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• Level: GCSE
• Subject: Maths
• Word count: 1607

# Beyond Pythagoras.

Extracts from this document...

Introduction

### Mathematics Coursework - Beyond Pythagoras

Sam Coates

In a right angled triangle, Pythagoras came to the conclusion that on a right angled triangle the addition of the squares of the two smallest sides are equal to the square of the longest square. Thus the equation - a²+b²=c²

Pythagorean Triples are numbers which are positive integers that comply with the rule.

For example, 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

Research

Testing the Theory

I will 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

## 2) Perimeter

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, the numbers increase by the odd numbers from 3 upwards. I have also noticed that the area is

½ (shortest side) x (middle side).

Middle

2

5

2x2=4

4+1=5 (correct)

From looking at my table of results and carrying out some testing of rules, I noticed that ‘an + n  = b’. So I took my formula for ‘a’ (2n + 1) multiplied it by ‘n’ to get ‘2n2 + n’. I then added my other ‘n’ to get ‘2n2 + 2n’. This is a parabola as you can see from the equation and also the graph

I will now test it using the first three terms.

2 x 1² + 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 2² + 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

2n² + 2n is the correct formula.

2 x 3² + 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 2n² + 2n is the correct formula for finding the middle side.

Middle side = 2n² + 2n

I now have the much easier task of finding a formula for the longest side.

Conclusion

1" rowspan="10">

## 4

2nd Difference is four

Difference between shortest and middle length sides generalisations.

## 220

1  7  17  31  49  71  97  127  161  199

6 10  14  18  22  26  30   34    38

2nd Difference is four

## 2310

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