• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1890

Fencing Problem

Extracts from this document...

Introduction

GCSE Maths Coursework:- By Sarah Wolfe, 10Ba2 A farmer has exactly 1000 metres of fencing and wants t fence off plot of level land. I am going to come up with a range of shapes which has a perimeter or circumference of 1000 metres. For each shape I draw I will draw different size ones and find their areas. The reason for finding their areas is because the farmer wants to find a shape with the maximum area. To get started on this exercise, I am first of all going to come with a shape which I think is easiest to use; this will be a rectangle. 1) Perimeter = 450 + 50 + 450 + 50 = 1000m Area = 450 x 50 = 22,500m 2) Perimeter = 400 + 100 + 400 + 100 = 1000m Area = 400 x 100 = 40,000m 3) Perimeter = 385 + 115 + 385 +115 = 1000m Area = 385 x 115 = 44,275m 4) Perimeter = 375 + 125 + 375+ 125 = 1000m Area = 375 x 125 = 46,875m 5) Perimeter = 350 + 150 +350 +150 = 1000m Area = 350 x 150 = 52,500m 6) Perimeter = 325 +175+325 +175 = 1000m Area = 325 x 175 =56,875m Whilst finding the areas of these shapes, I'm finding that when I decrease the size of the length and increase the width, the area seems to get bigger. ...read more.

Middle

I predict that, like the square, the equilateral triangle will give the biggest area. 333.3 = 166.6 + h 111,111.1 = 277,7.7+ h h = 111,111.1- 277,7.7 h = 108,333.4 h = 108,333.4 h = 329.1m Area = b x h 2 = 333.3 x 329.1 2 Area = 54,850 m (to 1d.p) I am going to do another table, so I can see clearly the areas I have already found. Base (m) Height (m) Area (m ) 100 450 22,350 200 400 38,700 300 350 47,400 350 325 47,950 33.3 33.3 54,850 400 300 44,800 My predication was right and the reason I know this is because when you increase the length higher 33.3m, then the area starts to decrease, as you can see in my table. I am not going to do a graph, as I know it will give me the same sort of graph when I did one for the square; the graph will be symmetrical. I think that this will be the pattern for all of the shapes I do; the one which is equilateral will give the biggest area of that family. It seems to me that the shapes with the smaller sides, give the smallest area, so I am now going to move on to using shapes with more sides; the first one I will start with will be a pentagon. I am going to stick to finding the area of a regular pentagon because, before when finding the areas of the rectangles and triangles, my conclusion has been that the shape which is equilateral, gives the biggest area. ...read more.

Conclusion

Here are a set of instructions of how I did it. 1) From the centre of 2) Then, find the central the shape, split it up angle. (360 the into equal triangles. number of sides) Then label the length of each side. (1000m the number of sides) 4) Once you have found 3) Now find the area the area of that triangle, of one of the triangles; multiply the area by the to do this, you need to number of sides. This find the height first, now gives the final area using trigonometry. of that shape. I could also use a formula; e.g if I were to find the area of a 20 sided shape. I could use 'n' as the number of sides 1) Split it up into 20 triangles 2) Find central angle - 360 = ? 20 3) Find area of each triangle- need height (trigonometry) 4) 'n' (20) x area of one triangle = total area of shape Now I have got worked out the area of the all possible shapes, I am now going to put the shapes with their biggest area, into a table. Shape Area m (best area of shape) Rectangle (square) 62,500m Triangle 54,850m Pentagon 68,800m Hexagon 72,150m Octagon 75,450m 10 sided shape 20 sided shape Circle 79,572m I am now going to increase the number of sides a lot higher. I am not going to do the ones in between as I know already that the more sides I have, the area will be bigger. I am going to do a few more examples ...read more.

The above preview is unformatted text

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.

    of H = 32500m2 A polygon shaped as a House (Rectangle + Triangle) The next shape that I shall be investigating is a rectangle and triangle that is shaped as a house. I have shown the shape that I shall be investigating below: I shall now divide the shape into two different segments.

  2. The Fencing Problem

    Area = {1/2[(1000 � 10000) x h]} x 10000 I have explored various regular polygons so far, and we can see now that as the number of sides is increased, the area is larger each time. I can now assume that a circle will finally give us the largest area

  1. t shape t toal

    - (7 x 6) 30 Up1 28 98 (5 x 28) - (7 x 6) 30 Up 2 from original 22 68 (5 x 22) - (7 x 6) 30 Up 3 from original 16 38 (5 x 16) - (7 x 6)

  2. Investigating different shapes to see which gives the biggest perimeter

    I can now use Pythagoras's theorem to work out the height of the triangle (a). a� = c� - b� = 4252 - 752 = 180625 - 5625 = 175000 a = V175000 = 418.3m So the height of the isosceles triangle is 418.3m because the height of the isosceles is the same as the height of the right-angled triangle.

  1. Geography Investigation: Residential Areas

    Using a scatter graph or something similar with this data would not give me accurate or clear results - so although I am repeating something I have done already in hypothesis one, I am doing it for a good reason.

  2. Maths GCSE Courswork

    360 x 264.57 / 2 = 95245.2 / 2 = 47622.6m2 OR Heron's Formula Area = = 500 (500 - 320)(500 - 360)(500 - 320) = 2268000000 = V2268000000 = 47622.6m2 The area of this isosceles triangle is 47622.6m2. The next triangle has two sides of length 330m and a

  1. The fencing problem 5-6 pages

    to gain the biggest areas. However I have also made another observation. The three-sided triangle has a smaller area than the four-sided square. It may just be possible that area increases with the number of sides. Consequently, this is my next hypothesis.

  2. The Fencing Problem My investigation is about a farmer who has exactly 1000 ...

    Firstly I am going to look at the isosceles triangle; this is a triangle in which 2 of the sides and 2 of the angles are the same. As I am using triangles I can use the equation: Area (A)

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