• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1641

The Fencing Problem.

Extracts from this document...

Introduction

Martin Jones 4M                Mr. Hogwood

The Fencing Problem

Target

My target is to try and find a shape that will give a farmer the largest possible plot of land that can be enclosed with a 1200-meter perimeter fence.

Hypothesis

I have studied many shapes and decided that a circle will give the largest area. I have come to this conclusion because shapes like polygons are made up from a number of triangles, the number of triangle is equal to the number of sides. Therefore as you increase the number of sides you increase the number of triangles.

A one sided shape would have no area at all; this therefore means that a circle should have the largest area as it has an infinite number of sides. My prediction is that the number of sides is relevant to the area as to say that the number of sides increases with the area, and visa versa.

Quadrilaterals

I will begin my investigation by investigating the areas of quadrilaterals; these are four sided shapes such as squares and rectangles.

...read more.

Middle

This shows that the equilateral triangle has the largest area. To check that this was true, I found the area of a triangle whose sides were close to the length of the sides of the equilateral triangle.

Base: 401m         Other Sides: 399m, 400m

image13.png

s (s – 401) (s – 399) (s – 400) = 69281m2

Area: 69281m2

From my calculations I have decided that the equilateral triangle is the triangle, which will provide the greatest area with a 1200m perimeter.

Polygons

        The next group of shapes that I will investigate are the polygons, there can be many, many different shaped polygons, with many sides, so I will make a chart using excel to calculate the areas of polygons with a great number of sides.  

I order to find the areas of polygons I will have to split them up into triangles. Each polygon is made from a large number of triangles, which is equal to the number of sides in the polygon. The formula I will use to calculate the polygon’s area will be:

360000/(n tan (180/n))

...read more.

Conclusion

p>

114172.3735

35

17.142857

5.142857

0.08975979

190.4727

3265.247021

114283.6457

40

15

4.5

0.078539816

190.5931

2858.896066

114355.8426

45

13.333333

4

0.06981317

190.6756

2542.340668

114405.3301

50

12

3.6

0.062831853

190.7345

2288.814458

114440.7229

60

10

3

0.052359878

190.8114

1908.113669

114486.8201

70

8.5714286

2.571429

0.044879895

190.8577

1635.923025

114514.6117

80

7.5

2.25

0.039269908

190.8877

1431.658101

114532.6481

90

6.6666667

2

0.034906585

190.9084

1272.722368

114545.0131

100

6

1.8

0.031415927

190.9231

1145.538574

114553.8574

150

4

1.2

0.020943951

190.958

763.8320225

114574.8034

200

3

0.9

0.015707963

190.9702

572.9106705

114582.1341

250

2.4

0.72

0.012566371

190.9759

458.3421084

114585.5271

300

2

0.6

0.010471976

190.979

381.9579007

114587.3702

400

1.5

0.45

0.007853982

190.982

286.4730071

114589.2028

500

1.2

0.36

0.006283185

190.9834

229.1801021

114590.0511

600

1

0.3

0.005235988

190.9842

190.9841864

114590.5118

700

0.8571429

0.257143

0.00448799

190.9846

163.7011281

114590.7897

800

0.75

0.225

0.003926991

190.9849

143.2387125

114590.97

900

0.6666667

0.2

0.003490659

190.9852

127.3234373

114591.0936

1000

0.6

0.18

0.003141593

190.9853

114.591182

114591.182

image04.png

Circle

A Circle has no sides or an infinite number of sides. So, if my theory is correct the circle should provide the largest area from all the shapes I have investigated.

Circumference: 1200m

image05.png

Diameter = 1200 = 382m

Π

Radius = 382 =191m

  2

Area = Π x 1912 = 114608m2

Area: 114608m2

.

114608m2 is the largest area that I have achieved using a perimeter/circumference of 1200m. This means that the circle would be the most effective shape to use in ‘The Fencing Problem’.

Conclusion

From the results I have gathered throughout my investigation I have come to the conclusion that the circle provides the largest surface area. In this investigation I tested quadrilaterals, triangles, polygons and a circle. Having tested all these shapes I came to the conclusion that the greater the amount of sides the greater the area.

The circle has the largest area because it has an infinite amount of sides. With 1200m of fence you can enclose a circle with a 1200m perimeter and 114608m2 area, this is the greatest area possible to enclose with that amount of fence.

...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

    Base Side A Side B Perimeter S Area (m�) 300 350 350 1000 500 47434.16 310 345 345 1000 500 47774.21 320 340 340 1000 500 48000.00 330 335 335 1000 500 48105.35 340 330 330 1000 500 48083.26 350 325 325 1000 500 47925.72 360 320 320 1000 500 47623.52 370 315 315 1000 500 47165.93 380 310

  2. Fencing problem.

    I shall now substitute this figure into the formula below: Area of the isosceles triangle = 1/2 � Base � Height Area of the isosceles triangle = 1/2 � 350m � 273.86m Area of the isosceles triangle = 47925.5m2 Right-angled triangle The third triangle that has to be explored is

  1. The Fencing Problem

    A graph which corresponds to the areas that I have just examined will also be shown following the drawings; the page which succeeds that will continue my investigation. 1) 475 475 50 h 475 25 2) h 450 50 3)

  2. Fencing Problem

    For me to get the results as accurate as possible I will be using the hero's formula. A B C A+B+C are the three sides of the scalene triangle that add up to give 1000m. To find the area of this triangle I will need the Hero's formula that gives me the area.

  1. Maths:Fencing Problem

    This is a Trapezium: This also becomes a rectangle so I know that its area will be less than 62500m This is a rhombus: This rhombus will give the maximum area of 62500m because it can be transformed into a square which has equal sides.

  2. The Fencing Problem

    when it is a rectangle. Therefore, a rectangle has a greater area than a parallelogram. I will investigate the areas of different rectangles, that each have the same perimeter of 1000 metres, because a rectangle is an easier shape for which to work out the area for than any other quadrilaterals.

  1. Geography Investigation: Residential Areas

    To back up what I see I will take photographs of eye-sores and the different types of residential areas I experience throughout my investigation. Methodology How do Basingstoke's residential areas change, improve and reflect different urban models? I have chosen this key question as the basis of my coursework because

  2. The Fencing Problem. My aim is to determine which shape will give me ...

    Shape Calculations Area (M2) EXAMPLE a c b VS-(S-A)(S-B)(S-C)=AREA S=PERIMETER/2 VS-(S-A)(S-B)(S-C) 4 400 250 350 V500(500-400)(500-250)(500-350)= 43301.27 43301.27 5 400 300 300 V500(500-400)(500-300)(500-300)= 44721.36 44721.36 6 350 400 250 V500(500-400)(500-350)(500-250)= 43301.27 43301.27 7 450 400 150 V500(500-450)(500-400)(500-150)= 29580.399 29580.399 Equilateral Triangle Shape No.

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