# The fencing problem.

Extracts from this document...

Introduction

## The fencing problemDaniel Briney

## The task

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.

## Solution

## 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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