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

1,000m of fence - what is the maximum area I can enclose?

Extracts from this document...

Introduction

Christopher Morfill        Page         5/2/2007

image00.png



1,000m of fence – what is the maximum area I can enclose?


Introduction:

In this project, I am trying to find out the maximum area of an enclosure I could make with 1000m of fencing.

I am going to attempt to tackle the problem by using different shapes, Including Quadrilaterals, Pentagons, Hexagons, Octogons, Isogons and circles. I already know that the mathematical solution (circles) may not be useful for the farmer because circles do not tessellate. The plots need to tessellate to make use of as much land as is possible – I have shown examples of tessellation below:

image10.png





image16.pngimage17.pngimage15.png

Historically, farmland in Britain was divided into hexagonal plots to use all available space. Since proper roads were built, this has no longer been possible, as space is at a minimum and roads usually run at the sides of fields.

Hypothesis:

My hypothesis is the more sides a shape has the more area it will contain and I believe the mathematical solution will be a circle because circles have infinite sides.

Logical Ideas for shapes:

  • Triangle
  • Quadrilateral
  • Pentagon
  • Hexagon

    - There is a pattern here, so I will jump to: -
  • Decagon (10 sides)
  • Isogons (20 sides)
  • n number of sides
...read more.

Middle

450

50 x 450 = 22500

60

440

60 x 440 = 26400

70

430

70 x 430 = 30100

80

420

80 x 420 = 33600

90

410

90 x 420 = 36900

100

400

100 x 400 = 40000

110

390

110 x 390 = 42900

120

380

120 x 380 = 45600

130

370

130 x 370 = 48100

140

360

140 x 360 = 50400

150

350

150 x 350 = 52500

160

340

160 x 340 = 54400

170

330

170 x 330 = 56100

180

320

180 x 320 = 57600

190

310

190 x 310 = 58900

200

300

200 x 300 = 60000

210

290

210 x 290 = 60900

220

280

220 x 280 = 61600

230

270

230 x 270 = 62100

240

260

240 x 260 = 62400

250

250

250 x 250 = 62500

Below I have compiled my own line graph, but I know I can make a more accurate one using ICT, so I have included a line graph compiled using ICT.

By looking at the line graphs, I can tell that continuing would make a mirror of the curve.

Triangle:

There are different types of triangles to use – Right Angle, Isoceles, Equilateral, Scalene, Acute and Obtuse. Although triangles are the easiest shape to use, it may not be easy to work out area – because the height is not always specified.

Finding the Area – if the perimeter is 1,000m

Equilateral

image19.png

To find the length of side x, I will have to use Pythagoras Theorem. The rule for Pythagoras theorem is a2 + b2 = c2.

a2 + b2 = c2

x2 + 166.6662 = 333.3332

x2 + 27777.77= 111111.11

83333.34 + 27777.77= 111111.11

image20.png

x = 288.6751461

x = 288.68

Now I know the height and base measurements, I will now continue to find the area of the triangle.

A = ½ bh

                                                                  = ½ (333.333x 288.6751288)
                                                                 = ½ 96224.94671

                                                                  = 48112.47335m2

                                                                  = 48112.48m2

Right Angled

A = ½ bh

                                                                  = ½ 300x200
                                                                 = ½ 60000

                                                                  =  30000m2

...read more.

Conclusion

image21.png

image22.png

image23.png

Tan  54 = . x .

                100

                                                                  1.3763819204711735382072095819109= . x .

                                                                                                                                      100

                                                        1.3763819204711735382072095819109 x 100 = x

                                                          137.63819204711735382072095819109 = x

                                                                                 137.64m = x(1dp)

                                         area of triangle = ½ bh

                                                                  = ½ 200 x 137.64
                                                                 = 0.5 x 200 x 137.64
image11.pngimage24.png

                                                                  = 100 x 137.64

                                                                  = 13764

                                Pentagon = 5 x Area of triangle

                                                = 5 x 13764

                                                   = 68820m2

Now that I have sampled regular Pentagons, I will now move on to irregular pentagons.

Irregular Pentagons

image12.png

image13.png

                Irregular pentagons still have the same amount of sides as a    regular pentagon, but usually a different area. image14.png

           To find out the area of the irregular pentagon on the left, I will

           Divide this into two shapes – an isosceles triangle and a square. I will then find the areas of both shapes and add them together.

Area of triangle

To find the length of side x, I will have to use Pythagoras Theorem. The rule for Pythagoras theorem is a2 + c2 = b2.

                  a2 + c2 = b2

1002 + c2 = 1502

    10000 + c2= 22500image13.png

c = sqrt 12500

c = 111.80image14.png

Now I know the height and base measurements, I will now continue to find the area of the triangle.

A = ½ bh

            = ½ (200x111.80)
= ½ 22360

   = 11180m^2

Now I need to work out the area of the square.

                                                     A = LW

     A = 200 x 250

A = 50000

So, the area of the whole shape:

A of Irregular Pentagon = 50000 + 11180

                                      = 61180m2

Hexagon

...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 Area Under A Curve

    The Trapezium Rule I will now use a formula to accurately find the area of trapeziums. This will increase accuracy even further by measuring more trapeziums and the formula ensures a closer degree of accuracy. The formula I will use is: (y0 + 2y1 + 2y2 + 2y3 + y4)

  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. Maths Fence Length Investigation

    This would also be a lot easier as many of the other shapes have millions of different variables. H h O The next shape that I am going to investigate is the pentagon. Because there are 5 sides, I can divide it up into 5 segments.

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

    = 1/2 of the base (b) x height (h) Which abbreviates to: A=1/2bh I do not know the measurement of the height so I will split my triangle into 2 by bisecting the angle as shown below to make this easier (all shapes drawn to scale 1cm:50m): To find the

  1. Fencing problem.

    the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular decagon. I shall half the interior angle to find this angle: Angle OGF = 1440 � 2 = 720 To find Angle GOF = 180 - (72 + 72)

  2. Investigate the shapes that could be used to fence in the maximum area using ...

    Height (m) Area (m2) 249 251 62499 249.5 250.5 62499.75 24975 250.25 6249993.75 250 250 62500 250.25 249.75 62499.9375 250.5 249.5 62499.75 251 249 62499 All of these results fit into the graph line that I have, making my graph reliable.

  1. I have to find the maximum area for a given perimeter (1000m) in this ...

    and a square is made up of triangles, what will it be when you have a triangle). This then applies to equilateral triangles being better than all other triangles because a) (which is why squares have more area than triangles, because of hypotenus, which if we think of right angled

  2. HL type 1 portfolio on the koch snowflake

    implies that the conjecture for is: "The conjecture suggests that as we move right along the x-axis (0 onwards), i.e. as the value of 'n' increases by 1 unit, the corresponding value on the y-axis ( 3 onwards)

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