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• Level: GCSE
• Subject: Maths
• Word count: 1839

# The Fencing Problem.

Extracts from this document...

Introduction

The Problem A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land. She is concerned about the shape of the plot, but must have a perimeter of 1000m. So it could be 400m 50m 450m 1000m Or anything else with a perimeter (or circumference) of 100m. She wishes to fence of the plot, which contains the maximum area. Investigate the shape, or shapes that could be used to fence in the maximum area using exactly 1000 metres of fencing each time. I am going to investigate different with shapes with the perimeter of 1000 m to find out the maximum area. I will start with rectangles as they have drawn some rectangles already. Then I will try Isosceles triangle and equilateral triangle. Then I will do some regular polygons. The I will try a circle. Rectangle and Square Length Width Area 0 500 0 10 490 4900 20 480 9600 30 470 14100 40 460 18400 50 450 22500 60 440 26400 70 430 30100 80 420 33600 90 410 36900 100 400 40000 110 390 42900 120 380 45600 130 370 48100 140 360 50400 150 350 52500 160 340 54400 170 330 56100 180 320 57600 190 310 58900 200 300 60000 210 290 60900 220 280 61600 230 270 62100 240 260 62400 ...read more.

Middle

310 190 60000 244.949 46540.31 1000 390 305 195 55000 234.5208 45731.55 1000 400 300 200 50000 223.6068 44721.36 1000 410 295 205 45000 212.132 43487.07 1000 420 290 210 40000 200 42000 1000 430 285 215 35000 187.0829 40222.82 1000 440 280 220 30000 173.2051 38105.12 1000 450 275 225 25000 158.1139 35575.62 1000 460 270 230 20000 141.4214 32526.91 1000 470 265 235 15000 122.4745 28781.5 1000 480 260 240 10000 100 24000 1000 490 255 245 5000 70.71068 17324.12 1000 500 250 250 0 0 0 1000 The formula for a triangle is 1/2 x base x height The formula used for column 3 which is a is =A4/2 Each time the coordinate changes. The formula used for column 4 is =POWER(B4,2)-POWER(C4,2) Each time the coordinate changes. We need to do this in order to work out the height. The formula for column 5 which is height is =POWER(D4,0.5) We need to work the height out in order to work out the area. The formula for column 6 which is area is =(A4*E4)/2 I then used my i.c.t skills to plot a graph between the relationship between base and height in order to help me find the maximum area within an isosceles triangle. ...read more.

Conclusion

I have used my i.c.t skills to plot a graph to find the relationship between the number of sides in a polygon and the area. I noticed from my graph Decagon has the largest area as it has most sides. My predication was right as the number of side's increases so does the area. I also noticed from triangle, rectangle, square, pentagon, heptagon, octagon, nonagon and decagon the area is increasing and the shape is becoming more like a circle therefore I decided to try circle next. Circle The formula for the area of a circle is ?rxr We need to find r We know circumference = 2?r Circumference = Perimeter We know the perimeter is = to 1000 Therefore 1000 = 2?r To find out r you do r = 1000/2? R= 159.1549431 Now we can work out the area. Area = ?rxr Area = ? x 159.159.1549431 x 159.1549.431 Area = 79577.47155m The largest possible area for a circle is 79577.47144m squared . Conclusion I conclude that the area that which gave the farmer the most area is a circle. This is because it is a regular shape that has infinite number of sides, the maximum area that the farmer could get would be 79577.47155m squared. I also found out as the number of sides increases so does the area. Regular shapes give the largest area. ...read more.

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# Related GCSE Fencing Problem essays

1. ## The Fencing Problem

that the area formula can also be seen as; Area = 400 x 100 x Sin ? = 40000 x Sin ? Below is the table as on the previous page, but in normal view. Base [a] (m) Side [b] (m) Angle (�) Height [h] (m) Perimeter (m) Area (m�)

2. ## My investigation is about a farmer who has exactly 1000 metres of fencing and ...

A = 5(1000/5)2/(4tan(180/5)) A = 68819.0960m2 This shows that my formula has given me the same answer as what I got when I worked this out the long way. I am now going to check my formula and my prediction by working out the area of regular heptagons (7 sides), octagons (8sides), nonagons (9 sides), decagons (10 sides)

1. ## Fencing problem.

I have discovered the area of five different quadrilaterals. Overall eight shapes have been discovered. The results have been shown below in a tabulated form: Name of quadrilateral Proportional Area (m2)

2. ## Fencing Problem

x (500 - 325.1) x (500 - 324.9) = 2296874250 m * 2296874250 m = 47925.71596 m� PERIMETER = 1000 m AREA = 47925.71596 m� I have conclude after testing the values of Scalene triangles that the closer the dimensions get to an Isosceles triangle the larger the area becomes, That means the closer the dimension are to an equilateral triangle the larger the area becomes.

1. ## Fencing Problem

Of Sides Interior Angle = 180�-Exterior Angle I will then use tangent of an angle, to work out the area of one of the forty segments of the shape. I will multiply this number by 40 because the 40 segments are all the same size.

2. ## The Fencing Problem

There are so many quadrilaterals with infinitely many different areas, that I will only work on the most basic ones; the trapezium and the parallelogram. Quadrilaterals: Comparing a rectangle with a Trapezium and Parallelogram Just by looking at this

1. ## The Fencing Problem

336.25 293.684 48090.674 328.0 336.00 293.258 48094.241 328.5 335.75 292.831 48097.493 329.0 335.50 292.404 48100.430 329.5 335.25 291.976 48103.050 330.0 335.00 291.548 48105.353 330.5 334.75 291.119 48107.338 331.0 334.50 290.689 48109.003 331.5 334.25 290.259 48110.347 332.0 334.00 289.828 48111.371 332.5 333.75 289.396 48112.072 333.0 333.50 288.964 48112.450 333.5 333.25 288.531

2. ## Fencing Problem

= 500 - 1/2 B We can therefore replace L to the above formula: A = 1/2 x B x V [(500 - 1/2 B) 2- (B2/4)] I will start the Base from 10 m and move upwards. I am hoping to reach to a point where I would obtain a maximum area after which the area starts to decrease.

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