• 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
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  • Level: GCSE
  • Subject: Maths
  • Word count: 1857

The fencing problem.

Extracts from this document...

Introduction

Maths Coursework

image00.png

 Yr 10

Maths Coursework

image73.png

image01.png

By Anonymous


Table Of Contents

Introduction

The fencing problem

Investigating the relationship between the length of sides and the area

Investigating Rectangles

Investigating Triangles

Investigating the relationship between the number of sides and the area

Pentagons

Hexagons

Heptagons

Octagons

Nonagons

Decagons

Results Table

Conclusion


Introduction

The fencing problem

A farmer has exactly 1000m of fencing.  She wishes to fence off a piece of land but does not know what shape will give her the largest area.  I will investigate different shapes to see if I can find out what the maximum area of land can be fenced off with a given length of fencing.


Investigating the relationship between the length of sides and the area

Investigating Rectangles

I will start off this section by testing a variety of different sized rectangles.  First I will try rectangles increasing or decreasing in 50’s, then in 10’s and finally in 5’s.  Then I will draw a table of the rectangles and finally draw a graph.

PREDICTION:        I think that none of the rectangles will have the same area because I have not used the same measurements for the length and breadth twice.

image11.png

image22.png

a = l x w

image28.png

Here are the rectangles and areas increasing and decreasing by 50.

l (m)

w (m)

a (m²)

1

50

450

l x w

50 x 450

22500

2

100

400

l x w

100 x 400

40000

3

150

350

l x w

150 x 350

52500

4

200

300

l x w

200 x 300

60000

5

250

250

l x w

250 x 250

62500

6

300

200

l x w

300 x 200

60000

7

350

150

l x w

350 x 150

52500

8

400

100

l x w

400 x 100

40000

9

450

50

l x w

450 x 50

22500


Here are the rectangles and areas increasing and decreasing by 10.

l (m)

w (m)

a (m²)

1

260

240

l x w

260 x 240

62400

2

320

180

l x w

320 x 180

57600

3

490

10

l x w

490 x 10

49000

4

460

40

l x w

460 x 40

18400

5

310

190

l x w

310 x 190

58900

Here are the rectangles and areas increasing and decreasing by 5.

l (m)

w (m)

a (m²)

1

345

155

l x w

345 x 155

53475

2

205

295

l x w

205 x 295

60475

3

325

175

l x w

325 x 175

56875

4

495

5

l x w

495 x 5

2475

5

425

75

l x w

425 x 75

31875

The point of this was to see which rectangle has the biggest area and whether I could see any relationships.  The following table shows the area of all the rectangles in order of smallest to largest.  I have also calculated the difference in metres between the length and width.

Length (m)

Width (m)

Area (m²)

Difference between length and width (m)

5

495

2475

490

10

490

4900

480

40

460

18400

420

50

450

22500

400

75

425

31875

350

100

400

40000

300

150

350

52500

200

155

345

53475

190

175

325

56875

150

180

320

57600

140

190

310

58900

120

200

300

60000

100

205

295

60475

90

240

260

62400

20

250

250

62500

0

...read more.

Middle

image108.pngimage04.png

image109.png

        area = 33541 m²


  1. Equilateralimage05.png

image06.pngimage06.png

image110.png

image111.pngimage07.png

        area = 48160 m²

4.        Right-angledimage08.png

image75.pngimage10.pngimage09.png

        area = 41625 m²image12.png

Base (m)

Side 1 (m)

Side 2 (m)

Area (m²)

450

200

350

33541

200

400

400

38729

250

333

417

41625

333.3

333.3

333.3

48160

Again the shape with the largest area is the one where all the sides are the same length.  Therefore my prediction was correct.


Investigating the relationship between the number of sides and the area

 I have now proved that regular shapes have the largest area.  Now I will try to find a relationship between regular shapes with up to ten sides.

Pentagons

I will start by drawing and finding the area of a regular pentagon and giving the answer to two decimal places.  I will find the area by using the SIN rule.

This is the SIN Rule

I will first split the shape into ten right-angled triangles, then find the angles, then height.  By then I will I will have enough information to use the formula:

image76.png


All I need to do then is multiply by ten then I will have the area for the pentagon.image13.png

image15.pngimage14.png

image16.png

image77.png

image78.png

image79.png

c = 137.64m  This is the height of the right-angled triangle in the regular pentagon.

The area of the right-angled triangle can now be calculated.

image76.png = image80.png = 6882.5m²

Therefore the area of the regular pentagon is 6882.5 x 10 = 68825m²


Hexagons

...read more.

Conclusion

number of sides = n

total length = 1000

length of sides = image99.png

I had to split the shape up into right-angled triangles and found that there was a relationship between the number of sides on the shape and the number of right-angles in the shape.  

number of right-angles = 2n

The area of a right-angled triangle is image76.png

        The base (b) is half of the length of the sides

        There for the base (b) = image101.pngx image99.png = image102.png

The height (h) is calculated by image103.pngimage72.png

        Where c = height (h)

Angle B is calculated by dividing the number of degrees in a circle (360) by the number of right-angled triangles in the shape.

        Angle B = image104.png

        Angle C = 180 – A - B

        Angle C = 180 – 90 - image104.png

        I can now substitute the information I have gained into the formula to calculate the height (h).

image105.png

I now have enough information to use the formula image76.pngto calculate the area of a right-angled triangle.

area = image106.png

I now have a formula that will calculate the area the right-angled triangle.  To find the area of the shape all I need to do is multiply this by the number of right-angled triangles in the shape.

Therefore the area of a shape with n sides can be calculated using the formula

image107.png

I have produce a graph using microsoft excel which shows the results of the formula for shapes with between 20 and 1000 sides.

©2002 Anonymous        Page  of

...read more.

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

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Fencing Problem

    that the closer the dimensions of the Isosceles triangles are to an Equilateral triangle the larger the area will be. The largest Isosceles triangle I have found with a perimeter of 1000 m has a base of 333.4 metres and sides of 333.3 metres, which is extremely close to the dimensions of an Equilateral triangle, which has sides of 333.3333333.

  2. Fencing problem.

    Four different types of triangles have been investigated. I have shown the results below in a tabulated form: Name of triangle Area (m2) Equilateral triangle 48095.2 Isosceles triangle 47925.5 Right angles triangle 41666 Scalene triangle 47160.14 I have created a graph, which represents the data that has been shown above: By looking at the graph above it is

  1. The Fencing Problem.

    The formula will be =(1000-B2)/2 where B2 is the base. It will be divided by 2 because 1000-B2 would give the sum of the two equal sides together. As previously , for the rectangles, there will be a formula to check that the perimeter is 1000m.

  2. Maths Coursework - The Fencing Problem

    Although the formula is different, the largest version (in terms of the area) of the shape is exactly the same as the simplest, like the square, or in this case the equilateral triangle. The Triangles The Conclusion As with the quadrilaterals, I have found that the most regular shape, in this case the equilateral, has the largest area.

  1. The Fencing Problem

    440 460 100 6) 450 450 100 7) 460 440 100 8) 470 430 100 9) 480 420 100 10) 490 410 100 11) 500 400 100 ^ Evidently, the results from the table of data earlier are much easily viewed through the graph. The graph is shaped as a wide "n" and has a smooth curve.

  2. Fencing Problem

    To do this I am going to split the triangle into two right-angled triangles so that the angle halves leaving me with 36� and 100m as the base. To work out the height I have to use trigonometry. Height = 100 = 137.64 (2 d.p)

  1. The Fencing Problem

    results table I can draw a graph of area against base of triangle. From the graph, I can see that the maximum area of a isosceles triangle with a range of bases, ranging from 10m to 490min in 10m steps, is a triangle with a base of 330m, with sides of 335m long and a perimeter of 1000m.

  2. The Fencing Problem

    This means that the area of the pentagon is 68,819cm2. Hexagons A hexagon can be split into six equilateral triangles. Here is one of them: I worked out the top angle using 360 / n and the base using 1000 / 6.

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