# The fencing problem.

Extracts from this document...

Introduction

The Fencing Problem

## Introduction

This coursework is a problem involving a farmer. She has 1000 metres of fencing and is looking to create a fenced off area that is as large as possible. She wants to know what shape to arrange the fencing in order to achieve this.

I hope to find out which shape would be most appropriate for this use. I will investigate the different formulas and areas of different shapes and compare results to come to a final conclusion.

The Triangles

### The Equilateral Triangle

The equilateral triangle is the almost ‘the square’ of the triangles in that all of the sides have to be the same length, and the angles must all be equal. As with the square, there is only one equatorial triangle with the perimeter of 1000m.

There is more than one way to calculate the area of an equilateral triangle so I will demonstrate each of the three different ways.

- A = x x x sin 60

= 48112.52 (to 2d.p.)

- A = x - 2 x h

h = sin 60 - - 3

= 48112.52 (to 2d.p.)

- A = ab sin c

= 48112.52 (to 2d.p.)

### The Right angle Triangles

There are three ratios that a right angle triangle must fit. I worked each of the areas out and calculated the largest of the three areas.

A = ½ b x h

##### Conclusion

Middle

In a way, this is a similar scenario as with the kite and the rhombus. Although the formula is different, the largest version (in terms of the area) of the shape is exactly the same as the simplest, like the square, or in this case the equilateral triangle.

Conclusion

I have found that the triangle with the largest area is an equilateral (regular) triangle.

The Quadrilaterals

### The Square

The only square with a perimeter of 1000 meters has four sides each being 250 metres in length.

Following the formula area = length x width, the areaworks out to be 62500 m².

##### Conclusion

As there is only one square, there is only one area that can be obtained, it is 62500 m².

The Rectangle

The rectangle is much more complicated, as there are many combinations in length and with that amount to many different areas.

Area of Rectangle = Base X Height

250 X 250 = 62500m

This is the biggest area that an rectangle can give

The table on the right shows all the areas of rectangle

### The Trapezium

Trapeziums

The formula to find the area of a trapezium is:

A = x h

h =

Example

Calculating the average of the parallel sides

=

=

=200

Calculating the distance between them (height)

h = 350 -150

=200

=100

250sq - 100sq

= 52,500

= 229.13

Area of Trapezium

A = x h

A = 200 x 229.13

A = 45825.76

The Polygons

Conclusion

### General Formula for regular polygons

I looked at formula for internal and external angles of polygons, and the trigonometry for finding the vertical height of the triangles. I was able to come up with this formula.

Area = ½ x base x tanθ x ½ x base x n

I can prove this by looking at the pentagon.

½ x 200 x tan 54 x ½ x 200 x 5 = 68819 m²

My earlier answer was 68800 m² when using a manual method to a greater decimal place.

### The Circle

By rearranging two equations that I already know I can say:

c = Pi x d

d =

By halving the diameter you get the radius

r =

Once you have the radius you are now able to find the area.

The formulae to find the area is:

A = Pi x r sq.

Here are my workings to show the area of a circle with a circumference of 1000m.

d =

d = 318.31

r =

r = 159.15

r sq. = 25330.30

A = 25330.30 x Pi

A = 79577.47 m sq.

The Conclusion

I have found that in order for the farmer to obtain the maximum possible area with her 1000 meters of fencing, she should arrange it in a circle.

I have also found that there is a relationship between the area and the number of sides. The more sides a shape has, the larger area it will have.

Also, I found out that regular shapes have larger areas than irregular shapes, and there is a general formula for the area of a polygon.

Area = ½ x base x tanθ x ½ x base x n

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

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