• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

Extracts from this document...

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.

...read more.

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

...read more.

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

...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. Investigate different shapes of guttering for newly built houses.

    Height Area of trapezium a�= h�- b� Area = 1/2 (a+b)?height = 5�- 1� = 0.5?(22+20)?4.89 a�= 24 = 102.69cm� a = ?24 = 4.89cm 2) Height Area of trapezium a�= h�- b� Area = 1/2 (a+b)?height = 6�- 1� = 0.5?(20+18)?5.92 a�= 35 = 112.1cm� a = ?35 = 5.92cm 3)

  2. t shape t toal

    18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Translation number T-number T-total Equation used t = 5x - 7g Difference by moving up 1 0 34 128 (5 x 34)

  1. t shape t toal

    The grid size I'll use will be 7 by 4. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Firstly here is the formula I'll need.

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

    To work out the area of an isosceles triangle, I will use the following formula: Area = 1/2 Base x Height So before I can work out the area, I need to work out the height of the isosceles triangle.

  1. Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

    In many ways the environment has improved, although some areas still could be better. Finally, judging by the reaction of many tourists and locals, the regeneration has had a generally good impact on the social side of things as well.

  2. Investigating heat loss in different sized animals.

    I will keep these the same: * All of the beakers will rest on the same surface so that heat loss by conduction will be the same. * None of the beakers will have lids on them. * All of the beakers are made of Pyrex.

  1. The aim of this investigation is to find what is the maximum area you ...

    It is now proven that when the triangle is an isosceles the area is larger. So now all we have to do to find the triangle that has the largest area out of the isosceles. To find the area of the triangles I will have to use Pythagoras' formula. E.g.

  2. Investigate the shapes that could be used to fence in the maximum area using ...

    Height (m) Area (m2) 249 251 62499 249.5 250.5 62499.75 24975 250.25 6249993.75 250 250 62500 250.25 249.75 62499.9375 250.5 249.5 62499.75 251 249 62499 All of these results fit into the graph line that I have, making my graph reliable.

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