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  • Level: GCSE
  • Subject: Maths
  • Word count: 2638

Beyond Pythagoras - Pythagorean Triples

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Introduction

Beyond Pythagoras Pythagorean Triples: Three integers a, b, and c that satisfy a2 + b2 = c2 are called Pythagorean Triples. a2 + b2 = c2 The numbers 3, 4 and 5 satisfy the condition because: 32 + 42 = 52 32 = 3 x 3 = 9 42 = 4 x 4 =16 52 = 5 x 5 = 25 32 + 42 = 9 +16 = 25 = 52 1. Each of the following sets of numbers satisfy a similar condition of (smallest number)2 + (middle number)2 = (largest number)2 a) 5, 12, 13. 52 + 122 = 132 52 = 5 x 5 = 25 122 = 12 x 12 = 144 132 = 13 x 13 = 169 52 + 122 = 25 +144 = 169 = 132 b) 7, 24, 25 72 + 242 = 252 72 = 7 x 7 = 49 242 = 24 x 24 = 576 252 = 25 x 25 = 625 72 + 242 = 49 +576 = 625 = 252 The numbers 3, 4 and 5 can be the lengths - in appropriate units - of the sides of a right-angled triangle. 3 5 4 The perimeter and area of this triangle are : Perimeter = 3 + 4 + 5 = 12 units Area = 1/2 x 3 x 4 = 6 square units The numbers 5, 12, and 13 can also be the lengths - in appropriate units - of a right-angled triangle : 5 13 12 The perimeter and area of this triangle are : Perimeter = 5 + 12 + 13 = 30 Area = 1/2 x 5 x 12 = 30 (c) ...read more.

Middle

Length of longest side (c) Perimeter Area 9 12 15 36 54 15 36 39 90 270 21 72 75 168 756 27 120 123 270 1620 33 180 183 396 2970 39 252 255 546 4914 45 336 339 720 7560 51 432 435 918 11016 From the above table I have noticed that there is a constant difference of 6 in the first column. I have also noticed that the third column is now obtained by adding 3 to column (b) this time, not 2. Column (b) can be again made using triangle numbers. Generalisation: (a) = Term no. 1 2 3 4 5 Sequence 9 15 21 27 33 Difference 6 6 6 6 The nth term for this is 6n + 3 (b)= Term no 1 2 3 4 5 Sequence 12 36 72 120 180 6n2 6 24 54 96 150 R2-R3 6 12 18 24 30 Difference 6 6 6 6 The nth term for this is 6n2 + 6n (c)= The formula for this term can be found using the nth term formula for column (b). It will simply be 6n2 + 6n + 3 Proof a2 + b2 = c2 ? (6n + 3) 2 + (6n2 + 6n) 2 = (6n2 + 6n + 3) 2 36n + 9 + 36n4 + 36n = 36n4 + 36n2 + 9 36n4 + 36n2 + 9 = 36n4 + 36n2 + 9 LHS = RHS Perimeter = To find the perimeter, you have to (a) + (b) + (c) = perimeter 6n2 + 6n + 3 6n2 + 6n + 6n + 3 12n2 + 18n + 6 Area = To find the area, you have to: 1/2 x (a) ...read more.

Conclusion

+ (b) + (c) = perimeter so... 16n2 + 16n + 8 16n2 + 16n + 16n + 8 32n2 + 48n + 16 Area = To find the area, you have to: 1/2 x (a) x (b) so... (16n2 + 16n) (16n + 8) 2 = 256n3 + 256n2 + 128n2 + 128n 2 = 256n3 + 384n2 + 128n 2 After investigating enlarging family of triples by a certain scale factor I can predict that the triples obtained by, for example, enlarging the odd family of triples by 7, will give the even family of triples enlarged by 7, when doubled. This is the same for all scale factors. I have also found that in the original odd family of triples, adding the middle and longest side and the square rooting the answer will give the shortest side. For example: 3,4,5 5,12,13 3 = (4 + 5) 5 = (12 + 13) A formula for generating all Pythagorean Triples is: a = t2 - s2 , b = 2st , c = t2 + s2 in which s and t are integer positive numbers, neither both even or both odd and in which s < t. by choosing values of s and t properly, Pythagorean Triples like these fore example are formed : s t (a) (b) (c) 1 2 3 4 5 2 3 5 12 13 1 3 8 6 10 Proving the formula The only triple that consists of consecutive integers is 3,4,5. Here are some properties that I found apply to all Pythagorean Triples : 1) Either (a) or (b) is divisible by 3. 2) Either (a) or (b) is divisible by 4. 3) Either (a), (b) or (c) is divisible by 5. 4) The product of (a), (b) and (c) is divisible by 60. 5) 6) 7) 8) ?? ?? ?? ?? ...read more.

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