• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Beyond Pythagoras

Extracts from this document...

Introduction

Maths Pure Coursework 1

Beyond Pythagoras

image00.jpg

Maths Pure Coursework 1

By: Ben Ingram

10R


Beyond Pythagoras

Pythagoras Theorem:

Pythagoras states that in any right angled triangle of sides ‘a’, ‘b’ and ‘c’ (a being the shortest side, c the hypotenuse): a2 + b2 = c2

image01.png

E.g. 1.    

32 + 42 = 52

9 + 16 = 25

52 = 25

2. 52 + 122 = 132                                                 3. 72 + 242 = 252

      25 + 144 = 169                                                      49 + 576 = 625

                         132  = 169                                                          252 = 625

All the above examples are using an odd number for ‘a’. It can however, work with an even number.

E.g. 1. 102 + 242 = 262

          100 + 576 = 676

                   262 = 676

N.B. Neither ‘a’ nor ‘b’ can ever be 1. If either where then the difference between the two totals would only be 1. There are no 2 square numbers with a difference of 1.

     32                      9

     42                      16

     52                      25

     62                      36  

     72                      49

     82                      64

     92                       81

     102                     100

     112                      121

As shown in the above table, there are no square numbers with a difference of anywhere near 1.


Part 1:

Aim: To investigate the family of Pythagorean Triplets where the shortest side (a) is an odd number and all three sides are positive integers.

...read more.

Middle

Investigation:

Patterns in ‘a’: The smallest numbers always increase by 2 in this family.

Relations between ‘n’ and ‘a’: The only pattern that I can see in these two sets of numbers is that ‘a’ is always double ‘n’ plus 1. E.g. 1 and 3

                                   1 x 2= 2

                                   2 + 1= 3

This works with all of the above pairs. It can be expressed algebraically as:  2n + 1=a

Patterns in ‘b’: The numbers in the ‘b’ column increase uniformly in a pattern that in itself increases by 4 every time.

8     12     16     20     24     28

                                            4      4      4      4       4

 This however cannot be used to form a formula as it could not be worked out for the nth term.

Relations between ‘a’ and ‘b’: The relationship here is that multiplying ‘a’ by ‘n’ then adding ‘n’ always leaves you with ‘b’. E.g. 1 and 5

                             5 x 1= 5

                              5 + 1= 6

This works for each one and can be expressed as the formulae: an + a= b

Relations between ‘b’ and ‘c’: The obvious relation here is that ‘c’ is always one more than ‘b’. E.g. 5 is one more than 4

     13 is one more than 12

     and so on…

Expressed algebraically this is: b + 1= c

However, these formulae only work if you do them in order. I.e.

...read more.

Conclusion

The fact that the two smaller numbers (when squared) add up to the same as the third number (when squared), proves my theory that multiples of Pythagorean triplets will also adhere to the rules of Pythagoras. My formulae can be changed to suit this:

y (2n + 1 = a)

y (2n2 + 2n = b)

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

y (4n2 + 6n + 2 = perimeter)

y (2n³ + 3n² + n = area)

In all the above cases, ‘y’ is a substitute for any number, when replaced it can be worked the same as the normal formula, multiplied and still produce a Pythagorean triplet.

E.g. 3 x ((2 x 4) + 1) = 27

     3 x ((2 x 4)² + 8) = 120

                 120 + 3 = 123

Once again the triplets from this family can be put into a table showing not only themselves, but also the fact that they conform with the rules of Pythagoras. In this case it is the multiples of my primary triplet; 3,4,5.

a

b

c

a² + b²

3

6

9

12

15

18

21

24

27

30

4

8

12

16

20

24

28

32

36

40

5

10

15

20

25

30

35

40

45

50

25

100

225

400

625

900

1225

1600

2025

2500

25

100

225

400

625

900

1225

1600

2025

2500

As you can see, when squared, the first two columns add up to the third number squared. This proves that my hypothesis is correct and multiples of triplets do conform with the rules of Pythagoras. This is true with both odd and even numbers being inserted into the ‘a’ column.

Ben Ingram        Page  of         10R

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Beyond Pythagoras essays

  1. Beyond Pythagoras

    +1)=13 It works for that one, as I can see on my table that with term 2, the longest side length is 13. (2x3(3+1) +1)=25 It works for that one, as I can see on my table that with term 3, the longest side length is 25.

  2. Beyond Pythagoras - I am investigating the relationships between the lengths of the three ...

    =12 2 x 3(3+1) =24 2 x 4(4+1) =40 2 x 5(5+1) =60 2 x 6(6+1) =84 2 x 7(7+1) =112 2 x 8(8+1) =144 2 x 9(9+1) =180 2 x 10(10+1) =220 etc. nth term for 'Length of longest side' 5, 13, 25, 41, 61, 85, 113, 145, 181, 221...

  1. Beyond Pythagoras

    Test that triangles are Pythagoras I will now test to see if the three new triangles I have got numbers for do comply with Pythagoras theorem. I will to this by adding side a2 and side b2 and see if I get the answer I get matches what I got for side C in my table.

  2. Beyond Pythagoras

    goes up by +4 every time from what you added to get your previous sequence. I also found that if you use the 2nd difference (the amount you always add on to your last total which gave you your previous number in the sequence)

  1. Beyond Pythagoras

    I am now going to find a general formula for the 2nd pythagorean family that I will name the b+2 family. From the values I have I know a = 2n+4 by using the differencing method I will now work out the 'b' and 'c' values.

  2. Pythagorean Theorem Coursework

    'n' 'a' 'b' 'c' Perimeter Area 1 2 0 2 4 0 2 4 3 5 12 6 3 6 8 10 24 24 4 8 15 17 40 60 5 10 24 26 60 120 Here are my formulae on how to get from n to all of the others:- 1.

  1. Beyond Pythagoras

    2 - (2n+1)2 Shortest (2n+1)* (2n+1) 4n2+ 2n 2n + 1 4n+ 4n2+1 Middle (2n2+2n) * (2n2+2n) 4n4+4n3 4n3+ 4n2 4n2 +8n3+4n4 Longest (2n2+2n+1) * (2n2+2n+1) 4n4+4n3+2n2 4n3+ 4n2+2n 2n2+2n+1 4n4+8n3+8n2+4n+1 (4n2 +8n3+4n4)= (4n4+8n3+8n2+4n+1)- (4n+ 4n2+1) Finally I will investigate the shortest Term. Mahmoud Elsherif Beyond Pythagoras P.7 Shortest Term2= Longest Term2- Middle Term2 (nth term)2= (nth term)2 + (nth term)2 (2n2+1)2 = (2n2+2n+1)

  2. Beyond Pythagoras ...

    I shall do this all over again but with an even short side. Shortest Side Middle Side Longest Side 6 8 10 10 24 26 14 48 50 18 80 82 22 120 122 26 168 170 Mahmoud Elsherif Beyond Pythagoras P.8 I shall find the prediction of the shortest side first.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work