• 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

image00.png

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.

...read more.

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

image17.png

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

...read more.

Conclusion

Circle

image13.png

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

Radius = 1000 / 2π

= 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        

...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. Geography Investigation: Residential Areas

    3 1 4 1 1 Cumberland Avenue 1 2 7 0 0 Vivaldi Close 10 0 0 0 0 The Beaches 10 0 0 0 0 For the above data I have created a percentage stacked bar chart, it is the easiest way to show this data.

  2. The Fencing Problem

    .5 ( 66.666 ( 188.679 = 6,257.864m� Now I will find the area for triangle B. But again I must first find the height of the triangle. Hypotenuse = 200m Base = 188.679m 200 � -188.679 = 4,400.234m V4,400.234 = 66.334m I can now find the area .5 ( 188.679

  1. The Fencing Problem

    Angle (�) Height [h] (m) Perimeter (m) Area (m�) 100 400 10 69.46 1000 6945.93 100 400 20 136.81 1000 13680.81 100 400 30 200.00 1000 20000.00 100 400 40 257.12 1000 25711.50 100 400 50 306.42 1000 30641.78 100 400 60 346.41 1000 34641.02 100 400 70 375.88 1000 37587.70 100 400 80 393.92 1000 39392.31 100 400

  2. Math Coursework Fencing

    We must spit the kite into two triangles as shown below: The area of a triangle is calculated using this formula: Since we are eventually going to double the area we get so we can derive this final formula to calculate the area of a kite.

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

    out what each angle is I need to divide the 180� between the 3 angles: 180/3 = 60� If I split my triangle in half I will find it easier to use the equation A=1/2bh. I do not know what the height is though so I will have to use trigonometry to come across this.

  2. Fencing Problem

    350 315 335 1000 500 47847.41 350 320 330 1000 500 47906.16 350 325.1 324.9 1000 500 47925.71596 333.3 333.2 333.5 1000 500 48112.50222 After analysing my second table of results for scalene triangles, the table has strengthened my assumption that the closer the dimensions are to an equilateral the larger the area is.

  1. Investigating different shapes to see which gives the biggest perimeter

    I can work this out because the dot at the centre of the pentagon equals 360�. So the size of each angle adjacent to the height of the triangle equals 360 � 10 = 36�. The following diagram helps explain this So to work out the area of each isosceles

  2. Fencing - maths coursework

    10594.49692 However, this is not the final bit of the formula, because there was 7 triangles in the heptagon so we have to times our answer (10594.49692) by 7 to get the area of the whole heptagon. So finally the last step to find out the area of the whole

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