Beyond Pythagoras Essay Example

Beyond Pythagoras Essay Example Beyond Pythagoras Essay Beyond Pythagoras Essay Mathematics Coursework BEYOND PYTHAGORAS By Asif Azam 1) The numbers 3, 4, and 5 satisfy the condition 3à ¯Ã‚ ¿Ã‚ ½ + 4à ¯Ã‚ ¿Ã‚ ½ = 5à ¯Ã‚ ¿Ã‚ ½ because 3à ¯Ã‚ ¿Ã‚ ½ = 33 =9 4à ¯Ã‚ ¿Ã‚ ½ = 44 = 16 5à ¯Ã‚ ¿Ã‚ ½ = 55 = 25 and so 3à ¯Ã‚ ¿Ã‚ ½ + 4à ¯Ã‚ ¿Ã‚ ½ = 9 + 16 = 25 = 5à ¯Ã‚ ¿Ã‚ ½ I will now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) à ¯Ã‚ ¿Ã‚ ½ + (middle number) à ¯Ã‚ ¿Ã‚ ½ = (largest number) à ¯Ã‚ ¿Ã‚ ½. a) 5, 12, 13 5à ¯Ã‚ ¿Ã‚ ½ + 12à ¯Ã‚ ¿Ã‚ ½ = 25 + 144 = 169 = 13à ¯Ã‚ ¿Ã‚ ½ b) 7, 24, 25 7à ¯Ã‚ ¿Ã‚ ½ + 24à ¯Ã‚ ¿Ã‚ ½ = 49 + 576 = 625 = 25à ¯Ã‚ ¿Ã‚ ½ 2) Perimeter 5 5 + 12 + 13 = 20 12 studentcentral.co.uk 7 7 + 24 + 25 = 56 24 wwfe few stfefeud efe fent cfe enfetral fecofe uk: b) Nth term Length of shortest side Length of middle side wwdd ddw stddddud edd ddnt cdd enddtral ddcodd uk. Length of longest side Perimeter Area +2 +1 1 wwbf bfw stbfbfud ebf bfnt cbf enbftral bfcobf uk. 3 4 5 12 6 2 5 12 13 30 30 +2 3 7 24 25 56 84 4 wwef efw stefefud eef efnt cef eneftral efcoef uk; 9 40 41 90 180 +2 5 11 60 61 132 330 +2 6 13 84 Tochv8Ey Visit studentcentral cb 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. I have also noticed that the area is 1/2 (shortest side) x (middle side). 3) In this section I will be working out and finding out the formulas for: * Shortest side * Middle side * Longest side In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2). But I dont know the formula so I will have to work that out. wwfe few stfefeud efe fent cfe enfetral fecofe uk. Firstly I will be finding out the formula for the shortest side. 3 5 7 9 11 2 2 2 2 The differences between the lengths of the shortest side are 2. This means the equation must be something to do with 2n. Lets see Nth term Length of shortest side 1 3 2n 2 x 1 = 2 (wrong) There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if Im correct I will now test this formula. 2n+1 Nth term Length of shortest side 1 3 21+1=3 (correct) Just in case I will test this formula in the next term: 2n+1 Nth term Length of shortest side 2 5 22=4 4+1=5 (correct) I now have to work out the formula for the middle side. I predict that the formula has got to do with something about the differences in lengths. (For instance in this case is 4), most likely 4n.However, because 4 is the difference, the formula must be nà ¯Ã‚ ¿Ã‚ ½. I now believe that the answer will have something to do with 4nà ¯Ã‚ ¿Ã‚ ½. So, I will now write the answers for 4nà ¯Ã‚ ¿Ã‚ ½. This coursework from www.studentcentral.co.uk (studentcentral.co.uk/coursework/essays/2255.html) Reproduction or retransmission in whole or in part expressly prohibited 4nà ¯Ã‚ ¿Ã‚ ½ work for the first term, but, it then collapses after this, as the difference between 4nà ¯Ã‚ ¿Ã‚ ½ gets larger, the thing you notice is that the difference in the 2nd term between 4nà ¯Ã‚ ¿Ã‚ ½ and the middle side is the middle side for the term before. This goes for all the other terms from the 2nd. This means that if I subtract the previous term, then I should in theory get the correct answer. 16-4 = 12 36-12 = 24 64-24 = 40 Etc. So, the equation I have so far is: 4nà ¯Ã‚ ¿Ã‚ ½- (previous middle side) = middle side All the previous terms is, (n 1), so if I put this into the above formula, then it should give me my middle side. 4nà ¯Ã‚ ¿Ã‚ ½ 4(1-1) à ¯Ã‚ ¿Ã‚ ½ = middle side This should in theory give me my middle side. I will test my theory with the first term. wweb ebw stebebud eeb ebnt ceb enebtral ebcoeb uk; 4 x 12 4(1-1) = 4 4 x 1 4 x 0à ¯Ã‚ ¿Ã‚ ½ = 4 4 4 x 0 = 4 4 0 = 4 4 = 4 My formula works for the first term. I will now check if this formula works for the next term. 4 x 2à ¯Ã‚ ¿Ã‚ ½ 4(2 1) à ¯Ã‚ ¿Ã‚ ½ = 12 4 x 4 4 x 1à ¯Ã‚ ¿Ã‚ ½ = 12 16 4 x 1 = 12 16 4 = 12 12 = 12 My formula also works for the 2nd term. You may be thinking that this is the correct formula but just to check, I will check this formula for the 3rd term. 4 x 3à ¯Ã‚ ¿Ã‚ ½ 4(3-1) à ¯Ã‚ ¿Ã‚ ½ = 24 4 x 4 4 x 2à ¯Ã‚ ¿Ã‚ ½ = 24 36 4 x 4 = 24 36 16 = 24 20 = 24 My formula did not work for the 3rd term. It now looks as if 4nà ¯Ã‚ ¿Ã‚ ½ 4(n 1) à ¯Ã‚ ¿Ã‚ ½ is not the correct formula after all. To check, I will look to see if the formula works using the 4th term. 4 x 4à ¯Ã‚ ¿Ã‚ ½ 4(4 1) à ¯Ã‚ ¿Ã‚ ½ = 40 4 x 16 4 x 3à ¯Ã‚ ¿Ã‚ ½ = 40 64 36 = 40 wwdb dbw stdbdbud edb dbnt cdb endbtral dbcodb uk. 28 = 40 My formula doesnt work for the 4th term either. I can now safely say that 4nà ¯Ã‚ ¿Ã‚ ½ 4(n 1) à ¯Ã‚ ¿Ã‚ ½ is definitely not the correct formula for the middle side. I believe the problem with 4nà ¯Ã‚ ¿Ã‚ ½ 4(n 1) was that 4nà ¯Ã‚ ¿Ã‚ ½, once you start using the larger numbers, it becomes far too high to bring it back down to the number I want for the middle side. Also, 4(n 1) à ¯Ã‚ ¿Ã‚ ½ is not as small when it gets larger so it doesnt bring the 4nà ¯Ã‚ ¿Ã‚ ½ down enough, to equal the middle side. I will now look at the difference to see if I can find a pattern there. 1 4 11 20 31 3 7 9 11 2 2 2 The difference here is 2, which means that the answer will involve 2 and nà ¯Ã‚ ¿Ã‚ ½. I will try 2nà ¯Ã‚ ¿Ã‚ ½. I can see that the difference between 2nà ¯Ã‚ ¿Ã‚ ½ and the middle number is the 2 times table. The 2 times table in the nth term is 2n. I now think 2nà ¯Ã‚ ¿Ã‚ ½ + 2n is the correct formula. I will now test it using the first three terms. 2 x 1à ¯Ã‚ ¿Ã‚ ½ + 2 x 1 = 4 2 x 1 + 2 = 4 2 + 2 = 4 4 = 4 My formula works for the first term; so, I will now check it in the next term. 2 x 2à ¯Ã‚ ¿Ã‚ ½ + 2 x 2 = 12 2 x 4 + 4 = 12 8 + 4 = 12 12 = 12 My formula works for the 2nd term. If it works for the 3rd term I can safely say that 2nà ¯Ã‚ ¿Ã‚ ½ + 2n is the correct formula. 2 x 3à ¯Ã‚ ¿Ã‚ ½ + 2 x 3 = 24 2 x 9 + 6 = 24 18 + 6 = 24 24 = 24 My formula also works for the 3rd term. I am now certain that 2nà ¯Ã‚ ¿Ã‚ ½ + 2n is the correct formula for finding the middle side. Middle side = 2nà ¯Ã‚ ¿Ã‚ ½ + 2n I now have the much easier task of finding a formula for the longest side. To start with, I am going to draw out a table containing the middle and longest sides. from www.studentcentral.co.uk Length of middle side Length of longest side +1 4 5 12 13 24 25 40 41 60 61 84 85 112 wwfa faw stfafaud efa fant cfa enfatral facofa uk. 113 144 145 180 181 220 221 From this table I know that there is only one difference, which is +1 between the middle and longest side. So: (Middle side) + 1 = longest side I predict that this formula will be the correct one for the longest side: 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 = longest side. wwbg bgw stbgbgud ebg bgnt cbg enbgtral bgcobg uk. I am very certain that this is the correct formula. I will check anyway using the first three terms: 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 = 5 2 x 1à ¯Ã‚ ¿Ã‚ ½ + 2 x 1 + 1 = 5 2 + 2 + 1 = 5 5 = 5 The formula works for the first term. 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 = 25 2 x 3à ¯Ã‚ ¿Ã‚ ½ + 2 x 3 + 1 = 25 18 + 6 + 1 = 25 wwcf cfw stcfcfud ecf cfnt ccf encftral cfcocf uk! 25 = 25 The formula also works for the second term 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 = 13 2 x 2à ¯Ã‚ ¿Ã‚ ½ + 2 x 2 + 1 = 13 8 + 4 + 1 = 13 13 = 13 The formula works for all three terms. So Longest side = 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 Now I will check that: 2n + 1 2nà ¯Ã‚ ¿Ã‚ ½ + 2n and 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 Form a Pythagorean triple or in other words aà ¯Ã‚ ¿Ã‚ ½ + bà ¯Ã‚ ¿Ã‚ ½ = cà ¯Ã‚ ¿Ã‚ ½ aà ¯Ã‚ ¿Ã‚ ½ + bà ¯Ã‚ ¿Ã‚ ½ = cà ¯Ã‚ ¿Ã‚ ½ This equals: (2n + 1) à ¯Ã‚ ¿Ã‚ ½ + (2nà ¯Ã‚ ¿Ã‚ ½ + 2n) à ¯Ã‚ ¿Ã‚ ½ = (2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1) à ¯Ã‚ ¿Ã‚ ½ If you then put these equations into brackets: (2n + 1)(2n + 1) + (2nà ¯Ã‚ ¿Ã‚ ½ + 2n)(2nà ¯Ã‚ ¿Ã‚ ½ + 2n) = (2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1)(2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1) If I work out this equation out by balancing them in each side and I end up with nothing, then 2n + 1, 2nà ¯Ã‚ ¿Ã‚ ½ + 2n and 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 is a Pythagorean triple. 4nà ¯Ã‚ ¿Ã‚ ½ + 4nà ¯Ã‚ ¿Ã‚ ½ = 4nà ¯Ã‚ ¿Ã‚ ½ + 2nà ¯Ã‚ ¿Ã‚ ½ + 2nà ¯Ã‚ ¿Ã‚ ½ 4n = 2n + 2n 8nà ¯Ã‚ ¿Ã‚ ½ = 4nà ¯Ã‚ ¿Ã‚ ½ + 4nà ¯Ã‚ ¿Ã‚ ½ 4n = 4n 1 = 1 I now end up with 0 = 0, so 2n + 1, 2nà ¯Ã‚ ¿Ã‚ ½ + 2n and 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 is a Pythagorean triple. I now have the nth term for each of the three sides of a right-angled triangle. I can now work out, both, the nth term for the perimeter and the nth term for the area. The perimeter of any triangle is just the length of the 3 sides added together. E.g. 1st term 3 + 4 + 5 = 12 So 12 is the perimeter for the first term 2nd term 5 + 12 + 13 = 30 3rd term 7 + 24 + 25 = 56 And so on. All I have to do is put all the 3 formulas together. Perimeter = (shortest side) + (middle side) + (longest side) f0KAU1K from f0KAU1K student f0KAU1K central f0KAU1K co f0KAU1K uk = 2n + 1 + 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 1 = 2nà ¯Ã‚ ¿Ã‚ ½ + 2nà ¯Ã‚ ¿Ã‚ ½ + 2n + 2n + 2n + 1 + 1 = 4nà ¯Ã‚ ¿Ã‚ ½ + 6n + 2 If I have done my calculations properly then I should have the right answer. To check this I am going use the 4th, 5th and 6th terms. 4th term: 4nà ¯Ã‚ ¿Ã‚ ½ + 6nà ¯Ã‚ ¿Ã‚ ½ + 2 = perimeter 4 x 4à ¯Ã‚ ¿Ã‚ ½ + 6 x 4 + 2 = 9 + 40 + 41 64 + 24 + 2 = 90 90 = 90 It works for the 4th term Lets see if it works for the 5th term: 4nà ¯Ã‚ ¿Ã‚ ½ + 6nà ¯Ã‚ ¿Ã‚ ½ + 2 = perimeter 4 x 5à ¯Ã‚ ¿Ã‚ ½ + 6 x 2 = 11 + 60 + 61 100 + 30 + 2 = 132 132 = 132 And it works for the 5th term And finally the 6th term: 4nà ¯Ã‚ ¿Ã‚ ½ + 6nà ¯Ã‚ ¿Ã‚ ½ + 2 = perimeter 4 x 6à ¯Ã‚ ¿Ã‚ ½ + 6 x 6 + 2 = 13 + 84 + 85 144 + 36 + 2 = 182 182 = 182 It works for all the terms so: Perimeter = 4nà ¯Ã‚ ¿Ã‚ ½ + 6n + 2 wwed edw stededud eed ednt ced enedtral edcoed uk! Like the area I know that the area of a triangle is found by: Area = 1/2 (b x h) b = base h = height Depending on which way the right angled triangle is the shortest or middle side can either be the base or height because it doesnt really matter which way round they go, as Ill get the same answer either way. Area = 1/2 (shortest side) X (middle side) = 1/2 (2n + 1) x (2nà ¯Ã‚ ¿Ã‚ ½ + 2n) = (2n + 1)(2nà ¯Ã‚ ¿Ã‚ ½ + 2n) I will check this formula on the first two terms: (2n + 1)(2nà ¯Ã‚ ¿Ã‚ ½ + 2n) = 1/2 (b x h) (2 x 1 + 1)(2 x 1à ¯Ã‚ ¿Ã‚ ½ + 2 x 1) = 1/2 x 3 x 4 3 x 4 = 1/2 x 12 12 = 6 6 = 6 2nd term: (2n + 1)(2nà ¯Ã‚ ¿Ã‚ ½ + 2n) = 1/2 b h (2 x 2 + 1)(2 x 2à ¯Ã‚ ¿Ã‚ ½ + 2 x 2) = 1/2 x 5 x 12 5 x 12 = 1/2 x 60 60 = 30 30 = 30 It works for both of the terms. This means: Area = (2n + 1)(2nà ¯Ã‚ ¿Ã‚ ½ + 2n)