# Beyond Pythagoras

Beyond Pythagoras

The aim of this investigation is to investigate Pythagoras theorem and to find a formula for the shortest side, middle length, hypotenuse, area and perimeter.

Because I have typed this up on a computer...

^ is squared

* is times

/ is divided

Pythagoras is

A2 + B2 = C2

I am going to prove this theory be finding out if the following numbers adhere to the rule.

Triangle 1

5 12 13

5^ + 12^ = 13^

5^ = 5*5 = 25

2^ = 12*12 = 144

3^ = 13*13 = 169

So

5^ + 12^ = 25 + 144 = 169 = 25^

The perimeter of the triangle is

All the lengths of the side added up

5 + 24 + 25 = 30

The area of the triangle is

/2 base * height

/2 * 12 * 5 = 30

Triangle 2

7 24 25

7^ + 24^ =25^

7^ = 7*7 = 49

24^ = 24*24 = 576

25^ = 25*25 = 625

So

7^ + 24^ = 49 + 576 = 625 = 25^

The perimeter of the triangle is

All the lengths of the side added up

7 + 24 + 25 = 56

The area of the triangle is

/2 base * height

/2 * 24 * 7 = 84

Length of shortest side

Length of middle side

Length of longest side

Perimeter

Area

3

4

5

2

6

5

2

3

30

30

7

24

25

56

84

9

40

41

90

80

1

60

61

32

330

3

84

85

82

546

5

12

12

239

840

7

44

44

305

224

To create this I used excel to find the Pythagorean triangles

Basically I created one horizontal line of numbers going up one at a time and another vertical line the same. I used the formula =B1+1 (the cell B1 has a 1 in it)

The formula in the white cells is

=IF(INT(SQRT(B\$1^2+\$A2^2))-SQRT(B\$1^2+\$A2^2)=0,SQRT(B\$1^2+\$A2^2), )

This means...

If the integer of the square root of B1 (1) squared by two plus A1 (1) squared by two is not a whole number print a zero, but if it is a whole number print the number.

The integer of a number is the bit after the point.

I took the square root because only a certain numbers two numbers each squared by two will produce a whole number.

Pythagorean triangles must have whole numbers.

(c)nick Peplow

By looking at the printout I can see, which three numbers, produce Pythagorean ...