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# Beyond Pythagoras

Extracts from this document...

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

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

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

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