The Fencing Problem  Math Coursework
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The task  A farmer has exactly 1000m of fencing; with it she wishes to fence off a level area of land. She is not concerned about the shape of the plot but it must have perimeter of 1000m. What she does wish to do is to fence off the plot of land which contains the maximun area. Investigate the shape/s of the plot of land that have the maximum area. Solution  Firstly I will look at 3 common shapes. These will be:  [IMAGE] A regular triangle for this task will have the following area: 1/2 b x h 1000m / 3  333.33 333.33 / 2 = 166.66 333.33Â²  166.66Â² = 83331.11 Square root of 83331.11 = 288.67 288.67 x 166.66 = 48112.52Â² [IMAGE]A regular square for this task will have the following area: Each side = 250m 250m x 250m = 62500mÂ² [IMAGE] A regular circle with a circumference of 1000m would give an area of: Pi x 2 x r = circumference Pi x 2 = circumference / r Circumference / (Pi x 2) = r Area = Pi x rÂ² Area = Pi x (Circumference / (Pi x 2)) Â² Pi x (1000m / (pi x 2)) Â² = 79577.45mÂ² I predict that for regular shapes the more sides the shape has the higher the area is. A circle has infinite sides in theory so I will expect this to be of the highest area. The above only tells us about regular shapes I still haven't worked out what the ideal shape is. Width (m) Length (m) Perimeter (m) Area (mÂ²) 500 0 1000 0 490 10 1000 4900 480 20 1000 9600 470 30 Need Writing Help?Get feedback on grammar, clarity, concision and logic instantly. Check your paper »How to Cite this Page
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1000
14100 460 40 1000 18400 450 50 1000 22500 440 60 1000 26400 430 70 1000 30100 420 80 1000 33600 410 90 1000 36900 400 100 1000 40000 390 110 1000 42900 380 120 1000 45600 370 130 1000 48100 360 140 1000 50400 350 150 1000 52500 340 160 1000 54400 330 170 1000 56100 320 180 1000 57600 310 190 1000 58900 300 200 1000 60000 290 210 1000 60900 280 220 1000 61600 270 230 1000 62100 260 240 1000 62400 250 250 1000 62500 240 260 1000 62400 230 270 1000 62100 220 280 1000 61600 210 290 1000 60900 200 300 1000 60000 190 310 1000 58900 180 320 1000 57600 170 330 1000 56100 160 340 1000 54400 150 350 1000 52500 140 360 1000 50400 130 370 1000 48100 120 380 1000 45600 110 390 1000 42900 100 400 1000 40000 90 410 1000 36900 80 420 1000 33600 70 430 1000 30100 60 440 1000 26400 50 450 1000 22500 40 460 1000 18400 30 470 1000 14100 20 480 1000 9600 10 490 1000 4900 0 500 1000 0 This table shows the results of every rectangle possible for the 1000m perimeter with a 10m gap between each reading. The results increase and then decrease symmetrically proving that the SQUARE 250 x 250 is the rectangle with the highest area because it is in the middle. This would produce a curve on a graph with the square at the top and in the middle. A square is a regular shape. The graph below is of the table on the previous page. [IMAGE] The graph has the curve that I explained on the previous page. Now I have completed the rectangles the next step I will choose is to try the same for triangles. I decided to use isosceles triangles: The base length (m) The 2 equal lengths (m) Perimeter (m) Area (mÂ²) 0.00 500.00 1000.00 0.00 20.00 490.00 1000.00 4898.98 40.00 480.00 1000.00 9591.66 60.00 470.00 1000.00 14071.25 80.00 460.00 1000.00 18330.30 100.00 450.00 1000.00 22360.68 120.00 440.00 1000.00 26153.39 140.00 430.00 1000.00 29698.48 160.00 420.00 1000.00 32984.85 180.00 410.00 1000.00 36000.00 200.00 400.00 1000.00 38729.83 220.00 390.00 1000.00 41158.23 240.00 380.00 1000.00 43266.62 260.00 370.00 1000.00 45033.32 280.00 360.00 1000.00 46432.75 300.00 350.00 1000.00 47434.16 320.00 340.00 1000.00 48105.35 333.33 333.33 1000.00 48000.00 340.00 330.00 1000.00 48083.26 360.00 320.00 1000.00 47623.52 380.00 310.00 1000.00 45731.55 400.00 300.00 1000.00 43487.07 420.00 290.00 1000.00 40222.82 440.00 280.00 1000.00 35575.62 460.00 270.00 1000.00 28781.50 480.00 260.00 1000.00 24000.00 500.00 250.00 1000.00 0.00 [IMAGE] This also shows that the shape with the largest area is the regular one. The graph isn't symmetrical this is because the isosceles triangles with the given perimeter can only have a base length from 0m  500m otherwise the triangle would be flat. This shows us that regular shapes are better. The triangle, square and circle tested make it appear that the amount of sides has something to do with the area. This is proved by the square having a greater area than that of the triangle but less than the circle. To prove this I will investigate regular shapes with more sides. A regular pentagon: [IMAGE] A regular pentagon has 5 sides and with the perimeter of 1000m it will therefore have 200m for each side. The way I will use to find the area of a pentagon will be too split it into triangles. To find the area we need to know the height, the values we know are: 54Ù’ [IMAGE] 72Ù’ The angle at the top of the triangle can be calculated by dividing 360Ù’ by 5 (the number of sides and triangles) this gives 72Ù’. Then to work out the other 2 angles 72Ù’ must be taken from 180Ù’ (the sum of angles in a triangle) to give 108Ù’. In an isosceles triangle the bottom two angles are the same so they will both be 108 / 2 = 54Ù’ 200m To then work out the area of the triangle we need to split it in half and use trigonometry. [IMAGE][IMAGE] 1 1 = Base: 100m 1 = Top angle: 36Ù’ 1 = Bottom angle: 54Ù’ To work out this are we will use TAN because we need the opposite and we have the adjacent. TAN 54 = opposite / 100 TAN 54 x 100 = opposite TAN 54 = 1.3764 Height of triangle = 137.64m Area of triangle = 137.64m x 100 = 13763.8 Area of pentagon = 13763.8 x 5 = 68819.10mÂ² This regular pentagon is as expected, the area is more than the triangle and the square but less than the circle. This once again proving that the more sides a regular shape has the more area there is. To work out the area for a regular shape I did the following: B = Base N = Number of sides P = Perimeter H = height Firstly I worked out the formula for the area of a triangle. This is = b x h / 2 Then I worked out the area for any shape = b x h x n / 2 This shows that the formula to work out the area of any regular shape is = 1/2 x P x P . n x 2 tan(180/n) This simplified is A = PÂ² . 4n Tan(180/n) 
