• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  • Level: GCSE
  • Subject: Maths
  • Word count: 2253

The Fencing Problem.

Extracts from this document...

Introduction

Maths Coursework: The Fencing Problem. Aim; To find different patterns of fencing that will make the maximum area and I am going to find different shapes that will make this. Working; Here are four different types of rectangle or squares that all have the same perimeter but will still give different amounts of area. All drawings are not to scale. In a rectangle there are two different length sides and they will add up to five hundred if we are looking for a perimeter of one thousand metres so each opposite side will have the same length. In the rectangle 450m by 50m the opposite side to 450m will have a length of 450m and the same goes for the opposite side for the 50m length all-adding up to 1000m in perimeter. In conclusion of this you can work out the area of any size rectangle if you have the perimeter and just one side of the rectangle. To work out the area of a rectangle with a width length of 150m I would subtract 150m from 500m, which would then leave me with 350m, and then I would multiply 150m by 350m giving me a total area of 52 500m2. From this evidence I can put this into an equation. (If x equals the length of a rectangle.) 1000=x(500-x) By using the equation you can make a prediction table. (overleaf). After 250m on each side I will not have to go any further as then I will be repeating myself. ...read more.

Middle

234.521 45731.554 400 300.0 223.607 44721.360 410 295.0 212.132 43487.067 420 290.0 200.000 42000.000 430 285.0 187.083 40222.817 440 280.0 173.205 38105.118 450 275.0 158.114 35575.624 460 270.0 141.421 32526.912 470 265.0 122.474 28781.504 480 260.0 100.000 24000.000 490 255.0 70.711 17324.116 500 250.0 0.000 0.000 As the regular square was the biggest in area of the rectangles I put in 333.3 as a triangle as that is the regular isosceles triangle. Here is a graph from the table above. You can see that the graph has the same positive trend in it as the first rectangle graph if I would have carried the rectangle graph on it would look identical to this one meaning it would be a parabola (an upward trend to start with then a downward trend in the middle.) Once again the regular shape has the most area. So again I am going to work the areas out at the regular shape triangle around 333, 333.25, 333.3, 333.5, 333.75 and 334. Base (m) Side (m) Height (m) Area (m2) 333 333.5 288.964 48112.450 333.25 333.4 288.747 48112.518 333.3 333.4 288.704 48112.522 333.5 333.3 288.531 48112.504 333.75 333.1 288.314 48112.410 334 333.0 288.097 48112.233 Once again this table has proved that all the results that I have recorded all fit on the line on the graph that I have produced proving that my results are correct and the regular isosceles triangle is the biggest area of triangles. As the last of the shapes that I have investigated have all been regular I am now going to just investigate regular shapes ...read more.

Conclusion

And it has been used in the table below. No. of sides Area (m2) 8 75444.174 9 76318.817 10 76942.088 As you can see the more amount of sides that there are the bigger the area is. And this has occurred on every type of shape that I have investigated so I am now going to research other shapes of a lot larger amount of sides consisting of 20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000 and 10000. No. of sides Area (m2) 20 78921.894 50 79472.724 100 79551.290 200 79570.926 500 79576.424 1000 79577.210 2000 79577.406 5000 79577.461 10000 79577.469 20000 79577.471 50000 79577.471 100000 79577.471 Here is the graph from the table above. As you can see from the graphs that I have produced the line starts to straighten as the number of sides increases as I am increasing the sides by large amounts the results are not changing so I am going to see the results from a circle. A circle has an unlimited amount of sides so the formula that I have used will not work on this shape like the others. In order to work out the area of a circle with a circumference of 1000m we need to find the diameter we know that the diameter is twice the radius. Diametre of a circle= Circumference�?=318.10�2=159.155 So the area of a circle is; ?r2=?x159.1552 =79577.472m2 If I place this result upon my graph it is on the same point as the last results I recorded from this I can conclude that a circle has the greatest area when using a similar circumference. Charlie Longuehaye 11da ...read more.

The above preview is unformatted text

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Fencing Problem

    I will use trigonometry again to figure out the height of the right-angled triangle. * I will use Tangent, as I have my opposite length to my angle and am trying to figure out my adjacent i.e. height. * Tan36 = 100 * I will change the subject or the formula to figure out the height.

  2. Fencing problem.

    I have shown the shape that I shall be investigating below: AB + BC + CD + DE + EF + FG + GH + HI + IA = 1000m Length of each side = Total perimeter � Number of sides Length of each side = 1000 � 9 =

  1. The Fencing Problem

    Due to the fact that to the find the height the operation involved is; Sin ? x 400 (because the sloping value is fixed to 400); and to find the area is to multiply the height by 100 each time (because the base value is fixed to 100), I deduce

  2. The Fencing Problem

    and I want to find out the opposite. height of triangle = 71.45m * tan64.3? = 148.5m Therefore, I can work out the area of the triangle because I know the height and base of it. Area of triangle = base/2 * height = 142.9m/2 * 148.5m = 10610.3m� Because

  1. t shape t toal

    where g is the grid size To get the number you take away, for example, on a 10 by 10 grid we can predict that if you make a translation of +1 you will have to take away 50 from the original T total.

  2. Math Coursework Fencing

    The internal angles in a scalene triangle are all different. The general formula for calculating an isosceles or equilateral triangle is half base times height or, The formula for calculating the area of a scalene triangle is Below a scalene triangle has been drawn and the angle measured in order to calculate the area.

  1. Geography Investigation: Residential Areas

    Back in hypothesis 4 I found the average area rating from all ten of my surveyed occupants and below is the table of results, I am going to use the statistical calculation of Standard Deviation to extrapolate from my data what the average area rating will be for Cyprus Road.

  2. Maths Fencing Coursework

    SIN 170�= x 200 200 SIN 170� = x x= 34.72963553 x 200 Area=6945.927107 18. SIN 180�= x 200 200 SIN 180� = x x= 0 x 200 Area=0 m2 If I do this working out it is zero because 180 � would be a straight line.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work