• 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. The Fencing Problem

    h 450 50 3) h 425 75 4) h 400 100 5) h 375 125 6) h 350 150 7) h 325 175 8) h 300 200 ^ The graph has a smooth curve,; however, unlike the scalene triangles, the isosceles triangles results aren't "symmetrical"; they ascend to a point, and then decline with greater magnitude.

  2. Fencing Problem

    the largest answer I have got is nowhere close to the area of a square. Trapezium I am not going to investigate the area of trapeziums as they share the same properties as parallelograms and I would get the same answers because if you split them up into shapes ypu

  1. The Fencing Problem

    multiply the area of the triangle by 6 because there are six triangles in the hexagon. area of hexagon = area of triangle * 6 = 12035.74m� * 6 = 72214.44m� Heptagon Heptagons have 7 sides therefore I can divide it up into 7 isosceles triangles, and the angle at the top an isosceles triangle would be 360/7=51.4?.

  2. t shape t toal

    If we substitute in the vales hopefully it will work. t = 5x + 7g t = (5 x 41) + (7 x 9) t = 205 + 63 t = 268 As we can see my prediction is correct.

  1. Geography Investigation: Residential Areas

    This is a personal view of the area, thus it is primary data. This can be found on my 'External Questionnaire' which I complete myself. (See Figure 8) I also have a hypothesis that requires me to have data about the appearance of the area - I have called this, Index of Decay.

  2. The fencing problem 5-6 pages

    The Area of the Pentagon is=68819.10m2 If I look at the result of this Area, I see that my theory was correct, as the Area of a Pentagon is bigger than the Area of a square. Therefore, I can say that the Area of a regular shape increases as the shape has more sides.

  1. Fencing Problem

    On the graph, the areas of the triangles go up and then start to decrease at a certain point. 1. This means when the difference between the sides is less, the area is larger. This graph is not symmetrical because the maximum is not in the middle.

  2. A farmer has exactly 1000m of fencing and wants to fence off a plot ...

    30 170 300 27658.63337 200 480 320 1000 500 20 180 300 23237.90008 200 490 310 1000 500 10 190 300 16881.94302 The maximum value in scalene is identical to an isosceles! This could mean that the more irregular shapes are the smaller the area, or this could be put

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