Maths GCSE Coursework Beyond Pythagoras Aim The aim of this coursework is to investigate the relationship between the lengths of the three sides of right angle triangles, the perimeters and the areas of the triangles. For any right-angled triangle the (shortest side) ^2+ (middle side) ^2 will equal to the (longest side) ^2. I have represented this in the simple formula below. a ^2+b ^2 = c ^2 The numbers 3, 4 and 5 satisfy the equation. This is because 3^2 + 4^2 = 5^2 = (3 x 3) + (4 x 4) = (5 x 5) 9 + 16 = 25 The overall formula for all Pythagorean triples is: – (shortest side) ^2+ (middle side) ^2 = (longest side) ^2 1.
a) I will use the set of numbers 5, 12 and 13 to check this formula. (shortest side) ^2+ (middle side) ^2 = (longest side) ^2 5^2 + 12^2 = 13^2 25 + 144 = 169 This set of numbers is a pythagorean triple as they follow the formula b) I am going to check if this set of numbers is also a triple 7, 24 and 25. (shortest side) ^2+ (middle side) ^2 = (longest number) ^2 7^2 + 24^2 = 25^2 49 + 576 = 625 These numbers follow the formula also and so they are a Pythagorean triple. The perimeter for this right angle triangle is to add all of the three sides together: – 3+4+5 = 12 units The formula for calculating the area of a right angle triangle is: – base x height 2 (3 x 4) = 6 sq units 2 2 a) I will now calculate the perimeter and area for the following triangles: – The perimeter for this triangle is: – The perimeter for this triangle is: – 5+12+13 = 30 units 7+24+25 = 56 units The area for this triangle is: – The area of this triangle is: – (5 x 12) = 30 sq units (7 x 24) = 84 sq units 2 2 2 b) These are the results of the perimeters and areas of the right angled triangles. Length of shortest side Length of middle side Length of longest side Perimeter (units) Area (square units) 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84 I am going to investigate the relationship between the perimeters, the areas, and the lengths of the sides. To start with I have placed the first seven pythagorean triples that I discovered with an odd number for the shortest side in a table with the value of n that will corresponds to the formulas worked out for each side length.
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BERMUDA: THE SATAN'S RING Introduction: A legendary triangle of Ocean lies between 3 countries upon the Atlantic ocean. The Cities are Bermuda, Puerto Rico and Fort Lauderdale. Ships, people and aeroplanes have been reported mysteriously disappearing off the face of the earth whilst travelling inside this triangle. It soon acquired the name "Devils Triangle" owing to peoples superstitions that the ...
The perimeters and area have also been calculated. I can use this table to check the answers for the formulas that I have worked out. n Shortest side Middle side Longest side 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 61 132 336 6 13 84 85 182 546 7 15 112 113 240 840 Firstly I am going to work out the formulas for each of the columns in the table. Shortest Side I worked out that the formula for the shortest side is 2 n+1 Method This is because there is a difference of two in each case. n 1 2 3 4 5 3 5 7 9 11 +2 +2 +2 +2 = 2 n + x when “n” is equal to three (2 x 3) +x = 7 x = 7 – (2 x 3) x = 7 – 6 x = 1 so the formula for the shortest side therefore is 2 n +1. Middle Side – Method I am going to work out the formula for the middle length side.
n 1 2 3 4 5 u 4 12 24 40 60 first difference 8 12 16 20 second difference 4 4 4 2 n^2 2 8 18 32 50 u – 2 n^2 2 4 6 8 10 2 2 2 2 The 2 n^2 in the method has been calculated from the second difference halved and squared. This means that the first part of the formula is 2 n^2. The second part is the difference in the last line which was 2 n. We can check to see of this is the final formula. E.
g Let n be equal to five. = 2 n^2 + 2 n = 2 x (5 x 5) + 2 x 5 = 2 x 25 + 10 = 60 The formula is correct as this answer matches the one in the table in the beginning. The final formula to work out the middle length is: – 2 n^2 + 2 n Longest Side To work out the longest side I found that the same method as calculating the middle side could be used. Method n 1 2 3 4 5 u 5 13 25 41 61 first difference 8 12 16 20 second difference 4 4 4 2 n^2 2 8 18 32 50 u – 2 n^2 2 4 6 8 10 2 2 2 2 The calculations are exactley the same as the ones for the middle length formula. The first part is 2 n^2 as the difference of four is halved and then the n is squared. The second part is 2 n because the difference of “u – n” is two.
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?? What is it like to transition from elementary to middle school? That is a question many kids ask each year. Having experienced both, I can tell you that there are plenty of similarities as well as some very big differences between the two types of schools. ?? ?? Elementary schools and middle schools have many traits in common. Typically, both are open five days a week for a set number of hours ...
This leaves us with the formula of 2 n^2 + 2 n but this on its own will not work. Another figure will have to be added to it. Using algebra I will now find out what the missing figure is. E.
g n is equal to 5 61 = 2 n^2 +2 n + x 61 = 2 x (5 x 5) + (2 x 5) + x 61 = 50+10 + x x = 61 – (50+10) x = 1 The formula has a “+1” added on the end so the term to define and calculate the longest length is: – = 2 n^2 + 2 n +1 I have found that the middle and longest sides are identical except from the +1 onto the end of the longest side. Perimeters I am now going to find a formula for the perimeter of the triangle. To find the perimeter the usual formula is: – (smallest length) + (middle length) + (longest length) In the diagram I have represented the lengths with letters. As I already have the formulas of the smallest, middle and longest sides I can substitute these terms into the perimeter formula. (smallest length) + (middle length) + (longest length) 2 n +1 2 n^2 + 2 n 2 n^2 + 2 n +1 = 2 n +1+ 2 n^2 + 2 n + 2 n^2 + 2 n +1 I have to collect “like” terms to simplify this formula.
For example all of the n^2 have been collected together. This is the formula when it has been simplified. = 4 n^2 + 6 n + 2 I am going to check that my formula works. when “n” is equal to five I will have to substitute five into all of the n’s. = 4 x (5 x 5) + (6 x 5) +2 = (4 x 25) + 30 +2 I have simplified these sums into: – = 100+32 = 132 This formula is correct as it matches the answer for the perimeter when n is equal to five in the table. Areas The formula to work out the area for all triangles is: – 0.
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5 x (base x height) or base x height 2 To work out a single formula for the area I can substitute the formulas from the middle side and the shortest side (base and height) into the above formula. base = middle side height = shortest side middle formula = 2 n^2 + 2 n shortest formula = 2 n+1 = 0. 5 x ( (2 n+1) x (2 n^2 + 2 n) ) For the moment I am going to ignore the 0. 5 part of the formula. = (2 n+1) x (2 n^2 + 2 n) After multiplying those numbers together the formula looks like this. = 4 n^3 + 4 n^2 + 2 n^2 + 2 n This formula has been simplified to – = 4 n^3 +6 n^2 +2 n At this point we must not forget to multiply by 0.
5 = 0. 5 x 4 n^3 +6 n^2 +2 n = 2 n^3 + 3 n^2 +n To check that the formula works I am going to take a result from the table and substitute it into the new formula to see of the right answer comes out. let n = 7 = 2 x (7) ^3 + 3 x (7) ^2 + 7 I multiply the two and the seven cubed then add that to the three multiplied by the seven squared and finally add those to the seven at the end. = (2 x 343) + (3 x 49) +7 = 686 + 147 + 7 = 840 840 is the figure in the table also showing that it is correct.
Therefore the formula is correct. Generalisations 1) Areas I have substituted the formulas from the sides into the equation for the area: – (2 n+1) (2 n^2+2 n) 2 This then simplifies down to: – 4 n^3 + 6 n^2 +2 n 2 and area = 2 n^3 + 3 n^2 + n 2) Perimeters The terms for the sides of the triangle have also been substituted into the equation for the perimeter. perimeter = a + b+ c = (2 n+1) + (2 n^2+2 n) + (2 n^2+2 n+1) This simplifies down to = 4 n^2 + 6 n + 2 3) Pythagoras’s Theorum I am substituting the side lengths into the rule for pythagoras’ Theorum. To find out any of the other side lengths: – a^2 + b^2 = c^2 (2 n+1) + (2 n^2+2 n) = (2 n^2+2 n+1) Now that I have worked out the formulas for the side lengths I will have to find out the value of n that the triple is. Without the n value I will not be able to work out any of the areas, lengths or perimeters. To find the value of n I have found that it can be done from using the shortest side length.
The formula for getting to the shortest side is 2 n + 1. To get from the length to the value of n I will have to inverse the process. This means that the 2 n + 1 now will be reversed to: – s – 1 2 s = shortest Example If I am given a right angled triangle with a shortest side of length 21 units and I would like to find out the area and the perimeter using a single formula I will have to work out the value of n that it equals to. To do this I will use the formula below: – n = s – 1 2 As the s means the shortest side this is the calculation that I will do: – n = 21 – 1 2 n = 20 2 n = 10 From this value you can work out the perimeter and the area. If the middle side and the longest side are not given then, using the formulas that I worked out earlier I will be able to calculate them aswell. I have come across pythagorean triples that do not fit in with these formulas.
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These triples have even numbers for the shortest side. e. g The numbers 6, 8 and 10 although a pythagorean triple do not work with the formulas I have figured out. Even Sides I am going to find the formulas for triples with even numbers for the shortest sides.
I have created a table including the triples that do not work. The sides, perimeters and areas all calculated accurately so that I can use the figures to check if my formulas are correct or not. n shortest side (units) middle side (units) longest side (units) perimeter (units) area (sq. units) 1 6 8 10 24 24 2 10 24 26 60 120 3 14 48 50 112 336 4 18 80 82 180 720 5 22 120 122 264 1320 Shortest Side To find out the shortest side value I figured out that the formula is 4 n + 2. This is because of the difference of four between each of the shortest side values.
The four then becomes 4 n and an extra number has to be added to it each time. This number is 2. E. g let n equal three. = (4 x 3) + 2 = 12 + 2 = 14 This formula is correct as it works and produces the right answer. Middle Side To work out the middle formula I will use the method: – n 1 2 3 4 5 u 8 24 48 80 120 first difference 16 24 32 40 second difference 8 8 8 4 n^2 4 16 36 64 100 u – 4 n^2 4 8 12 16 20 difference 4 4 4 4 As the second difference is eight it has to be halved and n^2.
This comes to 4 n^2 which will be the first part of the formula. The second part is the four which is the difference for u – 4 n^2. The four becomes 4 n. This makes the formula for the middle side 4 n^2 +4 n I have to check this formula and see if it is working correctly. E. g let n be equal to four.
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= 4 x (4 x 4) +4 x 4 The brackets will have to be multiplied out. = (4 x 16) +16 = 80 Eighty is the correct answer. Longest Side The formula for calculating the longest side for pythagorean triples with odd numbers as the shortest sides was exactley the same except the added one on the end. I am going to predict the formula for the longest side when the shortest side is an even number.
n shortest side (units) middle side (units) longest side (units) perimeter (units) area (sq. units) 1 6 8 10 24 24 2 10 24 26 60 120 The first part is the same as the middle formula 4 n^2 +4 n. The added number onto the end will this time be two as there is a difference of two between the middle side and the shortest side as you can see in the table. I will use the proper method to see if my prediction is correct.
n 1 2 3 4 5 u 10 26 50 82 122 first difference 16 24 32 40 second difference 8 8 8 4 n^2 4 16 36 64 100 u – 4 n^2 6 10 14 18 22 difference 4 4 4 4 The formula is 4 n^2 +4 n + x I will have to find out what the x stands for. E. g Let n be equal to two. 26 = 4 x (2 x 2) + (4 x 2) + x 26 = (4 x 4) + 8 +x 26 = 16+8 +x 26 = 24+x x = 26-24 x = 2 This makes the final formula: – 4 n^2 +4 n + 2 My prediction was correct but I had to go through it to check the right answer. Area The formula for area is: – base x height 2 I already have the formulas for the base (shortest side) and the height (middle side) so I can substitute these into the equation above.
Shortest side = (4 n+2) Middle side = (4 n^2 +4 n) When the terms have been substituted in, the formula should look like this: – (4 n+2) x (4 n^2 +4 n) 2 I am going to work out the top section of the formula first. = (4 n+2) x (4 n^2 +4 n) To work out this formula I will have to multiply all of the figures together. 16 n^3 + 24 n^2 + 8 n^2 + 8 n By collecting like terms this then simplifies down to: – 16 n^3 + 24 n^2 +8 n I am going to divide this by two as I left it out at the beginning. 16 n^3 + 24 n^2 +8 n 2 = 8 n^3 + 12 n^2 +4 n The formula for finding out the area is: – 8 n^3 + 12 n^2 +4 n Perimeter The formula for perimeter is a +b+c as shown below.
To work out the perimeter I am going to add together all of the terms to define each side of the triangle. a + b + c = (4 n+2) + (4 n^2 +4 n) + (4 n^2 +4 n +2) = 4 n+2 + 4 n^2 +4 n + 4 n^2 +4 n +2 To simplify this equation i am going to collect like terms. 8 n^2 +12 n +4 This means that the final formula for the perimeter for right angled triangles when the shortest side is odd is: – 8 n^2 +12 n +4 I am going to check if this formula works. E. g Let n equal to five The answer should equal to 264 264 = (8 x 25) + (12 x 5) + 4 264 = 200 + 60 + 4 264 = 264 This formula is working correctly. Generalisations The terms to define the length of each side has been placed in the diagram to show which term relates to which side clearly.
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For the area I have found a generalisation for the formula. This is the basic formula to find the area of a triangle: – base x height 2 This is the generalisation that I have made. (4 n+2) (4 n^2+4 n) 2 I have done the same for the perimeter. Perimeter = side a +side b +side c Perimeter = (4 n+2) + (4 n^2+4 n) + (4 n^2+4 n+2) To calculate the longest side using Pythagoras’ Theorem the generalisation that I have made is: – (shortest side) ^2+ (middle side) ^2 = (longest side) ^2 (4 n+2) + (4 n^2+4 n) = (4 n^2+4 n+2) I will have to find out the value of n that the triple is. Without the n value you will not be able to work out any of the areas, lengths or perimeters. To find the value of n I have found that it can be done from using the shortest side length.
The formula for getting to the shortest side from n is 4 n + 2. To get from the length to the value of n I will have to inverse the process. This means that the 4 n + 2 now will be reversed to: – s – 2 4 s = shortest Example If I am given a right angled triangle with a shortest side of length 26 units and I would like to find out the area and the perimeter using a single formula I will have to work out the value of n that it equals to. To do this I will use the formula below: – n = s – 2 4 As the s means the shortest side, this is the calculation that I will do: – n = 26 – 2 4 n = 24 4 n = 6 The n value will enable me to work out the perimeter and area using a single formula.
The lengths of the other sides can also be calculated using value n. This will make it a much quicker and effective way of calculating lengths and areas.
