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

The Fencing Problem

Extracts from this document...

Introduction

The Fencing Problem I am writing this investigation for a farmer who has 1000m2 of fencing and wants to find out how he can create the most area by using the fencing and rearranging the fence in different shapes. From only 3 sides to 99999 sides I have worked out the highest and lowest area for the farmers plot of land. ...read more.

Middle

(500-500) (500-300) =5916.079 Area = 500(500-400) (500-400) (500-200) =38729.833 Area = 500(500-450) (500-450) (500-150) =20916.501 Now I will try a different family of shapes a 4 sided shape but I think that the greatest area of a shape will come from a regular shape. Area = 250 x 250 = 62,500m2 Area = 200 x 300 = 60,000m2 Area = 150 x 350 = 52,500m2 Area = 100 x 400 = 40,000m2 Area = 50 ...read more.

Conclusion

I will now increase the number of sides on a shape and see if I can find a maximum area. I can see that as the number of sides increase the area also increases. The question now is does a circle have infinity sides, 1 single side or a peak number of sides n- sided Area = This formula shows that as n increases to infinity approaches tan 90 so that the angle between adjacent sides becomes 2x 90= 180o thus giving a circle. For a circle of a radius r = ...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

    For example: Equilateral triangle - 3 Equal Sides - 48112.52243 m� Square - 4 Equal Sides - 62500 m� By looking at this it strengthens my hypothesis. * 1000 m / 5 = 200 m * Then I will split up the entire pentagon into 5 Isosceles triangles.

  2. Fencing problem.

    I shall multiply this figure by the number of triangles present within the hexagon: Area of hexagon = Area of one triangle � Number of triangles Area of hexagon = 10617.5 � 7 = 74322.5m2 Octagon The fourth shape I shall be investigating is a nine-sided polygon also known as an octagon.

  1. The Fencing Problem.

    I can do this in Microsoft Excel by dragging the formula boxes down, thus duplicating them but allowing them to refer to a different number of sides. Hypothesis I predict that as you increase the number of sides the area increases because the maximum area for a rectangle is 62500m2, and the maximum area for an isosceles triangle is 48112.52243m2.

  2. The Fencing Problem

    As we can see, the highest value shown is 47925.72, before the area begins to decline to 44721.36. We can evaluate that an isosceles triangle with base = 350m, both sloping sides = 325m each accommodates the highest area whilst adhering to the set perimeter of 1000m.

  1. The Fencing Problem

    I used sine because I know the hypotenuse (width) and I want to work out the opposite (perpendicular height). I will now work out the area of the following parallelogram. h = 200m * sin45? = 141.4m a = 300m * 141.4m = 42420m� I will change the angle x

  2. Geography Investigation: Residential Areas

    This is true; there is in fact a weak correlation between them. This means that the higher the index of decay the more likely the occupants are to own their own homes. Over the course of this hypothesis I have found out two main things about Basingstoke.

  1. The fencing problem 5-6 pages

    I will start of with the pentagon. Each side= 200m (1000/5) All corresponding angles are the same Exterior angle = 360/N or 360/5 = 72 Interior angles= 180-72=1080 (angles on a straight line) Like all regular shapes, a pentagon can be split up into equal triangles, which will be isosceles as two sides, and the corresponding angles are equal.

  2. Fencing Problem

    H = V (c2- b2) Therefore if c = L and b = 1/2 B We can substitute H = V (c2- b2) into H = V (L2- B2/4) b B Therefore: A = 1/2 x B x H A = 1/2 x B x V (L2- B2/4)

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