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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.

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.

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

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

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