• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 3055

My investigation is about a farmer who has exactly 1000 metres of fencing and wants to use all of it to fence off the maximum area of land possible. I want to find the regular polygon that will have the maximum area with a perimeter

Extracts from this document...

Introduction

Mathematics Investigation

The Fencing Problem.

 By Waseem khalique

        My investigation is about a farmer who has exactly 1000 metres of fencing and wants to use all of it to fence off the maximum area of land possible.  I want to find the regular polygon that will have the maximum area with a perimeter (or circumference) of exactly 1000 metres (m).  I also want to look at compound shapes and see how their areas compare to those of the earlier shapes I looked at.

I assume that the land is flat and that there is no need to leave gaps in the fencing for gates or anything.  I also presume that the fence can be flexible i.e. will be able to curve into a circle.

First of all I am going to look at rectangles, also known as irregular quadrilaterals, quad meaning 4, as not all the angles and sides are the same.  I already know that to find the area (A) of a rectangle you multiply the length (L) by the width (W):

        Area = Length x Width

I am going to start with a rectangle with a length of is 475m and a width of 25m, the reason why I am not making the length 950m and the width 50m is because the perimeter will add up to 2000m and not 1000m.  It is now simple to find out the area of the rectangle because all I have to do is put the numbers into the equation:

        A = L x W

        A = 475 x 25

        A = 11875m2

...read more.

Middle

Base (b)/2 = 133 ⅓ / 2 = 166 ⅔m

Now I can work out the height of my triangle:

        Tan60 = h / 166 ⅔

h = 166 ⅔ x Tan60

h = 288.6751m (4 d.p)

        Now I have all of the information to work out the area of my triangle:

                A = ½bh

                A = 166 ⅔ x 288.67

                A = 48112.5224m2 (4 d.p)  

        To compare my work on triangles I will draw a table:

Name

Area (m2)

Isosceles triangle

44721

Equilateral triangle

48112.5224

I was correct in saying that the regular shapes have the biggest area; therefore I will only look at regular shapes from now on.  Now I will have a look at regular pentagons so I can see how their area’s compare to those which I have already looked at.

I will start with the regular 5 sided shape known as a pentagon.  From reading my Letts GCSE Mathematics Revision Guide, I know that the easiest way to find the area of a regular polygon is to split the shape into isosceles triangles starting from the centre and connecting this point to each of the vertices.

To find the length of each side, I use the same equation as before:

        1000/n = 1000/5 = 200m

As I will be using the formula A = ½bh, I need to divide the length into 2 as it will be the base of my triangle:

        b/2 = 200/2 = 100m

Now I need to find the size of the angles in each of these triangles, starting with the exterior.  You do this by dividing 360° by 5, the number of isosceles triangles:

        Exterior Angle = 360°/n = 360°/5 = 72°

I then need

...read more.

Conclusion

Although I will not be able to use my formula to work out the area of the circle, I can find it out by rearranging the formula to find the circumference of a circle which is:

Circumference (C) = Pi (3.14) x Diameter (d)

C = 3.14 x d

This can be rearranged to:

d = C/3.14

        This then works out as:

                d = 1000/3.14

                d = 318.3099m (4 d.p)

        I then divide this by two since I need the radius (r), which is half of the diameter:

                r = 318.3099/2 = 159.1550 (4 d.p)

        I can then use the formula, A = 3.14 x r2 to work out the area of the circle:

                A = 3.14 x r2

                A = 3.14 x 159.15502

                A = 79577.4716m2 (4.d.p)

        This proves that my prediction is correct as the area does in fact increase and the number of sides increases as shown below:

Number of Sides

Area in metres

squared to 4 d.p

3

48112.5224

4

62500.0000

5

68819.0960

6

72168.7837

7

74161.4785

8

75444.1738

9

76318.8172

10

76942.0884

11

77401.9827

12

77751.0585

20

78921.1894

2000

79577.4061

5000

79577.4611

20000

79577.4709

50000

79577.4714

Infinite (Circle)

79577.4716

         I think that the reason why circles have the biggest area is because they do not have any obvious vertices (corners) like a square does, therefore there is less wastage with the area because it all spreads out evenly from the one centre point where as in a square it starts to spread out but then has to go off into the corners.  This is why when more sides were added, the bigger the area became, basically because there was less wastage.

...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. A farmer has exactly 1000m of fencing and wants to fence off a plot ...

    495 505 249975 496 504 249984 497 503 249991 498 502 249996 499 501 249999 500 500 250000 501 499 249999 502 498 249996 503 497 249991 504 496 249984 505 495 249975 The results appear to be correct, but I will test values even closer to the maximum to make certain.

  2. Medicine and mathematics

    From the table above you can see that after every 6 hours the line decreases gradually because every hour the value is reduced to 60% of the value before, but then suddenly the line shoots back up, to averagely 314 grams, representing that the new dose has been administrated.

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

    So in my example x is 1/2 of 4 = 2. ?My formula will begin with 2r2. To find 'c' I have to find the value of t, when r = 0. So I am going draw another table. R 0 1 2 3 4 t -1 4 13 26

  2. Maths Investigation on Trays.

    To take the hypothesis a step further I have to test it on a rectangle tray. I don't think it will work to this because the rectangle is much more complicated it has two different measurements so it won't work as well.

  1. The sphere of influence in Brent Cross is bigger than that of North Finchley ...

    I asked my parents if they thought I was correct and they said that a good criminal record would make a good shopping area, because if it has a bad criminal record then people wouldn't be attracted there because they would think that their in danger but however if it

  2. The coursework problem set to us is to find the shape of a gutter ...

    at a time to get the maximum area. I predict it to be 45? as this is exactly between 40? and 50?. Side angle Height Top area 15 41 11.32064 9.840885 111.4052 15 42 11.14717 10.03696 111.8837 15 43 10.97031 10.22998 112.226 15 44 10.7901 10.41988 112.4315 15 45 10.6066 10.6066 112.5 15 46 10.41988 10.7901 112.4315 15

  1. Investigate the shapes that could be used to fence in the maximum area using ...

    Now that I have found that a square has the greatest area of the rectangles group, I am going to find the triangle with the largest area. Because in any scalene triangle, there is more than 1 variable, there are countless combinations, so I am only going to use isosceles triangles.

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

    Shape Calculations Area (M2) EXAMPLE a c d b b/2*d=AREA d= V(b/2)2*C2 b/2*d d= V(b/2)2*C2 1 450 450 100 100/2 = 50 50 x 50 = 2500 450 x 450 = 202500 202500 - 2500 = 20000 V20000 = 141.4 50 x 141.4 = 7070 7070m2 2 400 400 200 200/2=100 100 x 100=10,000

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