• 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
  10. 10
  11. 11
  12. 12
  13. 13
  • Level: GCSE
  • Subject: Maths
  • Word count: 1242

The Fencing Problem.

Extracts from this document...






The Fencing Problem

The fencing problem requires me to investigate the shape with the maximum area using 1000 metres of fencing. I have decided to start my search on triangles. A triangle being a shape with the minimum sides logically concludes that I begin there, as they are the most simple. From then on I will continue shape by shape as the number of sides increase; quadrilaterals, pentagons etc…

The area of a triangle is measured by calculating: ½ (base * height)

Therefore if I was to choose the length of the base then the area would be automatically decided by the height of the triangle. The type of triangle which I am going to investigate is the ‘isosceles’ triangle because as I vary the base, the other two sides would remain equal and this would give the maximum height for the triangle, thus giving the maximum area.


Formula to find out the length of the other two sides:image16.pngimage06.pngimage10.png

 ((1000 – (variable base))/2)

        So as I have chosen to do isosceles triangles, the main variable is the base.

...read more.



Now that parallelograms have been eliminated, I can just concentrate on finding the maximum area for a rectangular shape using different base sizes. The formula is easier to work out for quadrilaterals than that of triangles:  AREA = (base * height).  

        AREA = (variable base * ((1000- (2*variable base))/2)



From the last two investigations on triangles and quadrilaterals, the maximum area has been found by using the regular shape, i.e. an equilateral, a square; where all the sides are equal. Therefore for the remainder of my investigation I realise that it would be sensible to just concentrate on the regular shape. Pentagon is the name given to a five-sided shape and the area of this is found by splitting the shape into 5 regular triangles.  The area of one of these triangles is worked out by splitting it further in half, once this is worked out it can be multiplied by five and the area of the total shape can be calculated.

Please turn over image11.png

I have found the maximum area for a five-sided shape, so now I am going to try another shape.

...read more.



        I can now go further and investigate what shape would have the maximum area using 1000 metres of fencing.

         My analysis shows that as the number of sides increase so does the maximum area. However, with increasing number of sides, the rate of rise in area gets smaller and smaller. As the number of sides increase the shape becomes increasingly circular. Therefore, I am going to see what the area of a circle values at.

Circumference of circle = 2πr

                                 1000 = 2πr

500          = πr

                        Radius = 500/ π

                           Radius = 159.1549431 metres

Area of Circle = πr²

                          = π(159.1549431²)

Area of Circle = 79577.47155m²

        It seems to be that there is an ASYMPTOTE in this problem. Looking at my graph there seems to be a number which the graph is trying to reach but will never actually get there. This asymptote is the area of a circle; this can be proven mathematically using explanation involving radians. Firstly the area of a circle can be linked to the formula I worked out to find the maximum area of any sided shape; this is because as the number of sides in a shape increase the actual structure becomes more and more like a circle.

Please turn over  image11.png


Π * 500/π²                            ≈                           500²/ (n (tan (180/n))

500²/π                                   ≈                           500²/ (n (tan (180/n))



500²/π                                   ≈                           500²/ (n (tan (π/n))



500²/π                                   ≈                           500²/ (n (π/n)

500²/π                                   ≈                           500²/πimage31.png

...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. Fencing Problem

    x (500 - 425) x (500 - 150) = 9843750000 m * 9843750000 m = 31374.751 m� PERIMETER = 1000 m AREA = 31374.751m� * 1000 / 2 = 500 m * 500 x (500 - 400) x (500 - 400)

  2. Fencing problem.

    I have decided that shall use Pythagoras's theorem to discover the height. This theorem works with right hand triangles only. Therefore I shall have to half the triangle that has been shown above, to what has been shown below: Pythagoras's theorem can now be used.

  1. The Fencing Problem.

    I am also going to find out whether the number of sides affects the results and whether there are any similarities in results to a triangle. This will help me find the shape that gives the maximum area. As previously for rectangles I will test some different sized isosceles triangles that have an area of 1000m.

  2. Maths Coursework - The Fencing Problem

    The Decagon 360/10 = 36 180 - 36 = 144 144/2 = 72 1000/10 = 100 50 tan 72 = 153.8 50 x 153.8 = 7690 m� 7690 x 10 = 76900 m� Conclusion I found that the total area of a decagon with a perimeter of 1000 is 76900 m�.

  1. The Fencing Problem

    Base [a] (m) Side [b] (m) Side [c] (m) Perimeter (m) s s-a s-b s-c Area (m�) 100 500 400 1000 500 400 0 100 0.00 100 490 410 1000 500 400 10 90 13416.41 100 480 420 1000 500 400 20 80 17888.54 100 470 430 1000 500 400

  2. Fencing Problem

    Will I achieve the maximum area when the three sides are the same? I will be using Pythagoras theorem and trigonometry to find the side lengths and area. The perimeter for the whole isosceles triangle is 1000 metres. X X The formula for a triangle is BxH 2 1000-2X The

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

    far all my graphs prove that my statement was true and I have also realised that the increase between the shapes have decreased. Before the increase between the equilateral triangle and the square was 14,376 compared to the difference of the hexagon and heptagon, which is 1993.526.

  2. The Fencing Problem

    A simple diagram can explain this. In the diagram, the 10 x 6 rectangle is being multiplied step by step as well as the 8 x 8 square. 10 8 20 16 30 24 40 32 50 40 60 48 - 56 - 64 As you can clearly see, when

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