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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. Pythagoras Theorem

    To test that these were possible Pythagorean triples, I put them through the original Pythagoras theory: * 9� + 40� = 1,681 V1,681 = 41 * 11� + 60� = 3,721 V3,721 = 61 * 13� + 84� = 7,225 V7,225 = 85 * 15� + 112� = 12,769 V12,769 = 113 I had noticed earlier that 'a' was an odd number and I wanted to see if Pythagorean triples would work when 'a' was an even number. To make the value of 'a' even I decided to double the original three sets of triples: A B C A B

    • Word count: 1455
  3. Beyond Pythagoras

    1; 2(1)+1=3 If term number = 2; 2(2)+1=5 If term number = 3; 2(3)+1=7 This works for all Pythagorean triples that have an odd numbered SL. I have investigated another three Pythagorean triples and they are stated below. Serial No. Shortest side a Middle Side b Longest side c 1 3 4 5 2 5 12 13 3 7 24 25 4 9 40 41 5 11 60 61 6 13 84 85 I will now test the Pythagorean triples I have just found with the aid of Pythagoras theorem and a diagram.

    • Word count: 5018
  4. 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
  5. BEYOND PYTHAGORAS

    Here is a table showing the results of the 3 Pythagorean Triples: Triangle No. Shortest side Medium side Longest side Perimeter Area 1) 3 4 5 12 62 2) 5 12 13 30 302 3) 7 24 25 56 842 From the above table, I can see a few patterns emerging. Here they are: i. The shortest side is always an odd number. ii. The medium side is always an even number. iii. The medium side plus one equals to the Longest side. iv. The Longest side is always odd. 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.

    • Word count: 1172
  6. Investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers.

    This means that the equation I want has nothing to do with 3 sides squared, and I can eliminate this from my list of sides to investigate. I will now try 2 sides squared. (Middle)� + Largest number = (smallest number)� = 122 + 13 = 52 = 144 + 13 = 25 = 157 = 25 This does not work and I know that neither will 132, because it is larger than 122. I have decided that there is also no point in squaring the largest and the smallest or the middle number and the largest number.

    • Word count: 3070
  7. Beyond Pythagoras.

    They are always consecutive numbers. I shall now investigate this. I will assume that the hypotenuse has length b + 1 where b is the length of the middle side. (c=b+1) b+1 a b Using Pythagoras: a? + b? = (b+1) ? Expanding this: a? + b? = b? + 2b+ 1 therefore a? = 2b +1 This means that a? must be odd because it equals 2b + 1. Since a? is odd this means that a must be odd also. (even x even = even)

    • Word count: 4266
  8. Investigating families of Pythagorean triples.

    As a result, a = 2n + 1 The difference between the numbers in column b is 8 at first, then 12, then 16, then 20, and so on, adding 4 to the difference each time. Therefore, the second difference is 4. If we divide 4 by 2, we get 2. As a result, the first part of the formula for b is: 2n2 I then noticed that by adding double n to the above formula, I could get all the numbers in column b.

    • Word count: 2413
  9. I am going to investigate Pythagorean triples where the shortest side is an odd number and all 3 sides are positive integers - I will then investigate other families of Pythagorean triples to see if Pythagoras’ theorem (a²+b²=c²) works

    6 8 10 +1 +1 +1 +1 +1 2n+1 Middle side N 1 2 3 4 5 Sequence 4 12 24 40 60 1st differences +8 +12 +16 +20 2nd differences +4 +4 +4 a=4=2 2 2n� 2 8 18 32 50 +2 +4 +6 +8 +10 +2 +2 +2 +2 2n 2 4 6 8 10 +0 +0 +0 +0 +0 2n�+2n Longest side N 1 2 3 4 5 Sequence 5 13 25 41 61 1st differences +8 +12 +16 +20 2nd differences +4 +4 +4 a=4=2 2 2n� 2 8 18 32 50 +3 +5 +7 +9 +11 +2 +2 +2 +2 2n 2 4 6 8 10 +1 +1 +1 +1 +1 2n�+2n+1 Smallest side (2n+1)

    • Word count: 1528
  10. For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.

    They are always consecutive numbers. I shall now investigate this. I will assume that the hypotenuse has length b + 1 where b is the length of the middle side. (c=b+1) b+1 a b Using Pythagoras: a� + b� = (b+1) � Expanding this: a� + b� = b� + 2b+ 1 therefore a� = 2b +1 This means that a� must be odd because it equals 2b + 1. Since a� is odd this means that a must be odd also. (even x even = even)

    • Word count: 4280
  11. Beyond Pythagoras

    Area 1 6 8 10 24 24 2 8 15 17 40 60 3 10 24 26 60 120 4 12 35 37 84 210 5 14 48 50 112 336 6 16 63 65 144 504 7 18 80 82 180 720 8 20 99 101 220 990 9 22 120 122 264 1320 10 24 143 145 312 1716 To work out the area and perimeter, I used the known formulae; Perimeter = a + b + c Area = a x b 2 Odd Triples First, I need to work out `a' in terms of `n': Difference

    • Word count: 1741
  12. Beyond Pythagoras

    Then I will put my results in a table and look for a pattern that will occur. I will then try and predict the next results in the table and prove it. n Smallest ( Middle ( Longest ( 1 3 4 5 2 5 12 13 3 7 24 25 There is a clear pattern between the middle and longest side. There is also a sequence forming. n = 1 S = 3 M = 4 L = 5 n = 2 S = 5 M = 12 L = 13 n = 3 s = 7 M = 24 L = 25 2n +1 Sxn +n M+1 I have established a connection between n and S, S and M, and, M and L.

    • Word count: 1258
  13. 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
  14. Maths GCSE coursework: Beyond Pythagoras

    but this is a part of another pattern which is that the square of the shortest side is the same as the middle and longest sides added together, So... a� = b + c The theory works because: 12 +13 = 25 = 5� 5� = 12 + 13 This now allows to predict, odd starting, Pythagorean triples. It is correct! We have to, now, look at the smallest number. 3 5 7 9 11 13 2 2 2 2 2 1st difference The 1st difference is 2 so we must find the formula for the 'smallest number'.

    • Word count: 3537
  15. Beyond Pythagoras .

    One of the most important was Pherekydes who many describe as the teacher of Pythagoras. Another teacher of his was Thales who was said to have first introduced him to mathematical ideas. Although he created a strong impression on Pythagoras, he probably did not teach him a great deal. However he did contribute to Pythagoras's interest in mathematics and astronomy, and advised him to travel to Egypt to learn more of these subjects. Pythagoras became a frequent visitor to Egypt where he gave lectures about Astronomy, Mathematics and also Geology and Cosmology. Pythagoras founded a philosophical and religious school in Croton (on the east of the heel of southern Italy)

    • Word count: 1975
  16. Investigating Towers of Hanoi Mathematics coursework

    You have collection of disks and three piles into which you can place them. The left most pile is the starting pile you need to move the discs to the right most pile, never putting a disk on top of one, which is smaller. The middle pile is there for intermediate storage. The piles work like a triangle and a disc can be moved from one pile to any other pile. Looking for the Pattern. To figure out how many turns it will take for more than four disks, and to figure out how long it will take the monks to finish their task, you need to find a pattern relating to the number of disks to the minimum number of turns it takes to win the game.

    • Word count: 1252
  17. 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
  18. Beyond Pythagoras

    I also knew that c was probably one larger than b. I tried multiple triples until I found one that seemed to agree the theorem. I came up with 9, 40, 41. I proved that this was right: 92+402= 412 Because 92= 9x9= 81 402= 40x40= 1600 412= 41x41= 1681 81+1600= 1681 Next I needed to find the perimeter and area for this triple. I used the formula: Perimeter= a+b+c P= 9+40+41 P= 90 I found out the area using the formula: Area= (axb)?2 A= (9x40)?2 A=360?2 A=180 Next I updated my table: Length of Shortest Side (a)

    • Word count: 6029
  19. 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
  20. Beyond Pythagoras - Pythagorean Triples

    This is also true for the numbers 7, 24 and 25: 7 25 24 The perimeter and area for this triangle are : Perimeter = 7 + 24 + 25 = 56 Area = 1/2 x 7 x 24 = 84 Below is a table showing many Pythagorean triples: Length of shortest side (a) Length of middle side (b) Length of longest side (c) Perimeter Area 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84 9 40 41 90 180 11 60 61 132 330 13 84 85 182 546 15 112 113 240

    • Word count: 2638
  21. Beyond Pythagoras

    12 6 5 12 13 30 30 7 24 25 56 84 9 40 41 90 180 11 60 61 132 330 13 84 85 182 546 15 112 112 239 840 17 144 144 305 1224 Please find enclosed "sheet 1" To create this I used excel to find the Pythagorean triangles Basically I created one horizontal line of numbers going up one at a time and another vertical line the same. I used the formula =B1+1 (the cell B1 has a 1 in it)

    • Word count: 1605
  22. 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
  23. Dice Game

    Here are the results : ROLLABCROLLABC1WIN21WIN2WIN22WIN3WIN23WIN4WIN24WI N5WIN25WIN6WIN26WIN7WIN27WIN8WIN28WIN9WIN29WIN10WIN 30WIN11WIN31WIN12WIN32WIN13WIN33WIN14WIN34WIN15WIN35 WIN16WIN36WIN17WIN37WIN18WIN38WIN19WIN39WIN20WIN40WI N4115398 ABC72013 As you can see from this pie chart, in practice it suggests that B is the most likely player yo win the game. From looking at this I would predict that when I have worked out all of the probabilities B will have the largest chance of winning the game by far and A will have the smallest chance. ROUNDTHROWSROUND THROWS1122126212224123242338441124245132526626261371327128112 8259372913102530381126311112514322513383338144113439152535121611 364141712371318133861819253951420124013 This frequency graph shows that in practice the game is most likely to be won in the first round.

    • Word count: 1331
  24. Maths Number Patterns Investigation

    4 +5 = 32 9 = 9 And... 24 + 25 = 72 49 = 49 It works with both of my other triangles. So... Middle number + Largest number = Smallest number2 If I now work backwards, I should be able to work out some other odd numbers. E.g. 92 = Middle number + Largest number 81 = Middle number + Largest number I know that there will be only a difference of one between the middle number and the largest number. So, the easiest way to get 2 numbers with only 1 between them is to divide 81 by 2 and then using the upper and lower bound of this number.

    • Word count: 3143
  25. Beyond Pythagoras

    Therefore, the first odd Pythagorean triple to satisfy these criteria is the 3,4,5 triple, and it is subsequently the first term of the odd triples sequence. The first sequence of triples that will be investigated are those with the first number (i.e. 3 of 3,4,5) as an odd number. I will refer to these triples as "odd triples" throughout the investigation for convenience. The second sequence of triples that will be investigated are those with the first number (i.e. 6 of 6,8,10, although this triple should not be counted as a "true" triple, as we shall see later)

    • Word count: 4016

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