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

Find out the greatest area that can be enclosed with 1000m of fence

Extracts from this document...

Introduction

Fence

The exercise is to find out the greatest area that can be enclosed with 1000m of fence. My first impression is that the circle has the largest area, but this needs to be investigated.

Starting with rectangles, any 2 different length sides will add up to 500, because each side has an opposite with the same length. So 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. The equation to work out a rectangle is

1000 = width(500 –width)

Using 10m increments here is a table of areas For a graph of Height/Area  see graph 1.

Height (m)

Width

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

This shape is a parabola. If you look at the table and the graph, the rectangle with a height of 250m has the biggest area.

...read more.

Middle

This is also a parabola, therefore a square is the rectangle with a larger area, I can try this with other regular polygons

          I will try isosceles triangles because I can vary the base easily with this, I believe that the biggest area will belong to the equilateral triangle

The formula for working the isosceles triangle out is simple. Split the base in half giving two right angles. Use Pythagoras to work out the triangles height using the base length divided by two and the other side length.

...read more.

Conclusion

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

         From the first graph, 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 much, this indicates a graph of this would continue this trend.

 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 pi.

  • 1000/pi = diameter
  • diameter / 2 = radius
  • pi X radius squared = area
  • which for a 1000m circumference  79577.47155 metres squared

This is larger than any number of sides and would fit on the graph, but this difference can become almost imperceptible.

Therefore, I have proved and conclude that the polygon which will give the largest area, for a perimeter of 1000 metres is the circle. It is also quite safe to conclude that this is the same for any perimeter because the size of perimeter and scale in theory make no difference.

...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. Geography Investigation: Residential Areas

    I did this by using a Bi-Polar analysis amongst other things, they can all be found on my 'external questionnaire'. The results from my 'external questionnaire' could have not been improved by collecting more data because I can only analyse the street once with the same questionnaire when filled in by myself.

  2. The fencing problemThere is a need to make a fence that is 1000m long. ...

    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

  1. When the area of the base is the same as the area of the ...

    Area: Area of 4 sides: 1cm 7cm 49cm 49cm 1x7x4=28cm 2cm 5cm 50cm 25cm 2x5x4=40cm 3cm 3cm 27cm 9cm 3x3x4=36cm 4cm 1cm 4cm 1cm 4x1x4=16cm This table shows the results I got for the measurements of the 9cm by 9cm net.

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

    Environmental Biased Friday 24.9.04 11am-2:30pm Sunny MD survey Street Transect Biased Friday 24.9.04 11am-2:30pm Sunny MD SX 482542 Photographs Biased Friday 24.9.04 11am-2:30pm Sunny MD Sketches Biased Friday 24.9.04 11am-2:30pm Sunny MD SX 491539 Interview Biased Friday 17.9.04 1pm Spitting GD Originality Biased Saturday 13.11.04 12-2pm Sunny, cold MD Pedestrian

  1. Difference in Japans Two Biggest Regions: Kanto and Kansai

    Kobe is also famous for its Kobe beef, the Arima Onsen (hot springs), and the expensive and elite Ashiya district (the haunt of wealthy yakuza) ("City of Kobe.", n.d.). Kanto Region The Kanto region is a geographical area of Honshu, the largest island in Japan.

  2. There is a need to make a fence that is 1000m long. The area ...

    I will work out and record results which I feel are sufficient enough to prove my point. There are three columns one showing the height (m) another showing the base length and the final column shows the area (m2). In this case the base length will be the variable I will change.

  1. The Area Under A Curve

    Rectangle 1 - 1x2 - 2cm Rectangle 2 - 9x2 -18cm Rectangle 3 - 25x2 - 50cm Rectangle 4 - 49x2 - 98cm Rectangle 5 - 81x2 - 162cm 330cm When all the areas of the rectangles are totalled up, they are equal to 330cm .

  2. Find the biggest value of the ratio area / perimeter for some triangles.

    = square root of 90x(90-40)x(90-60)x(90-80) = square root of 90x50x30x10 = square root of 1350000 = 1161.90 Ratio = a/p = 1161.90 � 90 = 12.91 Triangle c: Perimeter = 60�Sin40 = 93.34 = 93.34+80+60 = 233.34 Area = s = p�2 = 233.34�2 = 116.67 = square root of s(s-a)(s-b)(s-c)

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