# Fencing problem.

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Introduction

## Problem specification

A farmer has exactly 1000 metres of fencing. With this 1000 meter of fencing she wishes to fence off a plot of level land.

She is not concerned with the shape of the plot, but must have a perimeter of 1000 m. her requirements are that the fence off the plot of land should contain the maximumarea.

## Plan

At the beginning of this experiment I shall begin experimenting with the simplest of all shapes. Triangles shall be investigated first. These triangles shall include equilateral triangle, an isosceles triangle, a right-hand triangle and a scalene triangle. I would find the area and plot the results upon a table.

After investigation triangles I shall start with the regular shapes with four sides, a square, rectangles, a trapezium, a parallelogram and a rhombus. Once again areas of these shapes shall be found and recorded on a table. Now I shall continue experimenting different shapes by increasing the number of sides of each shape. These will include: triangles, quadrilaterals, a pentagon, a hexagon, a heptagon, and octagon, a nonagon and a decagon.

I would take the whole investigation one step further and experiment with polygons shapes such as letters. For example a polygon shaped as an H, a polygon shaped as an E, and so on. These areas shall also be found and recorded.

Finally I shall scrutinize a circle. The results of the area shall also be observed and noted down on a table.

After doing all the above stages I shall then devise a term that would allow be to calculate any regular shape. The term would be entered into a spreadsheet program such as Microsoft Excel. I will do this so to make sure that my term is correct and also to list the area of many more shapes.

Middle

Rectangle 4

331m × 169m × 331m × 169m

55939

Trapezium

200m × 200m × 200m × 400m

51963

Parallelogram

100m × 400m × 100m × 400m

34640

Rhombus

250m × 250m × 250m × 250 m

62500

I have made a graph that displays the data above:

By observations it is obvious that the rhombus and the square have the most area coverage whereas rectangle 1 has the least area coverage.

#### Polygons

In the three division of my coursework I shall be investigating polygons. I shall be firstly investigating regular polygons. The following list shows the regular polygons that shall be investigated during this segment of my coursework:

- Pentagon: this is a regular five sided polygon
- Hexagon: this is a regular six sided polygon
- Heptagon: this is a regular seven sided polygon
- Octagon: this is a regular eight sided polygon
- Nonagon: this is a regular nine sided polygon
- Decagon: this is a regular ten-sided polygon.

#### Regular Pentagon

I will begin the first part of the investigation of regular polygons by discovering the area of a regular five-sided figure that is known as a regular pentagon. I have shown the shape that I shall be investigating during this part of my coursework:

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000m ÷ 5

Length of each side = 200m

The angles of the triangle from the centre of the pentagon can be found by dividing 3600 by the number of sides of the shape that has been shown above:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 5

Exterior angles = 720

I shall now find the interior angles of the above shape. This can be found by excluding 1800 from the exterior angle.

Interior angle = 1800 - Exterior angle

Interior angle = 1800 - 72 = 1080

Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular pentagon. I shall half the interior angle to find this angle:

Angle ODC = 1080 ÷ 2 = 540

To find Angle DOC = 180 – (54 + 54)

Angle DOC = 720

I shall now appraise to find the area of one of the five triangles that join to make a regular pentagon. The formula to find the area of a triangle has been shown below:

Area of a triangle = ½ × Base × Height

As we know that the base is known whereas the height is unknown. I shall now find the height of the triangle shown below:

∆ DOF = 72 ÷ 2 = 360

I shall use trigonometry to find the height of the above triangle. The trigonometry function TAN shall be used to find the height, which can also be known as length of opposite.

TAN Ø = Opposite ÷ Adjacent

TAN 540 = Opposite ÷ 100

Opposite = TAN 540 × 100

Opposite = 137.6m = Height.

I shall now substitute the height into the formula below:

Area of a triangle = ½ × Base × Height

Area of a triangle = ½ × 200 × 137.6 = 13760m2

I have now found the area of one triangle. I shall multiply this figure by the number of triangles present within the pentagon:

Area of pentagon = Area of one triangle × Number of triangles

Area of pentagon = 13760 × 5 = 68800m2

### Hexagon

The second polygon that I shall be investigating is a regular hexagon. I shall use the same method as I did with the previous shape. Firstly I will divide the polygon into individual triangles, and then I shall find the area of each triangle and then multiply this value by the number of triangles.

The length of each side of the shape above can be found by the following formula:

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000 ÷ 6 = 166.7m

The angles of the triangle from the centre of the pentagon can be found by dividing 3600 by the number of sides of the shape that has been shown above:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 6

Exterior angles = 600

I shall now find the interior angles of the above shape. This can be found by excluding 1800 from the exterior angle.

Interior angle = 1800 - Exterior angle

Interior angle = 1800 - 60 = 1200

Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular pentagon. I shall half the interior angle to find this angle:

Angle OED = 1200 ÷ 2 = 600

To find Angle EOD = 180 – (60 + 60)

Angle EOD = 600

∆ EOG = 60 ÷ 2 = 300

I shall use trigonometry to find the height of the above triangle. The trigonometry function TAN shall be used to find the height, which can also be known as length of opposite.

TAN Ø = Opposite ÷ Adjacent

TAN 600 = Opposite ÷ 83.35m

Opposite = TAN 600 × 83.35

Opposite = 144.4m = Height.

I shall now substitute the height into the formula below:

Area of a triangle = ½ × Base × Height

Area of a triangle = ½ × 166.7 × 144.4 = 12035.7m2

I have now found the area of one triangle. I shall multiply this figure by the number of triangles present within the hexagon:

Area of hexagon = Area of one triangle × Number of triangles

Area of hexagon = 12035.7 × 6 = 72214.2m2

### Heptagon

The third polygon that I shall be investigating is a regular heptagon. I shall construct the method as I have done before. The polygon shall be divided into triangles form the centre point. I shall calculate the area of one triangle then multiply the figure that will be obtained by the number of triangles present within the hexagon. The following is the shape that I shall be investigating:

The length of each side of the shape above can be found by the following formula:

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000 ÷ 7 = 142.9m

The angles of the triangle from the centre of the heptagon can be found by dividing 3600 by the number of sides of the shape that has been shown above:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 7

Exterior angles = 51.40

I shall now find the interior angles of the above shape. This can be found by excluding 1800 from the exterior angle.

Interior angle = 1800 - Exterior angle

Interior angle = 1800 – 51.4 = 128.60

Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular heptagon. I shall half the interior angle to find this angle:

Angle OED = 128.570 ÷ 2 = 64.290

To find Angle EOD = 180 – (64.29 + 64.29)

Angle EOD = 51.40

∆ OEH = 51.42 ÷ 2 = 25.170

I shall use trigonometry to find the height of the above triangle. The trigonometry function TAN shall be used to find the height, which can also be known as length of opposite.

TAN Ø = Opposite ÷ Adjacent

TAN 64.30 = Opposite ÷ 71.5m

Opposite = TAN 64.30 × 71.5m

Opposite = 148.6m = Height.

I shall now substitute the height into the formula below:

Area of a triangle = ½ × Base × Height

Area of a triangle = ½ × 142.9 × 148.6 = 10617.5m2

I have now found the area of one triangle. I shall multiply this figure by the number of triangles present within the hexagon:

Area of hexagon = Area of one triangle × Number of triangles

Area of hexagon = 10617.5 × 7 = 74322.5m2

### Octagon

The fourth shape I shall be investigating is a nine-sided polygon also known as an octagon. I shall be taking the same steps into consideration as I have done previously. The following shape is an octagon and I shall be investigating the shape during this segment of the coursework:

AB + BC + CD + DE + EF + FG + GH + HA = 1000m

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000 ÷ 8 = 125m

The angles of the triangle from the centre of the octagon can be found by dividing 3600 by the number of sides of the shape that has been shown above:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 8

Exterior angles = 450

Interior angle = 1800 - Exterior angle

Interior angle = 1800 – 45 = 1350

Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular octagon. I shall half the interior angle to find this angle:

Angle OFE = 1350 ÷ 2 = 67.50

To find Angle FOE = 180 – (67.5 + 67.5)

Angle FOE = 450

∆ OFE = 45 ÷ 2 = 22.50

TAN Ø = Opposite ÷ Adjacent

TAN 67.50 = Opposite ÷ 62.5m

Opposite = TAN 67.50 × 62.5m

Opposite = 150.9m = Height.

I shall now substitute the height into the formula below:

Area of a triangle = ½ × Base × Height

Area of a triangle = ½ × 125 × 150.9 = 9431.3m2

I have now found the area of one triangle. I shall multiply this figure by the number of triangles present within the octagon:

Area of octagon = Area of one triangle × Number of triangles

Area of octagon = 9431.3 × 8 = 75450.4m2

### Nonagon

The fifth shape I shall be investigating is a nine-sided shape this is also be known as a nonagon. I shall be taking the same procedures that I took before. I shall divide the shape into triangles, which shall give me nine triangles. I shall be retrieve the area of one triangle then I shall multiply the value obtained from discovering one triangle from the shape by the number of triangles within the polygon to provide the area of the whole shape. I have shown the shape that I shall be investigating below:

AB + BC + CD + DE + EF + FG + GH + HI + IA = 1000m

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000 ÷ 9 = 111.1m

The angles of the triangle from the centre of the nonagon can be found by dividing 3600 by the number of sides of the shape that has been shown above:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 9

Exterior angles = 400

Interior angle = 1800 - Exterior angle

Interior angle = 1800 – 40 = 1400

Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular nonagon. I shall half the interior angle to find this angle:

Angle OFE = 1400 ÷ 2 = 700

To find Angle FOE = 180 – (70 + 70)

Angle FOE = 400

∆ FOJ = 40 ÷ 2 = 200

TAN Ø = Opposite ÷ Adjacent

TAN 700 = Opposite ÷ 55.6m

Opposite = TAN 700 × 55.6m

Opposite = 152.8m = Height.

I shall now substitute the height into the formula below:

Area of a triangle = ½ × Base × Height

Area of a triangle = ½ × 111.1 × 152.8 = 8488m2

I have now found the area of one triangle. 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. I shall divide the shape into triangles, which shall give me ten triangles. Now I shall find the area of one triangle and then the multiply the value obtained form the discovery of the area of one triangle by the total number of triangles present within the shape. I have shown the shape that I shall be investigating during this part of the coursework below:

AB + BC + CD + DE + EF + FG + GH + HI + IJ + IA = 1000m

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000 ÷ 10 = 100m

The angles of the triangle from the centre of the nonagon can be found by dividing 3600 by the number of sides of the shape that has been shown above:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 10

Exterior angles = 360

Interior angle = 1800 - Exterior angle

Interior angle = 1800 – 36 = 1440

Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular decagon. I shall half the interior angle to find this angle:

Angle OGF = 1440 ÷ 2 = 720

To find Angle GOF = 180 – (72 + 72)

Angle GOF = 360

∆ KOF = 36 ÷ 2 = 180

TAN Ø = Opposite ÷ Adjacent

TAN 720 = Opposite ÷ 50m

Opposite = TAN 720 × 50m

Opposite = 153.9m = Height.

I shall now substitute the height into the formula below:

Area of a triangle = ½ × Base × Height

Area of a triangle = ½ × 100 × 153.9 = 7695m2

I have now found the area of one triangle. I shall multiply this figure by the number of triangles present within the decagon:

Area of decagon = Area of one triangle × Number of triangles

Area of decagon = 7695 × 10 = 76950m2

### Results

I have now completed the third segment of my coursework. I have discovered the area of six different polygons. These polygons that have been discovered are all regular. The results have been shown below in a tabulated form:

Name of regular polygon | Area (m2) |

Pentagon | 68800 |

Hexagon | 72217.2 |

Heptagon | 74322.5 |

Octagon | 75450.4 |

Nonagon | 76392 |

Decagon | 76950 |

Conclusion

Area =106 ÷ 4N2 * Tan [180 – (360 ÷ N) ÷2] N

Area =106 ÷ 4 * 52 * Tan [180 – (360 ÷ 5) ÷2] 5

Area =106 ÷ 4 * 25 * Tan [180 – (360 ÷ 5) ÷2] 5

Area =106 ÷ 100 * Tan [180 – (360 ÷ 5) ÷2] 5

Area =1000 * Tan [180 – (360 ÷ 5) ÷2] 5

Area =1000 * Tan [(180 – 72) ÷2] 5

Area =1000 * Tan [(180 ÷ 2] 5

Area =1000 * Tan [54] 5

Area =1000 * Tan 54 * 5

Area =1000 * 1.37638192 * 5

Area =13764 * 5

Area =68820 m2

The answer that I had obtained was only 20m off therefore I could say that the formula was correct enough for such a large scale.

Conclusion

I now am able to conclude that as the sides of the shape increases each side becomes smaller, each exterior angle becomes smaller but the height of each sub triangle of a polygon becomes larger resulting in a large triangle area and thus a larger area of the entire polygon.

After investigating many shapes, I have noticed that the shape, which occupies the most area, would most definitely have to consist of as many sides as possible so a shape with infinite shaped sides such as the circle would occupy the most area. A circle with a circumference of 1000m occupies a total area of 79,579m2 and since a circle has infinite sides it would obviously have the most area.

I have shown a bar graph that shows the area that was obtained from all the shapes that were investigated below. The graph below shows that the circle has the most area. A shape with more sided will have a larger area and conclusively the shape with the most area under the restriction of thousand meter perimeter is the circle.

The investigation is now complete and the farmer should use a circle shape with her fencing in order to accumulate the most area occupancy.

Anjum Kohli GCSE Mathematics Coursework Fencing Problem

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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