The highlighted row gives the biggest area; after I go past this row I start to repeat my self.
A square with the perimeter 1000m gives me the largest area. I will further investigate this because I was going up in increments of 10. I will go into the decimal widths and lengths.
Below is a table of results.
From this table I can see that the perfect square with the perimeter 1000m produces the largest area.
From this graph I can see that the highest point is on the 250m mark. The graph is a parabola and is symmetrical. At the centre point of the graph I reach the highest value for the area, after this I just repeat my self.
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 * 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 base length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.
Area = x(500 – x)
Key: x = width of rectangle
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 I will investigate triangles. I will first investigate isosceles triangles because triangles such as scalene have more than one different variable so there are millions of possible combinations. If I know the base length I can work out the length of the other two sides because they are the same.
For example if the base is 200m. I can take this away from 1000m and that answer I can divide by 2. Or more simply:
Side = (1000 – 200) ÷ 2 = 400
To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras’s theorem. Below is the formula and area when using a base of 200m.
B2 = C2 – A2
B2 = 4002 – 1002
B2 = 150000
B = 387.298
½ * 200 * 387.298 = 38729.833m.
On the next page there is a table of results for triangles.
From the table I can see that the regular shape of the family has given me the largest area.
I can see that the regular shapes of these families give 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.
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.
Tangent = opposite/adjacent
Tan54 = o / a
O = tan54 * a
O = tan54 *100
O = 137.638192m
137.638192m * 200 =27527.63841
27527.63841/2=13763.81921
13763.81921*5=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 * b * H = ½ * 83 1/3 * 144.338 = 6014.065
6014.065 * 12 = 72168.784m2
Heptagon:
1000 ÷ 7 = 142.857 ÷ 2 = 71.429m
360 ÷ 7 = 51.429 ÷ 2 = 25.714 degrees
Area = ½ * b * H = ½ * 71.429 * 148.323 = 5297.260
5297.260 * 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
To find the general formula for an n sided regular polygon I am going to construct a formula.
To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tangent
A = [(1000/n)/2]/ Tan (360/s)/2
Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it’s A * base *0.5 * number of sides.
As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.
Diameter = 318.47m
pR²
3.14 * 159.235 (2 d.p) * 159.235 (2 d.p) = 79617.16561m²
I can see that this produces the largest area.
