Beyond Pythagoras.

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Beyond Pythagoras

For this piece of coursework I am trying to find Pythagorean triplets (these are whole numbers that’s satisfies Pythagoras theorem).

Pythagoras Theorem is a2 + b2 = c2. (a) being the shortest side, (b) being the middle side and (c) being the longest side (hypotenuse) of a right angled triangle.

The numbers 3, 4 and 5 satisfy this condition

                        32 + 42 = 52

because            32 = 3 x 3 = 9

                        42 = 4 x 4 = 16

                        52 = 5 x 5 = 25

                     = 32 + 42 = 9 + 16 = 25 = 52

To find the perimeter we add all the sides together.

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Perimeter = 3 + 4 + 5 = 12

Finally to find the areas we times the smallest and middle side and then divide by to.

Area = ½ x 3 x 4 = 6

 

 

I will now put the first term in to a table and try to find the next terms up to 10.

 

  • s increases by +2 each term
  • s is equal to the term number times 2 then add 1
  • the square root of (m + l) = a  
  • ...

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