Beyond Pythagoras

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Beyond Pythagoras

In this investigation, I shall be studying the relationship between the lengths of the three sides of right angled triangles, their perimeters and their areas.

I aim to be able to:

-Make predictions about Pythagorean triples

-Make generalizations about the lengths of side

-Make generalizations about the perimeter and area of corresponding triangles

My Table of Results for the Triangles

The reason I used certain triples in my table (for example there is also another triple for 9 as the shortest side which has 12 as the middle side length and 15 as the hypotenuse) is because they followed the pattern I was looking at. Even though some of the other triple combinations might have worked, I chose the ones out of them which went best with the other combinations. The 9,40,41 triple has a difference of 1 between its middle and longest side lengths, which is what the other triples I’d chosen have had, so it would be more reliable if I used the result which matched with my others instead of something completely different. If we wish to work out a formula, a general rule and a relationship then I am going to use the triple which is most likely to help with that.

I have worked out the formulae, using the term number (n) ,which the properties of the triangles can be found from. To work out the first formula, I used the way you work out the nth term formula. I knew it was going up in 2s, so I put a 2 before the n, meaning that the term number was going to be timed by 2. I then took 2 away from 3 to work out how much I need to add on. To work out the formula of the middle one, I used trial and error and then for the hypotenuse, I just used the same formula but added on a plus 1 because I knew from my table of results that the longest side was one unit longer than the length of the middle side.

All my formulas are going to be tested three times, to make sure they are reliable and don’t only just work for one set of results.

Length of the shortest side: a=2n+1

I now need to test my formula to see if it works

so:

(2x1)+1=3

It works for that one, as I can see on my table that with term 1, the side length is 3.

(2x2)+1=5

It works for that one, as I can see on my table that with term 2, the side length is 5.

Join now!

(2x3)+1=7

It works for that one, as I can see on my table that with term 3, the side length is 7.

As you can see here:

Length of the middle side: b=2n (n+1)

I now have to test my formula to see if it works

(2x1(1+1))=4

It works for that one, as I can see on my table that with term 1, the middle side length is 4.

(2x2(2+1))=12

It works for that one, as I can see ...

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