# Beyond Pythagoras

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Introduction

Mathematics Coursework – Beyond Pythagoras May 2006

BEYOND PYTHAGORAS

MATHS COURSEWORK

INTRODUCTION

Pythagoras was a Greek mathematician and philosopher. He lived in 400 BC and was one of the first great mathematical thinkers. He spent most of his life in Sicily and southern Italy. He had a group of follows who went around and thought other people what he had taught them who were called the Pythagoreans.

Pythagoras himself is best known for proving that the Pythagorean Theorem was true. The Sumerians, two thousand years earlier, already knew that it was generally true, and they used it in their measurements, but Pythagoras proved that it would always be true. The Pythagorean Theorem says that in a right triangle, the sum of the squares of the two right-angle sides will always be the same as the square of the hypotenuse (the long side). A2 + B2 = C2

Pythagoras theorem can also help in real life. Here is an example:

Say you were walking though a park and wanted to take a short cut. With Pythagoras’s theorem you could work out exactly how long you would have to walk though the grass, rather then talking the long route by walking on the paths.

PLAN

I am going to investigate the three triangles I have been given. They are all right-angled triangles, with 3 sides, all different lengths.

The three triangles satisfy the Pythagoras theorem. The theorem states that the hypotenuse side (longest side) must equal the 2 shorter sides squared.

Here is the Pythagoras theorem:

PYTHAGORAS = a2 + b2 = c2

Here are the three triangles I have been given:

a) b) c)

I am now going to test if three triangles I have been given:

Triangle A = 32 + 42 = C2

9 + 16 = C2

25 = C2

5 = C

Middle

A = the first term

D = the first number in the changing sequence

C = the second, continuous difference

N = term

I will now substitute the numbers from the sequence into the formula.

Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c

4 + ( n – 1 ) 8 + ½ ( n – 1 ) ( n – 2 ) 4

4 + 8n – 8 + 2 [n2 – 3n + 2]

4 + 8n – 8 + 2n2 – 6n + 4

2n2 + 2n + 0

I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.

So when n = 7

2 x 72 + 2 x 7 + 0

2 x 49 + 14

98 + 14 = 112

In my sequence under the 7th term I have 112. This proves that my nth term is correct.

nth term for hypotenuse side

I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.

5 13 25 41 61 85 113

\ / \ / \ / \ / \ / \ /

8 12 16 20 24 28

\ / \ / \ / \ / \ /

4 4 4 4 4

Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c

5 + ( n – 1 ) 8 + ½ ( n – 1 ) ( n – 2 ) 4

5 + 8n – 8 + 2 [n2 – 3n + 2]

5 + 8n – 8 + 2n2 – 6n + 4

2n2 + 2n + 1

I will now again test this nth term using the same method as above.

So when n = 7

2 x 72 + 2 x 7 + 1

2 x 49 + 14 + 1

98 + 14 + 1 = 113

Both say the 7th term is 113, so it must be correct.

Testing the nth terms

I am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.

nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2

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

Iwill now expand the brackets for eachnth term.

( 2n + 1 )2 expanded:

( 2n + 1) ( 2n + 1)

= 4n2 + 4n + 1

( 2n2 + 2n )2 expanded:

( 2n2 + 2n ) ( 2n2 + 2n )

= 4n4 + 4n3 + 4n3 + 4n2

= 4n4 + 8n3 + 4n2

(2n2 + 2n + 1)2 expanded:

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

= 4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n +2n2 + 2n + 1

= 4n4 + 8n3 + 8n2 + 4n + 1

So: 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2= 4n4 + 8n3 + 8n2 + 4n + 1

8n2

Conclusion

nth term for hypotenuse side

15 39 75 123 183 255 339

\ / \ / \ / \ / \ / \ /

24 36 48 60 72 84

\ / \ / \ / \ / \ /

12 12 12 12 12

I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.

Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c

15 + ( n – 1 ) 24 + ½ ( n – 1 ) ( n – 2 ) 12

15 + 24n – 24+ 6 [n2 – 3n + 2]

15 + 24n – 24 + 6n2 – 18n + 12

6n2 + 6n + 3

I will now again test this nth term using the same method as above.

So when n = 7

6 x 72 + 6 x 7 + 3

6 x 49 + 45

294 + 45 = 339

In my sequence under the 7th term I have 336. This proves that my nth term is correct.

Both say the 7th term is 336, so it must be correct.

Testing the nth terms of ‘difference of 3’

I am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.

nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2

So: ( 6n + 3)2 + ( 6n2 + 6n )2 = (6n2 + 6n + 3)2

Iwill now expand the brackets for eachnth term.

( 6n + 3 )2 expanded:

( 6n + 3) ( 6n + 3)

= 36n2 + 36n + 9

( 6n2 +6 )2 expanded:

( 6n2 + 6n ) ( 6n2 + 6n )

= 36n4 + 36n3 + 36n3 + 36n2

= 36n4 + 72n3 + 36n2

(6n2 + 6n + 3)2 expanded:

(6n2 + 6n + 3) (6n2 + 6n + 3)

= 36n4 + 36n3 + 18n2 + 36n3 + 36n2 + 18n +18n2 + 18n + 9

= 36n4 + 72n3 + 72n2 + 36n + 9

So:

36n2 + 36n + 9 + 36n4 + 72n3 + 36n2= 36n4 + 72n3 + 72n2 + 36n + 9

72n2 + 36n + 9 + 72n3 + 36n4 = 36n4 + 72n3 + 72n2 + 36n + 9

36n4 + 72n3 + 72n2 + 36n + 9 = 36n4 + 72n3 + 72n2 + 36n + 9

This proves that the nth terms are all right and work with Pythagoras theorem.

Extension conclusion for ‘Difference of 3’ and final conclusion

Comparing the 3 nth terms I can conclude that when there is a different of 2 between the middle side and the hypotenuse side the nth term is doubled from when there was a difference of 1. When there is a different of 3 between the middle side and the hypotenuse side the nth term is tripled. I predict that when there is a difference of 4 between the middle side and the hypotenuse side and nth term will be quadrupled and so on.

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

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