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.
I have found that a square has the greatest area of the rectangles group, I will now find the triangle with the largest area. Because in any scalene or eight angled triangle there is always more than 1 variable side, there are countless combinations, so I am only going to use isosceles triangles. This is because if I know the base length, then I can work out the other 2 lengths, since 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.
To work out the area I need to recognise the height of the triangle. When the base is different heights, to work out the correct height I use Pythagoras' theorem:
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. 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 closest to 333 as I did whilst investigating rectangles:
This has proved that once again, the regular shape has the largest area.
The next shape that I am going to investigate is the pentagon. The last 2 shapes have had the largest areas when they are regular, so I will use regular shapes from now on which will be a lot easier as many of the other shapes have an infinite number of different variables.
There are 5 sides so I’ll divide it into 5 segments. Each segment possesses the properties of an isosceles triangle, with the top angle equalling 720. (A fifth of 3600). Now I can work out both the remaining angles by subtracting 72 from 180 and dividing the answer by 2. This results in two 540 angles. 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. Using SOH CAH TOA I’ve worked out that Tangent (Tan) will be required.
This has given me the length of H so I can work out the area:
A = 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 (5 X 2) and I get the area of the shape:
A = 6881.910 X 10
=68819.096m2.
All of the results that I have got so far have shown that as the number of sides increase, the area also 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
A = 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
A = ½ X b X H
=½ X 71.429 X 148.323
=5297.260
5297.260 X 14
=74161.644m2
My predictions were correct. As you can see, it is quite clear that as the number of sides increase, the area increases. Below is a table of the number of sides against area that I have calculated so far:
From the method that I used to find the area for the pentagon, hexagon and heptagon I can 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 but as I need to find half of that value I need to put 1000 / 7 / 2. The method that I used above has been put into an equation below.
1000 ÷ 7 = 142.857 ÷ 2 = 71.429
360 ÷ 7 = 51.429 ÷ 2 = 148.323
Area = ½ X b X H = ½ X 71.429 X 148.323 = X
5297.244 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:
These are exactly the same as the results in my previous results proving that they are correct.
Now that I’ve this equation, I will use it to work out the area for a regular octagon (8 sides), nonagon (9 sides), and decagon (10 sides):
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 a much greater amount of sides:
