• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  • Level: GCSE
  • Subject: Maths
  • Word count: 4002

Maths Coursework - Beyond Pythagoras

Extracts from this document...

Introduction

Maths Coursework - Beyond Pythagoras

        In this piece of coursework I will be looking at and investigating the relationship between the lengths of the three sides of right angled triangles and also the perimeter and area of these triangles. In particular I will be investigating triangles which satisfy the condition of a Pythagorean Triple. I will initially be looking at Pythagorean triples for which the shortest side is an odd number. I will begin my coursework by using the three triangles which I have already been given.

I will begin my investigation by checking to see if these triangles are actually Pythagorean triples. If they are they should satisfy the condition of the Pythagorean Triple shown below.

a2 + b2 = c2

The number 3,4 and 5 satisfy the condition…..

a2 + b2 = c2

This is because……  32 + 42 = 52

32 = 3 x 3 = 9            (Shortest Side)

42 = 4 x 4 = 16          (Middle Side)

52 = 5 x 5 = 25          (Hypotenuse)

32 + 42 = 9 + 16 = 25 = 52

The numbers 3, 4 and 5 are lengths, in appropriate units of the sides of a right angled triangle. This is also true for the other right angled triangles. The perimeter and area of this triangle.

Perimeter =   a + b + c

3 + 4 + 5 = 12 units

Area =   ½ ( a x b )

½ ( 3 x 4 ) =  6 square units

The above methods apply to all other right angled triangles.

The 3, 4, 5 triangle is the simplest triangle which satisfies the condition of Pythagoras’ theorem.

The next triangle with the shortest side as an odd number is the 5, 12, 13 triangle.

This triangle also satisfies the condition…..

a2 + b2 = c2

Because……   52 + 122 = 132

52 = 5 x 5 = 25                   (Shortest Side)

122 = 12 x 12 = 144           (Middle Side)

132 = 13 x 13 = 169           (Hypotenuse)

52 + 122 = 25 + 144 = 169 = 132

Perimeter

5 + 12 + 13 = 30 units

Area

½ ( 5 x 12 ) = 30 square units

The next triangle in the sequence is the 7, 24, 25 triangle.

This triangle also satisfies the condition……

a2 + b2 = c2

Because…….   72 + 242 = 252

72 = 7 x 7 = 49                   (Shortest Side)

242 = 24 x 24 = 576           (Middle Side)

252 = 25 x 25 = 625           (Hypotenuse)

72 + 242 = 49 + 576 = 625 = 252

Perimeter

7 + 24 + 25 = 56 units

Area

½ ( 7 x 24 ) = 84 square units

Now I will collect all of my results into a table for the first three triangles in the sequence.

Nth term

Shortest Side

Middle Side

Hypotenuse

Perimeter

Area

1

3

4

5

12

6

2

5

12

13

30

30

3

7

24

25

56

84

...read more.

Middle

3

7

24

25

56

84

4

9

40

41

90

180

5

11

60

61

132

330

6

13

84

85

182

546

7

15

112

113

240

840

Now that I have the rules for the sides of the triangle I will work out the rules for the perimeter and area using them.

Perimeter

Perimeter =   a + b + c

So if I let a, b, and c be the rules for each of the sides it should give me the rule for the perimeter. This way I will be able to work out the perimeter of any triangle just by using the nth term of the triangle.

Shortest Side  +  Middle Side  +  Hypotenuse  =  Perimeter

          a           +           b           +          c           =  Perimeter

  ( 2n + 1 )   +   ( 2n2 + 2n )+   ( 2n2 + 2n + 1 ) =  Perimeter      

          2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter

If I collect all the like terms I will end up with the rule for the perimeter.

          2n2  + 2n2  + 2n + 2n + 2n + 1 + 1 =  4n2 + 6n + 2

                                          4n2 + 6n + 2 = Perimeter

To see if this is the correct formula I will test it with the nth term. If it is correct I should get the perimeter of the triangles.

Nth Term        4n            6n           2           4n2 + 6n + 2          Perimeter

This proves that this is the correct formula for working out the perimeter for any triangle in the sequence.

The next thing I am going to do is to work out the rule of the area for these triangles.

Area

½ ( a x b )

Above you can see the equation that is used to work out the area of any triangle. The 2 values needed to do this are the lengths of the shortest and middle sides.

In a similar way to the method I used to work out the perimeter, I can substitute the rule for the shortest and middle side into the equation. This should give me the rule to work out the area of any triangle in the sequence.

The formula :   ½ ( a x b )       becomes…….     ½ x ( 2n + 1 ) x ( 2n2 + 2n )

...read more.

Conclusion

I will not try to go any further with this because the equations for the perimeter and area are even more complex than the equation for the middle side.

Conclusion

        In my coursework I have found out that there is a strong link between pythagorean triples. The sides, perimeter and area of the triangles which are made from these pythagorean triples showed a distinct pattern. Because of this I was able to find a rule for each property of the right-angled triangles. These rules allowed me to be able to find the properties of any pythagorean triple in the sequence. I could easily find many pythagorean triples beyond those give at the beginning of the coursework. I have also found some formulas that link the sides, perimeter and area of the triangles together. If I was not limited by my abilities I might have found all the equations and this would have allowed me to work out all the properties of the pythagorean triangle with either one of the sides, the perimeter or the area. I was unable to reverse the rule for the middle side. I needed to do this to find the links between the middle side and the other sides, perimeter and area. Either I do not have the knowledge to do this or it is impossible. I was able to do it for the shortest side so there must be a way for the middle side and I don’t know how to do it..

        If I did this piece of coursework again I would definitely spend a greater amount of time trying to figure out the formulas that I have given up on. By doing this I might have found them and I would have been able to make more links between the sides, perimeter and the area. I would also have gone into more depth with trigonometry and worked out the gradient of the curve created by the angles in the triangles.  

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Medicine and mathematics

    Therefore, we have to proceed using logs. 10/300=0.6x 0.03.=0.6x Log0.03./Log0.6 = x 6.86(3 s.f.) = x 6.86 Hours. 60 hours 0.86 = 51.6 minutes 60 minutes x 0.6 = 36 seconds Meaning that after 6 hours, 51.6 minutes, and 36 seconds after administration there will be less than 10mg amount of active penicillin in the bloodstream.

  2. Investigation to find out the number of matchsticks on the perimeter in a matchstick ...

    r = number of rows; P = number of matchsticks on perimeter; t = total number of matchsticks r P t 1 4 4 2 10 13 3 16 26 4 22 43 And I am going to make another table to find out the perimeter difference from the number of rows and the number of matchsticks on perimeter.

  1. Mathematics Gcse Coursework Tubes Investigation

    3. 4. 5. 6. 7. To in order to find the formula that work for every polygon I will substitute the letters in instead of the numbers. And I will follow the steps in the box above. Explanation of the formula: The first equation explains 1 - 3 steps in the first box on this page.

  2. Fencing Problem - Maths Coursework

    Area = 1/2 � b � H = 1/2 � 100 � 137.638 = 6881.910 I now have the area of half of one of the segments, so I simply multiply that number by 10 and get the area of the shape.

  1. Geography Investigation: Residential Areas

    So I can examine the correlation of externality features I created the table in Figure 8 which is based on my view of the area. Using this primary evidence and all of the streets I surveyed in my investigation I will examine this hypothesis.

  2. GCSE Maths Coursework

    It must be at least 1. * With rectangles there is an infinite number of equable shapes, it occurs for every size over 2 as my table proves.

  1. Fencing - maths coursework

    Now I am going to look at other four-sided shapes to work out that the square has the biggest area in the four-sided shape family. The area of a square is length x width, However the area of a parallelogram base x perpicardicular high.

  2. Fencing problem.

    3600 � 10 Exterior angles = 360 I shall now find the interior angles of the above shape. This can be found by excluding 1800 from the exterior angle. Interior angle = 1800 - Exterior angle Interior angle = 1800 - 36 = 1440 Now by using the help of

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work