• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Fencing problem.

Extracts from this document...

Introduction

Imy Jacobs         Math c/w        2004

Fences Coursework

2004

A farmer has exactly 1000m of fencing. With it she wishes to fence off a plot of level land. She isn’t concerned about the shape but states that it must be 1000m in perimeter.

What she wants to know is what is the maximum area that she can have with that amount of fencing.

By drawing scale outlines, I will examine some possible shapes for the plot of lad. In each case I will need to ensure that the perimeter is 1000m, and then I need to obtain the enclosed area.


Rectangles

The first shapes that I will be investigating are rectangles. I am going to draw three rectangles and see which of the three has the bigger area.

There are different types of rectangles that I could draw, including squares.

...read more.

Middle

Base/m

Sides/m

Area/m2

50

475

11 858.5

100

450

22 360.7

150

425

31 374.8

200

400

38 729.9

250

375

44 194.2

300

350

47 434.2

350

325

47 925.7

400

300

44 721.4

450

275

35 575.6

A rectangular field has a bigger area because it has more sides.

Step 3; Zooming In

So I can gain a more accurate peak in my first Triangle graph I am planning to zoom in and find the areas of some of the triangle between 300m and 350m along the base.

Three equilateral triangles

1) Area =  s(s-a)(s-b)(s-c)

            = 1000(1000 -1000/3)(1000 -1000/3)(1000 -1000/3)

            = 48 112.52m2

2)

...read more.

Conclusion

Formula;                                                                                Scale;

b x h             and Tan = Opposite                                                        1cm : 100m

    2                               Adjacent

Polygon 1 – Pentagon

Polygon 2; Hexagon

Polygon 3; Octagon

More Polygons

No. of sides/m

Length/m

Area/m2

5

200

68819.1

6

1000/6

72168.8

7

1000/7

74 161

8

125

75444.2

9

1000/9

76 319

10

100

76 942

20

50

78 922

50

20

79 472.7

100

10

79 551


As you can see by the table the area grows larger and larger as the polygons gain more sides. This means that the more sides the greater the area so we would want to build a field with as many sides as possible.

Looking at the diagrams you may notice that as the shapes gain more sides the lengths become shorter and they look more and more like a circle. Because of this I have decided to check out a circle and see how big an area that has.

Area = πr2

C = πd

1000 = πd

d = 1000 = 318.309

 π

r = 1000 = 159.15

        2π

Final Conclusion

We can now see that a circle has the greatest area of all shapes that we looked at, and therefore that is the best shape for the farmer to use when she builds her field.

...read more.

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. The Fencing Problem

    As a result, I will now analyse rectangles in the next stage of my investigation. Quadrilaterals - Rectangles I will proceed to show the maximum possible area for a rectangle with perimeter = 1000m; again, presenting my results in a graph.

  2. A length of guttering is made from a rectangular sheet of plastic, 20cm wide. ...

    =C23/2 =E23/TAN(RADIANS(D23)) =E23*F23 =G23*B23 30 6 =A24/B24 =180/(B24*2) =C24/2 =E24/TAN(RADIANS(D24)) =E24*F24 =G24*B24 30 7 =A25/B25 =180/(B25*2) =C25/2 =E25/TAN(RADIANS(D25)) =E25*F25 =G25*B25 30 8 =A26/B26 =180/(B26*2) =C26/2 =E26/TAN(RADIANS(D26)) =E26*F26 =G26*B26 30 9 =A27/B27 =180/(B27*2) =C27/2 =E27/TAN(RADIANS(D27)) =E27*F27 =G27*B27 30 10 =A28/B28 =180/(B28*2)

  1. Fencing problem.

    of triangles Area of octagon = 9431.3 � 8 = 75450.4m2 Nonagon The fifth shape I shall be investigating is a nine-sided shape this is also be known as a nonagon. I shall be taking the same procedures that I took before.

  2. The Fencing Problem

    The ratio of these sides will be 8:6:4. Perimeter = 1000m 8 + 6 + 4 = 18 1000m ( 18 = 55 5/9 I can now multiply this number by the ratio to find their length as part of the triangle with a perimeter of 1000m.

  1. Fencing Problem

    minus all of the exterior lengths I have discovered so far from 1000 m and then divide the answer by 2 to give me the length of one side. * 1000 - 100 - 100 - 70.71067812 - 70.71067812 = 658.5786438 * Now that I have the side length of

  2. The Fencing Problem

    I'm starting by going up by 50m each time: BASE(m) HEIGHT(m) AREA(m�) 50 450 22500 100 400 40000 150 350 52500 200 300 60000 250 250 62500 300 200 60000 Already the area has started to decrease somewhere around the 200m-300m spot so I will zoom in to find a closer answer: BASE(m)

  1. Geographical Inquiry into the proposed redevelopment plan of the Elephant and Castle.

    Two educational facilities that are also in the area are the South Bank University and the London College of Printing so these aspects need to be taken in to consideration during regeneration of the area. E D W A R D How to create the new vision Regeneration Proposals * 3000 retail and leisure jobs by 2007.

  2. The Fencing Problem. My aim is to determine which shape will give me ...

    I can also prove this with shape 1 + 2. In shape 1, which had a length of 10 and a width of 490, the area was 4900, while with shape 2, which had a length of 20 and width of 480, the area was 9600.

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