# Fencing problem.

Extracts from this document...

Introduction

Jonathan Rusbridge 11.2

Maths Coursework

All shapes that I will investigate can be divided into triangles. Shapes will all equal sides will all have the same number of , (equally sized) triangles as their sides. Shapes with different length sides will still be divided into the same number of triangles as sides, however these triangles will not all be the same size. I will measure the area of the overall shapes by using trigonometry to work out the area of one of the triangles and then multiplying that by the number of triangles there are in the shape.

I will begin with looking at Triangles.

This is an equilateral Triangle, (all sides and angles are equal).

To find the area of this shape I will make it into two equal right angled

Triangles, and then use Pythagoras’s theorem to find the height.

Square on the hypotenuse=sum of the squares on the other two sides.

(333.32=)111088.89 – (166.62=) 27755.56 = 83333.33

√83333.33=288.67 height= 288.67 area = 166.6x288.67=48092m2

This is a right angled triangle

The sides are only approximate. Area of this shape is (I will use

290m 410 Pythagoras’s theorem quite a lot so I will no longer say when I am)

290/2=145 area = 145x290=42050

290m

c

This in an isosceles triangle. Angles a and b = 75o

Middle

52500

300

200

60000

250

250

62500

200

300

60000

150

350

52500

100

400

40000

50

450

22500

Looking at the results I can see that the square has the largest surface area out of all the rectangles.

An interesting pattern that I noticed in these results, is that, not only do the surface areas increase as the sides become more equal, but as the surface area increases it does so in ever smaller increments. The first increment is 17500 while the last is only 2500. The difference between the increments is constant, the surface area increases by 5000 less than the previous area did. As soon as the length goes beyond 250m the difference between surface area’s increase by 5000 each time. This suggests to me that a graph of these results would be similar to if not actually a parabola. To check this I will draw the graph

(it is stuck on the back of this sheet). The graph is indeed similar to a parabola (however I cannot prove if it is one or not) it also shows that the square does have the highest possible surface area out of the different rectangles.

Rectangles are only regular quadrilaterals therefore I will now investigate irregular quadrilaterals:

e

both sides marked e are 225m, side h is 300m and side w is 250m.

Conclusion

At this point I have decided to work out my formulae (it took a long time and a LOT of scribbles so I have done it one rough paper).

Where: “n” is the number of sides.

“c” is the circumference (100m)

n c c / tan 360

2n 2n 2n m2

c

n c 2n

2n tan360

2n

Using this formulae I can find the area of any shape that has all sides equal length. Here are a few

## Number of sides | Area (m2) |

6 | 73040 |

7 | 74340 |

8 | 75483 |

9 | 76309 |

10 | 77160 |

As you can see as the number of sides on the shapes increase so does the area. This seems to suggest that the shape that has the most sides will have the highest area. The shape that has the most sides is a circle, because a circle has an infinite number of sides.

Using different formulas I can find the area of a circle with a circumference of 1000m.

Circumference is 1000m.

Diameter is 1000

╥

Area is ╥ D2

Area is 158845m2

Therefore the farmers 1000m of fence would be best put to use in fencing off a circle.

By

Jonathan Rusbridge

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

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