Beyond Pythagoras - Year 10 Maths Coursework

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Beyond Pythagoras - Year 10 Maths Coursework
By Nadia House 10G

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 now have to find out if the following sets of numbers satisfy a similar condition of:

Smallest number² + Middle number² = Largest number²
        (S² + M = L²)

a) 5, 12, 13

5² + 12² = 13²        ~   (25 + 144 = 169)

b) 7, 24, 25

7²+24² = 25²                ~   (49 + 576 = 625)

Here is a table containing the results:

I looked at this table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.

I already know that the Smallest number² + Middle number² = Largest number². So I know that there will be a connection between the numbers written above. The only problem is that it is
obviously not:

        Middle Side² + Large Side² = Small Side²

Because, 12² + 13² = 144 + 169 = 313
                                      and 5² = 25

     The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared.

I will now try 2 sides squared.

                M² + L = S²
                = 12² + 13 = 5²
                = 144 + 13 = 25
                = 157 = 25

     This does not work and neither will 13², because it is larger than 12². There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side squared.

12² + 13 = 5

     This couldn’t work because 12² is already larger than 5, this also goes for 13². The only number now I can try squaring is the smallest number.

                12 + 13 = 5²
                        25 = 25

     This works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles I know.

                4 +5 = 3²
                     9 = 9

                    And…

                24 + 25 = 7²
                        49 = 49

     It works with both of my other triangles. So…

M + L = S²

   If I now work backwards, I should be able to work out some other odd numbers.

E.g. 9² = M + L
       81 = M + L

     I know that there will be only a difference of one between the middle number and the largest number. So, the easiest way to get 2 numbers with only 1 between them is to divide 81 by 2 and then using the upper and lower bound of this number. So…

                81 = 40.5
                 2

Lower bound = 40, Upper bound = 41.
M = 40, L = 41.

To see if this works I’ll do it for a triangle that I already know.

                7² = M + L
                49 = M + L

                49 = 24.5
                 2

Lower bound = 24, Upper bound = 25.
M = 24, L =25.

This matches the answers that I already have with 7 being the short side, so I think that this equation works.
I now believe I can fill out a table containing the Shortest, Middle and longest sides, by using the odd numbers starting from 3. I already know that the middle and longest side with the shortest length being 3, 5, 7 or 9. So I will start with the shortest side being 11.

                11² = M + L
                121 = M + L

                121 = 60.5
                 2

Lower bound = 60, Upper bound = 61.
M = 60, L =61.
                        
                
               13² = M + L
                169 = M + L

                169 = 84.5
                 2

Lower bound = 84, Upper bound = 85.
M = 84, L =85.
                                        

                15² = M + L
                225 = M + L

                225 = 112.5
                 2

Lower Bound  = 112, Upper Bound = 113.
M = 112, L =113.
                                        

                17² = M + L
                289 = M + L

                289 = 144.5
                 2

Lower bound = 144, Upper bound = 145
M = 144, L =145.
                                        

                192 = M + L
                361 = M + L

                361 = 180.5
                 2

Lower bound = 180, Upper bound = 181.
M = 180, L =181.
                                                        

                212 = M + L
                441 = M + L

Join now!

                441 = 220.5
                 2

Lower bound = 220, Upper bound = 221.
M = 220, L =221.

 I now have 10 different triangles, which I think is easily enough to find a relationship between each side.

S² = M + L

     The way I mentioned above and on the previous pages, is quite a good way of finding the middle and longest sides. An easier and faster way to work out the sides would be by using the nth term. I will now try to work out the nth term for each side (shortest middle and longest).

The formula I will work ...

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