# Beyond Pythagoras

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Introduction

Beyond Pythagoras

The numbers 3, 4 and 5 satisfy the condition:

3² + 4²= 5²

Because 3 Because 3²= 3x3 =9

4²= 4x4 =16

5²= 5x5 =25

and so..... 3² + 4²=9 + 16= 25 = 5²

I will now find out each of 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

I also had to find the perimeter and area of (3, 4, 5) (5, 12, 13) (7, 24, 25)

Perimeter and area of (3, 4, 5) triangle:

Perimeter = 3+4+5= 12units

Area=3 x 4 ÷2=6

Perimeter and area of (5,12,13) triangle:

Perimeter = 5+12+13 = 30

Area = 5 x 12 ÷ 2 = 30

Perimeter and area of (7,24,25) triangle:

Perimeter = 7+24+25 = 56

Area = 7 x 24 ÷ 2 = 84

Here is a table containing the results:

N | Length of Smallest side | Length of Middle side | Length of Largest side | Perimeter | Area |

1 | 3 | 4 | 5 | 12 | 6 |

2 | 5 | 12 | 13 | 30 | 30 |

3 | 7 | 24 | 25 | 56 | 84 |

I looked at the table and noticed that the difference between the length of the middle side and the length of largest side is always 1. I know that the (smallest number) ² + (middle number) ² = (largest number) ². I knew that the there will be connection between the numbers written above. So I tried experimenting some:

(Middle number) ²+ (largest number) ²= (smallest number) ²

Because, 24² + 25² = 576+625 = 901

7²= 49

The difference between 49 and 901 is 852 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared.

Now I tried squaring the smallest Length

4 + 5= 9 = 3²

This work with 3² now I will try some other numbers:

12 + 13= 25 = 5²

24 + 25= 49 = 7²

It works for both of the other triangles.

Middle number + Largest number = Smallest number²

I can use this to work out some other odd numbers, if you turn Middle number + Largest number = Smallest number² backwards to Smallest number²= Middle number + Largest number

7²= Middle number + Largest number

Middle

So the rule should look like 2n²+1. The only thing was left was that I had to add another n to 2n²+1 to get my rule

So the rule is 2n²+1n or 2n²+n

I tried this rule 2n²+n

N=2 2 x2²+2=10

It didn’t work I try 3rd term

N=3 2 x3²+3= 21

I noticed that I did something wrong with my rule I should multiplied the n by 2 to get the answer because every time I did

2n²+n the difference between 2n²+n and middle side was N so I decided to multiply n by 2 this would give me 2 n

So the rule should be 2n²+2n

I will now try this rule out

N= 2 2 x2²+2 x 2=12

It works for 2nd term I will try it on 3rd term

N= 3 2 x3²+2 x 3=24

It works for 3rd term I will try it 4th term to make sure the rule is right

N= 4 2 x4²+2 x 4=40

The rule work

MIDDLE SIDE RULE: 2n²+2n

Another Rule for middle side:

I noticed that the Middle side numbers where all multiples of 4:

4 , 12 , 24 , 40 , 60 , 84 ,112

I decide to find out where in the 4’s time table did these multiples of 4 went:

Multiples of 4:

4 = 1 x 4=4

12 = 3 x 4= 12

24 = 6 x 4= 24

40 = 10 x 4= 40

60 = 15 x 4= 60.

84 = 21 x 4= 84

112 = 28 x 4= 112

I noticed that all the numbers underlined were triangle numbers. So this shows us that the rule has something to with triangle numbers.

So look at the triangle numbers rule:

½ n (n +1) (triangle number rule)

I tried this rule on the one of the odd number

Conclusion

2n+4 x N ² +4n + 3 =

2

This the rule for the area or I can simplify this

2n+4 x N ² +4n + 3

2

2n3 + 8n²+ 6n +4n² +16n + 12

2

2n3 +12n² +22n + 12

2

n3 +6n²+ 11n +6 = area

I will now try this rule out to see if it works

N= 2

2x 2 x 2 + 6 x 2² +11 x 2+6 =60

It works for 2nd being the nth term I will try 3rd nth term

N= 3

3x 3 x 3 + 6 x 3² +11 x 3+6 =120

This rule work

Area: n3 +6n²+ 11n +6

Now, I will check that 2n + 4, n² + 4n+3 and n² + 4n + 5 forms a Pythagorean triple (or a² + b² = c²).A = 2n + 4, b = n² + 4n+3, and c = n² + 4n + 5.

(2n + 4)² + (n² + 4n+3)² = (n² + 4n + 5)²

(2n + 4) (2n + 4)+ (n² + 4n+3) (n² + 4n+3) = (n² + 4n + 5) (n² + 4n + 5)

4n²+8n+8n+16+n4 +4n3 +3n²+ 4n3 +16n² +12n +3n² +12n+9 = n4 +4n3 +5n²+4n3 +16n²+ 20n+ 5n² +20n +25

n4 +8n3 +26n² +40n + 25 = n4 +8n3 +26n² +40n + 25

I now end up with 0 = 0, so 2n + 4, n² + 4n+3and n² + 4n + 5 has got to be a Pythagorean triple

Universal Rule

The Pythagoras is a set of three integers, ( a, b and c). Which apply to the Pythagoras' Theorem which is a2 + b2 = c2

The Pythagoras triple can be found using only two integers, U and V ( u is greater than v, u>v). One of the two numbers from u and v has to be and odd number and one has to be even number, still u has to be greater than v, u>v) e.g.

U= 6 and v= 3 or it can be u= 9 and v= 4

A = U2+ V2

B = 2UV

C = U2 +V2

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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