Beyond Pythagoras.

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Mathematics Coursework - Beyond Pythagoras

Mathematics Coursework BEYOND PYTHAGORAS

By Asif Azam


1)

The numbers 3, 4, and 5 satisfy the condition

3² + 4² = 5²

because 3² = 3x3 =9

4² = 4x4 = 16

5² = 5x5 = 25

and so

3² + 4² = 9 + 16 = 25 = 5²

I will now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) ² + (middle number) ² = (largest number) ².

a) 5, 12, 13

5² + 12² = 25 + 144 = 169 = 13²

b) 7, 24, 25

7² + 24² = 49 + 576 = 625 = 25²

2) Perimeter

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b)

I looked at the table and noticed that there was only 1 difference between the length of the middle side and the length of the longest side. And also if you can see in the shortest side column, it goes up by 2. I have also noticed that the area is

½ (shortest side) x (middle side).

3)

In this section I will be working out and finding out the formulas for:

  • Shortest side 
  • Middle side 
  • Longest side 

In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2). But I don’t know the formula so I will have to work that out. wwfe few stfefeud efe fent cfe enfetral fecofe uk.

Firstly I will be finding out the formula for the shortest side.

Join now!

3 5 7 9 11

2 2 2 2

The differences between the lengths of the shortest side are 2. This means the equation must be something to do with 2n.

Let’s see…

2n

2 x 1 = 2 (wrong)

There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I’m correct I will now test this formula.

2n+1

2x1+1=3 (correct)

Just in case I will test this formula in the next term:

2n+1

2x2=4

4+1=5 (correct)

I now have to work out ...

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