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  • Level: GCSE
  • Subject: Maths
  • Word count: 1065

Beyond Pythagoras

Extracts from this document...

Introduction


Contents:

Introduction                                                                        3,4
Satisfying the condition/Table 1                                        5,6

Calculations to find results in sequence                                7

Finding the nth term for Table 1                                                7,8

Perimeter and area for Table 1                                                9,10,11

Enlarging Triangles                                                        11,12

Conclusion                                                                        13

Table of notations:

S = Smallest side

M = Middle side

L = Longest side

A = Area of triangle

P = Perimeter of triangle

E = Enlargement

N = Position in table (used in nth term)


Introduction:

        The Greek mathematician and philosopher Pythagoras developed Pythagoras’ Theorem. Several philosophers who no doubt had a considerable influence on his future life had taught him from an early age.

Pythagoras’ Theorem:

                 C

                               A                                         B

Pythagoras’ Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

i.e.                                 BC2 = AB2 + AC2

(Refers to the diagram above)


In a triangle such as:

                                                5

                         3                

                                           4

32 + 42 = 52

        The sum of the lengths of the two short sides equals the length of the hypotenuse.

...read more.

Middle

60

61

132

330

6

13

84

85

182

546

7

15

112

113

240

840

8

17

144

145

306

1224

9

19

180

181

380

1710

10

21

220

221

462

2310


Calculations to find out the next results in the sequence:

I discovered that for the shortest side the difference was always +2 (3,5,7,9….). For the middle and longest side the difference was always the same, and followed the pattern of the 4 times tables (e.g. 8,12,16,20…).

Finding the nth term for each column in Table 1:

N

1

2

3

4

5

6

Short side

3

5

7

9

11

13

1st difference

2

2

2

2

2

Column A

Trial and error was my method to find the nth term.

When n=1

2n = 2 x (1) = 2

2n + 1 = 2 x (1) + 1

2 + 1 = 3

When n=2

2n + 1= 2(2) +1

           = 4 + 1

          = 5

This formula works, therefore for this column:

2n + 1 is the correct formula

For column B, the second differences were the same, so I knew it was a squared formula.

N

1

2

3

4

Middle side

4

12

24

40

1st Difference

8

12

16

2nd Difference

4

4

Column B

I then used trial and error

When n=1

2(12) + 2(1)

2 + 2 = 4

When n=2

2(32) + 2(3)

18 + 6 = 24

This formula works, therefore for this column:

2n2 + 2n is the correct formula


...read more.

Conclusion

th terms calculated, it would mean that they could be enlarged by a scale factor of ‘E’.

S = E (2n+1)        M = (2n2+2n)        L = E (2n2+2n+1)

For the area, I knew that the units are UNITS SQUARED, so….

If the ratio of lengths is 1:E

                                1:E2

    Then Ratio of areas is 1:E2

Therefore, this means that for the area column:

A = E2 n (n+1) (2n+1)

Perimeter is a length so he ratio of perimeter is 1:E

P = E (4n2 + 6n + 2)


Conclusion:

In this project I have investigated several possible Pythagorean triples, and by trial and error found formulas to express this.

I have found that for any whole number ‘n’ a Pythagorean triple is given by:

S = 2n + 1

M = 2n2 + 2n

L = 2n2 + 2n + 1

Further triples can be found by enlarging these by ‘E’.

S = E (2n + 1)

M = E (2n2 + 2n)

L = E (2n2 + 2n + 1)

...read more.

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