• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1274

Perimeter Investigation

Extracts from this document...

Introduction

I will be investigating the shape, or shapes, that could be used to fence a plot of land, which contains the maximum area, using exactly 1000 metres. To start I will be investigating the rectangle family as shown below:

image00.png

image01.png

image12.pngimage23.png

image31.pngimage40.png

image48.png

image55.png

image56.png

image57.png

image03.pngimage04.pngimage02.pngimage07.pngimage05.pngimage06.png

image08.pngimage10.pngimage09.png

image14.pngimage13.pngimage11.png

image16.pngimage15.pngimage17.png

image20.pngimage19.pngimage18.png

From these results, the maximum area using exactly 1000 metres of fencing is the rectangle that measures 250 by 250. Its area is 62500m², which is the biggest area and it is square in shape, which proves that the square is the best to use, when investigating four sided shapes.

I have plotted a graph from these results obtained:

image58.png

...read more.

Middle

image24.pngimage36.pngimage24.pngimage35.pngimage24.pngimage34.pngimage33.pngimage24.pngimage32.pngimage42.pngimage41.png

Base/m

Sides/m

Perpendicular Height/m

Area/m²

50

475

474.3

11857.5

100

450

447.2

22360

150

425

418.3

31372.5

200

400

387.3

38730

250

375

353.6

44200

300

350

316.2

47430

350

325

273.9

47932.5

400

300

223.6

44720

450

275

        158.1

35572.5

From these results, the maximum area using exactly 1000 metres of fencing is the triangle that measures, base 350m, sides 325m and an area of 47932.5m². This triangle is the best to use when investigating 3 sided shapes, because it has the biggest area and a perimeter of 1000 metres.

I decided to use regular shapes through-out my investigation because only regular shapes, i.e. shapes with equal sides, give the maximum area. So this is why I decided to use equilateral triangles.  

In my triangle investigation, I decided to start with a base of 50 metres and investigate the area using that base. I did not investigate triangles with a base above 450 metres because the area kept on decreasing. For example, the triangle with base 500 metres and sides 250 metres gave an area of 0 metres.

(Here will be a graph for Base/Area and another graph for sides/perpendicular height)

From the triangle exercise, I have determined that the shape with equal sides gives the maximum area.

...read more.

Conclusion

From these results, it appears that if we make a polygon of infinite sides, it would give me the maximum area. The polygon with maximum sides can only be a circle because each point on the circumference could be a side and the height will be the radius. If we make a polygon with 1000 sides of 1m each, the length of a side would become a dot and the shape would become a circle. Therefore it can be concluded that the circle would have the maximum area for a given perimeter.

...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. Investigation to find out the number of matchsticks on the perimeter in a matchstick ...

    difference 2nd difference In my table when b = 0, t = 0 so c = 0.

  2. Mathematics Gcse Coursework Tubes Investigation

    The width is three times longer and the length is three times shorter. Given the piece of paper is 8cm by 96cm. To calculate the volume of this cube then must follow the steps: 96 � 4 = 24cm 24 � 24 = 576cm2 576 � 8 = 4608cm3 The

  1. Fencing investigation.

    2002 = 1002 + x2 2002 - 1002 = x2 30000 = x2 V30000 = x 173.205m = x Now that the perpendicular height of the trapezium has been solved, working out the area is an easy procedure. Area = 1/2 (base + top)

  2. Borders Investigation

    Therefore the formula appears to work. As with perimeter, we can now look at the formula and obtain some visual understanding of why it works. We mentioned previously that the diameter of the shape could be represented by (2n for either side, 1 for the centre).

  1. Geography Investigation: Residential Areas

    Figure 7 Figure 8 Table 1 Hypothesis (page 2 & 3) Method Used How the method will provide evidence for the hypothesis How the data will be presented 1) Bi Polar Analysis The method will change words into numeric's to enable me to analyze the data.

  2. Fencing - maths coursework

    the number of sides This is a graph with all my results (the areas I have found.) My graph and table clearly shows that as the numbers of sides increase also the area increases. So to find out the maximum area for a farmer with perimeter of 1000m I will

  1. My investigation is about a farmer who has exactly 1000 metres of fencing and ...

    h = 83 ? x tan60 h = 144.3376m (4 d.p) Then I find out the area of my isosceles triangle: A = 1/2bh A = 83 ? x 144.3376 A = 12028.1306m2 (4 d.p) Then I multiply my answer by 6, the number of isosceles triangles in my hexagon: A = 12028.1306 x 6 = 72168.7837m2 (4 d.p)

  2. Fencing problem.

    � r2 Area of a circle = ? � (159.1549431)2 Area of a circle = ? � 25330.29591 Area of a circle = 79577.47155m2 Hence, we can also write the area of the circle to five significant figures, which looks more presentable.

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