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.
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.
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.
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.
The next shape that I am going to investigate is the pentagon.
Using SOHCAHTOA I can work out that I need to use Tangent.
O 100
T tan36
This has given me the length of H so I can work out the area.
Area = ½ X b X H = ½ 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:
1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.
360 ÷ 6 = 60 ÷ 2 = 30
Area = ½ X b X H = ½ x 83 1/3 X 144.338 = 6014.065
6014.065 X 12 = 72168.784m2
Heptagon:
1000 ÷ 7 = 142.857 ÷ 2 = 71.429
360 ÷ 7 = 51.429 ÷ 2 = 25.714
Area = ½ X b X H = ½ 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
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.
1000 ÷ 7 = 142.857 ÷ 2 =
360 ÷ 7 = 51.429 ÷ 2 =
Area = ½ X b X H = ½ 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.
Now that I have this equation, I am going to use it to work out the area for a regular octagon, nonagon and decagon.
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.
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.
1000/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
