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  • 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.

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