# The Fencing Problem

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Introduction

Nick Murphy The Fencing Problem Mathematics GCSE In this piece of coursework I will be addressing the Fencing Problem. This is: A farmer has exactly 1000 meters 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 1000m. What she does wish to do is fence off the plot of land, which contains the maximum area. Quadrilaterals I will start by looking at quadrilaterals. 50m 50 x 450 = 22500m� 45Om 400m 100m 100 x 400 = 40000m� 350m 150m 150 x 350 = 52500m� 200m 300m 200 x 300 = 60000m� 250m 250m 250 x 250 = 62500m� 250m As you can see I have drawn a large specturm of rectangles. I have noticed that the closer the size of the two measurements, the larger the area. Also, the closer the sizes are, the more like a square the shape becomes. In a rectangle, any two different length sides will add up to 500, because each side has an opposite with the same length. For example, in a 150 x 350 rectangle there are two sides opposite each other that are 150m long and two sides next to them that are opposite each other that are 350m long. ...read more.

Middle

+ h� 325m 325m 105625 - 30625 = 75000 h� = 75000 h = 273.9 350m 175m 1/2 base x height = 175 x 273.9 = 47932.5m� Hyp.� = h� + b� 333.3� = 166.7� + h� 111088.89 = 27788.89 + h� 111088.89 - 27788.89 = 83300 h� = 83300 h = 288.6 333.3m 333.3m 333.3m 333.3m 333.3m 166.7 1/2 base x height = 166.7 x 288.6 = 48109.6m� I have drawn here triangles with bases ranging from 50m to 350m going up in 50's. I have also drawn an equilateral triangle, as it is a regular triangle, and last time it was a regular quadrilateral that had the biggest area and it is the same in this case. Here is a base against area graph: As we can see the regular triangle has the largest area. Pentagon There are five sides to a pentagon, and it can be divided into five separate segments. The segments are isosceles triangles. We know the top angle is 72� by dividing 360 � 5, and therefore find the other two angles by 180 - 72 � 2 as the angles are equal. This equals 54� and because we can split an isosceles triangle into two right-angled triangles, I can use trigonometry to find the area. ...read more.

Conclusion

tan (90 - (180/n) = h 2 (1000/n) x h = Area of one segment 2 Area x n = Area of whole shape To prove this, here are some examples: Pentagon:- (1000/5) tan (90 - (180/5) = 138 2 (1000/5) x 138 = 13800 2 13800 x 5 = 69000m� Hexagon: - (1000/6) tan (90 - (180/6) = 144.3 2 (1000/6) x 144.3 = 12025 2 12025 x 6 = 72150m� Now that I have this equation, I can work out the area for an octagon and a nonagon. Octagon: (1000/8) tan (90 - (180/8) = 150.8 2 (1000/8) x 150.8 = 9425 2 9425 x 8 = 75400m� Nonagon: (1000/9) tan (90 - (180/9) = 152.6 2 (1000/9) x 152.6 = 8477.7 2 8477.7 x 9 = 76299.3m� Circle We have seen that as that as the number of sides increases, so does the area. Now I will find the area for a circle. Because a circle has an infinite number of sides, you cannot use any previous formula. To work out the area of a circle: 1000 = circumference pi 1000 = 318.3 � 2 = radius pi 318.3 � 2 = 159.15 pi x r� = Area pi x 159.15� = 79572.5 A = 79572.5m � In conclusion Here is a table and graph showing all my results for regular shapes: Therefore the circle has the greatest area with a 1000m circumference. ...read more.

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