I began by checking to see if these triangles fit into pythagoras’ triangle theorem:
a) 5, 12, 13 b) 7, 24, 25
Both fit into the pattern.
The numbers, 3, 4, 5 could be used to make a right angled triangle as shown below:
The perimeter and area of this triangles can be worked out as follows:
* Perimeter = 3 + 4 + 5 =12 units
or smallest length + middle length + largest length = perimeter in appropriate unit.
e.g. 3 cm + 4 cm + 5 cm = 12 cm perimeter
* Area = ½ x 3 x 4 = 6 square units
or ½ x smallest length x middle length = area in appropriate squared units
e.g. ½ x 3 cm x 4 cm = 6 cm
The other 2 examples of triangles that fit the pattern, 5, 12, 13 and 7, 24, 25 can also be the lengths of right angled triangles:
These are the results of the perimeter and area of all 3 triangles. Another 2 triangles have been added to the sequence in order to find a pattern and rule:
Code Name
n
1
2
3
4
5
To complete this table I used the first 3 in the table to get a basic rules:
a = n + 2
b = n + (n + 1) - 1
c = n + (n + 1)
P = 4n + 6n + 2
A = 2n + 3n + n
I worked all the above formulas out using this method:
e.g. b
n 1 2 3 4 5
4 12 24 40 60 <------- b
8 12 16 20
4 4 4 <------- meant n was involved
This meant that n was included in the formula. So I squared n and compared them:
1 4 9 16 25 <------- n
3 8 15 24 35 <-------- difference between b and n
At this point I noticed that the difference between b and n was n - 1. therefore I worked out the formula for b to be n + (n + 1) - 1.
I used the same method to find formulas for a, c, P, and A.
From these rules, I can estimate that the next 2 triangles in the pattern will be:
n
6
7
Next I plan to investigate even numbers for the shortest length side on the triangle. Then the triangles that fit into the Pythagorean Triples pattern, but don’t follow the rules that I’ve found.
First I investigated even numbers for the short side of the triangle. I put the results I found in a table:
Code Name
n
2
3
4
5
The first thing I noticed was that n5 didn’t fit in with the pattern of the other three. All the others progressed in numerical order, whereas n5 would have fitted in between n3 and n4 numerically.
Also, the only strand I could find a relevant formula for was a. Therefore I decided not to continue with this part of my investigation and move on to the next part.
Next I looked into the Pythagorean triples that didn’t fit the rules and formulas I have already found, some of them have been included in previous tables, but I used them in the table below because they fit in.
Code Name
n
1
2
3
4
5
6
7
These are the formulas u found and used to complete this table:
a = 3n
b = 4n
c = 5n
P = 12n
A = 6n
I worked the formulas out using the same method as before.
