• 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

Beyond Pythagoras Pythagoras Theorem is a� + b� = c�. 'a' being the shortest side, 'b' being the middle side and 'c' being the longest side (which is always the hypotenuse) of a right angled triangle. The numbers 3, 4 and 5 satisfy this condition: 3� + 4� = 5� because 3� = 3 x 3 = 9 4� = 4 x 4 = 16 5� = 5 x 5 = 25 and so 3� + 4� = 9 + 16 = 25 = 5� We also checked to see if similar sets of numbers also satisfy this condition: (smallest number)� + (middle number)� = (largest number)� The numbers 5, 12 and 13 also satisfy this condition: 5� + 12� = 13� because 5� = 5 x 5 = 25 12� = 12 x 12 = 144 13� = 13 x 13 = 169 and so 5� + 12� = 25 + 144 = 169 = 13� The numbers 7, 24 and 25 also satisfy this condition: 7� + 24� = 25� because 7� = 7 x 7 = 49 24� = 24 x 24 = 576 25� ...read more.

Middle

n To get these formulas I did the following: Take side 'a' for the first five sets of numbers; 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because they are consecutive odd numbers. From looking at my table of results, I noticed that 'an + n = b'. So I took my formula for 'a' (2n + 1) multiplied it by 'n' to get '2n� + n'. I then added my other 'n' to get: 2n� + 2n. Side 'c' is just the formula for side 'b' +1: 2n� + 2n + 1 The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1), my formula for 'b' (2n� + 2n) and my formula for 'c' (2n� + 2n + 1). Then I did the following: 2n + 1 + 2n� + 2n + 2n� + 2n + 1 This can be rearranged to equal: 4n2 + 6n + 2 The area = (a x b) ...read more.

Conclusion

= (a + 2d)� a� + a� + ad + ad + d� = (a + 2d)� 2a� + 2ad + d� = (a + 2d)� 2a� + 2ad + d� = (a + 2d)(a + 2d) 2a� + 2ad + d� = a� + 2ad + 2ad + 4d� 2a� + 2ad + d� = 4d� + a� + 4ad If you equate these equations to 0 you get the following: a� - 3d� - 2ad = 0 Change a to x: x� - 3d� - 2dx = 0 Factorise this equation to get: (x + d)(x - 3d) Therefore: x = -d x = 3d x = -d is impossible as you cannot have a negative dimension. a, a+d, a + 2d Is the same as: 3d, 4d, 5d This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc. Mathematics GCSE Coursework Beyond Pythagoras Luke Hopwood 11B Candidate number: 7484 The Mirfield Free Grammar ...read more.

The above preview is unformatted text

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

  1. Maths GCSE coursework: Beyond Pythagoras

    12 24 40 60 84 8 12 16 20 24 1st difference 4 4 4 4 2nd difference The 2nd difference is 4 so we must find the formula for the 'middle number'. 4 / 2 = 2 [4 is halved due to the use of n� So we must

  2. Maths Number Patterns Investigation

    It now looks as if "4n2 - 4(n - 1)2" is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.

  1. Pythagoras Theorem

    2a� + 1 = a4 + 2a� + 1 = a4 + 2a� + 1 = a4 +2a� +1 By cancelling the formula down, so that it is the same on both sides, I have proven that the formulae: o b = (a� - 1)/2 o c = (a� +

  2. Beyond Pythagoras

    = 145 Longest side = 145 I would like to check if the values found are accurate with the aid of a Pythagoras theorem. a2 + b2 = c2 172 + 1442 = 142 21025 = 21025 The results show that (17,144,145)

  1. BEYOND PYTHAGORAS

    M is always even. iii. L is always even. iv. M+2=L. To show that these patterns are correct and to see if there are any more patterns, I am going to extend this investigation and draw two more Pythagorean Triples.

  2. Beyond Pythagoras.

    However: 1x2=2 2x2=4 3x2=6 This is wrong. However I have noticed a pattern. Each of the numbers in this pattern is one less than the one I was aiming to get. This must mean that 2n+1 is the correct formula To check this I will pick a random number between

  1. Beyond Pythagoras

    M= Sxn +n M= (2n +1) xn +n = 2n�+n+n = 2n�+2n So the general rule for the middle side is: 2n�+2n Finally I am going to find the general rule for the longest side. L= M+1 L= (2n�+2n) +1 = 2n�+2n +1 So the general rule for the longest

  2. Beyond Pythagoras .

    to list all the Pythagorean triples but only the triples where the smallest number is an odd number. I came up with many triples but I decided to try and find patterns and observations by using only eight triples. Here are the triples.

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