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GCSE: Pythagorean Triples

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  1. GCSE Maths questions

    • Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
    • Level: GCSE
    • Questions: 75
  2. Math's Coursework: Pythagoras triples.

    * The lengths of the shortest sides are 2 more then the previous shortest side length. I'm going to extend the table with the wish that I find more patterns and figure out a formula. 9 40 41 90 180 11 60 61 132 330 Patterns of the Middle Value side. 4 8 12 4 12 24 4 16 40 4 20 60 I have found that the middle value has a difference of four. All of the middle values are devisable by 4; I'm going to divide all of middle values by four, then see what further patterns I can spot: 4 4 = 1 12 4 = 3 24 4 = 6 40 4 = 10 60 4 = 15 I have established that the middle values divided by four equal triangular numbers.

    • Word count: 881
  3. Beyond Pythagoras

    Must be one bigger than b I am now going to construct a table for these Pythagorean triples and I will try and work out patterns and formulas to help me work out the value of a, b and c and the perimeter and area of the triangles. (n) (a) (b) (c) (p) (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 5 11 60 61 132 330 6 13 84 85 182 546 7 15 112 113 240 840 8 17 144

    • Word count: 744
  4. Beyond Pythagoras

    = 25 x 25 = 625 and so 7� + 24� = 49 + 576 = 625 = 25� For the set of numbers 3, 4 and 5: Perimeter = 3 + 4 + 5 = 12 Area = 1/2 x 3 x 4 = 6 For the set of numbers 5, 12 and 13: Perimeter = 5 + 12 + 13 = 30 Area = 1/2 x 5 x 12 = 30 For the set of numbers 7, 24 and 25: Perimeter = 7 + 24 + 25 = 56 Area = 1/2 x 7 x 24 = 84

    • Word count: 937
  5. Beyond Pythagoras

    Further into my investigation I will also look at triangles that don't fit into the rules I've found, but whose smallest and middle length side when squared do add up to the longest length side squared. I began by checking to see if these triangles fit into pythagoras' triangle theorem: a) 5, 12, 13 b) 7, 24, 25 Both fit into the pattern. The numbers, 3, 4, 5 could be used to make a right angled triangle as shown below: The perimeter and area of this triangles can be worked out as follows: * Perimeter = 3 + 4 + 5 =12 units or smallest length + middle length + largest length = perimeter in appropriate unit.

    • Word count: 730
  6. 3 Digit Number - Maths Investigations

    Can I use Algebra to explain this? abc=100a+10b+c acb=100a+10c+b bac=100b+10a+c bca=100b+10c+a cba=100c+10b+a +cab=100c+10a+b 222a+222b+222c = 222(a+b+c) a+b+c =222 What if 2 of the 3 digits are the same? If 2 digits are the same : - 223 334 566 322 343 656 +232 +433 +665 777 ? 7=111 1110 ? 10=111 1887 ? 17=111 224 559 772 242 595 727 +422 +955 +277 888 ? 8=111 2109 ?19=111 1776 ? 16=111 It seems that when 2 of the 3 digits are the same the answer to the problem is 111.

    • Word count: 888

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