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  • Level: GCSE
  • Subject: Maths
  • Word count: 2220

Maths Fence Length Investigation

Extracts from this document...

Introduction

There is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.

100m

150m

400m

         I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. Below are 2 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.

                                                                         350m

In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side. To work out the area of a rectangle with a bas length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.

1000 = x(500 – x)

Below is a table of results, worked out using the formula. I have gone down by taking 10m off the base every time.

Height (m)

x

Area (m2)

0

500

0

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

500

0

0

         Using this formula I can draw a graph of base length against area.

         As you can see, the graph has formed a parabola.

...read more.

Middle

H2 = 150000

H = 387.298

½ X 200 X 387.298 = 38729.833m.

Below is a table of result for isosceles triangles from a base with 10m to a base with 500m.

Side =                   . Where X is the base length. I have used this formula to work out the area when the base is different heights. To do this I need to use Pythagoras’ theorem. Below is a diagram of the isosceles triangle.

Base (m)

 Side (m)

 Height (m)

 Area (m )

0

 500.0

 500.000

 0.000

10

 495.0

 494.975

 2474.874

20

 490.0

 489.898

 4898.979

30

 485.0

 484.768

 7271.520

40

 480.0

 479.583

 9591.663

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

.

Because the regular rectangle was the largest before, I added 333.

...read more.

Conclusion

No. of sides

Area (m2)

8

75444.174

9

76318.817

10

76942.088

         As you can see, the larger the number sides on the shape, the larger the area is. This pattern has carried on going for all of the shapes that I have investigated, so I am going to investigate shapes with the following amount of sides:

20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.

Below is a table showing the results that I got.

No. of sides

Area (m2)

20

78921.894

50

79472.724

100

79551.290

200

79570.926

500

79576.424

1000

79577.210

2000

79577.406

5000

79577.461

10000

79577.469

20000

79577.471

50000

79577.471

100000

79577.471

         On the next page is a graph showing the No. of sides against the Area.

         As you can see form the graphs, the line straightens out as the number of side’s increases. Because I am increasing the number of sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using o. To work out the circumference of a circle the equations is od. I can rearrange this so that the diameter is circumference/o. From that I can work out the area using the or2 equation.

1000/o = 318.310

318.310/2 = 159.155

o X 159.1552 = 79577.472m2

         If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference

...read more.

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