Beyond Pythagoras - Pythagorean Triples
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 . 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) 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
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
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. Smallest side Middle side Longest side 3 4 5 +1 (from middle side number) +2 +8 5 12 +4 13 +1 (from middle side number) +2 +8 7 24 +4 25 +1 (from middle side number) +2 +8 9 40 +4 41 +1 (from middle side number) +2 +8 11 60 61 +1 (from middle side number) a²+b²=c² a²+b²=c² a²+b²=c² 3²+4²=5² 7²+24²=25² 11²+60²=61² 9+16=25 49+576+625 121+3600=3721 25=25 625=625 3721=3721 There was only one pattern I noticed in the smallest side, which was the difference of two between each number. However, in the middle side the first difference was +8. This then increased by +4. Therefore, the difference between each number was +4. The longest side patterns were very easy to find. The number was one extra than the number before in the middle side i.e. 4(+1) =5. Smallest side N 1 2 3 4 5 Sequence 3 5 7 9 11 st differences +2 +2 +2 +2 2n 2 4 6 8 10 +1 +1 +1 +1 +1 2n+1 Middle side N 1 2 3 4 5 Sequence 4 12 24 40 60 st differences +8 +12 +16 +20 2nd differences +4 +4 +4 a=4=2 2 2n² 2
3 Digit Number - Maths Investigations
CWK1 3 Digit Number Take any 3 digit number, write down all possible numbers that can be made with the three digits, add them up, divide the total by the sum of the 3 digits. Investigate. If All Digits Are Different : - 123 456 789 132 465 798 213 546 879 231 564 897 321 654 987 +312 +645 +978 332 ? 24=222 3330 ? 15=222 5328 ? 24=222 147 258 369 174 285 396 471 528 639 417 582 693 741 825 936 +714 +852 +963 2664 ? 12=222 3330 ? 15=222 3996 ? 18=222 It seems that when all 3 digits are different the answer to the problem is 222. 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. Can I use Algebra to explain this? aab=100a+10a+b aba=100a+10b+a +baa=100b+10a+a 222a+111b = 111(2a+b) 2a+b =111 What if all 3 digits are the same number? If All Digits Are The Same : - 333
Maths Investigation: Number of Sides
The numbers 3, 4 and 5 satisfy the condition 32+42=52, Because 32= 3x3 =9 42= 4x4 =16 52= 5x5 =25 And so... 32+42=9+16=25=52 I now have to find out if 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 = 25+144 = 169 = 132. b) 7, 24, 25 72+242 = 49+576 = 625 +252 Here is a table containing the results: I looked at the table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side. I already know that the (smallest number) 2+ (middle number) 2= (largest number) 2. So I know that there will be a connection between the numbers written above. The problem is that it is obviously not: (Middle number) 2+ (largest number) 2= (smallest number) 2 Because, 122 + 132 = 144+169 = 313 52 = 25 The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared. I will now try 2 sides squared. (Middle)2 + Largest number = (smallest number)2 = 122 + 13 = 52 = 144 + 13 = 25 = 157 = 25 This does not work and neither will 132, because it is larger than 122. There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side squared. 22 + 13 = 5 This couldn´t work because 122 is already
Beyond Pythagoras
Beyond Pythagoras Aim I am conducting this investigation to discover formulae that will allow me to calculate many Pythagorean triples. I will first find formulae for odd Pythagorean triples and then even ones. For each, I will find formulae for `a', `b', `c', perimeter and area, all in terms of `n'. After this I hope to discover a general formula for all Pythagorean triples, although it is unlikely that I will. Contents Page Number 1 Aim and Contents 2 Odd and Even Tables of Pythagorean triples 3 Odd Formula for `a' in terms of `n' 4 Odd Formula for `b' and `c' in terms of `a' 5 Odd Formula for `b' and `c' in terms of `n' 6 Odd Perimeter and Area in terms of `n' 7 Even Formula for `a' in terms of `n' 8 Even Formula for `b' and `c' in terms of `a' 9 Even Formula for `b' and `c' in terms of `n' 10 Even Perimeter and Area in terms of `n' 11 Odd and Even Summary - All Formulae Odd Pythagorean Triples Table of first 10 odd triples 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 6 13 84 85 182 546 7 15 112 113 240 840 8 17 144 145 306 1224 9 19 180 181 380 1710 10 21 220 221 462 2310 Even Pythagorean Triples Table
Beyond Pythagoras
Beyond Pythagoras Pythagoras Theorem is a² + b² = c². 'a' being the shortest side, 'b' being the middle side and 'c' being the longest side (which is always the hypotenuse) of a right angled triangle. The numbers 3, 4 and 5 satisfy this condition: 3² + 4² = 5² because 3² = 3 x 3 = 9 4² = 4 x 4 = 16 5² = 5 x 5 = 25 and so 3² + 4² = 9 + 16 = 25 = 5² We also checked to see if similar sets of numbers also satisfy this condition: (smallest number)² + (middle number)² = (largest number)² The numbers 5, 12 and 13 also satisfy this condition: 5² + 12² = 13² because 5² = 5 x 5 = 25 12² = 12 x 12 = 144 13² = 13 x 13 = 169 and so 5² + 12² = 25 + 144 = 169 = 13² The numbers 7, 24 and 25 also satisfy this condition: 7² + 24² = 25² because 7² = 7 x 7 = 49 24² = 24 x 24 = 576 25² = 25 x 25 = 625 and so 7² + 24² = 49 + 576 = 625 = 25² 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 From these sets of numbers I have noticed the following: - * 'a' increases by +2 for each set of numbers * 'b' increases by +4 for each set of numbers * 'c' is
Beyond Pythagoras
Beyond Pythagoras This investigation is to study Pythagoras Theorem. I will try to find patterns and formulae to help predict Pythagorean Triples. About Pythagoras Pythagoras was a Greek Philosopher and Mathematician who is believed to have lived in the 6th century BC. He discovered many theorems but his most famous was: a2+b2= c2 What is a Pythagorean Triple? To answer this I first need to explain Pythagoras Theorem. Pythagoras States that in any right-angled triangle, a2+b2=c2. a is the shortest side, b the middle length side and c the hypotenuse (the longest side). A Pythagorean Triple is any set of integers that agrees this condition. For example 3, 4, 5 is a Pythagorean Triple because: 32+42=52 Because 32= 3x3= 9 42= 4x4= 16 52= 5x5= 25 9+16= 25 . The numbers 5, 12, 13 satisfy the condition: 52+122=132 Because 52= 5x5= 25 22= 12x12= 144 32= 13x13= 169 25+144= 169 The numbers 7, 24, 25 72+242= 252 Because 72= 7x7= 49 242= 24x24= 576 252= 25x25= 625 44+576= 625 2. (a) I found perimeter by using the formula: Perimeter= a+b+c P= 5+12+13 P= 30 Perimeter= a+b+c P= 7+24+25 P= 56 I found area using the formula: Area= (axb)?2 Area= 0.5x5x12 Area= 30 Area= (axb)?2 Area= 0.5x24x25 Area= 84 (b) I next put the results for perimeter and area into the table below. Length of Shortest Side (a) Length of Middle Side (b) Length of
Beyond Pythagoras.
BEYOND PYTHAGORAS Introduction For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician. Pythagoras lived on the island of Samos and was born around 569BC. He did not write anything but he is regarded as one of the world's most important characters in maths. His most famous theorem is named after him and is called the Pythagoras Theorem. It is basically a?+b?=c?. This is what the coursework is based on. I am going to look at the patterns, which surround this theorem and look at the different sequences that can be formed. The coursework The numbers 3, 4 and 5 satisfy the condition 3?+4?=5? because 3?=3x3=9 4?=4x4=16 5?=5x5=25 And so 3?+4?=9+16=25=5? I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ?+(middle number) ?=(largest number) ? a) 5, 12, 13 5?=5x5=25 2?=12x12=144 25+144=169 V169 = 13 This satisfies the condition as 5?+12?=25+144=169=13? b) 7, 24, 25 7?=7x7=49 24?=24x24=576 49+576=625 V625=25 This satisfies the condition as 7?+24?=49+576=625=25? The numbers 3,4 and 5 can be the lengths - in appropriate units - of the side of a right-angled triangle. 5 3 The perimeter and area of this triangle are: Perimeter = 3+4+5=12 units Area = ?x3x4=6 units ? The numbers 5,12,13 can also be the lengths - in appropriate units - of a right-angled triangle.
Beyond Pythagoras
BEYOND PYTHAGORAS By: Megan Garibian 0A What this coursework has asked me to do is to investigate and find a generalisation, for a family of Pythagorean triples. This will include odd numbers and even numbers. I am going to investigate a family of right-angled triangles for which all the lengths are positive integers and the shortest is an odd number. I am going to check that the Pythagorean triples (5,12,13) and (7,24,25) cases work; and then spot a connection between the middle and longest sides. The first case of a Pythagorean triple I will look at is: The numbers 5, 12 and 13 satisfy the connection. 5² + 12² = 13² 25 + 144 = 169 69 = 13 The second case of a Pythagorean triple I will look at is: The numbers 7, 24 and 25 satisfy the connection. 7² + 24² = 25² 49 + 576 = 625 625 = 25 There is a connection between the middle and longest side. This is that there is a one number difference. So if M= middle and L= longest L = M + 1 I am going to use the triples, (3,4,5), (5,12,13) and (7,24,25) to find other triples. 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 ( 3 4 5 2 5 2 3 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
Beyond Pythagoras
Beyond Pythagoras The aim of this investigation is to investigate Pythagoras theorem and to find a formula for the shortest side, middle length, hypotenuse, area and perimeter. Because I have typed this up on a computer... ^ is squared * is times / is divided Pythagoras is A2 + B2 = C2 I am going to prove this theory be finding out if the following numbers adhere to the rule. Triangle 1 5 12 13 5^ + 12^ = 13^ 5^ = 5*5 = 25 2^ = 12*12 = 144 3^ = 13*13 = 169 So 5^ + 12^ = 25 + 144 = 169 = 25^ The perimeter of the triangle is All the lengths of the side added up 5 + 24 + 25 = 30 The area of the triangle is /2 base * height /2 * 12 * 5 = 30 Triangle 2 7 24 25 7^ + 24^ =25^ 7^ = 7*7 = 49 24^ = 24*24 = 576 25^ = 25*25 = 625 So 7^ + 24^ = 49 + 576 = 625 = 25^ The perimeter of the triangle is All the lengths of the side added up 7 + 24 + 25 = 56 The area of the triangle is /2 base * height /2 * 24 * 7 = 84 Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 2 6 5 2 3 30 30 7 24 25 56 84 9 40 41 90 80 1 60 61 32 330 3 84 85 82 546 5 12 12 239 840 7 44 44 305 224 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.