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  • Level: GCSE
  • Subject: Maths
  • Word count: 2088

Beyond Pythagoras.

Extracts from this document...

Introduction

The Pythagoras tethered was found by a mathematician from Italy called Pythagoras who is said to be the first ever true mathematician.  He was also the first to teach that the Earth is a sphere that revolves around the Sun.

                    This how the Pythagoras thorium is defend:

A thorium stating that a square of a hypotenuse of a right angled triangle is equal to the sum of the squares of the other two sides.

            Formula that Pythagoras discovered:

          A² + B² = H²

                                                       H                                                                                      

To Check that each of the following sets of numbers satisfy a

Similar condition of the

(Smallest)²+ (middle)² = (largest number)²

1) a) 5,12,13                  

a²+b²=h²                  

5²+12²=h²        

25+144=169                    

 169= 13          

I put these figures in the formula for the Pythagoras theorem to

find that these numbers do satisfy the condition

(Smallest)²+ (middle)² = (largest number)²

l) b) 7,24,25                

a²+b²=h²                  

7²+24²=h²        

49+576=625                    

625=25

After finding that they are Pythagoras triples, I found the perimeter and area.

2)

(5,12,13)

Perimeter = 5 + 12 + 13

                 = 30

Area = 5 × 12

               2

         = 60

             2

         = 30

(7,24,25)                                                                

Perimeter = 7 + 24 + 25

                 = 56

Area = 7 × 24

               2

         = 168

              2

         = 84

I put everything that I did in to a table showing the unit of each side and the perimeter and area.

Shortest side

‘a’

Middle side

‘b’

Longest side (Hypotenuse)

‘h’

Perimeter

Area

3

4

5

12

6

5

12

13

30

30

7

24

25

56

84

To find the Perimeter I simply had to find the total of all three sides.  For the area I had to

P = a + b + h

find multiply side ‘a’ with side ‘b’ and half the total

A = ab

        2

Bellow is have shown in detail how I found the perimeter and area for each triple.

2)

(5,12,13)

Perimeter = 5 + 12 + 13

                 = 30

Area = 5 × 12

               2

         = 60

             2

         = 30

(7,24,25)                                                                

Perimeter = 7 + 24 + 25

                 = 56

...read more.

Middle

2n2 + 2n

      = 2 × 4² + 2 × 4

      = 2 × 16 + 2 × 4

      = 32 + 8

      = 40

Term 5:

‘b’  = 2n2 + 2n

      = 2 × 5² + 2 × 5

      = 2 × 25 + 2 × 5

      = 50 + 10

      = 60

Side ‘h’

It is obviouse that the set of results found for side ‘h’ are just one plus to the results of side ‘b’

Middle side

‘b’

Longest side (Hypotenuse)

‘h’

4         + 1

5

12       + 1

13

24       + 1

25

40       + 1

41

60       + 1

61

84       + 1

85

So I took the formula for side ‘b’ and simply added one onto it:

2n2 + 2n + 1

 I tested this formula by randomly selecting 2 terms.

Term 5:

‘h’ = 2n2 + 2n + 1

      = 2 × 5² + 2 × 5 + 1

      = 2 × 25 + 10 + 1

      = 50 + 11

      = 61

Term 4:

‘h’ = 2n2 + 2n + 1

      = 2 × 4² + 2 × 4 + 1

      = 2 × 16 + 8 + 1

      = 32 + 9

      = 41

Perimeter

To find the perimeter for any shape you must find the total of all the sides so I would do

a + b + c.  So I took the formulas that I fount for all 3 sides and added them together to make the formula for the perimeter.

2n+1 + 2n2 + 2n + 2n2 + 2n + 1

4n2  + 6n +2

I then tested this formula

Term 2:

‘p’ = 4n2  + 6n + 2

      = 4×22  + 6 × 2 + 2

       = 4×4 + 12 + 2

       = 16 + 14

       = 30

Term 5:

‘p’ = 4n2  + 6n + 2

      = 4×52  + 6 × 5 + 2

       = 4×25 + 30 + 2

       = 100 + 32

       = 132

Area

To find the area of a right angled triangle I would have to multiply the base and the height and half the answer as one would to find the area for a simple right angled triangles.  So I took both formulas from the base and height (‘a’’b’) and put them into the following formula:

½ (2n+1) (2n2 + 2n)

Then multiplied it out to get:

4n3 + 6n2 + 2n

          2

Then divide 4n3 + 6n2 + 2n by 2 to get

2n3 + 3n2 + n

I then looked at the relationship between the perimeter and area.

 P = 4n2  + 6n + 2

A = 2n3 + 3n2 + n

...read more.

Conclusion

b = 4

c = 0

I then used this to replace the letters in the formula to find the nth term

an² + bn + c = 0

               When replaced

4n² + 4n + 0

So the nth term formula would be : 2n² + 2n

It could see that everything double and I ended up with a double nth term equation

I then tested this formula for 2 randomly selected terms.

Term 4:

‘b’ = 4n2 + 4n

      = 4 × 4² + 4 × 4

      = 4 × 16 + 4 × 4

      = 64 + 16

      = 80

Term 5:

‘b’  = 4n2 + 4n

      = 4 × 5² + 4 × 5

      = 4 × 25 + 4 × 5

      = 100 + 20

      = 120

Side ‘h’ even

I only had to double the formula for for the hypotenuse

2n2 + 2n + 2

         4n2 + 4n + 2

But as the results have doubled, instead of there being a difference of 1 between side ‘a’ and ‘h’ there is a difference of 2

Middle side

‘b’

Longest side (Hypotenuse)

‘h’

8          + 2

10

24        + 2

26

48        + 2

50

80        + 2

82

60        + 2

122

168      + 2

170

I then tested this formula

Term 5:

‘h’ = 4n2 + 4n + 2

      = 4 × 5² + 4 × 5 + 2

      = 4 × 25 + 20 + 2

      = 100 + 22

      = 122

Term 4:

‘h’ = 4n2 + 4n + 2

      = 4 × 4² + 4 × 4 + 2

      = 4 × 16 + 16 + 2

      = 64 + 18

      = 82

Perimeter even

There are 2 ways to find the perimeter for even.  I can add all the formulas I have found together or I can multiply the whole equation by 2.  Either way this is the formula I got:

4n+2 + 4n2 + 4n + 4n2 + 4n + 2

8n2  + 12n +4

I then tested this formula

Term 2:

‘p’ = 8n2  + 12n + 4

      = 8×22  + 12 × 2 + 4

       = 8×4 + 24 + 4

       = 32 + 28

       = 60

Term 5:

‘p’ = 8n2  + 12n + 4

      = 8×52  + 12 × 5 + 4

       = 8×25 + 60 + 4

       = 200 + 64

       = 264

Area even  

The area for the even results would be

A = pn

If I multiplied the perimeter by n I should get the area.

P = 8n2  + 12n + 4

I will now see if this is correct by using the formulas for side a and b (simple area formula)

Area = ½ (4n + 1) (4n2  + 4n)

        = 16n³ + 10n  + 8n2 + 8n

                             2

        = 16n³ + 12n2  + 8n

        = 8n + 12n2 + 4n  

Area = p × n

This shows that if the perimeter is multiplied by n it well equal to the answer.

...read more.

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