• 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
• Level: GCSE
• Subject: Maths
• Word count: 2099

# The Fencing Problem.

Extracts from this document...

Introduction

The Problem

A farmer has exactly 1000 meters of fencing and wants to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000 meters. She wishes to fence off the plot of land, which contains the maximum area.

Investigate the shape or shapes, which could be used to fence in the maximum area using exactly 100 meters of fencing each time. In this investigation I am going to experiment with different shapes and I will find out which shape will give here the maximum area

Investigation

Squares

There are several shapes that I will be able to investigate for this coursework and to start with I will find the area of a square as this is the easiest shape to work out.

Area of a Square = Length x Width

Firstly to find the length of each side I will divide 1000m by 4 which equals 250m. Then to find the area we must square this number.

• 1000 / 4 = 250 meters
• 250 x 250 = 62500meters ²

Rectangle

The next shape that I will work out is a rectangle as this is almost as easy as a square to work out.

Area of Rectangle = Width x Length

In this table in the following page I will work out the area for a large number of rectangles.

Middle

430

1000

424

29698

150

425

1000

418

31375

160

420

1000

412

32985

170

415

1000

406

34527

180

410

1000

400

36000

190

405

1000

394

37402

200

400

1000

387

38730

210

395

1000

381

39983

220

390

1000

374

41158

230

385

1000

367

42254

240

380

1000

361

43267

250

375

1000

354

44194

260

370

1000

346

45033

270

365

1000

339

45781

280

360

1000

332

46433

290

355

1000

324

46985

300

350

1000

316

47434

310

345

1000

308

47774

320

340

1000

300

48000

330

335

1000

292

48105

340

330

1000

283

48083

350

325

1000

274

47926

360

320

1000

265

47624

370

315

1000

255

47166

380

310

1000

245

46540

390

305

1000

235

45732

400

300

1000

224

44721

410

295

1000

212

43487

420

290

1000

200

42000

430

285

1000

187

40223

440

280

1000

173

38105

450

275

1000

158

35576

460

270

1000

141

32527

470

265

1000

122

28782

480

260

1000

100

24000

490

255

1000

71

17324

500

250

1000

0

0

From the table above it is clear to see that when the base is 330 meter then the maximum area is obtained in 48105m².

To check all possibilities I then drew up another table for the figures in-between 332 and 337. This would give me a true reading as to which triangle would give me the greatest area.

 Base(meters) Side(meters) Perimeter(meters) Height(meters) Area(meters ²) 332 334.0 1000 290 48111.4 333 333.5 1000 289 48112.5 334 333.0 1000 288 48112.2 335 332.5 1000 287 48110.7 336 332.0 1000 286 48107.9 337 331.5 1000 285 48103.7

From this more detailed table it is clear that if she was to have a base of 333 meters then she would gain maximum area in 48122.5 meters².

From this I can see that a regular shape will have a greater area than an irregular form of a certain shape. Then next idea I would like to put forward is that all polygons are made up of a number of isosceles triangles. I can show this idea by investigating the area of a pentagon.

Pentagon

From the diagram we can see that we can divide the pentagon into 5 triangles which is also the number of sides it has.

Conclusion

Circle

The final shape that I will investigate in my coursework is a circle. I predict that this will have the largest area because of my previous calculations. I calculated that as the number of sides on a polygon increased then so would the area. With a circle there are an unlimited number of sides meaning it will have the biggest area.

As we have the perimeter (or circumference) of the circle to start with I will need to work back wards to find the radius then I can work out the area. My calculations begin on the following page.

Circumference = 2 x π x radius

So

= 159.1549431

and then to find the area we use the formula:

Area of a Circle = π x radius ²

= 159.1549431 ² x π

= 79577.47155 m²

So above is the area for a circle and that is the final shape I will work out. I will now put the areas of the different shapes I have investigated into a table to compare them.

Results

 Shape Area (m²) Square 62500 Rectangle N/A Isosceles Triangle 48112.5 Pentagon 68819.096 100 sided polygon 79551.28988443 1,000,000,000 sided polygon 79577.47154595 Circle 79577.47155

GCSE Maths Coursework                                                                    By James Nott

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

# Related GCSE Fencing Problem essays

1. ## Fencing Problem

the Circle * Circumference = x Diameter * I know my perimeter must be 1000 m so I will replace circumference by

2. ## Fencing problem.

I have discovered the area of six different polygons. These polygons that have been discovered are all regular. The results have been shown below in a tabulated form: Name of regular polygon Area (m2) Pentagon 68800 Hexagon 72217.2 Heptagon 74322.5 Octagon 75450.4 Nonagon 76392 Decagon 76950 I have represented the

1. ## The coursework problem set to us is to find the shape of a gutter ...

I went on to investigate shapes with more complex variations; the first was the isosceles triangle. The only variation here was in the angles at which the sides fell; the sides themselves were kept as 15cm long each, as the triangle was isosceles.

2. ## Maths Coursework - The Fencing Problem

The Quadrilaterals The Conclusion I have found that all of the quadrilaterals are connected to the area formula of a square. Because of this, they all have a maximum area of 62500m�. This is achieved in most cases using the formula, Area = 250m x 250m The Triangles The Equilateral

1. ## The Fencing Problem

400 100 2) 400 100 3) 400 100 4) 400 100 5) 400 100 6) 400 100 7) 400 100 8) 400 100 9) 400 100 10) 400 100 = ^ We can see that the "symmetrical" data is very clearly being portrayed on the graph, and I have

2. ## Math Coursework Fencing

The area of a parallelogram can be found by using the formula (Base multiplied by Height equals area). My prediction of the greatest area here, is that the closer the shape gets to a square the greater its area will be.

1. ## Geography Investigation: Residential Areas

Extrapolation I will calculate the average area rating from all of the other data I have collected. For each of the hypothesis I have just mentioned the main method I will use throughout the investigation as a breakdown of the last few pages.

2. ## Maths Fencing Coursework

This made me realise that a square is a rectangle which is something I have learnt in this section which I didn't know before. Parallelograms I have now done squares and rectangles; I have to move on to quadrilaterals. I have chosen to do Parallelograms.

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