# Maths Coursework - The Fencing Problem

Extracts from this document...

Introduction

Maths Coursework

The Fencing Problem

## Anthony Lynch

## Introduction

This coursework is a problem involving a farmer. She has 1000 metres of fencing and is looking to create a fenced off area that is as large as possible. She wants to know what shape to arrange the fencing in order to achieve this.

I hope to find out which shape would be most appropriate for this use. I will investigate the different formulas and areas of different shapes and compare results to come to a final conclusion.

The Quadrilaterals

### The Square

The only square with a perimeter of 1000 meters, has four sides each being 250 metres in length.

Following the formula area = length x width, the areaworks out to be 62500 m².

##### Conclusion

As there is only one square, there is only one area that can be obtained, it is 62500 m².

The Rectangle

The rectangle is much more complicated, as there are many combinations in length and with that amount to many different areas.

Length of Side | Width of Side | Area |

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 |

250 | 250 | 62500 |

260 | 240 | 62400 |

270 | 230 | 62100 |

280 | 220 | 61600 |

290 | 210 | 60900 |

300 | 200 | 60000 |

310 | 190 | 58900 |

320 | 180 | 57600 |

330 | 170 | 56100 |

340 | 160 | 54400 |

350 | 150 | 52500 |

360 | 140 | 50400 |

370 | 130 | 48100 |

380 | 120 | 45600 |

390 | 110 | 42900 |

400 | 100 | 40000 |

410 | 90 | 36900 |

420 | 80 | 33600 |

430 | 70 | 30100 |

440 | 60 | 26400 |

450 | 50 | 22500 |

460 | 40 | 18400 |

470 | 30 | 14100 |

480 | 20 | 9600 |

490 | 10 | 4900 |

Middle

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

250

250

62500

260

240

62400

270

230

62100

280

220

61600

290

210

60900

300

200

60000

310

190

58900

320

180

57600

330

170

56100

340

160

54400

350

150

52500

360

140

50400

370

130

48100

380

120

45600

390

110

42900

400

100

40000

410

90

36900

420

80

33600

430

70

30100

440

60

26400

450

50

22500

460

40

18400

470

30

14100

480

20

9600

490

10

4900

### The Parallelogram

I knew that the side length that produces the largest area would be 250 by 250. The area of a parallelogram is the base x vertical height. The vertical height can be worked out using ‘side length sin θ’.

I realised that the maximum value of θ was 90. When I worked out 250 sin θ,

###### I found that it produced the result 250. This therefore produced a square.

##### Conclusion

###### I found that again, the largest area produced was 62500m². This is the same as the parallelogram, the rectangle, the kite and especially the square.

### The Rhombus

The rhombus, like all quadrilaterals must have fours sides, these must all be equal. Therefore there can only be one value for any side, 250.

##### Conclusion

Although the angles of a rhombus can differ, I knew that regular shapes have larger areas. So, in this case 250 x 250 and angles of 90° form the largest possible area. Again, like the square, they produce an area of 62500m².

### The Trapezium

###### The trapezium can be worked out in almost the same way the same way as the parallelogram using the ‘vertical height = side sin 90’ formula. Using the area formula ‘ area = ½ x (a + b) x h’ I realised that, as in the parallelogram, the largest value of sin could be 90. Thus creating the largest area, and transforming the trapezium into a square.

Conclusion

Conclusion

I looked at formula for internal and external angles of polygons, and the trigonometry for finding the vertical height of the triangles. I was able to come up with this formula.

Area = ½ x base x tanθ x ½ x base x n

I can prove this by looking at the pentagon.

½ x 200 x tan 54 x ½ x 200 x 5 = 68819 m²

My earlier answer was 68800 m² when using a manual method to a greater decimal place.

### The Circle

###### There can only be one circle that has the circumference of 1000m.

###### Firstly to work out the diameter, I divided 1000 by ∏. This gave me 318.3.

The radius was simply half of 318.3 and therefore 159.1

Then I used the equation ‘∏R²’. This gave me a final area of 79522.5m².

Conclusion

I have found, using the formula ‘∏R²’ that the circle has the largest area possible with 79522.5m².

The Conclusion

I have found that in order for the farmer to obtain the maximum possible area with her 1000 meters of fencing, she should arrange it in a circle.

I have also found that there is a relationship between the area and the number of sides. The more sides a shape has, the larger area it will have.

Also, I found out that regular shapes have larger areas than irregular shapes, and there is a general formula for the area of a polygon.

Area = ½ x base x tanθ x ½ x base x n

Anthony Lynch – The fencing problem – Page

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