• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The Fencing Problem

Extracts from this document...

Introduction

The Fencing Problem A farmer has the need to enclose an area of land with 1,000 metres of fencing. He has to do so trying to make sure that he has enclosed the largest possible area of land. Therefore I will be investigating the shapes with the largest area that could be used to fence with 1000m of fencing. I will start by investigating different polygons. A polygon is a many sided shaped of strait lines which will be easy to measure, giving me more accurate results. These polygons will have a perimeter of 1000m. In this first section I will investigate the first set of polygons. Shape Equation Total area Perimeter Equilateral: 333.3+333.3+333.3 24,052� 1000�3= 333.3 288.64 x (333.3�2) ...read more.

Middle

The second shape of the polygon's family is the square: Shape Equation Total area Perimeter Quadrilateral: 250+250+250+250 62,500� 1000�4 250 x 250 = 62,500� To find the area of this square I have to: * Divide 1000 which is the perimeter by 3 which is the number of the polygon's sides * Multiply two of the polygon's sides By now I can see that between these polygons the square has the largest area. By the third shape, if this pattern is right the pentagon should have a larger area... But before continuing I will try two irregular polygons and investigate if by having different lengths of sides adding up to 1000m would affect the area. ...read more.

Conclusion

* Multiply the two shorter sides. I will have the area of a square so I divide it by 2 to make the area of an equilateral. Here I have proved myself wrong but I have stated something right: that irregular polygon's have smaller area than a regular polygon would have. With this investigation I have also shown again that the equilaterals have smaller area than the quadrilaterals. I have already shown the difference between irregular and regular polygons. I will continue investigating more areas of polygons and I think that as more lines the polygon has the larger the area. This document was downloaded from Coursework.Info - The UK's Coursework Database ?? ?? ?? ?? II ...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.

    The area of one isosceles triangle will be calculated using the formula =(B3*F3)/2 where B3 is one equal side and F3 is the height. Finally the total area will be calculated by multiplying the area of one isosceles triangle by the number of sides.

  2. The Fencing Problem

    Triangle 1: I will now look at an equilateral triangle. An equilateral triangle is a triangle with all sides and angles the same. This means that an equilateral triangle perimeter will be 1000m divided by 3. Perimeter = 1000m 1000m ( 3 = 333.333�m This means that each side will be 333.333�m long.

  1. Geography Investigation: Residential Areas

    4 15 4 6 3 1 1 18 5 10 6 1 1 23 6 6 3 3 9 31 7 6 3 4 16 34 8 7 4 4 16 41 9 10 6 3 9 52 10 4 2 8 64 ?d2= 124 Rs =1-(6? d2 / n3-n)

  2. 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.

  1. Fencing Problem

    249.8 1000 62499.96 After analysing my 2nd table of results I have come to the conclusion that the closer the dimensions of the rectangle are to the ones of a square the larger the area. The largest area for a rectangle that I have discovered has dimension of 249.9 by

  2. Math Coursework Fencing

    We must spit the kite into two triangles as shown below: The area of a triangle is calculated using this formula: Since we are eventually going to double the area we get so we can derive this final formula to calculate the area of a kite.

  1. The Fencing Problem

    I will work out the results using 249m, 249.5 and 249.75, as shown below. Height (m) Base (m) Area (m2) 251 249 62499 250.5 249.5 62499.75 250.25 249.75 62499.9375 250 250 62500 249.75 250.25 62499.9375 249.5 250.5 62499.75 249 251 62499 All of these results fit into the graph line that I have, making my graph reliable.

  2. The Fencing Problem

    This shape is a circle and I will try to work out the area of a circle with a circumference of 1000m. However, I will need to know the radius of the circle before I can calculate its area.

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