* See next page for Graph
Graph
Triangle (half × base × perpendicular height)
I am only going to use isosceles triangles. This is because I know the base length, so I can work out the other 2 lengths, because they are the same.
Example…
If the base is 200m long then I can subtract that from 1000 and divide it by 2.
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’ theorem. Below is the formula and area when using a base of 200m.
H2 = h2 – a2
H2 = 4002 – 1002
H2 = √150000
H = 387.298
(H=Height)
Area: ½ X 200 (B) X 387.298 (H) = 38729.833m
*NOT TO SCALE
Regular 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.
Graph
Conclusion
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.
Regular Pentagon
Because there are 5 sides, I can divide it into 5 segments. Each segment is an isosceles triangle, with a top angle of 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 (1000 divided by 5), so the base of the triangle is 100m.
Using Trigonometry I can work out the unlabeled side.
SOHCAHTOA
Tangent = Opposite side/Adjacent side
Opposite= 100cm and Adjacent= h
Tan36o= 100 _
h
h= 100
Tan36 o
h= 137.638m
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
Now I 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 now going to investigate this further with a regular hexagon (6 sides) and a regular heptagon (7 sides).
I am going to apply the same method as before to solve the areas of these two regular shapes.
Hexagon:
1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3
360 ÷ 6 = 60 ÷ 2 = 30
COS= Adjacent/Hypotenuse
Adjacent= 831/3 m and Hypotenuse= h
= 831/3 / h
=83.33333335/COS60
=166.6666667 = 166.66666672 – 83.33333352
= √20833.33332
h = 144.338
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
TAN= Opposite/Adjacent
= x/71.4285 = h= 148.323
Area = ½ X b X H = ½ X 71.429 X 148.323 = 5297.260
5297.260 X 14
= 74161.644m2
Octagon
TAN= Opposite/Adjacent
TAN 67.5o = h/62.5
h= 62.5xTan67.5 o
=Area= ½ x 125 x h
COS= Adjancent/Hypotenuse = 62.5/x
= 62.5/COS67.5=163.320m
TAN=62.5 x Tan67.5= 150.888m
Area= ½ x 150.888 x 125= 9430.3 x 8
=73444m2
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:
Circle
I have already come to the conclusion that: as the number of side’s increases, the area increases. Circles have no sides and are infinite. This should prove my prediction correct.
C= 2 π x r
1000= 2 x 3.14 x r
1000= 6.28 x r
1000
6.28 = r
r= 159.23
Area = πr2
= 3.14 x 159.23 x 139.23
=79612.2m2
From this I conclude that a circle has the largest area when using a similar circumference.
