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• Level: GCSE
• Subject: Maths
• Word count: 2083

# Beyond Pythagoras

Extracts from this document...

Introduction

Beyond Pythagoras

## Intro

I will be investigating Pythagoras triples using the Pythagoras theorem. A triple is a set of 3 whole numbers where the rule is a2 + b2  = c2.  I will be working out the values of the sides of the triangle.

The shortest side of the triangle will be represented by the letter ‘a’, the second longest side will be represented by the letter ‘b’, and the longest side, the hypotenuse, will be represented by the letter ‘c’.

## Odds

 nth Length of shortest side (a) Length of middle side (b) Length of longest side (c) Perimeter (a+b+c) Area(a x b)/2 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 61 132 330 6 13 84 85 182 546 7 15 112 113 240 840 8 17 114 145 306 1224

To begin with, I will draw up a table containing the first eight Pythagorean triples with ‘a’ being an odd number, and shows the lengths of the sides, their perimeters and their areas. Then I will be able to see if there are any connections or relationships between the numbers, and I will be able to find the nth term for each side, area and perimeter.

When I put the numbers into the table like this, I realised two things, firstly that ‘a’ increases by two, and secondly that ‘c’ is b+1 on both of the terms. As I currently only have two numbers in the sequence for ‘a’, my assumption about ‘a’ increasing by two each time is not necessarily correct, because the sequence’s 1st

Middle

This formula for ‘a’ is correct as 2n + 1 does equal ‘a’.

a = 2n + 1

The next formula I want to find is the nth term for ‘b’. I will use the same method as I used to find the nth term for ‘a’. I have written out the first 5 numbers in the sequence of ‘b’.

This sequence is different to the sequence for ‘a’ in that there is not a common difference until the second line. When this happens, you half the common difference, and multiply it by ‘n2’ so, in this case, I would get ‘2n2’. Then I will substitute ‘n’ for each of the first 5 numbers, and work out what it would give. I will call this new sequence ‘sequence 2’. Next, you compare it to the original sequence. So I will work out ‘2n2’ for the first 5 numbers, and compare it to the original sequence for ‘b’.

Here I have substituted ‘n’ for the numbers from 1 to 5, and found ‘sequence B’. Now I will compare it with the original sequence for ‘b’.

I have worked out the differences between the ‘sequence 2’ and the original sequence. The differences will form a new sequence, which I will call ‘sequence 3’.

Now, I work out the nth term for this sequence. There is a common difference on the first line, so I just multiply 2 by ‘n’, and because the previous number would be 0

Conclusion

th term for ‘c’. The nth term for ‘b’ is ‘n2 + 4n + 3’, and ‘c’ is 2 more than this, then the nth term for ‘c’ must be n2 + 4n + 5.  I will test my prediction using the number 1, which will replace ‘n’.

b = n2 + 4n + 5

b = 12 + 4 x 1 + 5

b = 1 x 1 + 4 x 1 + 5

b = 1 + 4 + 5

b = 10

This formula for ‘c’ is correct as 2n + 4 does equal ‘a’.

b = n2 + 4n + 3

## Perimeter

To find the perimeter you have to add a, b and c together.

If ‘a’ is 6, ‘b’ is 8 and ‘c’ is 10, like the 1st term, the perimeter will be 24 because 6+8+10=24.

If ‘a’ is 8, ‘b’ is 15 and ‘c’ is 17, like the 2nd term, the perimeter will be 40 because 8+15+17=40.

The formula to find the perimeter is correct as a + b + c does equal the perimeter

a + b + c

## Area

To find the area you have to multiply ‘a’ and ‘b’ together and then divide the answer by 2.

If ‘a’ is 6 and ‘b’ is 8, like the 1st term, the area will be 24 because 6x8=48 and 48÷2= 24

If ‘a’ is 8 and ‘b’ is 15, like the 2nd term, the area will be 60 because 8x15= 120 and 120÷2= 60

The formula to find the area is correct as (a x b)/2 does equal the area

(a x b)/2

I have worked out the nth terms for ‘a’, ‘b’ and ‘c’ and I have proved that they are correct

Conclusion table

 ‘a’ ‘b’ ‘c’ Perimeter Area Odds 2n + 1 2n2 + 2n 2n2 + 2n + 1 a + b + c (a x b)/2 Evens 2n + 4 n2 + 4n + 3 n2 + 4n + 3 a + b + c (a x b)/2

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