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Beyond Pythagoras

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Introduction

BEYOND PYTHAGORAS

Pythagoras theorem states that in a right angled triangle, a² + b² = c², this can also be interpreted as (shortest side) ² + (middle side) ²

= (longest side) ².

Two sets of numbers that satisfy the theorem are 5, 12, 13 and

7, 24, 25.

5² + 12² = 13²                                 7² + 24² = 25²

25 + 144 = 169                                     49 + 576 = 625

The perimeter of a triangle is calculated using a + b + c. The area of a triangle is calculated using ½ x a x b. The perimeters and areas for the triangles above are:

PERIMETER –  

        5 + 12 + 13 = 30

        7 + 24 + 25 = 56

AREA -

        ½ x 5 x 12 = 30

        ½ x 7 x 24 = 84

Below is a table of side length, area and perimeter for Pythagorean triples that start with ODD numbers.

TERM

LENGTH OF

LENGTH OF

LENGTH OF

PERIMETER

AREA

SHORTEST SIDE

MIDDLE SIDE

LONGEST SIDE

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

Using the sequence formula :

nth term = a + (n-1)d + ½ (n-1) (n-2)c

a = first number

d = first difference

c = second difference

...read more.

Middle

20

25

60

150

nth

3n 

4n 

5n 

12n 

6n²

TRIPLE: 5,12,13

TERM

LENGTH OF

LENGTH OF

LENGTH OF

PERIMETER

AREA

SHORTEST SIDE

MIDDLE SIDE

LONGEST SIDE

PERIMETER

AREA

1

5

12

13

30

30

2

10

24

26

60

120

3

15

36

39

90

270

4

20

48

52

120

480

5

25

60

65

150

750

nth

5n 

12n

13n 

30n 

30n²

TRIPLE: 7,24,25

TERM

LENGTH OF

LENGTH OF

LENGTH OF

PERIMETER

AREA

SHORTEST SIDE

MIDDLE SIDE

LONGEST SIDE

1

7

24

25

56

84

2

14

48

50

112

336

3

21

72

75

168

756

4

28

96

100

224

1344

5

35

120

125

280

2100

nth

7n 

24n 

25n 

56n 

84n²

I put the nth terms in a table against the number triple it was (t). I was then able to work

...read more.

Conclusion

Formula for Area:

(n² - m²) 2mn

2

Perimeter = Area:

(n² - m²) + 2mn + n² + m² = (n² - m²) 2mn

        2

I simplified the formula to give so that I was able to work out values for n and m.

(n² - m²) + 2mn + n² + m² = (n² - m²) 2mn

                                                        2

2mn + 2n² = (n + m)(n -m)mn

                                  2n(m + n) = (n + m)(n – m)mn

                                           2 = m x (n - m)

m = 2    (n – m) = 1    n = 3

                                               or            m = 1    (n – m) = 2   n = 3

When these two solutions for n and m are inserted into the general formulas the two Pythagorean triples with perimeter = area are generated.

m = 2   n = 3

n² - m ²        2mn                n² + m²

3² - 2²         2x2x3 = 12        3² + 2²

9 – 4 = 59 + 4 = 13

Pythagorean triple in which area = perimeter : 5, 12, 13

m = 1   n = 3

n² - m²        2mn        n² + m²

3² - 1²        2x1x3 = 6        3² + 1²

9 - 1 = 8        9 + 1 = 10

Pythagorean triple in which area = perimeter: 6, 8 , 10

...read more.

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