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• Level: GCSE
• Subject: Maths
• Word count: 2543

# The Fencing Problem

Extracts from this document...

Introduction

GCSE Mathematics Coursework 2 –

The Fencing Problem

## Scenario

A farmer has exactly 1000 meters of fencing and wants to fence off a plot of level land. So therefore there is a need to make a fence that is 1000m long. The actual shape itself has to have a maximum area. I will now investigate this and hopefully find a general formula that would make it easier to solve various problems relating to different shapes.

Firstly, to start the investigation of, I am going to investigate rectangles, all with a perimeter of 1000m. I have drawn various rectangles, to show how different shapes with the same perimeter can have different areas.

With a rectangle, two sides are parallel, which means that any two-side lengths will add up to 500m. For example in a 200 x 300 rectangle, two sides that are opposite each other that are 200m long and two sides next to them that are opposite each other that are 300m long.  With rectangles in this situation it is easy to work out the area, even with only one of the sides.

1000 = x(500 – x)

If you have a rectangle with a base of 200m, to work out the area, I take 200 and subtract it from 500, which equals 300 and then simply times 200 by 300 thus giving me the area. This method can be simplified significantly by putting the numbers into the formula above.

Middle

50

475.0

474.342

11858.541

60

470.0

469.042

14071.247

70

465.0

463.681

16228.832

80

460.0

458.258

18330.303

90

455.0

452.769

20374.617

100

450.0

447.214

22360.680

110

445.0

441.588

24287.342

120

440.0

435.890

26153.394

130

435.0

430.116

27957.557

140

430.0

424.264

29698.485

150

425.0

418.330

31374.751

160

420.0

412.311

32984.845

170

415.0

406.202

34527.163

180

410.0

400.000

36000.000

190

405.0

393.700

37401.537

200

400.0

387.298

38729.833

210

395.0

380.789

39982.809

220

390.0

374.166

41158.231

230

385.0

367.423

42253.698

240

380.0

360.555

43266.615

250

375.0

353.553

44194.174

260

370.0

346.410

45033.321

270

365.0

339.116

45780.727

280

360.0

331.662

46432.747

290

355.0

324.037

46985.370

300

350.0

316.228

47434.165

310

345.0

308.221

47774.209

320

340.0

300.000

48000.000

330

335.0

291.548

48105.353

333.3

333.3

288.675

48112.522

340

330.0

282.843

48083.261

350

325.0

273.861

47925.724

360

320.0

264.575

47623.524

370

315.0

254.951

47165.931

380

310.0

244.949

46540.305

390

305.0

234.521

45731.554

400

300.0

223.607

44721.360

410

295.0

212.132

43487.067

420

290.0

200.000

42000.000

430

285.0

187.083

40222.817

440

280.0

173.205

38105.118

450

275.0

158.114

35575.624

460

270.0

141.421

32526.912

470

265.0

122.474

28781.504

480

260.0

100.000

24000.000

490

255.0

70.711

17324.116

500

250.0

0.000

0.000

I added 333.3 in the table, as it is the base at which the triangle becomes regular. Below is a graph of base against area :

As can be seen from the graph, the regular triangle seems to have the largest area out of all the areas. To improve the accuracy of my results I am going to find the areas of the values around 333.

 Base (m) Side (m) Height (m) Area (m2) 333 333.5 288.964 48112.450 333.25 333.4 288.747 48112.518 333.3 333.4 288.704 48112.522 333.5 333.3 288.531 48112.504 333.75 333.1 288.314 48112.410 334 333.0 288.097 48112.233

The table above again shows that the regular triangle has the largest area.

For the past two shapes that I have looked at, the regular shapes of them both seem to have the largest areas; so from now on I am going to investigate regular shapes. I feel that this is the easiest and quickest solution, as the other shapes have many variables, which would need to be taken into account when investigating them.

After having investigated rectangles and triangles, I am going to move on to a slightly more complex shape and investigate it, pentagon.

In a pentagon there are 5 sides and each of those five sides can be divided up into 5 segments. Each segment is split up into an isosceles triangle, with the top angle being 720 (one fifth of 360o). This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each.

Conclusion

C = πd

1000 = πd

d = 1000/π

r = 500/π

d = 1000/π = 318.31

C = πd

= π x 318.31 = 1000

π x 159.152

= 79577.47

The calculation above shows that from the circumference the area can be found out. This can be investigated into further more.

One radian is the angle made when the arc length of a sector is the same as the radius.  Now that this is known it is possible therefore to state that there are 2π radians in a circle.  When looking at my formula for regular polygons, it can be seen that I require the equivalent of 180° not 360° and it is therefore easy to see that this would be equivalent to just π.  The new formula for working out the area of a regular polygon when using radians is therefore the following :

The polygon with the highest area would have infinity number of sides, thus the angles being very small. I can now remove the tan from the formula above, which gives me this formula :

Cancelling out n would leave just π on its own. Solving the equation would give the area of the circle with the circumference of 1000m as can be shown by the calculations below:

In conclusion I can say that the shape with the largest area and perimeter of a 1000m is a circle.

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

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