• 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. Math Coursework Fencing

    Circles are simple closed curves, dividing the plane into an interior and exterior. Sometimes the word circle is used to mean the interior, with the circle itself called the circumference(C). Usually, however, the circumference means the length of the circle, and the interior of the circle is called a disk.

  2. The Fencing Problem

    I will construct another table of data with the side lengths having a difference of 2 between them. 1) 400 500 100 2) 410 490 100 3) 420 480 100 4) 430 470 100 5) 440 460 100 6) 450 450 100 7)

  1. fencing problem

    side equal to 166.6m Perimeter 1000 =166.6m 6 360 = 60 6 60=30 2 tan30= 83.3 H H=83.3 = 144.27 0.57735 Area=166.6�144.27 = 12017.69 2 12017.69�6= 72106.14 m� = 72106.14 m� Heptagon Each side equal to 142.9m Perimeter 1000 =142.9m 7 360 = 51.42 7 51.42 = 25.71 2 Tan

  2. The Fencing Problem.

    Length (m) Width (m) Perimeter (m) Area (square m) 490 =500-B2 =(B2+C2)*2 =B2*C2 Having entered the correct information I will be able to calculate the areas of many different sizes of rectangles with a perimeter of 1000m. I can do this in Microsoft Excel by dragging the formula boxes down,

  1. Fencing Problem

    As an equilateral triangle has 3 equal sides and my perimeter is 1000 m I will divide 1000m (Perimeter) by 3 (Sides), which equals 333.3333333� m (recurring) per side. I will follow Hero's formula in order for me to find the area : * 1000 / 3 = 333.3333333� m (Lengths of Each Side)

  2. the fencing problem

    200 387.2 38729 1000 425 425 150 418.3 31374 1000 450 450 100 447.2 22360 1000 In this search I went up by 25 for the lengths and down by 50 for the base so that both lengths and base add up to 1000 meters.

  1. Fencing Problem

    I am now going to investigate an equilateral triangle to find out its area. 4. Base=333.3m Sides=333.3m Area: 1/2 x b x h 1/2 x 333.33 x 289 =48166m� Table Of results Base Sides Area 100 450 23850m� 200 400 38700m� 300 350 47400m� 400 300 44800m� 333.3 333.3 48166m�

  2. The Fencing Problem

    = 48,074.02cm2 This shows that shapes with all sides the same length have the highest formula. Why do regular shapes have the biggest area? All regular sided shapes have an n (n being the number of sides in that shape)

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