# Fencing Problem

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Introduction

Fencing Problem In this piece of coursework I shall try to obtain the biggest possible area enclosed by 1800m of fencing. I shall look at different types of sizes and shapes of fences to obtain the biggest area. First I shall look at rectangles as they are easiest to calculate. P=2L+2w 1800=2L+2w 900=L+w 900-w=L I shall use this formula to work out different lengths by substituting a different number for w. A=L x w A=(900-w) x w w l=900-w Area=w(900-w) 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900-50=850 900-100=800 900-150=750 900-200=700 900-250=650 900-300=600 900-350=550 900-400=500 900-450=450 900-500=400 900-550=350 900-600=300 900-650=250 900-700=200 900-750=150 900-800=100 900-850=50 50(900-50)=42500 100(900-100)=80000 150(900-150)=112500 200(900-200)=140000 250(900-250)=162500 300(900-300)=180000 350(900-350)=192500 400(900-400)=200000 450(900-450)=202500 500(900-500)=200000 550(900-550)=192500 600(900-600)=180000 650(900-650)=162500 700(900-700)=140000 750(900-750)=112500 800(900-800)=80000 850(900-850)=42500 I tabulated the results because it allows you to see the different areas with the different lengths. This table shows that a rectangle of both lengths and widths of 450 (square) gives the biggest area in terms of the rectangle. ...read more.

Middle

I have found that each time I have found the area the sides and angles are all the same. Use Trigonometry to find the height Use O to help you when you are trying to find A T A A=O/T A=(900/n)/tan(180/n) 1/2x1800/n x ((900/n)/tan(180/n)) 1x1800x900/n 2x n tan(180/n) (810000/n) ntan(180/n) nx (810000/n) ntan(180/n) Therefore, after all of my calculations my formula for an n-sided shape has come to 810000 ntan(180/n) I shall check it with a pentagon 810000 =222973.8711 5tan(180/5) Centre angle=72 Each side =360 72/2=36 (T) 360/2=180 (O) A =O/T We need to find the height A=180/tan36 A=247.7487457 Now to find the area of the triangle. (247.7487457x360)/2=44594.77422 Times the answer by 5 because there are 5 triangles in a pentagon 44594.77422x5=222973.8711 I shall now use my new formula to calculate 3-50 sided shapes instead of working everything out like a pentagon. I have also used excel to do this. N-sided shape 810,000/(n(tan(180/n))) 3 155885 4 202500 5 222974 6 233827 7 240283 8 244439 9 247273 10 249292 11 250782 12 251913 13 252792 14 253489 15 254050 16 254509 17 254889 18 255208 19 255477 20 255707 21 255905 ...read more.

Conclusion

With this vital piece of information I came to estimate that the area will be biggest in an equilateral triangle with all sides 600. After a couple of workings out of some triangles I came to discover another faster and easer way to calculate a triangle and this formula is called the semi perimeter. After a lot of working out I found that my estimate was right with an area of 155884.5727. I then moved onto other regular polygons and saw that each time the area increased however the amount it decreased by got smaller each time. I also noticed that each time you add a side onto a shape it looks more and more like a circle. After I had finished the regular polygons I guessed that because it was the only shape I had not done the shape with the most maximum area will be a circle. I was right in guessing that the shape with the largest areas was a circle with 257,831. ?? ?? ?? ?? ...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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