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  • Level: GCSE
  • Subject: Maths
  • Word count: 1815

The problem for this investigation is as follows: Farmer Jones has 1000 metres of fencing. What is the maximum area she can enclose?

Extracts from this document...

Introduction

Omar KadirMaths Coursework  : Farmer Jones’ Fencing Problem26/02/03

Problem:        The problem for this investigation is as follows:

Farmer Jones has 1000 metres of fencing. What is the maximum area she can enclose?

Mathematically, the problem can be interpreted to:

Which shape, with a perimeter of 1000 metres, has the largest area?

In order to go about answering the question, different types of shapes must be considered.

The original question of whether the shape will be a triangle, rectangle, polygon, or circle can be broken down into smaller questions. What will the lengths of the shapes be? If it is a triangle, will it be an Isosceles, an equilateral, or a Scalene? If it is a rectangle, what will the values of the width and length be? And if it is a polygon, will it be a regular or irregular shape?

I will take each shape separately, and investigate the different areas found within that particular shape. I will start with three-sided shapes and go on to four, five and six sided shapes, finishing with the circle.

Prediction:I predict that a circle with perimeter 1000metres will have the largest area.

I predict this because I also think the triangle will have a smaller area than the square (rectangle), which will be smaller than the pentagon, which will be smaller than the hexagon, and so on, until the final shape, the circle is reached.

...read more.

Middle

260

100.000

24,000.00

image11.png490

255

70.711

17,324.20

image12.pngimage13.png

image14.pngimage15.pngThe length of the base cannot go up to 500m,                 

because that would leave 250m for the other

two sides , therefore making it an ‘impossible triangle’,

image16.pngimage17.pngas the two con

The results from the spreadsheet can be shown on a graph:

image18.pngimage19.pngimage20.pngimage21.png

The formula to work out the area of any Isosceles triangle, by simply knowing the length of the base(x), can then be:

area = ½ base x height

     a = ½    x     (  c - b )

              a = ½    x        1000-x  -  x

                             2          2

Equilateral Triangles:

There can only be one equilateral triangle with a perimeter of 1000 metres. Each one of the sides will have a value of 1000m divided by three (333.3m).

To find the area, we can use the same equation:

                a = ½ b h

                   = ½ 333.3 x (  c - b )

                   = ½ 333.3 x 288.675

   = 48,112.50 m2

We can now see that the triangle with the largest

area is the Equilateral trianngle, and not the Isoceles.

Scalene Triangles:

After looking into Isosceles triangles, and the equilateral triangle, I do not see it necessary to carry on any further with scalene triangles.

I noticed that as the sides of the triangle became closer in length, the area increased, and therefore it makes sense that the triangle with equal sides has the maximum area.

Investigating Rectangles

To find out the areas of the rectangles with a perimeter of 1000m, I used a spreadsheet again. By filling in the values for the width, the length and the area of each triangle can be worked out.

image22.pngWe know that the perimeter of the rectangle will be 1000m, and that one length plus one width will be 500m, all the time.

image24.pngTo find the length, we minus the width value from 500m. The formula used in the spreadsheet is:

=500-A2

image25.pngimage26.pngimage26.pngFrom that, the area can easily be worked out, using a =wl, and on the spreadsheet using:

=B2*A2

Width(m)

Length(m)

Area(m )

20

480

9600

40

460

18400

60

440

26400

image27.png80

420

33600

100

400

40000

image28.png120

380

45600

140

360

50400

160

340

54400

180

320

57600

200

300

60000

220

280

61600

240

260

62400

image09.png250

250

62500

260

240

62400

280

220

61600

image30.png300

200

60000

320

180

57600

340

160

54400

360

140

50400

image31.png380

120

45600

400

100

40000

420

80

33600

440

60

26400

460

40

18400

480

20

9600

...read more.

Conclusion

theta, so we use TAN to find the adjacent.

image47.png

image48.pngimage50.png

Area of triangle = ½ b x h                                                         

                      a = ½ x 100 x     100                After finding the area of one triangle, I must

TAN 36o                multiply that figure by the number of triangles

a = 6,881.91 m2                found in the pentagon.

Area of Pentagon = no. of triangles  x  area of 1 triangle

                      a = 10  x  6881.91

a = 68,819.1 m2

Investigating Hexagons

image51.pngimage52.pngI will use the same method for finding the area of a regular hexagon, as I did for finding the area of a regular pentagon.

image53.png

Area of triangle = ½ b x h

   a = ½ x 83.33 x   83.33

          TAN 30o

   a = 6,013.58 m2image54.png

Area of Hexagon = no. of triangles  x  area of 1 triangle

                                                                           a = 12  x  6013.58

a = 72,163.01 m2

Investigating Octagons

image55.pngAgain, the same method is used.

image52.png

image56.pngimage58.png

Area of triangle = ½ b x h

   a = ½ x 62.5 x     62.5

image59.png       TAN 22.5o

   a = 4715.26 m2

image60.pngArea of Octagon = no. of triangles  x  area of 1 triangle

                                                                           a = 16  x  4715.26

a = 75,444.17 m2

Throughout investigating these polygons, I have been using the same formula over and over again. I have noticed that within the formula, the one thing that kept changing was the number of triangles in the shape ( t ). From that, I can form this equation to work out the area of any polygon:

a        =        t  ( ½ b  xb/2    )

TAN 0

Investigating Circles

image61.pngThere can only be one circle with a perimeter of 1000 metres. To find the area of this circle, we first need to know the radius.

image62.pngRadius = 1000

image63.png                2

r = 159.15

image64.png

Area =    r2

a = 3.142 x 159.152

a = 79,572.53

Conclusion

My prediction is correct, and it has been proven. There is a definite relationship between the number of sides and the area, as this graph shows:

image65.png

...read more.

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