#### 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

• Word count: 6029
• Level: GCSE
• Subject: Maths

#### Beyond Pythagoras

Beyond Pythagoras Pythagoras was a great mathematician who created theorems and one of his famous theorems was the "Pythagoras Theorem". You start with a right-angled triangle. The hypotenuse is labeled "c". The bottom of the triangle is "b" and the side of the triangle is labeled "a". Pythagoras Theorem says that in any right angled triangle, the lengths of the hypotenuse and the other two sides are related by a simple formula. So, if you know the lengths of any two sides of a right angled triangle, you can use Pythagoras Theorem to find the length of the third side: Algebraically: a2 + b2 = c2 The numbers 3, 4 and 5 satisfy the condition 9 + 16 = 25 Because 3x3=9 4x4=16 5x5=25 And so 9 + 16 = 25 I now have to find out if the following sets of numbers satisfy a similar condition of: (Shortest Side) 2 + (middle Side) 2 = (Longest side) 2 a) 5, 12, 13 a2 + b2 = c2 52 + 122 = 132 25 + 144 = 169 69 = 169 b) (7, 24, 25) a2 + b2 = c2 72 + 242 = 252 49 + 576 = 625 625 = 625 (3, 4, 5), (5, 12, 13) and (7, 24, 25) are called Pythagorean triples because they satisfy the condition, (Shortest side)2 + (Middle side)2 = (Longest Side)2 We know from the Pythagorean triples the shortest side is always an odd number. So far I have observed the following patterns: The shortest side length advances by two each time. Both the shortest and longest side lengths are

• Word count: 5018
• Level: GCSE
• Subject: Maths

#### For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.

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 a²+b²=c². 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 3²+4²=5² because 3²=3x3=9 4²=4x4=16 5²=5x5=25 And so 3²+4²=9+16=25=5² I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ²+(middle number) ²=(largest number) ² a) 5, 12, 13 5²=5x5=25 12²=12x12=144 25+144=169 V169 = 13 This satisfies the condition as 5²+12²=25+144=169=13² b) 7, 24, 25 7²=7x7=49 24²=24x24=576 49+576=625 V625=25 This satisfies the condition as 7²+24²=49+576=625=25² 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 = 1/2x3x4=6 units ² The numbers 5,12,13 can

• Word count: 4280
• Level: GCSE
• Subject: Maths

#### 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.

• Word count: 4266
• Level: GCSE
• Subject: Maths

#### Beyond Pythagoras

Beyond Pythagoras Introduction Believed to have been born in 582BC and died in 500BC approximately, Pythagoras was a Greek philosopher and mathematician. He discovered some of the most influential theories of number, geometry and proportion which are still frequently used in modern mathematics. Pythagorean triples, on which this investigation is based, are sets of three numbers as seen in the above diagram. Here it is the simplest triple; 3,4,5. They will be written in the following form throughout this paper; a,b,c. There are other Pythagorean triples such as 5,12,13; which is the 2nd odd triple and 361,65160,65161; which is the 180th odd triple. His theorem is this: the square of the hypotenuse (longest side (c (5 (25)))) of a right-angled triangle is equal to the sum of the squares of the other two sides (a (3 (9)) and b (4 (16))). This can be expressed using the following equation: a2+b2=c2 Aim The aim of the investigation is to explore and mathematically express the relationships between the number of the triple (n), the length of each side (a; the shortest), (b; the intermediate), (c; the longest), the perimeter (P) and area (A) of a right-angled triangle. Please note: Throughout this investigation, the sum on the bottom line is the final answer to each set of sums. For example: y=2+2 y=4 In this example, the bottom line (y=4) should be interpreted as

• Word count: 4016
• Level: GCSE
• Subject: Maths

#### Maths GCSE coursework: Beyond Pythagoras

Maths GCSE coursework: Beyond Pythagoras Within this investigation I will look at the relationships between the lengths, perimeters and areas of right-angled triangles. This looks at Pythagorean Triples, three numbers that satisfy the condition of: (smallest number)2 + (middle number)2 = (largest number)2 This can also be expressed as: a2 + b2 = c2 I will look first at an odd number as the 'smallest number' then continue to even numbers. The objective of this investigation is to be able to: Make predictions about Pythagorean triples. Make generalisations about the lengths of sides. Make generalisations about the perimeter and area of the corresponding triangles. To prove a Pythagorean triple you must check that the three numbers satisfy the condition. So... 5, 12, 13 52 + 122 = 25 + 144 = 169 = 132 It is a Pythagorean triple because 5² + 12² = 13². To try a second one: 7, 24, 25 7² + 24² = 49 + 576 = 625 = 25² This is also a Pythagorean triple 7² + 24² = 25² Pythagorean triples also can be used in right-angled triangles Perimeter: 5 + 12 + 13 = 30 Area: 0.5(5 x 12) = 30 13 5 12 To work out the perimeter of these triangles you add together all the sides, and to find out the area of the triangle you have to multiply the shortest sides and then you half the answer. A table showing the perimeter, area of Pythagorean triples: Short side

• Word count: 3537
• Level: GCSE
• Subject: Maths

#### Pythagorean Triples

Pythagorean

• Word count: 3196
• Level: GCSE
• Subject: Maths

#### 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

• Word count: 3143
• Level: GCSE
• Subject: Maths

#### Maths Number Patterns Investigation

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

• Word count: 3143
• Level: GCSE
• Subject: Maths

#### Investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers.

Beyond Pythagoras The aim of this piece of my coursework is to investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers. I will collate the relevant data and formulae for the nth terms by using a grid and from this data I should be able to make predictions on the nth terms of Pythagorean Triples. I will keep a narrative of what I am doing and discovering, referring within it to each process of my investigation. 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. * 5, 12, 13 52+122 = 25+144 = 169 = 132. * 7, 24, 25 72+242 = 49+576 = 625 +252 Here is a table containing the results: a b c area perimeter 3 5 7 9 1 4 2 24 40 60 5 3 25 41 61 6 3 84 80 330 2 30 56 90 32 2n +1 2n² + 2n 2n³ + 2n + n² +n 2n³ + 2n + n² + n (2n + 1) + (2n (n+1)) + (2n² + (2n + 1) 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

• Word count: 3070
• Level: GCSE
• Subject: Maths