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

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Introduction

Christopher Morfill Page 5/2/2007

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:

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

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

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

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

Conclusion

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

= 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

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

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= 22500

c = sqrt 12500

c = 111.80

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

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

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