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  • Level: GCSE
  • Subject: Maths
  • Word count: 2944

A farmer has exactly 1000m of fencing and wants to fence off a plot of level land. She is not concerned about the shape of the plot. She wishes to fence off the plot of land which contains the maximum area. I will be investigating the shape

Extracts from this document...

Introduction

A farmer has exactly 1000m of fencing and wants to fence off a plot of level land. She is not concerned about the shape of the plot. She wishes to fence off the plot of land which contains the maximum area. I will be investigating the shape, or shapes that could be used to fence in the maximum area using exactly 1000m of fencing each time.

I will attempt to work this problem out using a logical sequence of thought. So I will first investigate the shape that is simplest to work out the area of, rectangles.

The formula to work out rectangles is width*height

I will now show a number of rectangles to show how to work out the areas. I will fill the values for height width and area by hand.

Here is the table of rectangle area’s I made using excel. I’ve cut it down to show the first part of the table up to an area of 16,000 cm2.  

width (A)

Height (B)

Area [C]

10

990

9900

20

980

19600

30

970

29100

40

960

38400

50

950

47500

60

940

56400

70

930

65100

80

920

73600

90

910

81900

100

900

90000

110

890

97900

120

880

105600

130

870

113100

140

860

120400

150

850

127500

160

840

134400

170

830

141100

180

820

147600

190

810

153900

200

800

160000

From my set of result’s I’ve made a table which shows a interesting result, has a perfect line of symmetry down the middle and the results which we began with we also end with.

...read more.

Middle

image00.png

The graph of the scalene triangles, has a line of symmetry shown

As you can see the maximum value highlighted below, show’s the shape of an isosceles triangle.  Therefore I will be doing isosceles triangles next.

I’ll start my investigation of isosceles triangles by working out the area of a few triangles. Then I will show my results in the form of a table and analyse them.

Answer = 11858.54cm2

Answer = 22360.68cm2

Answer = 31374.75cm2

Answer = 38729.83cm2

Answer =44194.17cm2

Answer =47434.16cm2

Answer =47925.72cm2

Answer =44721.36cm2

Answer =35575.62cm2

I will be using hero’s formula to find the area of isosceles triangles. Here is my table.

BASE  [C]

SIDE 1 (A)

SIDE 2 (B)

Perimeter

S

S-A

S-B

S-C

AREA

200

400

400

1000

500

100

100

300

38729.83346

210

395

395

1000

500

105

105

290

39982.80881

220

390

390

1000

500

110

110

280

41158.23125

230

385

385

1000

500

115

115

270

42253.69806

240

380

380

1000

500

120

120

260

43266.61531

250

375

375

1000

500

125

125

250

44194.17382

260

370

370

1000

500

130

130

240

45033.321

270

365

365

1000

500

135

135

230

45780.72739

280

360

360

1000

500

140

140

220

46432.74706

290

355

355

1000

500

145

145

210

46985.37006

300

350

350

1000

500

150

150

200

47434.1649

310

345

345

1000

500

155

155

190

47774.20852

320

340

340

1000

500

160

160

180

48000

330

335

335

1000

500

165

165

170

48105.35313

340

330

330

1000

500

170

170

160

48083.26112

350

325

325

1000

500

175

175

150

47925.72378

360

320

320

1000

500

180

180

140

47623.5236

370

315

315

1000

500

185

185

130

47165.9305

380

310

310

1000

500

190

190

120

46540.30511

390

305

305

1000

500

195

195

110

45731.55366

400

300

300

1000

500

200

200

100

44721.35955

410

295

295

1000

500

205

205

90

43487.06704

420

290

290

1000

500

210

210

80

42000

430

285

285

1000

500

215

215

70

40222.81691

440

280

280

1000

500

220

220

60

38105.11777

450

275

275

1000

500

225

225

50

35575.62368

460

270

270

1000

500

230

230

40

32526.91193

470

265

265

1000

500

235

235

30

28781.50448

480

260

260

1000

500

240

240

20

24000

490

255

255

1000

500

245

245

10

17324.11614

...read more.

Conclusion

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        #

Now I can present the formula not in its finished state, but a usable one, the penultimate stage. I’ve cancelled out the like terms and I’m left with only a few more things to simplify.

I will now do the finished version of the formula and then test it out on 3 n-gon shapes.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

Now to test the general formula, I will at first use the shapes which areas I’ve found out.

Octagons – 25000/8TAN(180/8)=  75444.174cm2

Nonagons – 25000/9TAN(180/9)=  76318.817cm2

Decagons (10 sided) – 250000/10TAN(180/10)= 76942.088cm2

50 sided shape -  250000/50TAN(180/50)= 79472.724cm2

Now here is a table containing the areas

Triangle

48112.52243

Square

62500

Pentagon

68819.09602

Hexagon

72168.78365

Heptagon

74161.47845

Octagon

75444.17382

nonagon

76,318.817

decagon

76942.088

dodecagon

77751.05849

My graph show’s a general increase as the sides increase reaffirming my prediction that the area would increase as the sides went up. Although to show this theory working in a larger scope here is a table of areas increasing in 50’s.

50 sided shape

79,472.72

100 sided shape

79551.28988

150 sided shape

79565.83568

200 sided shape

79570.92645

250 sided shape

79573.28271

The graph of the shapes going up in 50’s shows the same trend as before

image01.png

After my research, I’ve concluded that the circle with a perimeter of 1000m is the largest area, and the farmer, if she wishes to have the largest area, should use this shape, although since circles do not tessellate, nor is it easy to make this shape on a large scale, she should use a simpler shape.

...read more.

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