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  • Level: GCSE
  • Subject: Maths
  • Word count: 1715

The Fencing Problem

Extracts from this document...

Introduction

THE FENCING PROBLEM

The problem:

A farmer has 1000m of fencing and wants to fence a plot of land.  It can be any shape but must have a perimeter of no more than 1000m.  I am going to investigate the effects of using different shapes to get the largest possible area using the 1000m of fence.

Triangles:

I will now investigate different types of triangles as they have the least number of sides.  Isosceles and scalene triangles will be my starting point.

Hypothesis:

The triangle’s area will increase as the triangle’s height increases.

Below are examples of different triangles.

Here is an equilateral triangle, a triangle with all angles and sides the same:

image00.png

Here is a right-angled triangle, a triangle with a right angle:

image01.png

image12.png

...read more.

Middle

48112 .52 m2.

Quadrilaterals:

Next, I will investigate simple four sided shapes called quadrilaterals.  I will investigate different rectangles and squares.

Hypothesis

I predict that the closer the length and width values get to equalling each other, the larger the area will be.

Below are a square and a rectangle:

                     100                                              150image40.pngimage02.png

                                                 50             RECTANGLE                50

100                SQUARE        100

                                                                       150

     100

The formula to work out the area is:

Area = Length x Height

E.g.

100 x 100 = 10000

50 x 150 = 7500

This shows just how different areas of shapes can be, even with the same perimeter.

QUADRILATERALS

Height

Width

Area

20

480

9600

40

460

18400

60

440

26400

80

420

33600

100

400

40000

120

380

45600

140

360

50400

160

340

54400

180

320

57600

200

300

60000

220

280

61600

240

260

62400

260

240

62400

280

220

61600

300

200

60000

320

180

57600

340

160

54400

360

140

50400

380

120

45600

400

100

40000

420

80

33600

440

60

26400

460

40

18400

480

20

9600

image43.png

Conclusion:

I have chosen different values of length and width and found that the biggest area is created when the lengths and widths of the sides are equal, at 250m.  This quadrilateral, with all sides the same, is a square.  
Therefore, the largest area for a four-sided shape is
62500m2.

Polygons:

The square is another regular shape, like the equilateral triangle, so from now on I will use regular polygons whilst finding out how the number of sides affects the area inside.

Hypothesis:

...read more.

Conclusion

POLYGONS:

n

Side

Angle

Radiens

1/2 side

1/2 angle

Height

Triangle area

Total area

5

200

72

0.628319

100

36

137.6382

13763.82

68819.1

6

166.6667

60

0.523599

83.33333

30

144.3376

12028.13

72168.78

7

142.8571

51.42857

0.448799

71.42857

25.71429

148.323

10594.5

74161.48

8

125

45

0.392699

62.5

22.5

150.8883

9430.522

75444.17

9

111.1111

40

0.349066

55.55556

20

152.6376

8479.869

76318.82

10

100

36

0.314159

50

18

153.8842

7694.209

76942.09

20

50

18

0.15708

25

9

157.8438

3946.095

78921.89

30

33.33333

12

0.10472

16.66667

6

158.5727

2642.879

79286.37

40

25

9

0.07854

12.5

4.5

158.8276

1985.344

79413.78

50

20

7.2

0.062832

10

3.6

158.9454

1589.454

79472.72

60

16.66667

6

0.05236

8.333333

3

159.0095

1325.079

79504.74

70

14.28571

5.142857

0.04488

7.142857

2.571429

159.0481

1136.058

79524.04

80

12.5

4.5

0.03927

6.25

2.25

159.0731

994.207

79536.56

90

11.11111

4

0.034907

5.555556

2

159.0903

883.835

79545.15

100

10

3.6

0.031416

5

1.8

159.1026

795.5129

79551.29

200

5

1.8

0.015708

2.5

0.9

159.1419

397.8546

79570.93

300

3.333333

1.2

0.010472

1.666667

0.6

159.1491

265.2485

79574.56

400

2.5

0.9

0.007854

1.25

0.45

159.1517

198.9396

79575.84

500

2

0.72

0.006283

1

0.36

159.1528

159.1528

79576.42

600

1.666667

0.6

0.005236

0.833333

0.3

159.1535

132.6279

79576.74

700

1.428571

0.514286

0.004488

0.714286

0.257143

159.1539

113.6813

79576.94

800

1.25

0.45

0.003927

0.625

0.225

159.1541

99.47133

79577.06

900

1.111111

0.4

0.003491

0.555556

0.2

159.1543

88.41905

79577.15

1000

1

0.36

0.003142

0.5

0.18

159.1544

79.57721

79577.21

image44.png

As you can see, the area increases as the number of sides go up.

Finding the area of the circle:

The circumference is 1000 (as this is the length of fencing available)

From this we can find the diameter of a circle with a 1000m circumference

Circumference = π x diameter

Diameter = circumference / πimage36.png

                  = 1000 / πimage37.png

                 = 318.3

Radius = ½ diameter

           = 159.15                                                        r = 159.15 m               c = 1000 m

Area = π x radius2

        = π x 25328.7

        = 79522.5 m2

Conclusion:

This proves my theory, as you can

see the circle has the largest area out

of all the shapes I have experimented with.  This is because a circle has an infinite number of sides and if the area size goes up as the number of sides goes up, an infinite number of sides will obviously have the largest area.

...read more.

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