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  • Level: GCSE
  • Subject: Maths
  • Word count: 1273

The fencing problem.

Extracts from this document...

Introduction

RANG BABATAHER11.1 – Maths Coursework

THE FENCING PROBLEM

Introduction

 In this investigation, I have to find out a farmers problem who needs to build a fence that is 1000m long. I am investigating which shape would give the maximum area with a 1000m perimeter. I will be investigating the properties of a 1000m perimeter fence.

I am going to start by drawing several regular and irregular rectangles, all with a perimeter of 1000m.

All drawings are not to scale.

m = Metres

4 Sided Shapes

Square

image00.png

image11.pngimage01.png

To find out an area of a square = base x length

In this case it will be                 = 250m x 250m

                                        = 62500m.

Rectangles

1.image26.pngimage20.png

image34.png

Area = 300 x 200 =image49.png

                                                = 60000m

2.image41.pngimage46.png

image47.png

Area = 350 x 150

    =52500m

3.

image48.pngimage02.png

image03.png

Area = 400 x 100

       =40,000m

4.

image04.pngimage05.png

image06.png

Area = 450 x 50

        =22,500m

In a rectangle, any two different length sides will add up to 500m, because each side has an opposite with the same length. That’s why when we look at the triangles above we can see its happening, for example if we look at the 1st which is  

200 x 300 when we add them up it will be 500.

...read more.

Middle

image15.png

I knew that the area of a triangle was image49.png

THE SQUARE ROOT OF: s(s-a)(s-b)(s-c)

Using this formula I could work out the area of an equilateral triangle shaped fence.

* = Multiplied

THE SQUARE ROOT OF (500*(500-333.333…)*(500-333.333…)*(500-333.333…)) = 48112.52245 m

I then tried this formula with an isosceles triangle.image17.pngimage17.pngimage18.png

image19.png

THE SQUARE ROOT OF (500*(500-400)*(500-400)*(500-200)) =  

38729.83346m

When I had worked out this formula I discovered that the two areas that had given the largest areas for 3 and 4 sided shapes had been regular shapes. I wondered if this meant that regular shapes gave the largest areas. I thought that it might have something to do with the lines of symmetry in a shape.

image49.png

5 Sided Shapes

Once I had worked out the largest areas of the squares and triangles I realised that a shape with the most lines of symmetry had the largest area. This meant that regular polygons would give the largest area, so from now on I will only look at regular shapes.image21.png

image19.pngimage19.png

image19.pngimage19.pngimage19.png

I then began to study pentagons. A pentagon is a five sided shape.

...read more.

Conclusion

This meant that my general formula was now 250000:  180

                                                   n               n

However, the formula for a circle involves multiplication, so I had to make my formula

250000 * n    

     n       180

I cAN now cancel out the two n’s, leaving me with the formula

250000 * 180

Then I realised that 250000 was my original r , as my radius was originally 500. This meant that my formula was now

r  * 180

My formula was now beginning to resemble the formula for a circle, but I still had to convert the 180 into π.

I realised that it must involve radians.

Radians

A radian is the angle subtended at the centre of the circle by an arc of length equal to the radius.

In radians 360° is equal to 2π. This means that 180° is equal to π, as 180 is half of 360 and π is half of 2π.

I now had my formula

r  * π

Or, in its familiar form

πr , which is the formula for a circle..

In my investigation I discovered that a circle would give the largest area for a fence of 1000m perimeter as it has an infinite number of sides.

I also discovered that the general formula for any n sided polygon is

250000:  tan (180)

n                   n

image42.pngimage43.png

...read more.

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