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# The Fencing Problem

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Introduction

The Fencing Problem I am writing this investigation for a farmer who has 1000m2 of fencing and wants to find out how he can create the most area by using the fencing and rearranging the fence in different shapes. From only 3 sides to 99999 sides I have worked out the highest and lowest area for the farmers plot of land. ...read more.

Middle

(500-500) (500-300) =5916.079 Area = 500(500-400) (500-400) (500-200) =38729.833 Area = 500(500-450) (500-450) (500-150) =20916.501 Now I will try a different family of shapes a 4 sided shape but I think that the greatest area of a shape will come from a regular shape. Area = 250 x 250 = 62,500m2 Area = 200 x 300 = 60,000m2 Area = 150 x 350 = 52,500m2 Area = 100 x 400 = 40,000m2 Area = 50 ...read more.

Conclusion

I will now increase the number of sides on a shape and see if I can find a maximum area. I can see that as the number of sides increase the area also increases. The question now is does a circle have infinity sides, 1 single side or a peak number of sides n- sided Area = This formula shows that as n increases to infinity approaches tan 90 so that the angle between adjacent sides becomes 2x 90= 180o thus giving a circle. For a circle of a radius r = ...read more.

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# Related GCSE Fencing Problem essays

1. ## Fencing Problem

* 1000 m / 4 = 250 m (Sides) * 250 x 250 = 62500 m� PERIMETER = 1000 m AREA = 62500 m� The largest area possible for a Square Side A Side B Side C Side D Perimeter Area (m�) 250 250 250 250 1000 m 62500 m� This table presents the largest possible area using 1000 metres of fence for a square.

2. ## Fencing problem.

Each side of the hexagon is 166.67 (1000m/6-number of sides) 2. I can divide the hexagon into six equal segments made up of isosceles triangles. The angle inside this triangle is 60�(360�/6). 3. I then place a perpendicular bisector to divide the isosceles triangles into right-angled triangles, this is when I introduce trigonometry into the calculation.

1. ## The Fencing Problem.

Isosceles Triangle x Number of Sides A = 9,430.52 x 8 A = 75,444.17 Nonagon and Decagon I decided not to calculate the nonagon and decagon because they followed the same principles as all the other regular polygons. By know I had started piecing together a universal formula that would work for all regular polygons.

2. ## The Fencing Problem

is basically the same except that height is replaced with a formula that equals height, 500 - Base. The formula for finding the area can be simplified Area = Base ( 500 - Base Area = Base (500 - Base)

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. ## The Fencing Problem

380 310 244.949 46540.305 390 305 234.521 45731.554 400 300 223.607 44721.360 410 295 212.132 43487.067 420 290 200.000 42000.000 430 285 187.083 40222.817 440 280 173.205 38105.118 450 275 158.114 35575.624 460 270 141.421 32526.912 470 265 122.474 28781.504 480 260 100.000 24000.000 490 255 70.711 17324.116 Using this

1. ## Fencing Problem

For me to get the results as accurate as possible I will be using the hero's formula. A B C A+B+C are the three sides of the scalene triangle that add up to give 1000m. To find the area of this triangle I will need the Hero's formula that gives me the area.

2. ## The Fencing Problem

/ n)) / n2, which can again be simplified to (250000 x Tan ((90 x (n - 2)) / n)) / n. We can simplify the fraction and it becomes: (250000 x Tan (90 - (180/n))) / n To check the equation, I will test it on the area of a square.

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