• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11

# 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.

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

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

Shape 1                                                                              shape 2

250m                                                                              240m

250m                                                                           2          260m

Area= length × width                                               Area= length × width

Area = 250×250                                                       Area = 240×260

= 62500m²                                                                 = 62400m²

Shape 3                                                                 shape 4

210m                                                                                  160m

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

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

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

Perimeter 1000 =142.9m

7

360 = 51.42

7

51.42 = 25.71

2

Tan 25.71 = 100

H

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

Perimeter 1000 = 125m

8

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 = 111.1m

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 = 100m

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

C= 1000

C=π*D

1000 m=3.14D

D=1000= 318.31

π

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

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Fencing Problem

1000 500 48112.23 334.5 332.75 332.75 1000 500 48111.64 335 332.5 332.5 1000 500 48110.71 After testing my fourth table of values I noticed the exact same pattern, and I also notice the area had increased quite fairly since I began narrowing down my results from my first table.

2. ## Fencing problem.

= the total perimeter of the scalene triangle � 2 Area of a scalene triangle (s) = 357 m + 349 m + 294 m � 2 Area of a scalene triangle (s) = 1000 � 2 Area of a scalene triangle (s)

1. ## The Fencing Problem

h 450 50 3) h 425 75 4) h 400 100 5) h 375 125 6) h 350 150 7) h 325 175 8) h 300 200 ^ The graph has a smooth curve,; however, unlike the scalene triangles, the isosceles triangles results aren't "symmetrical"; they ascend to a point, and then decline with greater magnitude.

2. ## Fencing Problem

This means that the angle in the middle for one of the triangles is 72�. Now that I have the angle for the triangle I have to work out the base length of the triangle.

1. ## The Fencing Problem

To work out the area, I will use the following formula: l = length of rectangle w = width of rectangle a = area of rectangle a = l * w The two examples I worked out are shown below.

2. ## Geography Investigation: Residential Areas

3 1 4 1 1 Cumberland Avenue 1 2 7 0 0 Vivaldi Close 10 0 0 0 0 The Beaches 10 0 0 0 0 For the above data I have created a percentage stacked bar chart, it is the easiest way to show this data.

1. ## Fencing Problem

Conclusion I have found that the three sided shape that had the biggest area overall when using 1000 meters of fencing, was an equilateral triangle with the measurement: Length=333.3m Sides=333.3m Height=289m and the area=48166m� Pentagons I am now going to look at a pentagon to find out its area when using 1000meters of fencing.

2. ## The fencing problem 5-6 pages

To find the Area we use the formula for a triangle Area=1 X base X height. We know that the base is 200m since it is the same as one side. Nevertheless, we need to find the height of the triangle to find the Area.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to