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  • Level: GCSE
  • Subject: Maths
  • Word count: 2134

Fencing Problem

Extracts from this document...

Introduction

image00.png

There is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.

100m

150m

400m

         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 bas 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 using the formula. I have gone down by taking 10m off the base every time.

Height (m)

x

Area (m2)

0

500

0

410

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

         Using this formula I can draw a graph of base length against area. 

...read more.

Middle

To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras’ theorem. Below is the formula and area when using a base of 200m.

H2 = h2 – a2

H2 = 4002 – 1002

H2 = 150000

H = 387.298

½ X 200 X 387.298 = 38729.833m.

Below is a table of result for isosceles triangles from a base with 10m to a base with 500m.

Side =                   . Where X is the base length. I have used this formula to work out the area when the base is different heights.

...read more.

Conclusion

No. of sides

Area (m2)

3

48112.522

4

62500.000

5

68819.096

6

72168.784

7

74161.644

         Now that I have this equation, I am going to use it to work out the area for a regular octagon, nonagon and decagon.

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 the following amount of sides:

20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.

Below is a table showing the results that I got.

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

Because I am increasing the number of sides by large amounts and they are not changing 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 o. To work out the circumference of a circle the equations is od. I can rearrange this so that the diameter is circumference/o. From that I can work out the area using the or2 equation.

1000/o = 318.310

318.310/2 = 159.155

o X 159.1552 = 79577.472m2

 From this I conclude that a circle has the largest area when using a similar circumference.

...read more.

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