# The Fencing Problem

Extracts from this document...

Introduction

The Fencing Problem The problem: A farmer has exactly 1000 metres of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000 metres. What she does wish to do is fence of f the plot of land, which contains the maximum area. The investigation: To solve this problem I will investigate two things, different shapes and the same shape but with different figures for its sides. I have decided to begin my investigation by looking at a four sided shapes, rectangles. Rectangles: A rectangle is a shape with four sides and four right angles. It has two parallel vertical sides and two parallel horizontal sides. I will construct four different rectangles. These shapes will all have a perimeter of 1000 metres. Square: As I have only 1000m of fence to use for its perimeter each side will need to be 250m in length. This is because a square has four sides equal in length. Perimeter = 1000m 1000m ( 4 = 250m The area of a square is found using the equation Base ( Height. Base = 250m Height = 250m 250 ( 250 = 62,500m� Rectangle 1: The height and base of a rectangle is of different height but each set of parallel lines is the same length. For rectangles there are many combinations I could make but I will start by looking at a rectangle with 300m and 200m sides. Perimeter = 1000m 2 ( 300 = 600m 2 ( 200 = 400m 600m + 400m = 1000m I will now find the area of the rectangle using these figures. To find the area of a rectangle the same formula is used as with the square, Base ( Height. Base = 200m Height = 300m 200 ( 300 = 60,000m� Rectangle 2: This rectangle will have a base of 100m and a height of. ...read more.

Middle

Opp = Sin ( Hyp Opp = Sin 48.59 ( 444.444 = 166.665m This means that I now have the necessary data to find the area of the original triangle. Base = 444.444m Height = 166.665m I will now find the area of the triangle using the following formula, half base multiplied by height. .5 ( 444.444 ( 166.665 = 37,036m� I have compiled a table using the data that I have found for the triangles to show the relation between area, base and height. Shape Height (metres) Base (metres) Area (Metres squared) Triangle 1 288.675 m 333.333m 48,112.5m� Triangle 2 166.666 m 333.333m 27,777.777m� Triangle 3 166.665 m 444.444 m 37,036m� This table shows again that the closer the height length is to the base height the larger the area. It shows that wider the gap between the height and base in length then the smaller its area is. It shows the same trend as that of the quadrilateral table, that is that a regular-sided shape has the largest area. Hexagons: Hexagons are six sided shapes. I will look at three hexagons to find which has the highest area. Hexagon 1: This first hexagon will be a regular hexagon, have all sides and angles the same. Perimeter = 1000m 1000 ( 6 = 166.666m I will now find the area of the hexagon. To find the area of the hexagon I will split it into triangles. To find the area of these triangles the equation is half Base ( Height. As the hexagon is regular all of the triangles will be equilateral and will be identical this means that I only need to find one which can be multiplied to find the rest of the triangles. It also means that all of the triangle sides equal the same as the base, 166.66m. However to find the height of the triangles it is necessary to split them in half to find their height. ...read more.

Conclusion

As my results show that the area of a regular polygon increases as the number of sides increases, I will investigate the maximum number of sides. I know that a circle has an infinite number of sides so I believe that it will have the largest area. I will now investigate the area of a circle. Circle: The area of a circle is found using the equation pi multiplied by the radius squared equals area. Area = ?r� However the only information that I have for the circle is its circumference which is the 1000m perimeter. To find the radius of the circle I will rearrange the formula circumference = 2?r. Circumference = 2?r Radius = C ( 2? I will now find the radius. 1000 ( 2? = 159.154m Radius = 159.154m With the information that I have just found I can now find the area of the circle. ? ( 159.154� Area = 79,577.471m� This area is the largest area that I have found out of all the shapes that I have investigated. It proves my earlier statement that a circle has a larger area. From my investigation I conclude that any regular polygon has a larger area than an irregular polygon with the same number of sides. I have found that the shape with the largest area is a circle. The table below illustrates this fact. Shape Number of Sides Area (m�) Equilateral Triangle 3 48,112.5 Square 4 62,500 Regular Pentagon 5 68,819.096 Regular Hexagon 6 72,166.662 Regular Heptagon 7 74,161.478 Regular Octagon 8 75,444.173 Regular Nonagon 9 76,318.817 Regular Decagon 10 76,942.088 Regular 15 sided polygon 15 78,410.501 Circle n 79,577.471 The graph below shows that the largest area is found when the 1000m perimeter forms a circle. It also shows that the increase in area lessens each time the regular polygon has an increased number of sides. For example from 3 sides to 4 sides the area increases by 14,387.5 whereas between 9 and 10 sides the area increases by 623.271. Craig Banfield - 1 of 33 ...read more.

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month