In a rectangle where the perimeter is 1000m long, any 2 different sides equal 500. Therefore in a rectangle 150x350m, there are two sides opposite each other that are 150m long and 2 sides opposite each other that are 350m long. This means that you can work out the area even if you have the length of only one side of the rectangle. For example, with a base of 300m, I subtract 200 from 500, giving 300. Then I can multiply 200 by 300. I can put this into a formula:
1000 = X (500-X)
The X in this formula, represents the value of a side that we know.
I constructed a spreadsheet using my formula: 1000 = X (500-X). I have gone down the table taking 10m of the base each time.
Using the data from this spreadsheet I can create a graph of base length of a quadrilateral against area.
As you can see the graph has formed a curve with a peak. According to the spreadsheet and the graph, the rectangle with a base of 250m has the greatest area. This shape is a square, which is a regular quadrilateral. I only measured to the nearest 10m so now I will use rectangles with bases around 250m to prove that the square has the largest area.
All of these results fit in with the graph I have made, therefore making it reliable.
This has proved once again that regular shapes give you the largest areas.
Since these last two shapes had the largest area when they were in their regular form, I will now be using only regular shapes from now on. This makes it easier since there are many different variables for each polygon.
The next shape I will be using is a polygon.
In a pentagon there are five sides, allowing me to split the shape into 5 segments like so.
Each segment is an isosceles triangle, with a top angle of 72º, this is because it is a fifth of 360 which is a full turn. From this I can find out the other angles must be 54º each because the angles in a triangle add up to 180º and (180-72) ÷ 2 = 54.
Since isosceles triangles can be split into two equal right angle triangles, I can work out the area of the triangle using trigonometry. I also know that each side of the pentagon is 200m long, so the base of the right angled triangle must be 100m.
Using SOHCAHTOA I can work out that I need to use tangent.
72 ÷ 2 = 36
tan 36 = 100 ÷ X so X = 100 ÷ tan 36 = 137.638
This has given us the length of H, so now I can work out the area of this right angled triangle.
Area = ½ × b × H = ½ × 100 × 137.638 = 6881.90960 m²
I now have the area of half of one of the segments on the pentagon, so I simply multiply the area by 10 to get the area of the whole shape.
Area = 6881.910 × 10 = 68819.096m²
My results so far show that as the number of sides increases, the area increases. I am now going to investigate this further with a regular hexagon and a regular septagon, which are 6 and 7 sided shapes.
I will work out the area of these shapes the same way I did for the pentagon.
Hexagon
1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3
360 ÷ 6 = 60 ÷2 = 30
X = 83 1/3 ÷ tan 30 = 144.3375673
Area ½ × b × H = ½ × 83 1/3 × 144.338 = 6014.065304
12 × 6014.065 = 72168.78364m²
Septagon
1000 ÷ 7 = 142.857 ÷ 2 = 71.429
360 ÷ 7 = 51.429 ÷ 2 = 25.714
X = 71.429 ÷ tan 25.714 = 148.3229569
Area = ½ × b × H = ½ × 71.429 × 148.323 = 5297.248461
14 × 5297.248 = 74161.47845m²
My prediction was correct, as the number of sides increases, so does the area of the shape. Here is a table showing the number of sides against area.
Here is the formula I used to work out the area of the pentagon, hexagon and septagon.
1000 ÷ n ÷ 2 = b
360 ÷ n ÷ 2 = tan a
H = b ÷ tan a
Area = b × H
Area × 2n = total area of shape
That is the full formula I used and it works with the equilateral triangle and square too.
Now that I have this formula, I will work out the area for a regular octagon, nonagon and decagon.
Octagon
1000 ÷ 8 = 125 ÷ 2 = 62.5
360 ÷ 8 = 45 ÷ 2 = 22.5
62.5 ÷ tan 22.5 = 150.8883476
½ × 150.888 × 62.5 = 4715.260864
4715.261 × 16 = 75 444.17382
Nonagon
1000 ÷ 9 = 111.111 ÷ 2 = 55.556
360 ÷ 9 = 40 ÷ 2 = 20
55.556 ÷ tan 20 = 152.6376344
½ × 152.638 × 55.556 = 4239.93429
4239.934 × 18 = 76318.81721
Decagon
1000 ÷ 10 = 100 ÷ 2 = 50
360 ÷ 10 = 36 ÷ 2 = 18
50 ÷ tan 18 = 153.8841769
½ × 153.884 × 50 = 3847.104421
3847.104 × 20 = 76942.08843
Here is a spreadsheet I made using the formula:
As you can see, the higher the amount of sides the larger the area. This pattern has carried on throughout my investigation. So now I will investigate shapes with much larger amounts of sides and see if it still carries on.
20 sided shape
1000 ÷ 20 = 50 ÷ 2 = 25
360 ÷ 20 = 18 ÷ 2 = 9
25 ÷ tan 9 = 157.8437879
½ × 25 × 157.844 = 1973.047348
1973.047 × 40 = 78921.89393
50 sided shape
1000 ÷ 50 = 20 ÷ 2 = 10
360 ÷ 50 = 7.2 ÷ 2 = 3.6
10 ÷ tan 3.6 = 158.9454484
½ × 10 × 158.945 = 794.7272422
794.727 × 100 = 79472.72422
100 sided shape
1000 ÷ 100 = 10 ÷ 2 = 5
360 ÷ 100 = 3.6 ÷ 2 = 1.8
5 ÷ tan 1.8 = 159.1025798
½ × 5 × 159.103 = 397.7564494
397.756 × 200 = 7955.28988
1000 sided shape
1000 ÷ 1000 = 1 ÷ 2 = 0.5
360 ÷ 1000 = 0.36 ÷ 2 = 0.18
0.5 ÷ tan 0.18 = 159.1544195
½ × 0.5 × 159.154 = 39.78860487
39.789 × 2000 = 79577.20975
10000 sided shape
1000 ÷ 10000 = 0.1 ÷ 2 = 0.05
360 ÷ 10000 = 0.036 ÷ 2 0.018
0.05 ÷ tan 0.018 = 159.1549379
½ × 0.05 × 159.155 = 3.978873446
3.979 × 20000 = 79577.46893
100 000 sided shape
1000 ÷ 100000 = 0.01 ÷ 2 = 0.005
360 ÷ 100000 = 0.036 ÷ 2 = 0.0018
0.05 ÷ tan 0.018 = 159.154943
½ × 0.005 × 159.155 = 0.397887357
0.398 × 200000 = 79577.47152
Here is a graph to show the number of sides against area. As you can see from the graph, the line straightens as the number of sides increases.
The last shape I will be using is a circle. A circle has an infinite amount of sides, so my formula cannot be used to find the area of the circle.
The formula for the circumference of a circle is:
Circumference = × diameter
We know the circumference so:
1000 = ×diameter
Now we have to rearrange the formula so we can use it fo find out the diameter. We need the diameter because the diameter ÷ 2 = radius. And with the radius we can work out the area. So now I shall rearrange the formula.
1000 ÷ = 318.3098862
Now to find the radius we divide the circumference by two.
318.310 ÷ 2 = 159.1549431
To work out the area we use the formula
so × 159.155² = 79577.47155
Conclusion
From my investigation I discovered that as the number of sides on a shape increases, so does the area.
I used the formula to work out the area of the shapes:
1000 ÷ n ÷ 2 = b
360 ÷ n ÷ 2 = tan a
H = b ÷ tan a
Area = b × H
Area × 2n = total area of shape
This worked for all my shapes apart from the circle. The circle gave the largest area because it has infinite sides.
In my investigation I also saw how that a regular shape gives a larger area too.
The area that the circle gave was 79577.472m²
