• 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
  17. 17
    17
  • Level: GCSE
  • Subject: Maths
  • Word count: 2941

The Fencing Problem.

Extracts from this document...

Introduction

The Fencing Problem A farmer has exactly 1000 metres of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000m. So it could be: Or anything else with a perimeter (or circumference) of 1000m. In this investigation all angles drawn to scale. All sides are to the scale of 50m:1cm What she does wish to do is fence off the plot of land which contains the MAXIMUM area. Part 1 By using scale drawings, or a closed loop of string, or otherwise examine some possible figures for the plot of land. In each case ensure that the perimeter is 1000m and obtain the enclosed area. Part 2 Investigate the shape, or shapes of the plot of land which have the maximum area. In order to find out the area of a given shape I need to know some formulae. Below is a range of formulae which could be useful in order to carry out this investigation. Area of a Triangle A = 1/2 x b x h Area = 1/2 x base x vertical height The alternative formula for a triangle is: A = 1/2 abSINC Area = 1/2 abSINC Area of a Parallelogram A = b x h Area = base x vertical height Area of a Trapezium A = 1/2 x (a+b) x h Area = average of x distance parallel sides between them Area of a Circle A = ? x r2 Area = Pie x (radius)2 Circumference = ? ...read more.

Middle

* From here it is a simple case of applying trigonometry to find out the height. TAN 540 = h / 100 h = TAN 540 x 100 1.37638192 x 100 = 137.638192 h = 137.638192 137.638192 x 100 = 13,763. 8192 x5 = 68,819.096 A = 68,819.096 The method used above to will be used to find the area of all the regular polygons from now on. I will go up to a ten sided shape (Regular Decagon). This will then , hopefully, be enough to extract a formula from. Regular Hexagon TAN 600 = h / 83.3 h = TAN 600 x 83.3 1.732050808 x 83.3 = 144.2798323 h = 144.2798323 144.2798323 x 83.3 = 12,028.13061 x6 = 72,168.78364 A = 72,168.78364 m2 Regular Heptagon TAN 64.285714290 = h / 71.42857143 h = TAN 64.285714290 x 71.42857143 2.076521397 x 71.42857143 = 148. 3229573 h= 148. 3229573 148. 3229573 x 71.42857143 = 10,594.49695 x7 = 74,161.47867 A = 74,161.47867 m2 Regular Octagon TAN 67.50 = h / 62.5 h = TAN 67.50 x 62.5 2.414213562 x 62.5 = 150.08883476 h = 150.08883476 150.08883476 x 62.5 = 9,430.521728 x8 = 75,444.17382 A = 75,444.17382 m2 Regular Nonagon TAN 700 = h / 55.55555556 h = TAN 700 x 55.55555556 2.747477419 x 55.55555556 = 152.6376344 h = 152.6376344 152.6376344 x 55.55555556 = 8,479.88858 x9 = 76,318.81722 A = 76,318.81722 m2 Regular Decagon TAN 720 = h / 50 h = TAN 720 x 50 3.077683537 x 50 = 153.8841769 h = 153.8841769 153.8841769 x 50 = 7,694.208843 x10 = 76,942.08843 A = 76,942.08843 m2 For all the Regular Polygons that I have worked out the maximum area for, I will display the information via tabulation. ...read more.

Conclusion

THE RELATIONSHIP BETWEEN MAXIMUM AREA AND NUMBER OF SIDES No. of Sides Maximum Area (m ) 30 79286.3704518549 40 79413.7796010919 50 79472.7242193263 60 79504.7361988674 70 79524.0359236919 80 79536.5611854907 90 79545.1480080987 100 79551.2898844352 110 79555.8340562759 120 79559.2902021367 130 79561.9798547583 140 79564.1139859050 150 79565.8356773256 160 79567.2447445089 170 79568.4125373509 180 79569.3911538321 190 79570.2193531608 200 79570.9264535885 Although the table has a lot of shape I need shapes which have even more sides No. of Sides Maximum Area (m ) 300 79574.5626425939 400 79575.8352930449 500 79576.4243456389 600 79576.7443240975 700 79576.9372607619 800 79577.0624839874 900 79577.1483365629 1,000 79577.2097463854 No. of Sides Maximum Area (m ) 2,000 79577.4060960806 3,000 79577.4424571250 4,000 79577.4551834840 5,000 79577.4610739747 6,000 79577.4642737380 7,000 79577.4662030858 8,000 79577.4674553697 9,000 79577.4683138474 10,000 79577.4689279207 11,000 79577.4693823211 12,000 79577.4697278827 13,000 79577.4699968068 14,000 79577.4702102085 15,000 79577.4703824103 No. of Sides Maximum Area (m ) 20,000 79577.4708914006 30,000 79577.4712550516 40,000 79577.4713823559 50,000 79577.4714411890 60,000 79577.4714733400 70,000 79577.4714925618 80,000 79577.4715047873 90,000 79577.4715131883 100,000 79577.4715193700 No. of Sides Maximum Area (m ) 200,000 79577.4715394123 300,000 79577.4715417683 400,000 79577.4715440107 500,000 79577.4715444446 600,000 79577.4715468591 700,000 79577.4715440847 800,000 79577.4715446172 900,000 79577.4715434230 1,000,000 79577.4715444541 Graphs for all these tables can be found on pages 18-23 The area keeps increasing as the number of sides does. The area is increasing very slightly but is still increasing. From 800 sides to 1,000,000 sides the area has stayed on 79,577.... Although the decimal place has moved fractionally each time. Maximum Area The shape with the most area is a circle. As the number of sides on each shape increase so does the area. As my prediction stated the fatter and rounded a shape became the larger the area is. The shapes which have lots of sides (1,000,000) have a large area because they are almost near a circle but not quite. ...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. When the area of the base is the same as the area of the ...

    when the height is 4cm and from the tables which I have presented, I have noticed that the ratio is always 1/6 when the square/net and length are divided between one another. I also predict that the ratio outcome for the 24cm by 24cm tray will also be 1/6, there

  2. Fencing Problem

    of the right angled triangle by 10, (number of sides x 2), to give me the area of the pentagon instead of multiplying it by 2 to give me the area of the isosceles triangle and then multiplying the answer of that by the amount of sides in the shape.

  1. Fencing problem.

    The trigonometry function TAN shall be used to find the height, which can also be known as length of opposite. TAN � = Opposite � Adjacent TAN 700 = Opposite � 55.6m Opposite = TAN 700 � 55.6m Opposite = 152.8m = Height.

  2. 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

  1. t shape t toal

    translation 2) enlargement 3) rotation and 4) reflection. I will be using two types of transformations firstly; these will be translation, which I have already been using, and rotation. I will also be using these transformations in combination. I will firstly start off by rotating the t-shape clockwise 90?

  2. t shape t toal

    Firstly here is the formula I need to use. T T+4 T+7 T+8 T+9 That is the converted version of the 4 by 6 grid size.

  1. t shape t toal

    3 x 7 = 21 Or g x 7 = 21 Grid size (g) Number that you take away Gained by 3 x 3 21 3 x 7 4 x 4 28 4 x 7 5 x 5 35 5 x 7 6 x 6 42 6 x 7 7

  2. Geography Investigation: Residential Areas

    I have evidence for this as Figure 12 shows that residents want to get away from the CBD which is known for its poor class housing and cramped conditions. Figure 13 shows that Basingstoke has developed according to the concentric model in a southwesterly direction.

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