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fencing problem

Extracts from this document...

Introduction

I have been given the fencing problem course work by my teacher. In this piece of course work I will be using ict (excel to help me draw graphs and word) and be drawing graphs to explain my statistics. This piece of course is about:

A farmer as exactly 1000 meter of fencing and wants 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 100m so it could be anything else with a perimeter (or circumference) of 1000m. She wishes to fence off the plot of land, which contains the maximum area.

The propose of this course is to work out all the maximum area for 1000 meter fence by drawing and calculating with 2D different shapes.

...read more.

Middle

33600

90

410

36900

100

400

40000

110

390

42900

120

380

45600

130

370

48100

140

360

50400

150

350

52500

160

340

54400

170

330

56100

180

320

57600

190

310

58900

200

300

60000

210

290

60900

220

280

61600

230

270

62100

240

260

62400

250

250

62500

image39.png

Now I will be drawing and calculating rectangles to find the maximum area of 1000meter

Shape 1                                                                              shape 2

       250m                                                                              240m       image12.png

image13.png

                         250m                                                                           2          260m

Area= length × width                                               Area= length × width

Area = 250×250                                                       Area = 240×260

= 62500m²                                                                 = 62400m²

Shape 3                                                                 shape 4                                                

     210m                                                                                  160mimage13.pngimage13.png

                               290m

                                                                                              340m

Area= length × width                                               Area= length × width

Area = 210×290                                         Area = 210×290

= 60900m²                                                                   = 54400m²

I have chosen 3 areas from the table above the rectangles .This table and the graph below  show the 3 rectangles and 1 square which I have draw area width of the rectangles. And I can see that the maximum area is a square.

shape

length

 width

area

1

160

340

54400

2

210

290

60900

3

240

260

62400

4

250

250

62500

This Graph will show the data above of area and width of the rectangles

image40.png

...read more.

Conclusion

aths/shape_space_and_measures/fencing_problem/819222/html/images/image15.png" style="width:24px;height:1.33px;margin-left:0px;margin-top:0px;" alt="image15.png" />

                                            2

                                       tan30= 83.3

                                                    H                                

image24.pngimage25.png

                                       H=83.3 = 144.27

                                            0.57735

        Area=166.6×144.27 = 12017.69

                    2

        12017.69×6= 72106.14

        = 72106.14

Heptagon

Each side equal to 142.9m

image41.png

Perimeter 1000 =142.9m

                    7

  360 = 51.42image28.pngimage16.pngimage27.pngimage26.png

    7

51.42 = 25.71

    2

Tan 25.71 = 100

                       Himage15.png

 h= 100 = 148.46

                                            0.48126         

                               Area= 142.9×148.46= 10607.46

                                                   2

                                      = 10607.46×7= 74252.22m²

Octagon

Each side equal to 125m

image31.pngimage32.pngimage29.pngimage30.png

                               Perimeter 1000 = 125m

                                                   8image16.png

image33.png

360 = 45

                                    8

45 = 22.5    

                                    2  

                                 Tan 25.7= 62.5        

                                          H

                               H= 32.5 = 150.8

                                     0.41421

                                Area= 125×150.8= 9625        

                                                2        

                                 =9625×8=75400m²

Nonagon

Each side equal to 111.1m

Perimeter 1000 = 111.1mimage42.png

                    9

360 = 40

         9

 40 = 20

           2

     Tan 20= 55.5

                     H                          

                                    H = 55.5 = 152.14

                                0.3639

                        Area= 111.1×152.14= 8451.37

                                                      2            

                                         8451.37×9= 76062.33m²

Decagon

Each side equal to 100m

   Perimeter =1000 = 100mimage16.pngimage43.png

                        10

360= 36                

                     10                            

36=18

                   2

                                              Tan 18= 50

                                                              H

                                            H= 50 = 1592.35

                            0.0314

           Area= 100×1592.35 = 79617.5

                 2

           = 79617.5×10= 796175m²

shapes

perimeter

 area (m² )

Pentagon

200

6881.5

Hexagon

166.6

72106.14

Heptagon

142.9

74252.22

Octagon

125

75400

Nonagon

111.1

76062.33

Decagon

100

796175

image44.png

image34.png

        C= 1000

                                           C=π*D

image35.png

                                          1000 m=3.14D

                                          D=1000= 318.31

π

        Radius= 318= 159

                                                          2

                                           Area=π×R²

                                           = 3.14×159×159

                                           = 79382

In conclusion in this piece of course I have achieved in my investigating on different shapes of perimeter of 1000m to which shape gives the maximum area, I found that the circle with a circumference of 1000m gives me the maximum area than other shapes.

krishan 10H

...read more.

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