He lived in 400 BC and was one of the first great mathematical thinkers. He spent most of his life in Sicily and southern Italy. He had a group of follows who went around and thought other people what he had taught them who were called the Pythagoreans.Pythagoras himself is best known for proving that the Pythagorean Theorem was true. The Sumerians, two thousand years earlier, already knew that it was generally true, and they used it in their measurements, but Pythagoras proved that it would always be true. The Pythagorean Theorem says that in a right triangle, the sum of the squares of the two right-angle sides will always be the same as the square of the hypotenuse (the long side). A2 + B2 = C2Pythagoras theorem can also help in real life. Here is an example:Say you were walking though a park and wanted to take a short cut. With Pythagoras’s theorem you could work out exactly how long you would have to walk though the grass, rather then talking the long route by walking on the paths.PLANI am going to investigate the three triangles I have been given. They are all right-angled triangles, with 3 sides, all different lengths.The three triangles satisfy the Pythagoras theorem. The theorem states that the hypotenuse side (longest side) must equal the 2 shorter sides squared.Here is the Pythagoras theorem:PYTHAGORAS = a2 + b2 = c2Here are the three triangles I have been given:a) b) c)I am now going to test if three triangles I have been given:Triangle A = 32 + 42 = C29 + 16 = C225 = C25 = CTriangle A is a Pythagorean triplet because when I put in the shortest and middle side into the formula, it gave me the answer that the hypotenuse was 5. 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 = C225 + 144 = C2169 = C213 = CThis is also correct and matches with that I have been given.Triangle C = 72 + 242 = C249 + 276 = C2625 = C225 = CAll these triangles are Pythagorean triplets.ACTIONI am going to put all the numbers I have been given into a table. I want to investigate how I cold work out what the next numbers in the next triangle would be. I am putting it into a table because it is easier to see if there is any pattern.TrianglesABCPerimeterArea1345126251213303037242556844940419018051160611323306128485182548nth term2n + 12n2 + 2n2n2 + 2n + 1Everything that I add to my table after this point will be in blue.I will now work out what the next three triangles in the family will be.Sequence for shortest side3 5 7 9 11 13 / / / / /2 2 2 2 2Every time 2 is being added to the previous number. I can work out that the next numbers will be 9, 11, and then 13.Sequence for middle side4 12 24 40 60 84 / / / / /8 12 16 20 24 / / / /4 4 4 4Here there is a continuous second difference of 4. I can tell that the number after 24 will be 40. I worked this out by finding the difference between 12 and 24 (12), adding 4 to it (16) then adding it on to 24.Sequence for hypotenuse side5 13 25 41 61 85 / / / / /8 12 16 20 24 / / / /4 4 4 4On this sequence the second difference is 4, and by adding 4 every time to the first non continuous difference I can tell that the next numbers will be 41, 61 then 85.Test that triangles are PythagorasI will now test to see if the three new triangles I have got numbers for do comply with Pythagoras theorem. I will to this by adding side a2 and side b2 and see if I get the answer I get matches what I got for side C in my table.Triangle 4A2 + B2 = C292 + 402 = C281 + 1,600 = C21681 = C2V1681 = C241 = Side CThis answer matches with what I predicted with my table. Therefore it is defiantly a Pythagoras triangle.Triangle 5A2 + B2 = C2112 + 602 = C2121 + 3600 = C23721 = C2V3721 = C261 = Side CAgain this is a Pythagoras triangle.Triangle 6A2 + B2 = C2132 + 842 = C2169 + 7056 = C27225 = C2V3721 = C285 = Side CI now can be sure that all the new triangles are Pythagoras.Perimeter and AreaI am going to work out the perimeter and area of each triangle.PerimeterI will work out the perimeter by adding all the sides of the triangles and see what they total.Triangle 1: 3 + 4 + 5 = 12Triangle 2: 5 + 12 + 13 = 30Triangle 3: 7 + 24 + 25 = 56Triangle 4: 9 + 40 + 41 = 90Triangle 5: 11 + 60 + 61 = 132Triangle 6: 13 + 84 + 85 = 182nth term for perimeter12 30 56 90 132 182 / / / / /18 26 34 42 50 / / / /8 8 8 8Now to work out the nth term I will use the formula explained on page 10.Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c12 + ( n – 1 ) 18 + 1/2 ( n – 1 ) ( n – 2 ) 812 + 18n – 18 + 4 [n2 – 3n + 2]12 + 18n – 18 + 8n2 – 12n + 84n2 + 6n + 2AreaTo work out the area of the six triangles I will be using the formula:Base x Height2Triangle 1: 4 x 3 = 62Triangle 2: 12 x 5 = 302Triangle 3: 24 x 7 = 842Triangle 4: 40 x 9 = 1802Triangle 5: 60 x 11 = 3302Triangle 6: 84 x 13 = 5462nth termNext, I am going to find out the nth term for the shortest side, the middle side and the hypotenuse side. I will do this because it makes it much easier to find out the length of a side for triangle n.To work out the nth term I will use the formula: Tn = d n + ( a – d )Where: a = the first termd = common differencen = termTn = nth termnth term for shortest side3 5 7 9 11 13 15 / / / / / /2 2 2 2 2 2Tn = d n + ( a – d )= 2 n + ( 3 – 2 )= 2 n + 1I will now test this nth term to see if it is correct. I will do this by finding out the 7th term using the nth term then checking it against my sequence.So when: n = 72 x 7 + ( 3 – 2)14 + 1 = 15This proves that the nth term is correct because it matches with what I worked out what would be the 7th term in my sequence.nth term for middle side4 12 24 40 60 84 112 / / / / / /8 12 16 20 24 28 / / / / /4 4 4 4 4Because this has a second difference I will use a second formula to work out the nth term.Forumla: Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) cWhere: Tn = nth termA = the first termD = the first number in the changing sequenceC = the second, continuous differenceN = termI will now substitute the numbers from the sequence into the formula.Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c4 + ( n – 1 ) 8 + 1/2 ( n – 1 ) ( n – 2 ) 44 + 8n – 8 + 2 [n2 – 3n + 2]4 + 8n – 8 + 2n2 – 6n + 42n2 + 2n + 0I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.So when n = 72 x 72 + 2 x 7 + 02 x 49 + 1498 + 14 = 112In my sequence under the 7th term I have 112. This proves that my nth term is correct.nth term for hypotenuse sideI will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.5 13 25 41 61 85 113 / / / / / /8 12 16 20 24 28 / / / / /4 4 4 4 4Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c5 + ( n – 1 ) 8 + 1/2 ( n – 1 ) ( n – 2 ) 45 + 8n – 8 + 2 [n2 – 3n + 2]5 + 8n – 8 + 2n2 – 6n + 42n2 + 2n + 1I will now again test this nth term using the same method as above.So when n = 72 x 72 + 2 x 7 + 12 x 49 + 14 + 198 + 14 + 1 = 113Both say the 7th term is 113, so it must be correct.Testing the nth termsI am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2So: ( 2n + 1)2 + ( 2n2 + 2n )2 = (2n2 + 2n + 1)2I will now expand the brackets for each nth term.( 2n + 1 )2 expanded:( 2n + 1) ( 2n + 1)= 4n2 + 4n + 1( 2n2 + 2n )2 expanded:( 2n2 + 2n ) ( 2n2 + 2n )= 4n4 + 4n3 + 4n3 + 4n2= 4n4 + 8n3 + 4n2(2n2 + 2n + 1)2 expanded:(2n2 + 2n + 1) (2n2 + 2n + 1)= 4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n +2n2 + 2n + 1= 4n4 + 8n3 + 8n2 + 4n + 1So: 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2 = 4n4 + 8n3 + 8n2 + 4n + 18n2 + 4n + 1 + 8n3 + 4n4 = 4n4 + 8n3 + 8n2 + 4n + 14n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1This proves that the nth terms are all right and work with Pythagoras theorem.CONCLUSIONI have now generalized the problem given to me at the beginning. I do not think I can go any further with it. I have proved and tested all my formulas.However, I do think I could extent this though by looking at a different family of Pythagorean triplets. In this family there was a difference of 1 between the values of the middle side and longest side. I will now look at a family that has a difference of 2. I want to see if there if the nth terms will be linked or have something in common.I can now produce any triplet in this family with my nth terms easily.EXTENSIONIn this extension I will do the same with what I did with the first family but with the new family and sides.To work out the new family I will double all the values from the last Pythagorean triplets.So Triangle 1 will be:3, 4, 5 doubled to 6, 8, 10Triangle 25, 12, 13 doubled to 10, 24, 26Triangle 37, 24, 25 doubled to 14, 48, 50Here is the new family:I will now check to see if these triangles are Pythagorean triplets by putting the shortest side and middle find into the formula and see if it gives me the answer for the hypotenuse.Triangle 1A2 + B2 = C262 + 82 = C236 + 64 = C2100 = C2C = 10This triangle is a member of a Pythagorean triplet.Triangle 2A2 + B2 = C2102 + 242 = C2100 + 576 = C2676 = C2C = 26Again, this triangle is a member of a Pythagorean triplet.Triangle 3A2 + B2 = C2142 + 482 = C2196 + 2304 = C22500 = C2C = 50All these triangles are Pythagorean triplets.Like last time I will now put these triangle into a table for the same reasons.TrianglesABC16810210242631448504188082522120122626168170Nth term4 n + 24n2 + 4n + 04n2 + 4n + 2Everything that I add to my table after this point will be in blue.I will now work out what the next three triangles in the family will be.Sequence for shortest side6 10 14 18 22 26 / / / / /4 4 4 4 4Every time 4 is being added to the previous number. I can work out that the next numbers will be 18, 22, and then 26.Sequence for middle side8 24 48 80 120 168 / / / / /16 24 32 40 48 / / / /8 8 8 8Here there is a continuous second difference of 8. I can tell that the number after 48 will be 80. I worked this out by finding the difference between 24 and 48 (24), adding 8 to it (32) then adding it on to 48.Sequence for hypotenuse side10 26 50 82 122 170 / / / / /16 24 32 40 48 / / / /8 8 8 8On this sequence the second difference is 8, and by adding 8 every time to the first non continuous difference I can tell that the next numbers will be 82, 122 then 170.Test that the three new triangles are in the Pythagorean familyI will again, like I did with the last family, test to see if the three new triangles I have got numbers for do comply with Pythagoras theorem. I will to this by adding side a2 and side b2 and see if I get the answer I get matches what I got for side C in my table.Triangle 4A2 + B2 = C2182 + 802 = C2324 + 6400 = C23360 = C2V6724 = C282 = Side CThis answer matches with what I predicted with my table. Therefore it is defiantly a Pythagoras triangle.Triangle 5A2 + B2 = C2222 + 1202 = C2484 + 14400 = C214884 = C2V14884 = C2122 = Side CAgain this is a Pythagoras triangle.Triangle 6A2 + B2 = C2262 + 1682 = C2676 + 28224 = C228900 = C2V28900 = C2170 = Side CI now can be sure that all the new triangles are Pythagoras.nth term of new familyNext, like last time, I am going to find out the nth term for the shortest side, the middle side and the hypotenuse side. I will do this because it makes it much easier to find out the length of a side for triangle n. I will use the same formula I describe when investigating the last family.nth term for shortest side6 10 14 18 22 26 30 / / / / / /4 4 4 4 4 4Tn = d n + ( a – d )= 4 n + ( 6 – 4 )= 4 n + 2I will now test this nth term to see if it is correct. I will do this by finding out the 7th term using the nth term then checking it against my sequence.So when: n = 74 x 7 + ( 6 – 4)28 + 2 = 30This proves that the nth term is correct.nth term for middle side8 24 48 80 120 168 224 / / / / / /16 24 32 40 48 56 / / / / /8 8 8 8 8Because this has a second difference I will use a second formula to work out the nth term that I have noted above when working out the nth term for the middle side on the first family.Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c8 + ( n – 1 ) 16 + 1/2 ( n – 1 ) ( n – 2 ) 88 + 16n – 16 + 4 [n2 – 3n + 2]8 + 16n – 16 + 4n2 – 12n + 84n2 + 4n + 0I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.So when n = 74 x 72 + 4 x 7 + 04 x 49 + 28196 + 28 = 224In my sequence under the 7th term I have 224. This proves that my nth term is correct.nth term for hypotenuse sideI will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.10 26 50 82 122 170 226 / / / / / /16 24 32 40 48 56 / / / / /8 8 8 8 8Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c10 + ( n – 1 ) 16 + 1/2 ( n – 1 ) ( n – 2 ) 810 + 16n – 16 + 4 [n2 – 3n + 2]10 + 16n – 16 + 4n2 – 12n + 84n2 + 4n + 2I will now again test this nth term using the same method as above.So when n = 74 x 72 + 4 x 7 + 24 x 49 + 28 + 2196 + 28 + 1 = 226Both say the 7th term is 226, so it must be correct.Testing the nth terms of new familyI am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2So: ( 4n + 2)2 + ( 4n2 + 4n )2 = (4n2 + 4n + 2)2I will now expand the brackets for each nth term.( 4n + 2 )2 expanded:( 4n + 2) ( 4n + 2)= 16n2 + 16n + 4( 4n2 + 4n )2 expanded:( 4n2 + 4n ) ( 4n2 + 4n )= 16n4 + 16n3 + 16n3 + 16n2= 16n4 + 32n3 + 16n2(4n2 + 4n + 2)2 expanded:(4n2 + 4n + 2) (4n2 + 4n + 2)= 16n4 + 16n3 + 8n2 + 16n3 + 16n2 + 8n +8n2 + 8n + 4= 16n4 + 32n3 + 32n2 + 16n + 4So:16n2 + 16n + 4 + 16n4 + 32n3 + 16n2 = 16n4 + 32n3 + 32n2 + 16n + 432n2 + 16n + 4 + 32n3 + 16n4 = 16n4 + 32n3 + 32n2 + 16n + 416n4 + 32n3 + 32n2 + 16n + 4 = 16n4 + 32n3 + 32n2 + 16n + 4This proves that the nth terms are all right and work with Pythagoras theorem.Extension conclusion for ‘Difference of 2’Comparing the 2 nth terms I can conclude that when there is a different of 2 between the middle side and the hypotenuse side the nth term is doubled from when there was a difference of 1. I predict that when there is a difference of 3 between the middle side and the hypotenuse side and nth term will be tripled.Difference of 3I will now look at a family with a difference of 3 to see if my prediction will be correct.TrianglesABC19121521536393217275427120123533180183639252255Nth term6 n + 36n2 + 6n + 06n2 + 6n + 3To get these triangle sides I have just trebled the numbers from the first family. To make sure that these are correct I will pick a triangle and check whether it is a Pythagorean triplet.Triangle 4 = A2 + B2 = C2272 + 1202 = C2726 + 14,400 = C215,126 = C2123 = CThis proves that this triangle and all of the new ones are Pythagorean triplets of a new family.nth term of the new familyI am now going to find out the new nth terns for all the sides using the same techniques as I have used before and for the same reasons I have stated above.nth term for shortest side9 15 21 27 33 39 45 / / / / / /6 6 6 6 6 6Tn = d n + ( a – d )= 6 n + ( 9 – 6 )= 6 n + 3I will now like last time test to see if this nth term is correct by finding out the 7th term using the nth tern then checking it against my sequence.So when: n = 76 x 7 + ( 9 – 6 )42 + 3 = 45This proves that the nth term is correct.nth term for middle side12 36 72 120 180 252 336 / / / / / /24 36 48 60 72 84 / / / / /12 12 12 12 12Because this has a second difference I will use a second formula to work out the nth term that I have noted above when working out the nth term for the middle side on the first family.Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c12 + ( n – 1 ) 24 + 1/2 ( n – 1 ) ( n – 2 ) 1212 + 24n – 24+ 6 [n2 – 3n + 2]12 + 24n – 24 + 6n2 – 18n + 126n2 + 6n + 0I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.So when n = 76 x 72 + 6 x 7 – 06 x 49 + 42294 + 42 = 336In my sequence under the 7th term I have 336. This proves that my nth term is correct.nth term for hypotenuse side15 39 75 123 183 255 339 / / / / / /24 36 48 60 72 84 / / / / /12 12 12 12 12I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.Tn = a + ( n – 1 ) d + 1/2 ( n – 1 ) ( n – 2 ) c15 + ( n – 1 ) 24 + 1/2 ( n – 1 ) ( n – 2 ) 1215 + 24n – 24+ 6 [n2 – 3n + 2]15 + 24n – 24 + 6n2 – 18n + 126n2 + 6n + 3I will now again test this nth term using the same method as above.So when n = 76 x 72 + 6 x 7 + 36 x 49 + 45294 + 45 = 339In my sequence under the 7th term I have 336. This proves that my nth term is correct.Both say the 7th term is 336, so it must be correct.Testing the nth terms of ‘difference of 3’I am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2So: ( 6n + 3)2 + ( 6n2 + 6n )2 = (6n2 + 6n + 3)2I will now expand the brackets for each nth term.( 6n + 3 )2 expanded:( 6n + 3) ( 6n + 3)= 36n2 + 36n + 9( 6n2 +6 )2 expanded:( 6n2 + 6n ) ( 6n2 + 6n )= 36n4 + 36n3 + 36n3 + 36n2= 36n4 + 72n3 + 36n2(6n2 + 6n + 3)2 expanded:(6n2 + 6n + 3) (6n2 + 6n + 3)= 36n4 + 36n3 + 18n2 + 36n3 + 36n2 + 18n +18n2 + 18n + 9= 36n4 + 72n3 + 72n2 + 36n + 9So:36n2 + 36n + 9 + 36n4 + 72n3 + 36n2 = 36n4 + 72n3 + 72n2 + 36n + 972n2 + 36n + 9 + 72n3 + 36n4 = 36n4 + 72n3 + 72n2 + 36n + 936n4 + 72n3 + 72n2 + 36n + 9 = 36n4 + 72n3 + 72n2 + 36n + 9This proves that the nth terms are all right and work with Pythagoras theorem.Extension conclusion for ‘Difference of 3’ and final conclusionComparing the 3 nth terms I can conclude that when there is a different of 2 between the middle side and the hypotenuse side the nth term is doubled from when there was a difference of 1. When there is a different of 3 between the middle side and the hypotenuse side the nth term is tripled. I predict that when there is a difference of 4 between the middle side and the hypotenuse side and nth term will be quadrupled and so on.

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