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  • Level: GCSE
  • Subject: Maths
  • Word count: 3279

Fencing problem.

Extracts from this document...

Introduction

Introduction A farmer has 1000m of fencing and he needs to know which shape encloses the largest possible area inside this 1000m of fencing. I will be experimenting with a variety of different shapes, eventually I will be able to observe a trend or pattern and then I can find out which shape gives the largest area. Plan It will be necessary for me to begin with the simplest shape so that I can work out the area and then I can gradually build upon that with other shapes. The formulae for working out the area of a particular shape is the length * Width. I have decided to start with the rectangle. I chose to begin with the rectangle because in my opinion it is the simplest shape to work out its area. Whilst I am experimenting with the rectangle, I can notice what the area of a square will give; and obviously, this will result with the largest possible area (1000,perimeter/4-number of sides *250) for the quadrilaterals. Next, I will probably experiment with the area of the triangle because it has the least number of sides. I am going to be working out both the isosceles and equilateral triangles, however I am going to exclude the scalene triangle out of this investigation. I decided this mainly because it will be very difficult for me to come up with different possible combinations of lengths of the scalene triangles. It will be very time consuming if I did so, therefore it would be much more appropriate to only work with the isosceles and equilateral triangles. The following shapes will be the regular shapes. These are shapes with the same length all the way round (pentagon, hexagon, heptagon, octagon, nonagon, decagon etc). I need to consider that to work out the areas of these regular shapes it is vital that I apply trigonometry because all regular shapes can be divided into equilateral triangles and therefore using these triangles, I can eventually calculate the area of the regular shape. ...read more.

Middle

Calculations Height (m) Calculations Result for height (m) Area (m�) 10 495 245025 25 245000 494.9747468 2474.874 20 490 240100 100 240000 489.8979486 4898.979 30 485 235225 225 235000 484.7679857 7271.52 40 480 230400 400 230000 479.5831523 9591.663 50 475 225625 625 225000 474.341649 11858.54 60 470 220900 900 220000 469.041576 14071.25 70 465 216225 1225 215000 463.6809248 16228.83 80 460 211600 1600 210000 458.2575695 18330.3 90 455 207025 2025 205000 452.7692569 20374.62 100 450 202500 2500 200000 447.2135955 22360.68 110 445 198025 3025 195000 441.5880433 24287.34 120 440 193600 3600 190000 435.8898944 26153.39 130 435 189225 4225 185000 430.1162634 27957.56 140 430 184900 4900 180000 424.2640687 29698.48 150 425 180625 5625 175000 418.3300133 31374.75 160 420 176400 6400 170000 412.3105626 32984.85 170 415 172225 7225 165000 406.2019202 34527.16 180 410 168100 8100 160000 400 36000 190 405 164025 9025 155000 393.7003937 37401.54 200 400 160000 10000 150000 387.2983346 38729.83 210 395 156025 11025 145000 380.7886553 39982.81 220 390 152100 12100 140000 374.1657387 41158.23 230 385 148225 13225 135000 367.4234614 42253.7 240 380 144400 14400 130000 360.5551275 43266.62 250 375 140625 15625 125000 353.5533906 44194.17 260 370 136900 16900 120000 346.4101615 45033.32 270 365 133225 18225 115000 339.1164992 45780.73 280 360 129600 19600 110000 331.662479 46432.75 290 355 126025 21025 105000 324.0370349 46985.37 300 350 122500 22500 100000 316.227766 47434.16 310 345 119025 24025 95000 308.2207001 47774.21 320 340 115600 25600 90000 300 48000 330 335 112225 27225 85000 291.5475947 48105.35 333.33 333.33 111108.8889 27777.22223 83331.66668 288.6722478 48111.56 340 330 108900 28900 80000 282.8427125 48083.26 350 325 105625 30625 75000 273.8612788 47925.72 360 320 102400 32400 70000 264.5751311 47623.52 370 315 99225 34225 65000 254.9509757 47165.93 380 310 96100 36100 60000 244.9489743 46540.31 390 305 93025 38025 55000 234.520788 45731.55 400 300 90000 40000 50000 223.6067977 44721.36 410 295 87025 42025 45000 212.1320344 43487.07 420 290 84100 44100 40000 200 42000 430 285 81225 46225 35000 187.0828693 40222.82 440 280 78400 48400 30000 173.2050808 38105.12 450 275 75625 50625 25000 158.113883 35575.62 460 270 ...read more.

Conclusion

eg1) 72168.72m� = 0.5 * 144.3375673 * 12 * 83.33 = 72168m� eg2) 74165.49m� = 0.5 * 148.32 * 71.43 * 14 = 74165.48m� The two examples prove that my equation is correct, therefore I am going to see weather the octagon, nonagon and decagon follow the trend (as number of sides increase the area increases) by substituting them into the area equation. The table of results below proves that my hypothesis is correct. To work out the areas of the regular polygons I am going to apply the area formulae which is A = 0.5 * h * S * l Number of Sides Area (m�) 3 48111.56 4 62500 5 68820 6 72168.72 7 74165.49 8 75444.17 9 76318.82 10 76942 This table of results proves that my hypothesis was correct, as the number of sides, increase the area becomes greater. Below I have constructed a graph to show this theory. I observed from the graph that as the number of sides increase, the line, which is the area, becomes steeper; eventually this will become a flat line of area as the numbers of sides increase. I expect that a circle will have the maximum area because it has infinite number of sides, therefore I cannot apply the area formula of the regular polygons to the circle. I am going to calculate the area of the circle by using the formula ?r� or ?d, where d is the diameter and r is the radius, in this case the diameter is the circumference of the circle which is 1000m C (circumference): 2? This is how I calculated the area of the circle, 1000 = 2?r r = 1000/2? = 79577.5m� I conclude that the area of the circle has the greatest area therefore, my hypothesis was correct. I stated that the circle would probably result with the greatest area because it has infinite amount of sides. In my hypothesis I stated that as the number of sides increase the area will also increase , the graphs and observations prove that this is true. Doha Salameh Maths 11JCU Mr McEvoy ...read more.

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