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  • Level: GCSE
  • Subject: Maths
  • Word count: 1242

The Fencing Problem.

Extracts from this document...

Introduction

MATHS GSCE COURSEWORK – EDEXCEL 2003 DISPATCH 1

THE FENCING PROBLEM

ADNAN YOUNIS

YEAR 4-LB


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.

image00.png

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.

Middle

image03.pngimage05.pngimage04.png

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)

image09.png

image08.png

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.

Conclusion

image34.png

        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


AREA OF CIRCLE       ≈         AREA OF ANY SIDED SHAPE!

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

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

image25.pngimage26.png

CONTINUING:image27.pngimage28.png

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

image30.pngimage29.png

THERFORE:

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

500²/π                                   ≈                           500²/πimage31.png

...read more.

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