Beyond Pythagoras

Perimeter = 5 + 12 + 13

5 + 12 + 13 = 30 units.

Area of 5,12 and 13

                 13

        5

12

Area = 0.5 or ½ * height * length

Area = ½ * 5 * 12

½ * 5 * 12 = 30 units

The perimeter of the numbers 5,12 and 13 is 30 units.

The area of the numbers 5.12 and 13 is 30 units.

Numbers 7, 24 and 25

Perimeter of 7,24 and 25

        25

        7

24

Perimeter = smallest number + middle number + largest number

Perimeter = 7 + 24 + 25

7 + 24 +25 = 56 units

Area of 7,24 and 25

        25

        7

24

Area = ½ * height * length

Area = ½ * 7 * 24

½ * 7 * 24 = 84 units 

The perimeter of the numbers 7,24 and 25 is 56 units.

The area of the numbers 7,24 and 25 is 84 units.

Now that I know this I can put what I know into a table so everything looks clearer and it is easy to read.

From now on I am going to change the smallest side to the letter A

                                             The middle side to the letter B

                                            The largest side to the letter C

I am now going to estimate the next line in the table.

I have estimate this because I can see that in the A column, the numbers all go up by 2 so I have made a guess that the next one is going to be 9.

I have put 40 in the B column as the middle side because I could see a pattern in the number before, 4 went up to 12, which shows there is a difference of 8 and between 12 and 24 there is a difference of 12. If you put 8 and 12 together there is a 4 difference in between the 8 and 12 so I added a difference of 4 onto 12 to get 16 then I added that 16 onto the 24 to get 40. That was my guess.

In the C column, I could see there was an obvious pattern in that every number was the one after the number in the B column so I estimated that the largest side was going to be 41. With these numbers I worked out the perimeter and the area.

Testing 9,40,41: -

Left Hand Side = 9 + 40

                    =81+1600

                    =1681

Right Hand Side= 41

                      =1681

Left Hand Side = Right Hand Side

This means that 9,40,41 is a Pythagorean Triple.

Perimeter of 9,40 and 41

   

40

        9

41

Perimeter = A + B + C

Perimeter = 9 + 40 + 41

9 + 40 + 41 = 90 units

Area of 9, 40 and 41

40

        9

41

Area = ½ * A * B

Area = ½ * 9 * 40

½ * 9 * 40 = 180 units

The perimeter of the numbers 9,40 and 41 is 90 units.

The area of the numbers 9,40 and 41 is 180 units.

All of these numbers (3,4,5), (5,12,13), (7,24,25) and (9,40,41) are all called Pythagorean Triple because they all satisfy the condition.

A  + B  = C    or 3  + 4  = 5

         

        3                        5        

        4

I am now going to investigate the relationships between the lengths of all three sides of the right – angled triangles. I am now going to find a formula for A.

The difference between the lengths of the shortest side is 2. This means the equation must be something with 2n. So, there is only a difference of +1 between 2n and the shortest side, s this means the formula should be 2n+1

A=2n+1

I will now test this formula: -

N=1

A=2*1+1

A=3

This means that my first term is right.

N=2

A=2*2+1

A=5

This shows that my second term is right.

N=3

A=2*3+1

A=7

This shows that my third term is right.

I will now try and find a method for B.

From this table I can see if I *1 and then +1 from A, I get B.

Example

N=1

A=3

B=3*1+1

B=4

This means this way works for the first term. Now I will try this for the next line.

N=2

A=5

B=5*1+1

B=6

I am half the number off so this means that if I double the *1 to *2 and the +1 to +2 then I will see if it works.

N=2

A=5

B=5*2+2

B=12

This means this way works for the second term. I will now try this formula for the next line.

N=3

A=7

B=7*2+2

B=16

This means that I was 8 off but I have seen a pattern that could work for this term. I am going to try *3 instead of *2 and +3 instead of +2.

N=3

A=7

B=7*3+3

B=24

This means that this way is right for the third term. I can see a pattern

I can see that the times and the add number is the same as the number for N.

So I now can find the algebraic way for this pattern.

I know that if I times A with N then add N I get B.

B=A * N + N

I know that A=2n+1.

If I put that into the equation instead of A, I get

B=2n+1* N + N

I will now put brackets around the 2n+1 so I can factorise this equation.

(2n+1)* N + N

2n*N = 2n

1*N = N

 

So now the formula reads: -

2n + n + n

I can now simply this formula,

2n + 2n

B= 2n + 2n

I will now test this formula to see if it works.

N=1

B= 1* 1= 1

B= 1*2 = 2

B= 2 + 1 = 3

So far the equation is: -

B= 3 + 2n

B= 2 * 1 = 2

B= 3 + 2

B=5

This means that this equation works for the first term in B. I will now try this formula for the second term.

N=2

B=2*2 =4

B=4+2=8

B=2*2=4

B=4+8

B=12

This means that the second term is right.

I can see that C is one more than B so if the equation for B is 2n + 2n. So if I add a +1 to that formula I should get C.

C= 2n + 2n + 1

I will now test this formula to see if this is right.

N=2

C=2*2=4

C=2*2=4

C=2*2=4

C=4+4+4+1

C=13

This means that this formula is right for the second term but now I will now try and do it for the third term.

N=3

C=3*3=9

C=9*2=18

C=2*3=6

C=18+6+1

C=25

This shows that this formula works for C.

Other Methods of finding B.

The second difference between the B numbers are all +4

                                                                   /2

This means that the rule involves 2n

When I tried 2n.  All of the numbers were out by twice the number of N.

Rule for B must be = 2n + 2n.

Other methods for finding B.

From observation I can see that if you add B and C together you get A.

For example: -

A=3 B=4 C=5

B and C = 4 + 5 = 9

A = 3

A = 9

B and C = 9 and A = 9

This shows that both sides are the same. I will now find this equation.

Equation is B+C=A

I know the equation for C and that is: -

C=B+1

If I now change the part of C on the formula I get,

B+B+1=A

I can simplify that equation.

2B+1=A

I want to put the 2B on its own so I must takes of the +1 and move this to the other side.

2B+1=A

2B=A –1

I now want B on its own so I need to change sides of the 2.

B=A –1

       2

I can now add the formula of A to the equation to get,

B=(2n+1) -1

         2

I need to lose the brackets around 2n+1. I will do this by factorising the 2n+1.

(2n+1)  (2n+1)

B=4n +4n+1-1

          2

The +1 and –1 cancel each other out to make,

B=4n +4n

        2

B=2n +2n

Other Methods of finding B.

From looking at the original table I noticed a pattern.

 

To get B I have to use a triangle number then times it by 4.

The triangle numbers are 1,3,6,10,15…………

The algebraic way for this is what I am going to show you.

I know that a triangle is,

Nth triangle number = n  (n+1)

                              2

B=triangle number *4

If I change the triangle number to,

B=N  (n+1) *4

     2

I need to lose the n

                        2

I am going to do this by putting the 4 with N to make 4N

                                                                     2

B=4n

      2

B=2n

The equation now is,

B=2n(n+1)

I need to lose the brackets around n+1. I will do this by multiplying everything inside the brackets with 2n.

B=2n*n=2n

B=2n*1=2n

The answer is,

B=2n +2n

This shows that the formula for B is 2n +2n and there cannot be any other formula for B.

Other Pythagorean Triples Predicted

I have predicted these numbers for A because I know that for A the numbers go up +2 so I predicted the next two to be 11 and 13. To make sure this is right I will use the formula for A to make sure.

A=2n+1

A=2*5+1

A=11

This shows that the fifth term is actually 11. I will now find the sixth term.

A=2n+1

A=2*6+1

A=13

This shows my predicted were correct. For B I have predicted that the next two numbers will be 60 and 84. I have predicted these because I know that the second difference between the two numbers is always +4 so I used that and came out with 60 and 84.

To see if I am right I will use the formula B to see if it true.

B=2n +2n

B=5*5=25

B=25*2=50

B=2*5=10

B=50+10

B=60

This shows that this works for the fifth term. I will now see if it works for the sixth term.

B=2n +2n

B=6*6=36

B=2*36=72

B=2*6=12

B=72+12

B=84.

This means that my predictions for B were correct and now I am going to see if my prediction for C were correct. I know that C is always +1 to B so I just added 1 to B but I have to see if 61 and 85 are right.

C=2n +2n+1

C=5*5=25

C=25*2=50

C=2*5=10

C=50+10+1

C=61

I can see that my prediction for the fifth term is correct. I will now try the sixth term.

C=2n +2n+1

C=6*6=36

C=36*2=72

C=2*6=12

C=72+12+1

C=85

I can my predictions for C is both correct.

Advanced Algebra

At the beginning of this coursework I said the formula to test a Pythagorean triple is that the left hand side of the formula is equal to the right hand side of the formula. Now that I have all the formula for A, B and C, I will now see if this formula is right.

A=2n+1

B=2n +2n

C=2n +2n+1

Left Hand Side

A=(2n+1)

I have to factorise this formula to get rid of the brackets.

A=(2n+1) (2n+1)

A=2n*2n=4n

A=2n*1=2n

A=1*2n=2n

A=1*1=1

A=4n +2n+2n+1

I can simplify this to,

A=4n +4n+1

I now have to do the same to the formula of B.

B=(2n +2n)

B=(2n +2n) (2n +2n)

B=2n +2n =4n

B=2n +2n=4n

B=2n+2n =4n

B=2n+2n=4n

B=4n +4n +4n +4n

This can be simplified to,

B=8n +8n

I now know the left hand side equation is: -

A and B=4n +4n+1+8n +8n

 

Now that I found the left Hand Side of the formula, I must find the right hand side of the formula. I have to do to C’s formula exactly what B’s formulas is and what A formula is.

C=2n +2n+1

I now have to factorise this.

C=(2n +2n+1) (2n +2n+1)

C=2n *2n =4n

C=2n *2n=4n

C=2n +1=2n

C=2n*2n =4n

C=2n*2n=4n

C=2n*1=2n

C=1*2n =2n

C=1*2n=2n

C=1*1=1

C=4n +4n +2n +4n +4n +2n+2n +2n+1

This can be simplified to,

C=4n +4n+1+8n +8n  

I got this answer by adding the two 4n together and then the other two 4n  together.