• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1764

The Fencing Problem

Extracts from this document...


The Fencing Problem


 The aim of this project is to find the largest possible area of containment by a fence of 1000 meters. This is achieved by experimenting with different shapes.


    I predict that the largest area of containment made by the fence possible would be in the shape of a circle. This is because there are no edges:


                                                             Area of circleimage01.png

                                                              Area of Hexagonimage12.png

The area shaded is the possible area that can be contained by a circle. As you can see the edges cut possible areas of containment, which the circular shape can hold. Therefore I predict that the maximum area from a shape with a perimeter of 1000 would be produced in a circle.


To carry out this investigation I started with a polygon with the least amount of side’s possible that could contain an area. This was a triangle.

To go about investigating the possible areas for an Isosceles triangle I made a table containing all the possible side lengths adding to 1000 and proceeded to work out each of their areas. I did this on a computer programme called excel. It allowed me to produce formulas to save me working out every individual triangle area.

The first column in my table determined which triangle I was finding the area of. I numbered these to save confusion from 1-50.

...read more.


                                                                     Area of Quadrilateral

                                                                     Area in Triangle                image16.pngimage14.pngimage15.pngimage13.png

As shown by this diagram the area for the Quadrilateral is larger than the triangle. This is the first sign of a trend that will be apparent as I proceed with the investigation. This is the fact that the more sides the polygon has the larger the area will be.

        I found the maximum area of the quadrilateral to be a square, with lengths of 250 on each side creating an area of 62500 meters squared as shown on the table.


Once I found out that there appeared to be a pattern the next test would be to do the same experiment on the next polygon. A pentagon. A regular pentagon however cannot have a series of different combinations for the side lengths, as there is only one combination where all sides (the perimeter) add up to 1000 meters or else the pentagon would be irregular. This is 1000 divided by 5, which is 200.


          200m                      200m                         Perimeter = 1000


    200m                                 200m



There for the area found is the maximum area for a regular pentagon with a perimeter of 1000.

To find this, the pentagon has to be divided up into 5 triangles each exactly the same. The area of one of these triangles is found and then multiplied by 5 to get the overall area of the pentagon.

...read more.


Base, which is 1000/the n’th term and height of the triangles within the polygons, which is found using the same method as before. Dropping the perpendicular and calculating the height using Tan. All this information is enough to find the area for every triangle within the polygons using the same methods as before. Finally to find the maximum areas in the last column a simple formula is needed which is the triangle area multiplied by the n’th term this will give you the maximum area for every polygon up to infinity.

    I conclude that this experiment agrees with firstly my prediction, which was that the maximum area would increase as the polygon sides increased. The results table and graphs show this clearly. Also I conclude that the best shape to use to contain this plot of land is a circle, as it will give the largest area with a perimeter of 1000. I think that this investigation went well and I managed to prove my theory and discover that although perimeters in a shape can stay the same increasing the shapes sides alters the area dramatically.

I could expand on this experiment by finding the formula in a shortened version, as one fraction so it would be faster and more practical to discover maximum areas.

...read more.

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

    to find the height, and then substituting the values into the conventional triangle formula (A = 1/2bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.

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

    had to divide the opposite by the adjacent when I used tan to find the height which equals: (b/2)/h This cancels down to: b/2h I need to put the tan into this part of my workings to find out the measurement of the height in my triangle, and this needs

  1. Fencing Problem

    350 320 330 1000 500 47906.16 350 325.1 324.9 1000 500 47925.71596 From looking at these results and referring to my past results for my Isosceles triangles I will now try and test the values that are the closest to Equilateral triangles as they have shown the largest area possible with a perimeter of 1000 m.

  2. Fencing Problem

    Of Sides Interior Angle = 180�-Exterior Angle I will then use tangent of an angle, to work out the area of one of the sixty segments of the shape. I will multiply this number by 60 because the 60 segments are all the same size.

  1. The Fencing Problem

    As we know, in algebraic terms, half the base is '500 � n'. Now I have everything that I need to put the area of a triangle into a formula in terms of 'n': Area of triangle = (500 � tan 180)

  2. Fencing problem.

    AB2 = AD2 + DB2 (325) 2 = AD2 + (157) 2 105625 = AD2 + 30625 AD2 = 105625 - 30625 AD2 = 75000 AD = V75000 AD = 273.86m Now we know that the height is equal to 273.86m.

  1. Pythagoras Theorem.

    1st difference 0 1 4 2 2 8 2 2 12 4 8 4 12 2 3 24 4 18 6 16 2 4 40 4 32 8 20 2 5 60 4 50 10 24 2 6 84 72 12 New sequence = Original sequence - 2n2 Quadratic equation

  2. Fencing Problem

    are not given the height therefore we have to use Pythagoras' Theorem a2+ b2= c2 (in this case a is the height) H2= c2- b2 rearranging it. (Just a reminder, b = 1/2 B and c = L) H = V (c2- b2)

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