• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1496

Maths Coursework: The Fencing Problem

Extracts from this document...

Introduction

Maths Coursework: The Fencing Problem A farmer has exactly 1000 metres of fencing, with it the farmer wishes to fence off a level plot of land. What the farmer does wish to do is fence off the plot of land which contains the maximum area. Introduction I have been given an investigation to examine which shape with a perimeter of 1000m will have the biggest area. By using scale drawings, or a closed loop of string, I'm going to look at some possible figures for the plot of land. In each case I'm going to ensure that the perimeter is 1000m and obtain the enclosed area. I'll then go onto further investigate the shape or shapes which have the maximum area and see whether there is a pattern within the concept of the tangent of an angle and the area of the shape. Triangles I'll begin by investigating the three - sided shapes. They consist of triangles of different shapes and sizes. Equilateral Triangle Area = 1/2 x 333? x 333? x Sin60? Area = 48112.52m2 (2.d.p) From this I can state that an equilateral triangle with all sides equal will have an area of 48112.52m2 (2 d.p.). Table to show the area of 6 isosceles triangles. Triangles Shape Side A Side B Side C Area 1 290 290 420 42000.00 2 300 300 400 44721.35 3 310 310 380 46540.31 4 320 320 360 47623.52 5 330 330 340 ...read more.

Middle

60? 180 - 60 = 120? 120 / 2 = 60? Tan 60? = H / 83? H = 144.3375673 144.3375673 x 83? = 12028.13061 / 2 = Area of right - angled triangle. 6014.065304 x 2 = 12028.13061m2 (Area of isosceles triangle). 12028.13061 x 6 = 72168.78366m2 Area of Hexagon = 72168.78366m2 Regular Octagon Now I'll aim to find the area of a regular octagon within a circle, therefore each length of the octagon = 125m. 45? 180 - 45 = 135? 135 / 2 = 67.5? Tan 67.5? = H / 62.5 H = 150.8883476 150.8883476 x 62.5 = 9430.521728 / 2 = 4715.260864 (Area of right - angled triangle). 4715.260864 x 2 = 9430.521728m2 (Area of isosceles triangle). 9430.521728 x 8 = 75444.17382m2 Area of Octagon = 75444.17382m2 Using a formula it's possible to find out the area of a regular polygon is as long as you know the number of sides. Shape Number of Sides Area of Regular Polygon 1 3 48112.52243 2 4 62500.00000 3 5 68819.09602 4 6 72168.78365 5 7 74161.47845 6 8 75444.17382 7 9 76318.81721 8 10 76942.08843 9 12 77751.05849 10 15 78410.50183 11 30 79286.37045 12 50 79472.72422 13 60 79504.7362 14 75 79530.92399 15 80 79536.56119 16 90 79545.14801 17 100 79551.28988 18 200 79570.92645 19 300 79574.56264 20 400 79575.83529 21 500 79576.42435 22 600 79576.74432 23 700 79576.93726 24 800 79577.06248 25 900 79577.14834 26 1000 79577.20975 ...read more.

Conclusion

7694.208843 x 10 = 76942.08843m2 Area of Decagon = 76942.08843m2 Regular Dodecagon Now I'll aim to find the area of a regular dodecagon within a circle, therefore each length of the dodecagon = 83?m. Tan 75? = H / 41? H = 155.502117 155.502117 x 41? = 6479.254874 / 2 = 3239.627437 (Area of right - angled triangle). 3239.627437 x 2 = 6479.254874m2 (Area of isosceles triangle). 6479.254874 x 12 = 77751.05849m2 Area of Dodecagon = 77751.05849m2 I've created a table to show the number of sides and also the tangent of the angle. Creating this table helps me in finding whether there is a relationship between the tangent of an angle and the number of sides. Shape Number of Sides Angle Tangent of the Angle 1 5 54 1.37638192 2 6 60 1.732050808 3 7 64.28571429 2.076521397 4 8 67.5 2.414213562 5 9 70 2.747477419 6 10 72 3.077683537 7 11 73.63636364 3.405687239 8 12 75 3.732050808 I've transformed the results in the table to a scatter graph to show the relationship in a better perspective. Reading of the scatter graph shows that there is some strong positive correlation between the number of sides and the tangent of the angle, meaning that there is a relationship. As the tangent of the angle increases the area of the regular polygon increases yet as the graph continues it looks like it will stabilise ?? ?? ?? ?? Rushan Perera 10D 20/04/2007 ...read more.

The above preview is unformatted text

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. Mathematics Gcse Coursework Tubes Investigation

    And divided the length of one side by 2 then you can find the length of half a side.

  2. Fencing Problem - Maths Coursework

    / 2 = 400 To work out the area I need to know the height of the triangle. Tow ork out the height I can use Pythagoras' Theorem. Below is the formula and area when using a base of 200m.

  1. Geography Investigation: Residential Areas

    To increase the accuracy of my results would be to increase my sample size as a number one priority. I have explained many times why this would be of help to me.

  2. The Fencing Problem

    ( 66.334 = 6,257.916m� I can now add up all these areas to find the total area of the hexagon. 6,257.916 + 6,257.864 + 17,777.777 = 30,293.609m� Hexagon 3: This hexagon will have sides of 250m, 200m, 100m, 150m, 100m and 200m.

  1. The Fencing Problem

    90 400.00 1000 40000.00 100 400 100 393.92 1000 39392.31 100 400 110 375.88 1000 37587.70 100 400 120 346.41 1000 34641.02 100 400 130 306.42 1000 30641.78 100 400 140 257.12 1000 25711.50 100 400 150 200.00 1000 20000.00 100 400 160 136.81 1000 13680.81 100 400 170 69.46

  2. Math Coursework Fencing

    The area of a parallelogram can be found by using the formula (Base multiplied by Height equals area). My prediction of the greatest area here, is that the closer the shape gets to a square the greater its area will be.

  1. Fencing Problem

    I will divide 1000 m (perimeter) by 4 (sides), to give me the length of each of the sides - 250 m. Then I will multiply the Length by Length to give me the area. * 1000 m / 4 = 250 m (Sides)

  2. GCSE Maths Coursework Growing Shapes

    I notice that with pattern 4 in columns when n = 4: 2 lots of (1 + 3 + 5) = 2 x 9 1 lot of 7 = 1 x 7 2 x 9 = 2(n-1)2 7 = 2n - 1 Ts = 2(n-1)2 + 2n - 1 Number of Lines Pattern no.

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