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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
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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
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
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.
This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.
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