GCSE: Beyond Pythagoras
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 Level: GCSE
 Questions: 75

Beyond Pythagoras  I am investigating the relationships between the lengths of the three sides of right angled triangles, the perimeters and areas of these triangles.
is equal to the sum of the squares of the two sides, opposite and adjacent. A a + b = c AC = AB + BC B C Pythagoras' thereom helps us to solve the lights on a rightangled triangle. Method Step 1: Square them Square the two numbers, the sides of the triangle, you are given. Step 2: Add or Subtract To find the longest side, you add the two squared number. To find the shorter side, you subtract the smaller squared number from the larger one. Step 3: Square root After adding or subtracting, take the square root.
 Word count: 1600

Pyhtagorean Theorem
+ n I have also added 2 more terms using what I think are my formulae. 'n' 'a' 'b' 'c' Perimeter 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 Here are my formulas. They are formulas on how to get from n to all of the others:  1. n to a 2n+1 2. n to b 2n�+2n 3.
 Word count: 1272

Beyond Pythagoras
Checking this against what I have been given, I can verify that it was correct. I will now test the other two triangle I have been given. Triangle B = 52 + 122 = C2 25 + 144 = C2 169 = C2 13 = C This is also correct and matches with that I have been given. Triangle C = 72 + 242 = C2 49 + 276 = C2 625 = C2 25 = C All these triangles are Pythagorean triplets.
 Word count: 4003

Beyond Pythagoras
The 9,40,41 triple has a difference of 1 between its middle and longest side lengths, which is what the other triples I'd chosen have had, so it would be more reliable if I used the result which matched with my others instead of something completely different. If we wish to work out a formula, a general rule and a relationship then I am going to use the triple which is most likely to help with that. I have worked out the formulae, using the term number (n)
 Word count: 2758

Beyond Pythagoras
As you can see in the table below there are 5 sets of Pythagorean triples for family 1. I was given 3 Pythagorean triples to begin with from family 1 but I have worked out and added two additional Pythagorean triples to the table (which are in green) Family 1 S M L 3 4 5 5 12 13 7 24 25 9 40 41 11 60 61 13 84 85 As you can see I have spotted patterns in family 1 I should be able to find a formula if I separate each side of family 1.
 Word count: 3204

Beyond Pythagoras
By using the differencing method I can see that 'a' has a difference of 2 between each pythagorean triplet I predict the next two 'a' values to be 9 and 11 From the table I can also see that there is a quadratic sequence for the 'b' value so I predict the next two values to be 40 and 60 Also from my table, I can see the 'c' value is b+1 so my two next predictions I'm going to prove are: a b c 9 40 41 11 60 61 a� = 9� =81 b� = 40� =1600 c� = 41� =1681 a�+b� =81+1600 =1681 a�+b�=c� so Pythagoras's theory holds for (9,40,41)
 Word count: 1692

Pythagorean Theorem Coursework
+ n I have also added 2 more terms using what I think are my formulae. 'n' 'a' 'b' 'c' Perimeter 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 Here are my formulas. They are formulas on how to get from n to all of the others:  1. n to a 2n+1 2. n to b 2n�+2n 3.
 Word count: 1272

Beyond Pythagoras
The difference is 2 Next I shall find the prediction of the middle side next. 4,12,24 It goes up by 4,8,12. So in my conclusion I think it will become 4,8,12,16,20,24 So it will be 4,12,24, 40, 60 84. The difference is 4,8,12. Now I shall find the difference and it is n*4 Next I shall find the prediction of the longest side next. 5,13,25 It goes up by 4,8,12. So in my conclusion I think it will become 4,8,12,16,20, 24 So it will be 5,13,25,41,61 85. The difference is 4,8,12. Now I shall find the difference and it is n*4.
 Word count: 2399

Beyond Pythagoras P.1 Pythagoras Theorem is a2+b2= c2 'a' is being the shortest side, 'b' being the middle side and 'c' being
4,12,24 It goes up by 4,8,12. So in my conclusion I think it will become 4,8,12,16,20,24 So it will be 4,12,24, 40, 60 84. The difference is 4,8,12. Now I shall find the difference and it is n*4 Next I shall find the prediction of the longest side next. 5,13,25 It goes up by 4,8,12. So in my conclusion I think it will become 4,8,12,16,20, 24 So it will be 5,13,25,41,61 85. The difference is 4,8,12. Now I shall find the difference and it is n*4.
 Word count: 2424

Beyond Pythagoras
Part 1: Aim: To investigate the family of Pythagorean Triplets where the shortest side (a) is an odd number and all three sides are positive integers. By putting the triplets I am provided with in a table, along with the next four sets, I can search for formulae or patterns connecting the three numbers. Pythagorean Triplet (n) 1st Number (a) 2nd Number (b) 3rd Number (c) Area (cm) Perimeter (cm) 1 2 3 4 5 6 7 3 5 7 9 11 13 15 4 12 24 40 60 84 112 5 13 25 41 61 85 113 6 30 84 180 330 546 840 12 30 56 90 132 182 240 Investigation: Patterns in 'a': The smallest numbers always increase by 2 in this family.
 Word count: 1313

Pythagoras of Sámos.
Indeed, Pythagoras is said to have been driven from S�mos by his disgust for the tyranny of Polycrates and in 530 BC Pythagoras settled in Crotona, a Greek colony in southern Italy, where he founded a movement with religious, political, and philosophical aims, known as Pythagoreanism. It was a difficult choice for him to make, as his leaving the guidance of his instructors likely slowed his progress. Clearly that was not true; after settling in Crotona, it was there that he created some of his greatest works.
 Word count: 1357

Spoleèná dopravní politika.
V tomto duchu formuluje �koly spolecn� dopravn� politiky tak� R�msk� smlouva o EHS, kter� mluv� o zaveden� spolecn�ch pravidel mezin�rodn� dopravy, zaji�ten� voln�ho pr�stupu k poskytov�n� dopravn�ch slu�eb uvnitr ka�d�ho clensk�ho st�tu pro dopravce z dal��ch clensk�ch st�tu, o tom, �e ��dn� odvetv� dopravy nem� b�t st�ty preferov�no. Od toho se ocek�valo, �e se do sektoru vnesou konkurencn� podm�nky, tak�e spolecn� postup v dopravn� politice mel sn�it n�klady. Smlouva se omezovala na koordinaci postupu ve vybran�ch oborech dopravy: na �eleznic�ch, na silnic�ch a vnitrn�ch vodn�ch cest�ch (tedy bez n�morn� a leteck� dopravy).
 Word count: 1906

Towers of Hanoi.
I will now confirm that with four discs it is possible to get from the start (A) to the finish (B) or (C) in a minimum of 15 moves. This is the position I will begin my challenge from. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) Moves: 1B 2C 1C 3B 1A 2B 1B 4C 1C 2A 1A 3C 1B 2C 1C As I confirmed, it is possible to complete this task in a minimum of 15 moves. Investigation 2 In my second investigation I am going to try to successfully move one disc from the start (A)
 Word count: 1594

Was Maths invented or discovered?
If we first analyse 'Pythagoras Theorem' which is a concept 'discovered' by a man called Pythagoras of Samos. The theorem is that in a rightangled triangle the square of the length of the hypotenuse equals the sum of the squares of the other two sides. It would be incorrect to define Pythagoras Theorem as an invention. Before Pythagoras himself found out about this theorem, a� already equalled to b� + c�, therefore he simply discovered this fact. Another example of a mathematical discovery is Pi, which is used as the symbol for the ratio of the circumference of a circle to its diameter.
 Word count: 1149

Research on Pythagoras and his work.
It is probable that he had two brothers although some sources say that he had three. Certainly he was well educated, learning to play the lyre, learning poetry and to recite Homer. There were, among his teachers, three philosophers who were to influence Pythagoras while he was a young man. One of the most important was Pherekydes who many describe as the teacher of Pythagoras. The other two philosophers who were to influence Pythagoras, and to introduce him to mathematical ideas, were Thales and his pupil Anaximander who both lived on Miletus.
 Word count: 5475

Beyond Pythagoras.
co cb uk cb for more cb Do not cb redistribute Tochv8Ey 85 182 546 7 15 112 113 240 840 +2 8 17 144 145 306 1224 9 19 180 181 380 1710 wwdd ddw stddddud edd ddnt cdd enddtral ddcodd uk. 10 21 220 221 462 2310 I looked at the table and noticed that there was only 1 difference between the length of the middle side and the length of the longest side. And also if you can see in the shortest side column, it goes up by 2.
 Word count: 2107

Investigation into a driving test.
Fri 17.00 M 19 6 A Fri 14.00 F 26 9 A Fri 16.00 F 31 7 A Tue 12.00 F 14 17 A Tue 11.00 M 18 19 A Thur 14.00 M 21 20 A Fri 17.00 F 11 14 A Wed 15.00 F 6 27 A Wed 10.00 M 13 17 A Fri 16.00 M 24 9 B Fri 13.00 M 13 28 B Mon 12.00 M 32 B Fri 11.00 F 10 22 B Fri 16.00 F 12 33 B Thur 14.00 F 17 19 B Fri 12.00 F 23 3 B Wed 11.00 F 13 19
 Word count: 5853

The Die Investigation.
I first set up a tree diagram to show what is happening. A B C W 1/6 * W 10/36 W 60/216 L L L # W  Wins L  Loses At # the diagram then starts again at * and follows this pattern continuously except the final probabilities will be different. Next I wrote down in a table the probabilities of A, B and C winning on their first, second, third and fourth attempts. Multiplying down through the probability tree's branches did this. Attempt 1 Attempt 2 Attempt 3 Attempt 4 A 1/6 60/1296 3600/279936 216000/60466176 B 10/36 600/7776 36000/1679616 2160000/362797056 C 60/216 3600/46656 216000/10077696 12960000/2176782336 From this I noted
 Word count: 1276

Beyond Pythagoras
This is because: Triangle Number 1 = 2 x 2 2 = 3 x 4 3 = 4 x 6 4 = 5 x 8 = 2n2 + 2n 5 = 6 x 10 6 = 7 x 12 7 = 8 x 14 8 = 9 x 16 Longest side = 2n2 + 2n + 1 Perimeter = A + B + C Area = A x B x 0.5 Box Methods Shortest� = (2n + 1) � 2n +1 2n 4n� 2n +1 2n 1 = 4n� + 4n + 1 Middle� = (2n� + 2n)� 2n� +2n
 Word count: 717

Maths Dice Investigation
This is obviously because we used 10 counters. After carefully looking at the results table it becomes apparent that "A Wins" "B Wins" Win = 1 Win = 6 Draw = 3 Draw = 3 Lose = 6 Lose = 1 We can write this as a probability equation: P(A Wins) = 1/10 P(B Wins) = 6/10 P(Draw) = 3/10 And P(A Wins) = 41/100 P(B Wins) = 59/100 Conclusion: From the results I have produced it becomes evident that my prediction was correct.
 Word count: 904

Pythagorean triplets
To work out the perimeter we use the condition: a + b + c = units 1st triplet  3 + 4 + 5 = 12 units 2nd triplet  5 + 12 + 13 = 30 units 3rd triplet  7 + 24 + 25 = 56 units To work out the area we use the condition: 1/2 * a * b = square units 1st triplet  1/2 * 3 * 4 = 6 square units 2nd triplet  1/2 * 5 * 12 = 30 square units 3rd triplet  1/2 * 7 * 24 = 84
 Word count: 767

Dice Maths Investigation
Seeing as the probability of the round passing is exactly equal to the probability of C winning in the first round, this value can only be 5/18. Therefore a player's winning probability is decreased by means of multiplication by a fraction less than one (5/18). When I come to create a formula, this will be very useful as I now know that it must involve 5/18. Formulae to find the probability of either A, B or C winning in the nth round.
 Word count: 2900

Mathematics Coursework  Beyond Pythagoras
4 +5 = 32 9 = 9 And... 24 + 25 = 7 49 = 49 It works with both of my other triangles. So this means that the Middle number + Largest number = (Smallest number) 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.
 Word count: 3132

Dice Game Maths Investigation
Method To do this investigation  first I played the game. We were told by our teacher to play it 30 times and to take down the results with ultimate care  which we did successfully. Results After carrying out the experiment in a careful and strategic way I found some interesting results. I found that A won just twice (2/30), B won thirteen time (13/30) and C won a total of fifteen times (15/30). This shows that at this stage that C is the more likely winner.
 Word count: 1716

Beyond Pythagoras  Year 10 Maths Coursework
The only number now I can try squaring is the smallest number. 12 + 13 = 5� 25 = 25 This works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles I know. 4 +5 = 3� 9 = 9 And... 24 + 25 = 7� 49 = 49 It works with both of my other triangles. So... M + L = S� If I now work backwards, I should be able to work out some other odd numbers.
 Word count: 3461