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• Level: GCSE
• Subject: Maths
• Word count: 1263

# The fencing problem.

Extracts from this document...

Introduction

## The fencing problemDaniel Briney

A farmer has exactly 1000m of fencing; with it she wishes to fence off a level area of land.  She is not concerned about the shape of the plot but it must have perimeter of 1000m.

What she does wish to do is to fence off the plot of land which contains the maximun area.

Investigate the shape/s of the plot of land that have the maximum area.

## Firstly I will look at 3 common shapes.  These will be:

A regular triangle for this task will have the following area:

1/2 b x h

1000m / 3 – 333.33

333.33 / 2 = 166.66

333.33² - 166.66² = 83331.11

Square root of 83331.11 = 288.67

288.67 x 166.66 = 48112.52²

A regular square for this task will have the following area:

Each side = 250m

250m x 250m = 62500m²

A regular circle with a circumference of 1000m would give an area of:

Pi x 2 x r = circumference

Pi x 2 = circumference / r

Circumference / (Pi x 2) = r

Middle

1000

62400

230

270

1000

62100

220

280

1000

61600

210

290

1000

60900

200

300

1000

60000

190

310

1000

58900

180

320

1000

57600

170

330

1000

56100

160

340

1000

54400

150

350

1000

52500

140

360

1000

50400

130

370

1000

48100

120

380

1000

45600

110

390

1000

42900

100

400

1000

40000

90

410

1000

36900

80

420

1000

33600

70

430

1000

30100

60

440

1000

26400

50

450

1000

22500

40

460

1000

18400

30

470

1000

14100

20

480

1000

9600

10

490

1000

4900

0

500

1000

0

This table shows the results of every rectangle possible for the 1000m perimeter with a 10m gap between each reading.

The results increase and then decrease symmetrically proving that the SQUARE 250 x 250 is the rectangle with the highest area because it is in the middle.  This would produce a curve on a graph with the square at the top and in the middle.

A square is a regular shape.

The graph below is of the table on the previous page.

The graph has the curve that I explained on the previous page.

Now I

Conclusion

1 = Base: 100m

1 = Top angle: 36ْ

1 = Bottom angle: 54ْ

To work out this are we will use TAN because we need the opposite and we have the adjacent.

TAN 54 = opposite / 100

TAN 54 x 100 = opposite

TAN 54 = 1.3764

Height of triangle = 137.64m

Area of triangle = 137.64m x 100 = 13763.8

Area of pentagon = 13763.8 x 5 = 68819.10m²

This regular pentagon is as expected, the area is more than the triangle and the square but less than the circle.  This once again proving that the more sides a regular shape has the more area there is.

To work out the area for a regular shape I did the following:

B = Base

N = Number of sides

P = Perimeter

H = height

Firstly I worked out the formula for the area of a triangle.  This is = b x h / 2

Then I worked out the area for any shape = b x h x n / 2

This shows that the formula to work out the area of any regular shape is =

1/2  x P x P                 .

n x 2 tan(180/n)

This simplified is A = P²        .

4n Tan(180/n)

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

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# Related GCSE Fencing Problem essays

1. ## Fencing Problem

triangle I can multiply it by the amount of sides in the pentagon (5)to give me the total area of the pentagon. * 13763.8192 x 5 = 68819.096 m� An observation I made while figuring out the area of the entire pentagon was that I could of multiplied the area

2. ## Fencing problem.

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Angle (�) Height [h] (m) Perimeter (m) Area (m�) 100 400 10 69.46 1000 6945.93 100 400 20 136.81 1000 13680.81 100 400 30 200.00 1000 20000.00 100 400 40 257.12 1000 25711.50 100 400 50 306.42 1000 30641.78 100 400 60 346.41 1000 34641.02 100 400 70 375.88 1000 37587.70 100 400 80 393.92 1000 39392.31 100 400

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2. ## Fencing Problem

170 34527.16 333.44 48112.515 190 37401.54 334 48112.233 210 39982.81 335 48110.712 230 42253.70 336 48107.879 250 44194.17 337 48103.725 270 45780.73 350 47925.724 Observation From the results of the table, the area increases as the base increase. However, there reaches a point where the areas stops increasing and in fact starts to decrease.

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