• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Fencing problem

Extracts from this document...

Introduction

Maths Coursework                                           SSMC                               Nicholas Thorburn

Fencing problem

I realise intuitively that a circle will cover the largest area, but it is hard to prove it. I decided to do this logically, according to the sheet, by looking at the area that can be covered by a rectangle. I decided to produce a table, and then a graph, showing varying lengths and breadths in a logical manner, decreasing one by 50, and increasing the other by 50.

These are my preliminary results:

Width (m)

150

175

200

225

250

275

300

325

350

Length (m)

350

325

300

275

250

225

200

175

150

AREA (m2)

52500

56875

60000

61875

62500

61875

60000

56875

52500

I produced a graph showing area from this:

image00.png

You can see that this graph is a parabola, so the highest point will indicate the greatest area, which is 62,500. I have put the base along the x axis because you can work out the other two

...read more.

Middle

225

212.5

200

Length (m)

200

212.5

225

237.5

250

262.5

275

287.5

300

AREA (m2)

60000

61093.75

61875

62343.75

62500

62343.75

61875

61093.75

60000

Triangles

        I am only going to use isosceles triangles. This is because if know the base length, then I can work out the other 2 lengths, because they are the same. If the base is 200m long then I can subtract that from 1000 and divide it by 2. This means that I can say that

Side = (1000 – 200) ÷ 2 = 400.

        I will then follow on by investigating the other type of triangle, scalene.

This is how I worked out the area:

b2 = a2 - c2

    = 4502 - 502

    = 200,000

    = 447.2135955

    = ½ X 25 X 447.2135955 = 5590.169944m

I produced a graph from this, below:

image02.png

        This again gives good evidence that that a regular sided shape (an equilateral triangle) covers more are than an isosceles triangle, and that scalene triangles cover less area than an isosceles triangle.

As you can see this graph is a-symmetrical.

...read more.

Conclusion

Using SOHCAHTOA I can work out that I need to use Tangent.

O100

T   =  tan36

This has given me the length of H so I can work out the area.

Area = ½ X b X H = ½ x 100 X 137.638 = 6881.910

I now have the area of half of one of the 5 segments, so I simply multiply that number by 10 and I get the area of the shape

Area = 6881.910 X 10 = 68819.096m2.

I am going to work out the area of the 2 shapes using the same method as before.

Hexagon:

1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.

360 ÷ 6 = 60 ÷ 2 = 30

Area = ½ X b X H = ½ x 83 1/3 X 144.338 = 6014.065

6014.065 X 12 = 72168.784m2

Heptagon:

1000 ÷ 7 = 142.857 ÷ 2 = 71.429

360 ÷ 7 = 51.429 ÷ 2 = 25.714

Area = ½ X b X H = ½ X 71.429 X 148.323 = 5297.260

5297.260 X 14 = 74161.644m2

         My prediction was right, so as the number of sides increases, the area increases. Below is a table of the number of sides against area. I have not gone far with this, or produced a graph, as all that proves is the area increase tails off, towards the higher-sided shapes.

No. of sides

Area (m2)

3

48112.522

4

62500.000

5

68819.096

6

72168.784

7

74161.644

...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. Operation Cedar Falls.

    All civilians were to be evacuated from the area which would be cleared and the tunnels destroyed. Phase II of Operation Cedar Falls was planned to last from two to three weeks. Cedar Falls was to be the largest and most significant operation to this point in the war.

  2. The Fencing Problem

    x h]} x 100 Regular Polygons - Chiliagon (1000-sided-shape) As you can see, I have not incorporated a diagram of a Hectogon because of its close resemblance to a circle (at this scale factor). However, the calculations of the base and height of the triangles are shown, and of course the calculations.

  1. Fencing Problem

    * To figure out the lengths of the sides of the Polygon I divided the perimeter,"1000 m" by "n" - the number of sides. * Then divided the interior angle, 360�, by "n" - the number of sides, to give me the angle of each of the Isosceles triangles �

  2. Geography Investigation: Residential Areas

    I personally enjoy green space and can't see lots, although there is some I may give a higher penalty point than someone who is indifferent on green space. In my key question I wanted to find out how Basingstoke changes and reflect different urban models.

  1. Fencing Problem

    Rectangles I have found out that regular shapes seem to have larger area but I am going to confirm that by looking at a rectangle. Since the perimeter of each rectangle is suppose to be 1000m. The formula for perimeter of a rectangle is: 2L + 2W = Perimeter 2L + 2W = 1000 2(L+W)

  2. Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

    In contrast to this, the Mayflower Steps, and even more so on Sutton Road, were recorded as having very high traffic counts. Gashouse Lane has just 1 van over a five-minute period. This is 3 lower than Artillery Flats which had 4, and Discovery Wharf had 5 vehicles altogether.

  1. Fencing problem.

    333.3 m We know that the exterior angle of the triangle would add up to 3600, this figure is obtained, as the sum of interior angles is always equal to 1800. To find out the exterior angles we would as following: Exterior angles = 3600 � Number of sides Exterior

  2. Fencing Problem - Math's Coursework.

    Tow ork out the height I can use Pythagoras' Theorem. Below is the formula and area when using a base of 200m. H� = h� - a� H� = 400� - 100� H� = 160000 - 10000 H� = 150000 H = 387.298 1/2 � 200 � 387.298 = 38729.833m wweb ebw stebebud eeb ebnt ceb enebtral ebcoeb uk.

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