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The Fencing Problem.

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Introduction

GCSE Mathematics

The Fencing Problem

The Problem

        A farmer has 1000 meters of fencing and wants to fence off a plot of level land. She is not troubled about the shape of the plot but it must have a perimeter of 1000 m. She needs to fence of a plot of land that contains the maximum area. I am going to investigate which shape will give the biggest area and show my working.

To start of I will use the simplest shapes and find the largest possible area of each shape. A simple shape is a shape, which has an easy formula to finding its area. For example a rectangle has the formula of length x width to find its area.

Rectangle/Square

I am going to start by investigating rectangles, that all have a perimeter of 1000 meters. Below are 2 rectangles showing how different shapes with the same perimeter can have different areas.

image00.png

image04.png

Length x Width

Perimeter = 1000m          

...read more.

Middle

=(1000-A3*2)

=A3*B3

Using this table I made a graph on excel of one width (A) against Area (AxB). This is on the next page.

The line of results on the graph has made a plane curve (parabola). From the table and the graph, we can see that the rectangle with a Length (A) of 250m has the greatest area. This shape is called a square and is the regular shape of rectangles.

Triangle

I then found the triangle with the largest area. I only used isosceles triangles because if I know the base I can work out the other 2 lengths because they are equal.image06.pngimage05.png

image02.pngimage01.png

image03.png

For example

If B = 200

1 Side = (1000 – 200) / 2 = 400

To work out the area I need to work out H (the height). To work out H (height) I can use Pythagoras’ Theorem. The formula and area a triangle with B (base) of 200m is shown below.

H² = h² - a²

H² = 400² - 100²

H² = 160000 – 10000

H² = 150000

H = 387.298

½ × 200 × 387.298 = 38729.

...read more.

Conclusion

Then using SOH CAH TOA (trigonometry) you can work out that you need to use Tangent.

H = 100 tan54 = 137.638

O = 100

T = tan 36

This gives me the length of H so I can work out the area.

Area = ½ x b x H = ½ x 100 x 137.638 = 6881.910

Now I have the area of half of one of the segments, all you need to do is multiply that number by 10 and get the area of the shape.

Area = 6881.910 × 10 = 68819.096m²

So from this I can be sure that the more sides a shape has the larger its area will be I can also be sure that the regular of each shape gives the largest area. Using the same method of working out as before I then worked out the area of a regular Hexagon and Heptagon.

I used the same method as before to work out the area of the 2 shapes.

Hexagon

1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.

360 ÷ 6 = 60 ÷ 2 = 30

Area = ½ * b * H = ½ * 83 1/3 * 144.338 = 6014.065

6014.065 * 12 = 72168.784m2

Heptagon

1000 ÷ 7 = 142.857 ÷ 2 = 71.429m

360 ÷ 7 = 51.429 ÷ 2 = 25.714 degrees

Area = ½ * b * H = ½ * 71.429 * 148.323 = 5297.260

5297.260 * 14 = 74161.644m2

...read more.

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