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• Level: GCSE
• Subject: Maths
• Word count: 3772

# Fencing Problem - Math's Coursework.

Extracts from this document...

Introduction

### Fencing Problem – Math’s Coursework

Fencing Problem – Math’s Coursework

A farmer has exactly 1000 meters of fencing and wants to fence of a plot of level land. She is not concerned about the shape of the plot but it must have a perimeter of 1000 m. She wishes to fence of a plot of land that contains the maximum area. I am going to investigate which shape is best for this and why.

I am going to start by investigating the different rectangles; all that have a perimeter of 1000 meters. Below are 2 rectangles (not drawn to scale) showing how different shapes with the same perimeter can have different areas.

### Text Box: 200

Below is a table of different rectangles.

Height

## Length

Area

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 studentcentral.co.uk

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 wwbg bgwstbgbgud ebg bgntcbg enbgtral bgcobg uk;

280

61600

230

270

62100

240

260

62400

250

250

62500 wwbb bbwstbbbbud ebb bbntcbb enbbtral bbcobb uk!

260

240

62400

270

230

62100

280

220

61600

290

210

60900

300

200 wweg egwstegegud eeg egntceg enegtral egcoeg uk!

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 wwga gawstgagaud ega gantcga engatral gacoga uk!

110

42900

400

100

40000

410

90

36900

420

80

33600

430

70

30100

440

60

26400

450

50

22500

460

40

18400

470

14100

480

20

9600

490

10

4900

Using this table I can draw a graph of height against area. This is on the next sheet.

As you can see, the graph has formed a parabola. According to the table and the graph, the rectangle with a base of 250m has the greatest area. This shape is also called a square.

Middle

410

295

212.132

43487.060

420 wwae aewstaeaeud eae aentcae enaetral aecoae uk.

290

200.000

42000.000

430

285

187.083

40222.845

440

280

173.205

38105.100

450

275

158.114

35575.650

460

270

141.421

32526.830

470

265

122.474

28781.390

480

260

100.000

24000.000

490

255

70.711

17324.195

Because the last two shapes have had the largest areas when they are regular, I am going to use regular shapes from now on.

The next shape that I am going to investigate is the pentagon.

Because there area 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle with the top angle being 72º. This is because it is a fifth of 360º. This means that I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 54º each. Because every isosceles triangle can be split into 2 right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle will be 100m. GnJr27sHS from GnJr27sHS student GnJr27sHS central GnJr27sHS co GnJr27sHS uk

Using SOH CAH TOA I can work out that I need to use Tangent.

H = 100 tan54 = 137.638

O = 100

T = tan 36

This gives me the length of H so I can work out the area.

Area = ½ × b × H = ½ × 100 × 137.638 = 6881.910

I now have the area of half of one of the segments, so I simply multiply that number by 10 and get the area of the shape.

Area = 6881.910 × 10 = 68819.096m²

Conclusion

5297.260 * 14 = 74161.644m2

My predictions were correct and as the number of side’s increases, the area increases. Below is a table of the number of sides against area

 No. Of sides Area (m2) 3 48112.522 4 62500.000 5 68819.096 6 72168.784 7 74161.644 wwga gawstgagaud ega gantcga engatral gacoga uk;

To find the general formula for an n sided regular polygon I am going to construct a formula.

To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tanq

A = [(1000/n)/2]/ Tan (360/s)/2

Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it’s A * base *0.5 * number of sides.

As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.

 Diameter 318.47m

pR²

3.14 * 159.235 (2 d.p) * 159.235 (2 d.p) = 79617.16561m²

I can see that this produces the largest area.

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

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