• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 2025

The Fencing Problem.

Extracts from this document...

Introduction

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.

Middle

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.

Conclusion

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. Maths investigation - The Fencing Problem

    A hexagon has 6 sides.... You get the idea. The shape with infinite sides must have infinite lines of symmetry. The only shape that has infinite lines of symmetry is the circle.

  2. Fencing Problem

    At the start of my table I have purposely made "angle a" smaller then "angle b" to see whether of not the area will be larger or smaller. I learnt that the area is smaller as I tested the same angles but made "angle a" the larger one and "angle

  1. Fencing problem.

    I shall multiply this figure by the number of triangles present within the hexagon: Area of hexagon = Area of one triangle � Number of triangles Area of hexagon = 12035.7 � 6 = 72214.2m2 Heptagon The third polygon that I shall be investigating is a regular heptagon.

  2. The Fencing Problem.

    The formula entered will be =G3*A3 where G3 is the area of one triangle and A3 is the number of sides. The formulas and headings will be entered in as shown in the table below. Number 1 Equal Side Perimeter Internal Angle Half Angle Height Area of 1 Triangle Total Area of Sides (m)

  1. The Fencing Problem

    =(A12*C12)/2 The table below is the same table as above but in normal view. Base (m) Sloping Height (m) Perpendicular Height (m) Perimeter (m) Area (m�) 50 475 474.34 1000 11858.54 100 450 447.21 1000 22360.68 150 425 418.33 1000 31374.75 200 400 387.30 1000 38729.83 250 375 353.55

  2. The Fencing Problem

    To work out the area of a parallelogram, I will use the following formula: l = length of parallelogram w = width of parallelogram h = perpendicular height of parallelogram a = area of parallelogram a = l * h However, because I only know the length and width of

  1. t shape t toal

    This proves my equation correct as the correct translated T-total is generated. t = (5x - 7g) - b(5g) + (a5) Only if you split the diagonal translation into two parts the horizontal movement and a vertical movement you can use this formula to generate you new T total It

  2. The Fencing Problem. My aim is to determine which shape will give me ...

    (152.6377* 31.25) * 18 4239.9361*18=76318.85 76318.85 Shape No. Shape Type Area (M2) 1 Equilateral Triangle 48126.3 2 Square 62500 3 Regular Pentagon 68823 4 Regular Hexagon 71999.99r 5 Regular Heptagon 74162.32683 6 Regular Octagon 75444.17389 7 Regular Nonagon 76318.85 Analysis I can tell from this graph that the hypothesis I

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