• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  • Level: GCSE
  • Subject: Maths
  • Word count: 2025

The Fencing Problem.

Extracts from this document...


Phil Bavister The Fencing Problem A farmer has exactly 100 metres of fencing and wants to fence off a plot of land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000 m. She wishes to fence off the plot of land, which contains the maximum area. Investigate the shape, or shapes, that could be used to fence in the maximum area using exactly 1000 m of fencing each time. Firstly I thought about some of the shapes the possible and decided to calculate their areas in a logical order. So I decided to start with triangles before moving onto quadrilaterals, pentagons, hexagons and so on. Triangles To calculate the area of a triangle you must times half the base by the vertical height. Formulae A = b x h Equilateral Triangle There is more than one way to calculate the area of an equilateral triangle so I will demonstrate each of the three different ways. 1. A = x x x sin 60 = 48112.52 (to 2d.p.) 2. A = x - 2 x h h = sin 60 - - 3 = 48112.52 (to 2d.p.) 3. A = ab sin c = 48112.52 (to 2d.p.) Isosceles Triangle A = b x h To find the vertical height you must split the isosceles triangle into two separate right-angled triangles and use Pythagoras' Theorem. Pythagoras can only be used in right-angled triangles and uses two sides to find the third side. ...read more.


Parallelograms To find the area of a parallelogram you would do: A = b x h But I have found out that a rectangle with the same dimensions has a larger area. This is because a parallelogram is like a rectangle squashed on its side, this means it has a smaller height than rectangle, therefore producing a smaller area. Trapeziums The formula to find the area of a trapezium is: A = x h h = Example Calculating the average of the parallel sides = = =200 Calculating the distance between them (height) h = 350 -150 =200 =100 250sq - 100sq = 52,500 = 229.13 Area of Trapezium A = x h A = 200 x 229.13 A = 45825.76 Regular Polygons A regular polygon is a any shape made up of straight lines with each of its sides and angles being the same. To find the area of a regular polygon you must divide your polygon into several (number of sides) isosceles triangles. You can do this by drawing lines outwards from the centre point to each of the points on the perimeter. From here draw a perpendicular line vertically from top of the triangle to the base. In doing so you will have created two identical right angled triangles. From here using Pythagoras and Trigonometry you will be able to find the area of the right angled triangle. By simply multiplying this by two you will generate the area of the isosceles triangle. ...read more.


For this investigation we consider a circle having infinite sides. Here is a graph to show my results: There is also a spreadsheet to show all my results with regular polygons. After calculating each of the areas I tabulated them, this was so that I didn't lose them and it helped me identify any trends that were appearing. I found that it was easier to calculate areas using numerical values first and then translate these into and algebraic equation. When performing this investigation I made a number of mistakes. One of which was to make sure all my triangles were possible. Example But I learnt from this that the sum of sides a and c must greater than the base (in this case over 500m). I noticed from my table of results that as the area increased, the area of the internal triangles decreased. This can be explained because as the area increased so did the number of sides but the perimeter remained the same, meaning the triangles were getting smaller but they were having to be multiplied by larger numbers to find the total area. Evaluation I a satisfied with my findings because I have gathered good results and developed formulas. I believe that my results are valid and well presented. I feel I have used formulas and computers well to aid my investigation. Ways to extend this investigation in the future would be to: 1. Note : All figures in this document have been rounded to two decimal places but in the calculations were given their full numerical value. I couldn' t find a squared sign so instead I have used the notation - sq. ...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. The Fencing Problem

    1000 44194.17 300 350 316.23 1000 47434.16 350 325 273.86 1000 47925.72 400 300 223.61 1000 44721.36 450 275 158.11 1000 35575.62 Naturally, the perimeter must always equal 1000, and the measurements of the triangles have all met this criterion.

  2. My investigation is about a farmer who has exactly 1000 metres of fencing and ...

    I will plot this information as well as the information from the other graph onto a new graph showing number of sides against area: This graph shows that my prediction is still correct as the area is getting larger as the amount of sides increase.

  1. Beyond Pythagoras

    Area (A) The formula for area is just 1/2 ab, so the formula is 0.5 � (10n + 5) � (an + 5n) = 0. 5 � 10an2 + 50n2 + 5an + 25n Perimeter (P) P = a + b + c P = 10n + 5 + 10n2

  2. Maths investigation - The Fencing Problem

    = 68794.7 Hexagon Using trigonometry 1/2 x 6 x 166 2/3 x 166 2/3 x sin 60 Area = 72168.8 Septagon Each side = 142.9 142.9/sin 51.4 = Y/sin 64.3 Y = 164.8 Area = 7x( 1/2 x 164.8 x 164.8 x sin 51.4)

  1. Beyond Pythagoras.

    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

  2. Beyond Pythagoras.

    210 5 14 48 50 112 336 6 16 63 65 144 504 7 18 80 82 180 720 8 20 99 101 220 990 To find the middle length and the long length I have been using the shortest length to help me.

  1. The Fencing Problem.

    Regular Polygons Having tested isosceles triangles and rectangles I found that regular sided shapes give the maximum area. I know this because the maximum area of an isosceles triangle is given when the sides are each 333.33m. The maximum area given by a rectangle is give by a square with 250m sides.

  2. The Fencing Problem

    right-angled triangles using trigonometry and then multiply it by 2 to get the area of the triangle above. This is what a general right-angled triangle looks like for each shape: So to find out the height, I will need to do the equation Tan (180 - ((180/n)

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