GCSE Maths questions

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• Level: GCSE
• Questions: 75

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

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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

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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

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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

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