• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  • Level: GCSE
  • Subject: Maths
  • Word count: 2223

The fencing problemThere 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

Extracts from this document...

Introduction

The fencing problem

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. Below are 2 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.

350m


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

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

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

As you can see, the graph has formed a parabola.

...read more.

Middle

Side = (1000 - 200) ÷ 2 = 400. This can be rearranged to the following formula.


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)

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

On the next page is a graph showing the No. of sides against the Area.

As you can see form the graphs, 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 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

If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference.

...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. A farmer has 1000 metres of fencing. She wants to use it to fence ...

    So it follows a sequence: 3, 4, and 5. General formula The general formula in this case will work out the area of any regular shape with the perimeter of 1000m, I will follow the method that I used to work out the area of a pentagon and hexagon because

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

    results I got for the measurements of the 7cm by 7cm net. The formulas that I used were the same to what I used for the others. The maximum volume is when the height is 1cm, but the area of the base does not equal to the area of the 4 sides.

  1. Fencing Problem

    out the area of the up coming regular polygons, using the exact same method as I used to figure out the area of the pentagon but changing the lengths / angles. Shape Sides Side Length (1000 / Sides) Angle Of Each Isosceles (360 / Sides)

  2. To investigate the effects of a parachutes shape and surface area, on it time ...

    In the end I found it easiest to get the length and area of the triangle and the just manipulate the area formulas for the square and circle to make them both equal the triangle and then work out the length and radius accordingly.

  1. A length of guttering is made from a rectangular sheet of plastic, 20cm wide. ...

    =E3*F3 =G3*B3 20 6 =A4/B4 =180/(B4*2) =C4/2 =E4/TAN(RADIANS(D4)) =E4*F4 =G4*B4 20 7 =A5/B5 =180/(B5*2) =C5/2 =E5/TAN(RADIANS(D5)) =E5*F5 =G5*B5 20 8 =A6/B6 =180/(B6*2) =C6/2 =E6/TAN(RADIANS(D6)) =E6*F6 =G6*B6 20 9 =A7/B7 =180/(B7*2) =C7/2 =E7/TAN(RADIANS(D7)) =E7*F7 =G7*B7 20 10 =A8/B8 =180/(B8*2) =C8/2 =E8/TAN(RADIANS(D8)) =E8*F8 =G8*B8 20 11 =A9/B9 =180/(B9*2) =C9/2 =E9/TAN(RADIANS(D9))

  2. Fencing problem.

    I shall multiply this figure by the number of triangles present within the nonagon: Area of nonagon = Area of one triangle � Number of triangles Area of nonagon = 8488 � 9 = 76392m2 Decagon The final polygon that shall be investigated during this segment of the coursework is a ten-sided figure known as a decagon.

  1. The coursework problem set to us is to find the shape of a gutter ...

    8.09017 5.877853 10 128.4545 10 37 7.986355 6.01815 10 127.9266 10 38 7.880108 6.156615 10 127.3159 10 39 7.77146 6.293204 10 126.622 After looking at the results in more detail I have noticed that the gutter with sides at 10cm and an angle of 30?

  2. The Fencing Problem

    and the resulting value is the area. Area = {1/2[(1000 � 5) x h]} x 5 1000 � 5 = 200 200 200 360 � 5 = 72� 200 200 200 h 200 100 Regular Polygons - Hexagon As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan)

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