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  • Level: GCSE
  • Subject: Maths
  • Word count: 1293

The fencing problem

Extracts from this document...

Introduction

The fencing problem

The objective of this task is to experiment with shapes to try and find the maximum area possible. The given problem is that a farmer has exactly 1000m of fencing. With it, she wants to fence off a plot of level land. She doesn’t mind what the shape is but it must have a perimeter of 1000 meters so we now have to find the best shape that the farmer can use.

Part one – By using scale drawings, I will examine some possible figures for the plot of land. In each case I must ensure that the perimeter is 1000m and I will obtain the enclosed area.

Part two – I will investigate the shape or shapes of the plot of land, which have the maximum area.

I am going to start the investigation by using a few quadrilaterals. I will experiment with these shapes, changing the lengths and widths and recording possible outcomes in a table.

Quadrilaterals

Key:        = 20m

Area = length x width

Area = 250 x 250

        = 62500m²

Area = length x width

Area = 200 x 300

       = 60,000m²

...read more.

Middle

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

When the length and width are the same, the area is the largest, and the areas before and after this are the same, and it continues up and down the rest of the table. Now that I have found that a square has the greatest area of the quadrilateral group, I am going to find the triangle with the largest area. Because there is more than 1 variable, there are countless combinations, so I am only going to use isosceles triangles. This is because if know the base length, then I can work out the other 2 lengths, because they are the same. I did this by going up in 10’s and discovered a pattern occurring in the table. I have used the results from the quadrilaterals I have drawn with lengths of 50m, 100m, 150m, 200m and 250m to create a graph.

...read more.

Conclusion

Area = 8660.25404 x 10 = 86,602.5m2 (1dp)

All of the results that I have got so far have shown that as the number of side’s increases, the area increases. I am going to investigate this further with a regular hexagon (6 sides) and a regular heptagon (7 sides).

Hexagon

1000 ÷ 6 = 166.666… ÷ 2 = 83.333…

360 ÷ 6 = 60 ÷ 2 = 30

Tan = opp ÷ adj  

Tan 60 = 83.3333333 ÷ adj

Tan 60 x 83.3333333 = 144.3375673 = 144.338 (3 d.p)

Area = ½ x b x h = ½ x 83.333… x 144.338 = 6014.065

6014.065 X 12 = 72168.784m2

No. of sides

Area (m2)

3

48112.522

4

62500.000

5

68819.096

6

72168.784

I have made a table of the area of triangles, quadrilaterals, pentagons and hexagons. As you can see, the more sides there are, the bigger the area becomes. So my conclusion is that the circle will have the largest area, as it has an infinite number of sides. I will prove this theory by investigating a circle.

Circle

Not to scale

Circumference =  pi x diameter

1000 ÷ pi = 318.3098862 ÷ 2

= 159.154931

= 159.2 (1.d.p)

Area = pi x radius ²

Area = pi x 159.1549431²

Area = pi x 25330.29591

Area = 79,577.5m² (1.d.p)

As you can see – my prediction was correct considering that the circle with an infinite number of sides has a larger area than the rest of the shapes I have investigated.

...read more.

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