# The Fencing problem.

Extracts from this document...

Introduction

The Fencing problem A farmer has exactly 1000 metres of fencing to fence of a piece of land. She wishes to fence off a plot of land, which contains the maximum area. I will investigate which shape will provide the maximum amount of area, whilst still maintaining a perimeter of 1000m. I shall start by looking at the shape with the least number of sides, the triangle The equilateral triangle The equilateral triangle is a set of special triangles in which all of the sides have are the same length, and the angles are equal. The isosceles triangle These, as well as the aforementioned equilateral triangle are the only triangles I shall be investigating. This is due to scalene and other irregular triangles have more than one variable, and so it is very difficult to calculate the area for them, as there are infinite combinations. With the isosceles triangle, if I know the base length, then I can work out the other two lengths, because they will be equal. If the base is 300m long then I can subtract that from 1000 and divide it by 2. ...read more.

Middle

The area of a parallelogram is equal to the base x vertical height. The vertical height can be worked out using 'side length sin ?'. I realised that the maximum value of ? was 90, as the graph for sine peaks at this point. I then worked out that this produced the result 250. The largest possible outcome, using a parallelogram is therefore the same as a square. The Rhombus The rhombus, like all quadrilaterals must have fours sides, but unlike others, these must all be equal, although the angles can change. Therefore, as I have found that regular shapes produce the greatest area, there can only be one value for any side, 250, and so the same outcome is produced as that of a square, producing an area of 62500m2. The Trapezium The trapezium can be worked out in almost the same way the same way as the parallelogram using the 'vertical height = side sin 90' formula. Using the area formula 'area = 1/2 x (a + b) x h' I realised that, as in the parallelogram, the largest value of sin could be 90. ...read more.

Conclusion

Using this formula, I worked out the area to be; 68819.1 Whilst, previously I found it to be; 68820 (when using a manual method to a lesser decimal place) Using this formula, I also worked out the area of shapes with large numbers of sides. No. of sides Area (m ) 50 79472.724 100 79551.29 200 79570.926 500 79476.424 1000 79577.21 And its graph; I have now come thoroughly proved that as the number of sides increases, so will the area. The results seem to be reaching a limit though, as they are getting closer and closer to one point. This point, or limiting factor, is the case of the circle, as it has an infinite number of sides. Hence, it will have a larger area, due to there being shorter vertices in a circle than any other shape, hence there is more area enclosed within the fence. To work out the area of a circle, I have to use a different formula, as I cannot put the number "infinity" into my formula. Instead, I shall use the formulae, A= ?r2, and C= ?d; C= ?d 1000 = ?d d = 1000 / ? d=318.31 (5 s.f) Hence, I have proved that the most efficient shape which the farmer should use with her fencing is a circle of perimeter 1000m. Will Witt The Fencing problem ...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