• 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

Mathematics Coursework        

Beyond Pythagoras

Introduction

There once lived a Greek Mathematician/Philosopher called Pythagoras.  He discovered many interesting theories and I will be looking at one in particular.  Pythagoras’ theorem of right-angled triangles, which states one clear fact:

a2 + b2 = c2 or "The square of the hypotenuse is equal to the sum of the squares of the other two sides".  Other mathematical formulas I will be using to aid me in my tasks are:  Perimeter = a + b + cand

Area = ½ x a x b

I have decided to investigate the relationship between different aspects of right-angled triangles.  I will study the connections between all three sides, the perimeter and the area of the triangles.  To do this, I will begin by constructing a table of various measurements taken from some right-angled triangles.  The triangles I have chosen, all increase by one unit on the shortest side.  In my table, s=a, m=b and l=c:

Term Number (n)

Length of

shortest side (s)

Length of

middle side (m)

Length of

longest side (l)

Perimeter

Area

1

3

4

5

12

6

2

5

12

13

30

30

3

7

24

25

56

84

4

9

40

41

90

180

Initially, I calculated the relationship between the shortest side and the other two sides until

...read more.

Middle

By constructing another table, I discovered that the expression is:  

m = 2n2 + 2n

To check this, I can use the formula (n x s) + n.

Length of Longest Side

By referring back to my table, I noticed that the longest side is 1 unit larger than the length of the middle side, which makes them consecutive.  This means that the formula for the length of the longest side would be exactly the same as the formula for the length of the middle side, with an addition of 1 unit.  So my new formula is as follows: l = 2n2 + 2n + 1.  

The reason for this is that the difference between the length of the shortest side and the length of the longest side, and the actual length of the shortest side, are equal to the length of the longest side.

To check that my formula worked, I decided to find the formula using a different method.  I realized that the difference between s and l was 2n2 by calculating that the third difference was 4 and then added this to my formula for s, and, to prove my formula correct, the expression was the same.

Proof

To prove my formulas are correct, I referred back to the initial formula: a2 + b2 = c2

...read more.

Conclusion

2 + 12n + 4

Area:                a = (n + 3n2 + 2n3) x 2

                        a = 2n + 6n2 + 4n3

After applying my new formulas to the new table, I discovered that my prediction is correct.

Investigating Other Pythagorean Triples

I decided to delve deeper into Pythagorean triples (sides of a right-angled triangle that are integers).  I began by drawing a table with s = 2 and used my formulas to calculate the lengths of the other sides, to check if they were integers.  

S

m

l

 2

2

2.93

2

3

3.61

2

4

4.47

2

5

5.39

2

6

6.32

2

7

7.28

After looking at the table, I realised that there could not possibly be a Pythagorean triple with shortest side 2, because, the longest side would need to be an integer and it would obviously never reach a whole number.  We can see this because, in the l column the number is always l = m + 0.x.  Because of this fact, and the fact that in the first possible s and m row, x is only 93 and from then on x decreases, we know that l can never reach a whole number that is larger than m.

I then tried a table for s = 4 but achieved the same outcome.  As it would be impractical and extremely time consuming to conduct every possible table for every possible s, I believe that there are no Pythagorean triples where s is an even number, except those that are enlargements of the initial Pythagorean triple.  

Asha Lancaster Thomas                11JL

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem 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 Fencing Problem essays

  1. Medicine and mathematics

    5 11.66400 6 156.99840 7 94.19904 8 56.51942 9 33.91165 10 20.34699 11 12.20820 12 157.32492 13 94.39495 14 56.63697 15 33.98218 16 20.38931 17 12.23359 18 157.34015 19 94.40409 20 56.64245 21 33.98547 22 20.39128 23 12.23477 24 157.34086 25 94.40452 26 56.64271 27 33.98563 28 20.39138 29 12.23483

  2. Pythagoras Theorem is a2 + b2 = c2. 'a' being the shortest side, 'b' ...

    the perimeter is equal to the area To find this out I have been to the library and looked at some A-level textbooks and learnt 'Polynomials' 4n2 + 6n + 2 = 2n3 + 3n2 + n 6n + 2 = 2n3 - n2 + n 2 = 2n3 -

  1. Beyond Pythagoras

    This means that the formula is 2n. Middle side (b) I can tell just by looking at the numbers in this column that the numbers are the squares of the triangle numbers minus 1. This means that the formula for b is n2 - 1 Longest side (c)

  2. Beyond Pythagoras.

    5th term: 4n� + 6n� + 2 = perimeter 4 x 5� + 6 x 2 = 11 + 60 + 61 100 + 30 + 2 = 132 132 = 132 And it works for the 5th term And finally the 6th term: 4n� + 6n� + 2 =

  1. Beyond Pythagoras.

    25 40 + 1 41 60 + 1 61 84 + 1 85 So I took the formula for side 'b' and simply added one onto it: 2n2 + 2n + 1 I tested this formula by randomly selecting 2 terms.

  2. Beyond Pythagoras.

    I am going to use the first length of the middle side to use as an example to see if this formula works. By looking at my results table it tells me that the answer should be 4. Lets see.

  1. Beyond Pythagoras

    I will now test the formula to see if it works: N=8 2 x 8=16 + 1 = 17 N= 5 2 x 5=10 + 1 = 11 The formula works for the shortest side so the formula is: Shortest Side: 2n+1 MIDDLE SIDE: By looking at my table I

  2. Beyond Pythagoras

    I calculated the squares putting the equations in the brackets and then multiplying them out. c2 = a2 + b2 with odd integers. The formulae: c2 = (2n2 +2n+1) (2n2 +2n+1) The terms= 4n^4+4n�+2n�+4n�+4n�+2n+2n�+2n+1 Simplified=4n^4+8n�+8n�+4n+1 a2= (2n+1) (2n+1) The terms= 4n�+1+2n+2n Simplified=4n�+4n+1 b2=(2n2 +2n)

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