. I now think 2nÂ² + 2n is the correct formula. I will now
test it using the first three terms.

2 x 1Â² + 2 x 1 = 4

2 x 1 + 2 = 4

2 + 2 = 4

4 = 4

My formula works for the first term; so, I will now check it in the
next term.

2 x 2Â² + 2 x 2 = 12

2 x 4 + 4 = 12

8 + 4 = 12

12 = 12

My formula works for the 2nd term. If it works for the 3rd term I can
safely say that

2nÂ² + 2n is the correct formula.

2 x 3Â² + 2 x 3 = 24

2 x 9 + 6 = 24

18 + 6 = 24

24 = 24

My formula also works for the 3rd term. I am now certain that 2nÂ² + 2n
is the correct formula for finding the middle side.

Middle side = 2nÂ² + 2n

I now have the much easier task of finding a formula for the longest
side. To start with, I am going to draw out a table containing the
middle and longest sides.

Length of middle side

Length of longest side

+1

[IMAGE]4

5

[IMAGE]12

13

24

25

40

41

60

61

84

85

112

113

144

145

180

181

220

221

From this table I know that there is only one difference, which is +1
between the middle and longest side. So:

(Middle side) + 1 = longest side

I predict that this formula will be the correct one for the longest
side:

2nÂ² + 2n + 1 = longest side.

I am very certain that this is the correct formula. I will check
anyway using the first three terms:

2nÂ² + 2n + 1 = 5

2 x 1Â² + 2 x 1 + 1 = 5

2 + 2 + 1 = 5

5 = 5

The formula works for the first term.

2nÂ² + 2n + 1 = 25

2 x 3Â² + 2 x 3 + 1 = 25

18 + 6 + 1 = 25

25 = 25

The formula also works for the second term

2nÂ² + 2n + 1 = 13

2 x 2Â² + 2 x 2 + 1 = 13

8 + 4 + 1 = 13

13 = 13

The formula works for all three terms. Soâ€¦

Longest side = 2nÂ² + 2n + 1

Now I will check that: -

2n + 1

2nÂ² + 2n and

2nÂ² + 2n + 1

Form a Pythagorean triple or in other words aÂ² + bÂ² = cÂ²

aÂ² + bÂ² = cÂ²

This equals:

(2n + 1) Â² + (2nÂ² + 2n) Â² = (2nÂ² + 2n + 1) Â²

If you then put these equations into brackets:

(2n + 1)(2n + 1) + (2nÂ² + 2n)(2nÂ² + 2n) = (2nÂ² + 2n + 1)(2nÂ² + 2n + 1)

[IMAGE]

[IMAGE]

[IMAGE]

If I work out this equation out by balancing them in each side and I
end up with nothing, then 2n + 1, 2nÂ² + 2n and 2nÂ² + 2n + 1 is a
Pythagorean triple.

4nÂ² + 4nÂ² = 4nÂ² + 2nÂ² + 2nÂ²

4n = 2n + 2n

8nÂ³ = 4nÂ³ + 4nÂ³

4n = 4n

1 = 1

I now end up with 0 = 0, so 2n + 1, 2nÂ² + 2n and 2nÂ² + 2n + 1 is a
Pythagorean triple.

I now have the nth term for each of the three sides of a right-angled
triangle. I can now work out, both, the nth term for the perimeter and
the nth term for the area.

The perimeter of any triangle is just the length of the 3 sides added
together. E.g.

1st term 3 + 4 + 5 = 12

So 12 is the perimeter for the first term

2nd term 5 + 12 + 13 = 30

3rd term 7 + 24 + 25 = 56

And so on. All I have to do is put all the 3 formulas together.

Perimeter = (shortest side) + (middle side) + (longest side)

= 2n + 1 + 2nÂ² + 2n + 2nÂ² + 2n + 1

= 2nÂ² + 2nÂ² + 2n + 2n + 2n + 1 + 1

= 4nÂ² + 6n + 2

If I have done my calculations properly then I should have the right
answer. To check this I am going use the 4th, 5th and 6th terms.

4th term:

4nÂ² + 6nÂ² + 2 = perimeter

4 x 4Â² + 6 x 4 + 2 = 9 + 40 + 41

64 + 24 + 2 = 90

90 = 90

It works for the 4th term

Let's see if it works for the 5th term:

4nÂ² + 6nÂ² + 2 = perimeter

4 x 5Â² + 6 x 2 = 11 + 60 + 61

100 + 30 + 2 = 132

132 = 132

And it works for the 5th term

And finally the 6th term:

4nÂ² + 6nÂ² + 2 = perimeter

4 x 6Â² + 6 x 6 + 2 = 13 + 84 + 85

144 + 36 + 2 = 182

182 = 182

It works for all the terms so:

Perimeter = 4nÂ² + 6n + 2

Like the area I know that the area of a triangle is found by:

Area = Â½ (b x h)

b = base

h = height

Depending on which way the right angled triangle is the shortest or
middle side can either be the base or height because it doesn't really
matter which way round they go, as I'll get the same answer either
way.

Area = Â½ (shortest side) X (middle side)

= Â½ (2n + 1) x (2nÂ² + 2n)

= (2n + 1)(2nÂ² + 2n)

I will check this formula on the first two terms:

(2n + 1)(2nÂ² + 2n) = Â½ (b x h)

(2 x 1 + 1)(2 x 1Â² + 2 x 1) = Â½ x 3 x 4

3 x 4 = Â½ x 12

12 = 6

6 = 6

2nd term:

(2n + 1)(2nÂ² + 2n) = Â½ b h

(2 x 2 + 1)(2 x 2Â² + 2 x 2) = Â½ x 5 x 12

5 x 12 = Â½ x 60

60 = 30

30 = 30

It works for both of the terms. This means:

Area = (2n + 1)(2nÂ² + 2n)