# I am going to start investigating different shape rectangles, all which have a perimeter of 1000m.

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Introduction

Paul Dunn

I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. 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 base 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 by using the above formula.

Height (m) | Width (x) | Area (m2) |

0 | 500 | 0 |

20 | 480 | 9600 |

40 | 460 | 18400 |

60 | 440 | 26400 |

80 | 420 | 33600 |

100 | 400 | 40000 |

120 | 380 | 45600 |

140 | 360 | 50400 |

160 | 340 | 54400 |

180 | 320 | 57600 |

200 | 300 | 60000 |

220 | 280 | 61600 |

240 | 260 | 62400 |

250 | 250 | 62500 |

260 | 240 | 62400 |

280 | 220 | 61600 |

300 | 200 | 60000 |

320 | 180 | 57600 |

340 | 160 | 54400 |

360 | 140 | 50400 |

380 | 120 | 45600 |

400 | 100 | 40000 |

420 | 80 | 33600 |

440 | 60 | 26400 |

460 | 40 | 18400 |

480 | 20 | 9600 |

500 | 0 | 0 |

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

Middle

Area (m )

0

500.0

500.000

0.000

20

490.0

489.898

4898.979

40

480.0

479.583

9591.663

60

470.0

469.042

14071.247

80

460.0

458.258

18330.303

100

450.0

447.214

22360.680

120

440.0

435.890

26153.394

140

430.0

424.264

29698.485

160

420.0

412.311

32984.845

180

410.0

400.000

36000.000

200

400.0

387.298

38729.833

220

390.0

374.166

41158.231

240

380.0

360.555

43266.615

260

370.0

346.410

45033.321

280

360.0

331.662

46432.747

300

350.0

316.228

47434.165

320

340.0

300.000

48000.000

333.3

333.3

288.675

48112.522

340

330.0

282.843

48083.261

360

320.0

264.575

47623.524

380

310.0

244.949

46540.305

400

300.0

223.607

44721.360

420

290.0

200.000

42000.000

440

280.0

173.205

38105.118

460

270.0

141.421

32526.912

480

260.0

100.000

24000.000

500

250.0

0.000

0.000

Because the regular rectangle was the largest before, I added 333.3 as a base length. This is the length of the base of a regular triangle. Below

is a graph of the base against area. The regular triangle seems to have the largest area out of all the areas but to make sure I am going to

Conclusion

1000 ÷ 7 = 142.857 ÷ 2 = 71.429

360 ÷ 7 = 51.429 ÷ 2 = 148.323

Area = ½ X b X H = ½ X 71.429 X 148.323 = X

5297.244 X 14 =

That is the full equation and it works on all of the shapes that I have already done, giving the same answers as before. Below is a table showing the answers I got whet I used the equation:

No. of sides | Area (m2) |

3 | 48112.522 |

4 | 62500.000 |

5 | 68819.096 |

6 | 72168.784 |

7 | 74161.644 |

These are exactly the same as the results in my previous results proving that they are correct.

Now that I’ve this equation, I will use it to work out the area for a regular octagon (8 sides), nonagon (9 sides), and decagon (10 sides):

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 a much greater amount of sides:

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 |

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

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