The fencing problem
There is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.
00m
50m
400m
I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. Below are 2 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.
350m
In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side. To work out the area of a rectangle with a bas length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.
000 = x(500 - x)
Below is a table of results, worked out using the formula. I have gone down by taking 10m off the base every time.
Height (m)
x
Area (m2)
0
500
0
0
490
4900
20
480
9600
30
470
4100
40
460
8400
50
450
22500
60
440
26400
70
430
30100
80
420
33600
90
410
36900
00
400
40000
10
390
42900
20
380
45600
30
370
48100
40
360
50400
50
350
52500
60
340
54400
70
330
56100
80
320
57600
90
310
58900
200
300
60000
210
290
60900
220
280
61600
230
270
62100
240
260
62400
250
250
62500
260
240
62400
270
230
62100
280
220
61600
290
210
60900
300
200
60000
310
90
58900
320
80
57600
330
70
56100
340
60
54400
350
50
52500
360
40
50400
370
30
48100
380
20
45600
390
10
42900
400
00
40000
410
90
36900
420
80
33600
430
70
30100
440
60
26400
450
50
22500
460
40
8400
470
30
4100
480
20
9600
490
0
4900
500
0
0
Using this formula I can draw a graph of base length against area.
As you can see, the graph has formed a parabola. According to the table and the graph, the rectangle with a base of 250m has the greatest area. This shape is also called a square, or a regular quadrilateral. Because I only measured to the nearest 10m, I cannot tell whether the graph is true, and does not go up just to the sides of 250m. I will work out the results using 249m, 249.5 and 249.75
Base (m)
Height (m)
Area (m2)
249
251
62499
249.5
250.5
62499.75
24975
250.25
6249993.75
250
250
62500
250.25
249.75
62499.9375
250.5
249.5
62499.75
251
249
62499
All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.
Now that I have found that a square has the greatest area of the ...
This is a preview of the whole essay
Base (m)
Height (m)
Area (m2)
249
251
62499
249.5
250.5
62499.75
24975
250.25
6249993.75
250
250
62500
250.25
249.75
62499.9375
250.5
249.5
62499.75
251
249
62499
All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.
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 or eight angled triangle, there is more than 1 variable, there are countless combinations, so I am only going to use isosceles triangles. This is because if know the base length, then I can work out the other 2 lengths, because they are the same. If the base is 200m long then I can subtract that from 1000 and divide it by 2. This means that I can say that
Side = (1000 - 200) ÷ 2 = 400. This can be rearranged to the following formula.
To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras' theorem. Below is the formula and area when using a base of 200m.
H2 = h2 - a2
H2 = 4002 - 1002
H2 = 150000
H = 387.298
/2 X 200 X 387.298 = 38729.833m.
Below is a table of result for isosceles triangles from a base with 10m to a base with 500m.
Side = . Where X is the base length. I have used this formula to work out the area when the base is different heights. To do this I need to use Pythagoras' theorem. Below is a diagram of the isosceles triangle.
Base (m)
Side (m)
Height (m)
Area (m )
0
500.0
500.000
0.000
0
495.0
494.975
2474.874
20
490.0
489.898
4898.979
30
485.0
484.768
7271.520
40
480.0
479.583
9591.663
50
475.0
474.342
1858.541
60
470.0
469.042
4071.247
70
465.0
463.681
6228.832
80
460.0
458.258
8330.303
90
455.0
452.769
20374.617
00
450.0
447.214
22360.680
10
445.0
441.588
24287.342
20
440.0
435.890
26153.394
30
435.0
430.116
27957.557
40
430.0
424.264
29698.485
50
425.0
418.330
31374.751
60
420.0
412.311
32984.845
70
415.0
406.202
34527.163
80
410.0
400.000
36000.000
90
405.0
393.700
37401.537
200
400.0
387.298
38729.833
210
395.0
380.789
39982.809
220
390.0
374.166
41158.231
230
385.0
367.423
42253.698
240
380.0
360.555
43266.615
250
375.0
353.553
44194.174
260
370.0
346.410
45033.321
270
365.0
339.116
45780.727
280
360.0
331.662
46432.747
290
355.0
324.037
46985.370
300
350.0
316.228
47434.165
310
345.0
308.221
47774.209
320
340.0
300.000
48000.000
330
335.0
291.548
48105.353
333.3
333.3
288.675
48112.522
340
330.0
282.843
48083.261
350
325.0
273.861
47925.724
360
320.0
264.575
47623.524
370
315.0
254.951
47165.931
380
310.0
244.949
46540.305
390
305.0
234.521
45731.554
400
300.0
223.607
44721.360
410
295.0
212.132
43487.067
420
290.0
200.000
42000.000
430
285.0
87.083
40222.817
440
280.0
73.205
38105.118
450
275.0
58.114
35575.624
460
270.0
41.421
32526.912
470
265.0
22.474
28781.504
480
260.0
00.000
24000.000
490
255.0
70.711
7324.116
500
250.0
0.000
0.000
.
Because the regular rectangle was the largest before, I added 333.3 as a base length. This is the length of the base of a regular triangle. Below is a graph of the base against area.
As you can see this graph is a-symmetrical. The regular triangle seems to have the largest area out of all the areas but to make sure I am going to find out the area for values just around 333.
Base (m)
Side (m)
Height (m)
Area (m2)
333
333.5
288.964
48112.450
333.25
333.4
288.747
48112.518
333.3
333.4
288.704
48112.522
333.5
333.3
288.531
48112.504
333.75
333.1
288.314
48112.410
334
333.0
288.097
48112.233
This has proved that once again, the regular shape has the largest area.
Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on. This would also be a lot easier as many of the other shapes have millions of different variables.
H
h
O
The next shape that I am going to investigate is the pentagon.
Because there are 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle, with the top angle being 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each. Because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle is 100m.
b
Using SOHCAHTOA I can work out that I need to use Tangent.
H= = =137.638
O 100
T tan36
This has given me the length of H so I can work out the area.
Area = 1/2 X b X H = 1/2 x 100 X 137.638 = 6881.910
I now have the area of half of one of the 5 segments, so I simply multiply that number by 10 and I get the area of the shape
Area = 6881.910 X 10 = 68819.096m2.
All of the results that I have got so far have shown that as the number of side's increases, the area increases. I am going to investigate this further with a regular hexagon (6 sides) and a regular heptagon (7 sides).
I am going to work out the area of the 2 shapes using the same method as before.
Hexagon:
000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.
360 ÷ 6 = 60 ÷ 2 = 30
Area = 1/2 X b X H = 1/2 x 83 1/3 X 144.338 = 6014.065
6014.065 X 12 = 72168.784m2
Heptagon:
000 ÷ 7 = 142.857 ÷ 2 = 71.429
360 ÷ 7 = 51.429 ÷ 2 = 25.714
Area = 1/2 X b X H = 1/2 X 71.429 X 148.323 = 5297.260
5297.260 X 14 = 74161.644m2
My predictions were correct and as the number of side's increases, the area increases. Below is a table of the number of sides against area
No. of sides
Area (m2)
3
48112.522
4
62500.000
5
68819.096
6
72168.784
7
74161.644
From the method that I used to find the area for the pentagon, hexagon and heptagon I ca work out a formula using n as the number of sides. To find the length of the base of a segment I would divide 1000 by the number of sides, so I could put , but as I need to find half of that value I need to put . All
The method that I used above has been put into an equation below.
000 ÷ 7 = 142.857 ÷ 2 =
360 ÷ 7 = 51.429 ÷ 2 =
Area = 1/2 X b X H = 1/2 X 71.429 X 148.323 = X
5297.260 X 14 =
That is the full equation and it works on all of the shapes that I have already done, giving the same answers as before. Below is a table showing the answers I got whet I used the equation.
No. of sides
Area (m2)
3
48112.522
4
62500.000
5
68819.096
6
72168.784
7
74161.644
Now that I have this equation, I am going to use it to work out the area for a regular octagon, nonagon and decagon.
No. of sides
Area (m2)
8
75444.174
9
76318.817
0
76942.088
As you can see, the larger the number sides on the shape, the larger the area is. This pattern has carried on going for all of the shapes that I have investigated, so I am going to investigate shapes with the following amount of sides:
20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.
Below is a table showing the results that I got.
No. of sides
Area (m2)
20
78921.894
50
79472.724
00
79551.290
200
79570.926
500
79576.424
000
79577.210
2000
79577.406
5000
79577.461
0000
79577.469
20000
79577.471
50000
79577.471
00000
79577.471
On the next page is a graph showing the No. of sides against the Area.
As you can see form the graphs, the line straightens out as the number of side's increases. Because I am increasing the number of sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using o. To work out the circumference of a circle the equations is od. I can rearrange this so that the diameter is circumference/o. From that I can work out the area using the or2 equation.
000/o = 318.310
318.310/2 = 159.155
o X 159.1552 = 79577.472m2
If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference.
