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• Level: GCSE
• Subject: Maths
• Word count: 3047

# The fencing problem.

Extracts from this document...

Introduction

INTRODUCTION

In this coursework, I have being given a problem on what type of shape of farm a farmer should use for her farm. Here is a briefing of the problem, the farmer has exactly 1000 metres of fencing and wants to fence off a plot of land. So she is looking for a shape with a perimeter (or circumference) of 1000 metres and she wishes to fence off the plot of land, which contains the maximum area.

In other to these, I will be looking at the following shapes; Quadrilaterals, triangles, regular shapes (polygons) and circles. I will be working on finding the areas of these different shapes while changing the sides (measurement) progressively. I will then present the results in graphs, tables and also some help from the “generated formula”.

I will draw a table of results showing the areas of the following shapes I’ve done, and look at the shape with largest area and mark it down, I’ll also draw a graph showing the results in a more logical order. About this logical order (systematic frequency), I’ve also produced all my coursework in a systematic order, to produce a professional and neat work. That is; the splitting of the shapes into triangles to allow neat calculations, the tables and the graphs, the use of other helpful formulas.

In this section, I will be looking at 4-sided shapes (Quadrilaterals).

Middle

418.3

150

31,372.5

387.3

200

38,730

353.6

250

44,200

316.2

300

47,430

273.9

350

47,932.5

After careful calculation I have discovered that as the base increases and the height decreases, the area of an isosceles triangle increases. The graph above shows this. To do the calculation I used the formula ½ x base x height (Area of a triangle). To carry out this formula I was required to find the height, using “Pythagoras theorem” and the remainder two sides, how I did this is shown in the calculations. The answers presented are rounded up to 1 decimal point but the actual figures were used in the calculation. Also, the results show that to have the greatest area the base must be a length between 300 and 350. And in this graph it shows that the base with the length of 350 has the largest area.

THREE REGULAR SHAPES (POLYGONS)

In this section, I will be looking at three regular shapes (polygons). The shapes I will be looking at are; the regular pentagon, hexagon, and octagon. I will be finding out how they might be the apposite shape to solve the problem. I will ensue this by finding their areas, in other to this; I would have to break down the shapes into “Logical Order” (systematic reference) by breaking down into triangles, finding the area of the triangles and then carrying on from there.

Pentagon

32

200        200

360 ÷ 50 = 72

==           ==             H

A

200

200

200        100

200

In this pentagon, I have split it into five triangles in other to find the area. I will then do the same thing I did for the triangles above; find their height in other to find their area.

Therefore, I’ll be finding the height of the right-angled triangle; unfortunately, I have got just one side of the right-angled triangle that is known. So I cannot find height using “Pythagoras theorem”, so I would be using trigonometry (SOH-CAH-TOA).

I am going to use TOA for the triangle because it is the appropriate formula to use now.

TOA = Tangent =   Opposite

Tan 36 =   100

A

a =       100

Tan 36

a = 137.64

So, the height of the triangle = 137.6cm²

Therefore, to find out the Area of the triangle,

½ BH

½ x Base x Height

½ x 200 x 137.6

½ x 27,520

Area of triangle = 13,760m²

I will now find out the area of the pentagon. Since I divided the pentagon into five triangles; I’ll now multiply the area of the triangle by five in other to get the area of the pentagon.

Therefore;

Area of pentagon = 13,760 x 5

= 68,800m²

Hexagon _

166.7                                   360 ÷ 6 = 60

166.7

166.7

166.7                 166.7

166.7

166.7

A        H

83.35

In this hexagon, I have split it into six triangles in other to find the area. I will then do the same thing I did for the triangles above; find their height in other to find their area.

Therefore, I’ll be finding the height of the right-angled triangle; unfortunately, I have got just one side of the right-angled triangle that is known. So I cannot find the height using “Pythagoras theorem”, so I would be using trigonometry (SOH-CAH-TOA).

I am going to use TOA for the triangle because it is the appropriate formula to use now.

TOA = Tangent =   Opposite

Tan 30 =   83.35

A

A =  83.35

Tan 30

A = 173.19

So, the height of the triangle = 173.2cm²

Therefore, to find out the Area of the triangle,

½ BH

½ x Base x Height

½ x 166.7 x 173.2

½ x 22,872.4

Area of triangle = 22,872.4m²

I will now find out the area of the hexagon. Since I divided the hexagon into six triangles;

I’ll now multiply the area of the triangle by six in other to get the area of the hexagon.

Therefore;

Area of hexagon = 22,872.4 x 6

= 137,234.4m²

OCTAGON

125        360 ÷ 8 = 45

125        125

125                                                                       125

125        125

125        125

H

A

62.5

In the octagon above, I have split it into eight triangles in other to find the area. I will then do the same thing I did for the triangles above; find their height in other to find their area.

Therefore, I’ll be finding the height of the right-angled triangle; unfortunately, I have got just one side of the right-angled triangle that is known. So I cannot find the height using “Pythagoras theorem”, so I would be using trigonometry (SOH-CAH-TOA).

I am going to use TOA for the triangle because it is the appropriate formula to use now.

TOA = Tangent =   Opposite

Tan 22.5 =   62.5

A

A =    62.5

Tan 22.5

A = 150.89

So, the height of the triangle = 150.9cm²

Therefore, to find out the Area of the triangle,

½ BH

½ x Base x Height

½ x 125 x 150.9

½ x 18,862.5

Area of triangle = 18,862.5m²

I will now find out the area of the octagon. Since I divided the octagon into eight triangles; I’ll now multiply the area of the triangle by eight in other to get the area of the octagon.

Therefore;

Area of octagon = 18,862.5 x 8

= 150,900m²

 Number of sides Area 5 68,800 6 137,234.4 8 150,900

Conclusion

Therefore;

Formula for circumference of a circle = 2πr

= 2 x π x radius

Thus;

= 1000m =2πr

= 1000m = r

2π

r = 1000

2π

r =    1000          (π = 22/7 == 3.142)

2 x 3.142

r =  1000

6.284

Therefore;

To find the area of the circle;

= πr²

The above is the formula for finding the area of the circle.

Area of circle = πr²

A = π159.13²

= π x 159.13²

= π x 25,322.4

= 3.142 x 25,322.4

Area of circle = 79,562.10m²

 Number of Sides Area 3 47,932.5 4 62,000 8 150,000 99 79,550.8 Circle 79,562.10

Surprisingly, according to the graph above and all the other shapes I’ve studied, it seems that the Octagon is the shape with the largest area, according to statistics, the circle should have the largest area, but according to my own investigation, the octagon has the largest area.

CONCLUSION

In conclusion, the results in my graph show that an 8- sided shape or an octagon has the largest area out of all the shapes studied. This shows that the farmer will fence off the plot in the shape of an octagon of 1000 metres to fence off the plot of the land which contains the maximum area. The illustrations and calculations above show the different sided shapes that could have been a possibility. These include quadrilaterals, triangles, regular shapes and the circle. But the octagon shows us that out of all the shapes I have mentioned the octagon still has the largest area.

The reason for my result could still be the reason that the octagon is almost close to a circle, so the shape with largest area could therefore be octagon.

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

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