= 4 x 3 = 2 x 3 = 6
2
Perimeter of a triangle = side + side + side
= 3 + 4 + 5 = 12
2) 5,12,13
The formula for Area of a triangle is: ½ x b x h
Area = ½ x 12 x 5
= 12 x 5 = 6 x 5 = 30
2
Perimeter of a triangle = side + side + side
= 5 + 12 + 13 = 30
3) 7,24,25
The formula for Area of a triangle is: ½ x b x h
Area = ½ x 24 x 7
= 24 x 7 = 12 x 7 = 84
2
Perimeter of a triangle = side + side + side
= 7 + 24 + 25 = 56
From the above data I can construct a table to make the identification of patterns easier.
From this table I have observed the following patterns:
- The shortest side length advances by two each time.
- All the middle numbers are even.
- The middle side is always a multiple of four.
- The longest side is always one unit more than the middle side.
- Both the shortest and longest side lengths are always odd.
- Both the area and perimeter are always even numbers.
Now I am going to draw another triangle to show the formula -
(Shortest side)² + (Middle side)² = (Longest side)² by using letters a,b,c. This will make it easier for me to illustrate the formula for all three – Pythagoras Theorem, Area and Perimeter by using letters.
Key:
Shortest side = a
Middle side = b
Longest side = c
Using these letters:
- Pythagoras theorem formula is a² + b² = c²
- Area formula is ½ x base x height
= ½ x b x a
- Perimeter formula is a + b + c
I am going to find three more Pythagorean triples. The shortest side has been given to me as 3, 5, 7, and I am supposed to find out the middle side and the longest side. To do so I am going to use the pattern as explained in point 4 for the first three triangles with sides 3,4,5; 5,12,13 and 7,24,25.
To work this out easier and faster, I have constructed a table to make the identification of patterns for the new three triples easier.
Note: The sides of the new three Pythagorean triples are highlighted in red.
The table above explains that for the terms 1,2,3, the difference between the lengths of the shortest sides 3,5,7 is 2. This means that to get the new shortest sides for the new triples, I will add 2 each time. By doing so I have got: 7 + 2 = 9; 9 + 2 = 11 and 11 + 2 = 13.
Similarly, looking at the lengths of the Middle sides 4,12,24, I can see that to get from 4 to 12, I have to add 8; then from 12 to 24, I have to add 12. So moving further I have seen that the number increases by 4 each time, indicating the difference of 4 as we move further. Hence, the next lengths of the three middle sides would be: 24 + 16 = 40; then 40 + 20 = 60 and 60 + 24 = 84.
The same pattern used to find out the middle sides was used to find the longest sides. To get from 5 to 13, I have to add 8; then from 13 to 25, I have to add 12. So moving further I have seen that the number increases by 4 each time, indicating the difference of 4 as we move further. Hence, the next lengths of the three longest sides would be: 25 + 16 = 41; then 41 + 20 = 61 and 61+ 24 = 85.
Also, we can see that there is another way to find the longest sides besides as mentioned above. We can see that the longest side for the first 3 terms is 1 more than the middle side. This indicates that to find the next three longest sides I just have to add one more to the middle number. I have constructed a table as shown below to understand this pattern better:
Now I am going to show that the new triples also obey the same relationship as I have shown under point above, which is that they satisfy the condition
(Shortest side) ² + (Middle side) ² = (Longest side) ² of the Pythagoras theorem.
1) The numbers 9,40,41 satisfy this condition because -
9² + 40² = 41²
So shortest side 9² = 9 x 9 = 81
Middle side 40² = 40 x 40 = + 1600
--------
1681
And Longest side 41² = 41x 41 = 1681
2) 11, 60, 61
11² + 60² = 61²
Shortest side 11² = 11 x 11 = 121
Middle side 60² = 60 x 60 = + 3600
--------
3721
Longest side 61² = 61 x 61 = 3721
3) 13, 84, 85
13² + 84² = 85²
Shortest side 13² = 13 x 13 = 169
Middle side 84² = 84 x 84 = + 7056
--------
7225
Longest side 85² = 85 x 85 = 7225
FAMILIES OF TRIPLES:
With the help of the above set of triples, I am now going to prove, as done earlier that all the above three sets of new triples also satisfy the Pythagoras Theorem and obey the rule, i.e. (Shortest side) ² + (Middle side) ² = (Longest side) ²
This I have down by multiplying the triples by 2 and 3. This is shown as below:
Therefore, 18² + 80² = 82² and 27² + 120² = 123²
324 + 6400 = 6724 729 + 14400 = 15129
6724 = 6724 15129 = 15129
They can also be called as Pythagorean triples.
Now I am going to work out the area and the perimeter of each of the above triangles.
1) 9, 40, 41
The formula for Area of a triangle is: ½ x b x h
Area = ½ x 40 x 9
= 40 x 9 = 20 x 9 = 180
2
Perimeter of a triangle = side + side + side
= 9 + 40 + 41 = 90
2) 11, 60, 61
The formula for Area of a triangle is: ½ x b x h
Area = ½ x 60 x 11
= 60 x 11 = 30 x 11 = 330
2
Perimeter of a triangle = side + side + side
= 11 + 60 + 61 = 132
3) 13, 84, 85
The formula for Area of a triangle is: ½ x b x h
Area = ½ x 84 x 13
= 84 x 13 = 42 x 13 = 546
2
Perimeter of a triangle = side + side + side
= 13 + 84 + 85 = 182
Now to fully understand the above pattern better, I am going to input all the data of first three triangles and the new ones that I have found along with the area and the perimeter in a table as shown below, so that it will be easy for me to find out an expression for all the three sides.
Based on the above table, I can work out the expressions for:
- How to get shortest side from ‘n’
- How to get middle side from ‘n’
- How to get longest side from ‘n’
To start with, now I am going to find an expression which gives the length of the shortest side, by using the expression for the term number as ‘n’. To find the expression for the shortest side, I have to find the difference between successive values of the shortest side, which I have shown in the table below:
As we can see above that the difference between successive values of the shortest sides is 2. Hence from these numbers, I can see that the expression will be: SS = 2n + 1
‘SS’ meaning length of the short side.
‘n’ meaning number of terms.
I can prove that this works by testing it against any one length of the SS and the ‘n’ term.
The triangle for the second term has a SS of 5 (output value) and the ‘n’ term against it is 2 (input value).
2n + 1 = SS
Therefore, 2 x 2 + 1 = 5
4 + 1 = 5
So, 5 = 5
Now I am going to find an expression for the size of the middle side. The procedure will be same as above, which is to find the difference between successive values of the middle side. This is shown in the table below:
From looking at the table of results, I noticed that SSn + n = b. So I took my formula for ‘SS’ as (2n + 1) multiplied it by ‘n’ to get ‘2n² + n. I then added my other ‘n’ to get 2n² + 2n.
Hence the expression for the middle side will be: MS = 2n² + 2n.
‘MS’ meaning length of the middle side.
‘n’ meaning number of terms.
I can prove that this works by testing it against any one length of the MS and the ‘n’ term.
The triangle for the third term has a MS of 24 (output value) and the ‘n’ term against it is 3 (input value).
2n² + 2n = MS
Therefore, 2 x 3² + 2 x 3 = 24
18 + 6 = 24
So, 24 = 24
Finding the values of the expression:
Looking at the repetition of 4 at the second difference between the set of numbers, I can now find the general form of expression, which is: an² + bn + c
With the help of this expression, I can now:
- Find ‘a’ by dividing the second difference 4 by two (2 being the input value), I am left with ‘a’, the co-efficient of n².
So 4/2 = 2.
Hence a = 2
-
Find ‘c’ by substracting 4 (second difference) from the first difference between the 1st and 2nd term of the middle side (that is 8) and then subtracting 4 again from the 1st side, I am left with 0, which is ‘c’
This is done like this: 8 – 4 = 4
4 – 4 = 0
Hence c = 0
- I can calculate ‘b’ by substituting ‘a’ and ‘c’ for the values I have found.
Substituting it - an² + bn + c
Input value output value
2 x (2) ² + b(2) + 0 = 12
2 x 4 + 2b = 12
8 + 2b = 12
2b = 12 – 8
2b = 4
b = 4/2 = 2
Hence b = 2
Now I am going to find the expression for the size of the longest side, using the same procedures as I have used to find the expression for the middle side.
However, this is straightforward, because longest side is just the formula for the middle side, which is ‘MS’ +1. However, looking at the table of results, I noticed that 2n² + 2n + 1= LS
Hence the expression for the longest side will be: LS = 2n² + 2n + 1
‘LS’ meaning length of the longest side.
‘n’ meaning number of terms.
I can prove that this works by testing it against any one length of the LS and the ‘n’ term.
The triangle for the second term has a LS of 13 (output value) and the ‘n’ term against it is 2 (input value).
2n² + 2n + 1 = LS
Therefore, 2 x 2² + 2 x 2 + 1 = 13
8 + 4 + 1 = 13
So, 13 = 13
Finally, to prove my formulas for ‘SS’, ‘MS’ and ‘LS’ are correct, I decided to incorporate them formulas into a2 + b2 = c2: -
a2 + b2= c2
(2n + 1) 2 + (2n2 + 2n) 2 = (2n2 + 2n + 1) 2
(2n + 1)(2n + 1) + (2n2 + 2n)(2n2 + 2n) = (2n2 + 2n + 1)(2n2 + 2n + 1)
4n2 + 2n + 2n + 1 + 4n4 + 4n2 + 4n3 + 4n3= 4n4 + 8n3 + 8n2 + 4n + 1
4n2 + 4n + 1 + 4n4 + 8n3 + 4n2= 4n4 + 8n3 + 8n2 + 4n + 1
4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1
I now have the nth term for each of the 3 sides of a right-angled triangle. This proves that my ‘SS’, ‘MS’ and ‘LS’ formulas are correct.
