Maths Coursework: Fencing Problem
Investigate the shape, or shapes that could be used to fence in the maximum area using exactly 1000 meters of fencing each time.
Introduction
I will attempt to solve this problem by investigating different shapes with different numbers of sides to find the maximum area that can be gained using 1000 meters of fence. The perimeter of the shapes, no matter how many sides it has, must be 1000m. I will begin by considering three sided shapes then I will move on to other shapes with various number of sides e.g. rhombus, trapezium, parallelogram, pentagon, hexagon etc.
Triangle
I will look at different types of triangles these are: Right angle triangle, scalene, isosceles and equilateral.
Right Angle Triangle
Considering the fact that I have to use exactly 1000m to construct a right angle triangle, it would be mathematically incorrect to assume the values of the sides because Pythagoras Theorem states that a2 + b2 = c2 . Therefore, I will have to make a model and construct a right angle triangle with a perimeter of 1000m. I will use a scale of 1: 10,000 using a string. I managed to find that the base and height are 3.75cm & 2.00cm, using Pythagoras Theorem the hypotenuse is 4.25cm. This would all add up to 10cm. using the scale; the sides would be as shown on the diagram below.
425m
375m
200m
I have found that the only right angle triangle that can be formed using 1000m as total length without any decimals would have the above lengths.
Area of a triangle is:
A = 1/2 x B x H
A = 1/2 x 200 x 375
A = 37500 m2
However, I can also have other lengths on the sides but this would have decimals:
a2 + b2 = c2
32 + 42 = 52
However, this would add up to 12 in total, which exceeds the normal perimeter by 2. Therefore, to get the sides it would be:
A = 1000 x 3 = 250 250
12
B = 1000 x 4 = 1000/3
12
C = 1000 x 5 = 1250/3 1000/3 , 12
I will move on to look at scalene triangles. In this case, it is safe to assume the sides and put any lengths so long as they add up to 1000m.
Scalene Triangles
To find the area of these triangles I can use Hero's Formula since I have all the sides given:
A = V[s(s-a) x (s-b) x (s-c)] where s = (a + b + c)/2
Therefore putting in the values in the formula, the results for both triangles will be:
A = 23473.39 m2 and 35355.34 m2
As you can see, there is a big different between the two areas. However, I cannot see a good explanation for this at this early stage but one thing I can notice is that the triangle with an obtuse angle seems to have smaller area than the one with no obtuse angle.
Isosceles Triangles
I am now moving on to isosceles triangles and look for a pattern.
The perimeter of each triangle is 1000 meters, therefore the equation for it is:
2L + B = 1000m (where L= Length of the sides and B= base)
We have to remember that an isosceles triangle has two sides equal in size.
Therefore, to find the value of B or L we rearrange the formula:
2L + B = 1000 to
B = 1000 - 2L and L = 500 - 1/2 B
I can find the area of a triangle using the formula = 1/2 x B x H
However, we are not given the height therefore we have to use Pythagoras' Theorem
a2+ b2= c2 (in this case a is the height)
H2= c2- b2 rearranging it. (Just a reminder, b = 1/2 B and c = L)
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)
Having to work with two different lengths i.e. L and B makes it difficult to find the area therefore if I can substitute L to B, I can easily work out the area.
Looking back at the previous rearranging of the formula 2L + B = 1000, which is:
L = 500 - 1/2 B
We can therefore replace L to the above formula:
A = 1/2 x B x V [(500 - 1/2 B) 2- (B2/4)]
I will start the Base from 10 m and move upwards. I am hoping to reach to a point where I would obtain a maximum area after which the area starts to decrease.
Using the formula the table below shows the areas of triangles with different bases.
Base
Area
Base
Area
0
2474.87
290
46985.370
30
7271.52
310
47774.209
50
1858.54
330
48105.353
70
6228.83
331
48109.003
90
20374.62
332
48111.371
10
24287.34
333
48112.450
30
27957.56
333.3
48112.522
50
31374.75
333.33
48112.522
70
34527.16
333.44
48112.515
90
37401.54
...
This is a preview of the whole essay
Using the formula the table below shows the areas of triangles with different bases.
Base
Area
Base
Area
0
2474.87
290
46985.370
30
7271.52
310
47774.209
50
1858.54
330
48105.353
70
6228.83
331
48109.003
90
20374.62
332
48111.371
10
24287.34
333
48112.450
30
27957.56
333.3
48112.522
50
31374.75
333.33
48112.522
70
34527.16
333.44
48112.515
90
37401.54
334
48112.233
210
39982.81
335
48110.712
230
42253.70
336
48107.879
250
44194.17
337
48103.725
270
45780.73
350
47925.724
Observation
From the results of the table, the area increases as the base increase. However, there reaches a point where the areas stops increasing and in fact starts to decrease. To be precise, I added decimal places to see exactly where the area starts to decrease. I then found out that the maximum area that can be formed from an isosceles triangle is 48,112.5 m2 with the base value of 333 1/3 m.
Evaluating the Results
Since the maximum value is obtained with a base measuring 333 1/3 then the remaining value is 666 2/3. Therefore, since an isosceles triangle has two sides equal then it means that all the three sides will have 333 1/3. This indicates that an equilateral triangle has the largest area. I will now plot the results into a graph
The graph also shows that the maximum area is from an equilateral triangle with sides 333.33m.
Observations
I entered the bases of 0m and 350 m in the graph. On the graph, the areas of the triangles go up and then start to decrease at a certain point.
. This means when the difference between the sides is less, the area is larger. This graph is not symmetrical because the maximum is not in the middle.
2. The largest area is 48112.52m2. This is when all the sides are 333/ 1/3 m (equilateral triangle).
To check that it is the maximum area, I inserted values just before it and after it. The areas of those values were less, so 48112.52m2is the largest area for a triangle of perimeter 1000m.
Four sided Figures
Trapezium
The first four-sided shape would be trapezium. I will take two examples:
Finding the area of a trapezium is:
A = 1/2 x (a + b) x h
However, we do not have height; therefore, I will have to use Pythagoras Theorem
a2 = c2 - b2
h2 = 1502 - [(400 - 300)/2]2
h = 141.42
A = 1/2 x (300 + 400) x 141.42
A = 49,497 m2
Applying the same procedure to the second trapezium, we would get:
A = 54,661.5 m2
Observations
From the above results, I can see that the second trapezium has a larger area than the first. The reason for this might be the fact that the difference between the values of the sides. As the difference becomes smaller, the area increases. We saw the same trend when looking at triangles.
I will now look other parallelogram.
Parallelogram & Rhombus
I chose to find the area of both parallelogram and rhombus at the same time because both have slanting sides. I had some difficulties in finding their area since I did not have the height or diagonal. However, I can mathematically prove that it would have less area then a rectangle or square.
We can change the rhombus to look like a rectangle by cutting the triangular edge, and put it on the other side it would look like this:
As you can see the slanting height is no longer there as a side because it has occupied the other side to make it form like a rectangle. This forms a new length, which is smaller, then the slanting height hence making it have a smaller area when compared to a square or rectangle.
Rectangles
I have found out that regular shapes seem to have larger area but I am going to confirm that by looking at a rectangle.
Since the perimeter of each rectangle is suppose to be 1000m. The formula for perimeter of a rectangle is:
2L + 2W = Perimeter
2L + 2W = 1000
2(L+W) = 1000
L+W = 500
This shows that the length plus the width should exactly be 500.
The value of L will vary from 1m to 499m it with width. Of course, I could have included 0 and 500 but if the length were 500m then it would mean that the width is 0 m, which is not realistic, and vice versa.
Area of a rectangle would be:
L x W = Area (m2)
e.g.
The table next page shows the results I got for the area of rectangles with different Length and Width.
Length (m)
Width (m)
Area (m2)
Length (m)
Width (m)
Area (m2)
0
490
4900
290
210
60900
30
470
4100
310
90
58900
50
450
22500
330
70
56100
70
430
30100
350
50
52500
90
410
36900
370
30
48100
10
390
42900
390
10
42900
30
370
48100
410
90
36900
50
350
52500
430
70
30100
70
330
56100
450
50
22500
90
310
58900
470
30
4100
210
290
60900
490
0
4900
230
270
62100
250
250
62500
270
230
62100
Evaluating the results.
From the table I can observe that the rectangle with the largest area has identical lengths and width of 250m and is in fact a square. The largest area that can be made from a four-sided figure is 62,500m2and it come from a square.
I can plot the results on a graph to show these results.
It will form a parabola shape because the curve reaches its peak, which is 62,500 with the length measuring 250, it goes back down again. When entering values for the length I could not have entered 0m or 500m because the shape would not be a rectangle then, it would have been a straight line. The areas in the table increases and then decreases. This proves that the areas of the rectangles get larger as the difference between the length and width get smaller. This type of curve is a quadratic graph, which comes from quadratic equations, as you will see next page. This can mathematically be explained.
The formula for perimeter we deduced earlier was L + W = 500
If we make W the subject, it will look like this: W= 500- L
Now putting this in the area formula which is A = L x W
This will change to L x (500-L)
Simplifying this would be 500L - L2
Rearranging this equation would look like this: -L2+ 500L making it a quadratic equation.
Using the completing the square method this could be factorized to look like this:
- (L2- 500L)
- (L-250)2+ 62,500
This indicates that the maximum value obtainable for the area of a rectangle with a perimeter of 1000 will be 62,500 and it will fall under the length value of 250.
Pentagon
I am first going to look at an irregular pentagon, the house pentagon even when I know a regular pentagon will give me a larger area. I am doing this just to prove that regular shapes give larger areas.
Now I am going to look at the regular pentagon of all sides = 200m.
The perimeter of the regular pentagon is 1000m so the equation is:
5L = 1000 (where L = Length)
This can be simplified as:
L = 200 m
I am using a regular pentagon to prove my statement; a regular pentagon has a bigger area then an irregular area both with 1000m, Even though we have seen that the rectangles and triangles on the previous pages were equilateral and produced the greatest area.
The formula for the area of a pentagon is simple; I will just divide the pentagon into five triangles. I will then find area of the triangle and multiply it by five.
I cannot use the previous method for finding area of triangle because we do not know the length of hypotenuse. Therefore, Pythagoras' Theorem is not applicable in this case.
The only way to find the height of the triangle in this case is therefore use:
Tan X = Opposite
Adjacent
Where opposite is half the base (L) and the adjacent is the height of the triangle.
X is half the interior angle.
In the case of the regular pentagon, the interior angle can be found by:
= 360 (degrees)
5
= 72 (degrees)
Tan X = 1/2 L x 1/H
Tan 72 = 200
2 2 x H
Tan 36 = 100
H
H = 100
Tan 36
H = 137.6 meters
Area of triangle = 1/2 x B x H
= 1/2 x 200 x 137.6
= 13,763.8 m2
Area of Pentagon = 13,763.8 x 5
= 68,819.1 m2
Evaluating Results
From the results I got, the largest area that can be got from a pentagon is 68,819.1 m2
In addition to that, this area was achieved from a regular pentagon.
Hexagon
I have now proven that regular polygon produce greatest area, so I will now move on to look at hexagon (6-sided polygon).
The perimeter will as usual be 1000m. Therefore, to we need to get obtain base (L) by 6L = 1000 m
L = 166.7 m
To get the area of the hexagon, I will use the same formula for getting the area of a triangle then multiply it by 6.
The interior angle is = 360
6
= 60 degrees
To get the height I would use the formula:
Tan 60 = 166.7
2 2 x H
H = 3.3
Tan 30
H = 144.3 m
Area = 1/2 x 166.7 x 144.3
= 12,028.1m2
Area of hexagon = 12,028.1 x 6
= 72,168.8 m2
Evaluating Results
From the above results, I have noticed an increase in the area. This implies that as the number of sides increase, the area also increases.
n Sided Polygon (n stands for any integer)
Now that I have discovered the fact that as the number of sides of a regular polygon increases, so does the area of that polygon. I will try to derive a general formula to find n-sided polygon.
Let us say a regular polygon has n sides.
To find the area of the polygon we need to get area of one triangle multiplied by n.
To find the base of each triangle (L) we use the formula:
L = 1000
n
To find area of triangle, we need to find the height.
To find height we use:
Tan X =__L__
2 x H
However, X is:
X = _360_
2 x n
Substituting L and X, we get:
Tan 180 = 1000
n 2Hn
Rearranging the above formula to make H the subject:
H = ___500___
n tan 180
n
A = (1/2 x L x H) x n
Area = (1/2 x 1000 x __500__) x n = __5002__
n tan 180 n tan 180
n n
However, the programme (Excel) I will be using to make this calculation will not use degrees. Instead, I will use radians. I found out from an internet research, that one radian is a unit of angle, equal to the central angle inscribed in a circle the arc of which is equal in length to the radius of the circle. So if the circumference of a circle is 2Pir and the angle is 360o, 2PI radians is equal to 360o (r subtends one radian). Therefore, 180o is equal to PI.
Therefore General formula for n-sided polygon (regular) is:
Area
Evaluating Results
I have now found a general formula to find the area of any regular polygon. The table below shows the areas of different types of regular polygon.
The table on the below shows that as the number of sides of a regular polygon increases, so does the area,
No. of sides
Area
No. of sides
Area
5
68819.10
3
78022.30
6
72168.78
4
78237.25
7
74161.48
5
78410.50
8
75444.17
6
78552.18
9
76318.82
7
78669.52
0
76942.09
8
78767.80
1
77401.98
9
78850.94
2
77751.06
20
78921.89
The areas as plotted on a graph looks like this:
This graph is not a continuous graph and the data is discreet because you cannot have an area for 3 1/2 sides. I have kept the line though so I can show how it evens out.
The values in the table are correct. I checked the 5 and 6 sided shape with the answers I got for the pentagons and the hexagons.
No. of sides
Length Difference
No. of sides
Length Difference
5
21.52
3
3.11
6
4.82
4
2.68
7
0.83
5
2.33
8
8.27
6
2.05
9
6.52
7
.82
0
5.27
8
.62
1
4.35
9
.45
2
3.65
20
.31
Length difference = Difference between height of triangle in the polygon and the
radius of circle.
No. of sides
Area difference
No. of sides
Area difference
5
0758.37
3
555.17
6
7408.69
4
340.22
7
5415.99
5
166.97
8
4133.30
6
025.29
9
3258.65
7
907.95
0
2635.38
8
809.67
1
2175.49
9
726.53
2
826.41
20
655.58
Area Difference = Difference between area of polygon and area of a circle.
No. of sides
Height
No. of sides
Height
5
37.64
3
56.04
6
44.34
4
56.47
7
48.32
5
56.82
8
50.89
6
57.10
9
52.64
7
57.34
0
53.88
8
57.54
1
54.80
9
57.70
2
55.50
20
57.84
No. of sides
Angle of Interior (Degrees)
No. of sides
Angle of Interior (Degrees)
5
72.00
3
27.69
6
60.00
4
25.71
7
51.43
5
24.00
8
45.00
6
22.50
9
40.00
7
21.18
0
36.00
8
20.00
1
32.73
9
8.95
2
30.00
20
8.00
Height = This is the height of the triangle inside the polygon.
Angle of Interior = This is the angle at the top of the triangle inside the polygon. It is the same as the interior angle of the polygon.
No. of sides
Base Length
No. of sides
Base Length
5
200.00
3
76.92
6
66.67
4
71.43
7
42.86
5
66.67
8
25.00
6
62.50
9
11.11
7
58.82
0
00.00
8
55.56
1
90.91
9
52.63
2
83.33
20
50.00
Base Length = This is the base length of the triangle and can be obtained dividing 1000 by number of sides of the polygon.
From the tables I made, there are several patterns. As the number of sides increased, the base, the angle at the top of the triangle and area of each triangle get smaller. In addition, as the number of sides increased, the height and area of the polygon increased. These can all be seen from the tables.
Adding to that, as you increase the sides, the area is tending towards a certain number. This can be seen on the graph. To find that certain number I am going to calculate the area of a circle.
As the number of sides increase, the area of the polygon draw nearer to the area of the circle, which is 79577.47m2, This is also shown on the column, which is the difference between the area of the circle and the area of the polygon. This decreases as n increases. The area of the circle and the area of the n-sided polygon when n draws nearer to infinity are almost the same because a circle has an infinite number of sides. So as the number of sides for the polygon tends towards infinity, the area becomes more like the area of a circle.
Even when you keep increasing the number of sides in the polygon, it will never have the exact area as the circle. This is because the graph the polygon forms is an asymptote. It will never touch the exact value, which is the area of the circle.
In addition, the circle has an infinite number of sides and you cannot write infinity down as a number, so there is no way of calculating the polygon with infinite number of sides. Therefore, the circle will always have the largest area of perimeter 1000m.
To explain this I will look at the table. As the number of sides gets bigger, the base gets smaller, the angle at the top gets smaller, the height gets bigger, and the triangle becomes more and more like a straight line, which is like the radius of the circle (159.15m). This is also shown on the column on the table, which is the difference between the height and the radius. The difference gets smaller as n increases. To explain this even better, I have researched an A-level book to provide rigorous proof.
In the A-Level book, it says applying Tan to a very small angle is the same as the angle itself. Therefore, when n comes closer to infinity thus PI/n becomes very small, Tan (PI/n) would be the same as PI/n. To test this we can have a look at the table. As n gets larger, the column PI/n becomes the closer or even same as Tan (PI/n).
Tan Pi/n
Pi/n
Tan Pi/n
Pi/n
0.727
0.628
0.246
0.242
0.577
0.524
0.228
0.224
0.482
0.449
0.213
0.209
0.414
0.393
0.199
0.196
0.364
0.349
0.187
0.185
0.325
0.314
0.176
0.175
0.294
0.286
0.167
0.165
0.268
0.262
0.158
0.157
After finding this out, I can compare the formulas to find the area of a circle area of an n-sided shape and try to find the connection. I have used radians because the area for the circle has PI in it and the area of the polygons has 180o in it. Therefore, I can change the 180o into PI so it is easier to compare the formula.
Area of Circle Area Of n-sided Polygon
Therefore, as you can see the formulas for the height of the triangle in the polygon when n tends to infinity and the radius of the circle are equal. In addition, the formulas for the area of the circle and the area of the polygon when n tend to infinity are equal. That is why the values are almost equal in the table.
Conclusion
I therefore conclude that the largest shape of 1000m perimeter the farmer can use is a circle of radius 159.15m that gives an area of 79577.47m2.
Hamza Sharif Ms. Butt 10X1
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