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• Level: GCSE
• Subject: Maths
• Word count: 2092

Fencing problem

Extracts from this document...

Introduction

Syllabus 1387/1388

Higher tier

Abdulbasit Asif

Edexcel 2003 Coursework Maths GCSE

A farmer has exactly 1000 metres 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 1000m. So it could be:

The Investigation

Firstly I will begin with triangles because they are the simplest polygon.

To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras' theorem. Below is the formula and area when using a base of 450m. I will not be using scalene triangles, since there are countless combinations that can be used.

H- height, h- hypotenuse, b- half of base

H² = h² - b²

H² = 275² - 225²

H² = 25 000

H = 158.113 883

½ × 200 × 158.113 883 = 35 575.623 68

Now I will try with different sized bases and see what areas I get.

The area for this last triangle is lower than the area of the one before, so the triangle with maximum area must have a base between 300 and 350m. I will investigate an equilateral triangle because it is in the range we are looking for.

The equilateral triangle has a larger area than all the other triangles I have found. I will now find areas of triangle around 333.

Middle

230

62100

260

240

62400

250

250

62500

240

260

62400

230

270

62100

220

280

61600

210

290

60900

200

300

60000

190

310

58900

180

320

57600

170

330

56100

160

340

54400

150

350

52500

140

360

50400

130

370

48100

120

380

45600

110

390

42900

100

400

40000

90

410

36900

80

420

33600

70

430

30100

60

440

26400

50

450

22500

40

460

18400

30

470

14100

20

480

9600

10

490

4900

Using the data from this spreadsheet I can create a graph of base length of a quadrilateral against area.

As you can see the graph has formed a curve with a peak. According to the spreadsheet and the graph, the rectangle with a base of 250m has the greatest area. This shape is a square, which is a regular quadrilateral. I only measured to the nearest 10m so now I will use rectangles with bases around 250m to prove that the square has the largest area.

 Base (m) Height (m) Area (m²) 249 251 62499 249.5 250.5 62499.75 249.75 250.25 62499.9375 250 250 62500 250.25 249.75 62499.9375 250.5 250.5 62499.75 251 249 62499

All of these results fit in with the graph I have made, therefore making it reliable.

This has proved once again that regular shapes give you the largest areas.

Since these last two shapes had the largest area when they were in their regular form, I will now be using only regular shapes from now on. This makes it easier since there are many different variables for each polygon.

The next shape I will be using is a polygon.

In a pentagon there are five sides, allowing me to split the shape into 5 segments like so.

Each

Conclusion

1000 ÷ 10000 = 0.1 ÷ 2 = 0.05

360 ÷ 10000 = 0.036 ÷ 2 0.018

0.05 ÷ tan 0.018 = 159.1549379

½ × 0.05 × 159.155 = 3.978873446

3.979 × 20000 = 79577.46893

100 000 sided shape

1000 ÷ 100000 = 0.01 ÷ 2 = 0.005

360 ÷ 100000 = 0.036 ÷ 2 = 0.0018

0.05 ÷ tan 0.018 = 159.154943

½ × 0.005 × 159.155 = 0.397887357

0.398 × 200000 = 79577.47152

Here is a graph to show the number of sides against area. As you can see from the graph, the line straightens as the number of sides increases.

The last shape I will be using is a circle. A circle has an infinite amount of sides, so my formula cannot be used to find the area of the circle.

The formula for the circumference of a circle is:

Circumference =  × diameter

We know the circumference so:

1000 =  ×diameter

Now we have to rearrange the formula so we can use it fo find out the diameter. We need the diameter because the diameter ÷ 2 = radius. And with the radius we can work out the area. So now I shall rearrange the formula.

1000 ÷  = 318.3098862

Now to find the radius we divide the circumference by two.

318.310 ÷ 2 = 159.1549431

To work out the area we use the formula

• r² = area

so  × 159.155² = 79577.47155

Conclusion

From my investigation I discovered that as the number of sides on a shape increases, so does the area.

I used the formula to work out the area of the shapes:

1000 ÷ n ÷ 2 = b

360 ÷ n ÷ 2 = tan a

H = b ÷ tan a

Area = b × H

Area × 2n = total area of shape

This worked for all my shapes apart from the circle. The circle gave the largest area because it has infinite sides.

In my investigation I also saw how that a regular shape gives a larger area too.

The area that the circle gave was 79577.472m²

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

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