• 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...


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.


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.


* 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.

    of 1 Triangle (rad.) (rad.) (m) (square m) (square m) 5 =1000/A3 =B3*A3 =2*PI()/A3 =D3/2 =(B3/2)/TAN(E3) =(B3*F3)/2 =G3*A3 Having entered the correct information I will be able to calculate the areas of many regular polygons with different numbers of sides and with a perimeter of 1000m.

  2. The Fencing Problem

    I can now work out the height of the triangle using Pythagorean theorem. This states that the square of the length of the hypotenuse of a right-angled triangle equals the sum of the squares of the lengths of the other two sides.

  1. Geography Investigation: Residential Areas

    d2 / n3-n) Rs =1-(6? d2 / n3-n) ?d2 = 24 therefore 6?d2 = 744 n = 10 therefore n3-n = 990 Rs = 1-(744/990) Rs = 0.25 (correct to 2 significant figures) Null Hypothesis: Accepted. My null hypothesis said that there is a correlation between the index of decay and whether or not the occupants own their own home.

  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

    I will use trigonometry again to figure out the height of the right-angled triangle. * I will use Tangent, as I have my opposite length to my angle and am trying to figure out my adjacent i.e. height. * Tan36 = 100 * I will change the subject or the formula to figure out the height.

  2. Math Coursework Fencing

    Worked example: a=400 b=100 C=150 It is clear that a table I necessary as it will allow me to work out the best kite in the quickest manner.

  1. The Fencing Problem

    If the base is 200m long then I can subtract that from 1000 and divide it by 2. This can be rearranged to the following formula: Side = (1000 - 200) ?= 400 I have used this formula to work out the area when the base is at different heights.

  2. The Fencing problem.

    41158.231 230 385.0 367.423 42253.698 240 380.0 360.555 43266.615 250 375.0 353.553 44194.174 260 370.0 346.410 45033.321 270 365.0 339.116 45780.727 280 360.0 331.662 46432.747 290 355.0 324.037 46985.370 300 350.0 316.228 47434.165 310 345.0 308.221 47774.209 320 340.0 300.000 48000.000 330 335.0 291.548 48105.353 333.4 333.3 288.675 48113.3 340

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