Looking at this area it is clear now that the area of a regular (equilateral) triangle is greater than the area of an isosceles triangle. The third type of triangles is a scalene.
-
Scalene: no sides are equal
B
A C
AB = BC = AC
This type of triangle has no equal sides or angles. There are countless combinations of numbers that add up to give the perimeter of 1000 meters. This makes it very hard to get the maximum area of a scalene triangle with a perimeter of 1000m. If we look back at the other two types of triangles we notice that as the sides get closer in length to each other, the area of the shape increases as well. As a conclusion regular 3 sided shapes (equilateral triangles) have the maximum area of all three types of triangles.
In the next step of my investigation I am going to try shapes with 4 sides (rectangles).
Rectangles:
Area of any rectangle = base x height
Looking at the table above I have noticed that the area keeps increasing as the length and width get closer. When the length and width were 250 the area was the maximum. This shape is called a square. A square is a shape with four equal sides. I am going to draw a graph of these results so that the relationship between the length of the rectangle and its area is clearer.
By looking at the graph it is now clear that the area keeps increasing until the shape is a square and then decreases at the same rate that it has increased by.
A square: all sides are equal
A B
D C
Area of a square= length2
= 2502
= 62500
Comparing the area of a square with that of an equilateral triangle, the square has a bigger area. Therefore 4-sided regular shapes with a perimeter of 1000 meters have a bigger area than 3-sided regular shapes with the same perimeter.
62500 > 48113
The next step in my investigation is to look at the area of five-sided regular shapes (regular pentagons) with a perimeter of 1000 meters.
Pentagons: five sides
A
E B
D N C
AB = BC = CD = DE = 200meters
By dividing the pentagon ABCDE into five equal triangles I can get its area.
In triangle DMC:
MN is the perpendicular bisector of DC.
DN = CN = 200 / 2
= 100m
Angle DMC = 360 / 5
= 72°
Therefore angle NMC = 72 / 2
= 36°
Since angle MNC = 90°
Therefore angle NCM = 180 – (90 + 36)
= 54°
tan = opp / adj
tan 36 = 100 / adj
adjacent = 100 / tan 36
= 137.6387192
MN = 137.62387192 meters
Area of triangle MCN = ½ x base x height
= ½ x 100 x 137.6387192
= 6881.909602 m2
Therefore area of triangle DMC = area of triangle MCN x 2
= 6881.909602 x 2
= 13763.8192 m2
Area of ABCDE = area of triangle DMC x 5
= 68819.096 m2
= 68819 (nearest d.p.)
Comparing the area of a regular 5-sided shape with a perimeter of 1000 meters with the area of a regular 4-sided shape with the same perimeter, I can see that the area of a regular pentagon is greater than the area of a square.
68819 > 62500
The next step in my investigation I am going to look at the area of regular shapes with 6 sides (Hexagons) and a perimeter of 1000 meters.
Hexagon: 6 sides
A B
F C
E N D
AB = BC = CD = DE = EF = 1662/3 meters
By dividing the hexagon ABCDEF into six equal triangles I can get its area.
In triangle EMD:
MN is the perpendicular bisector of ED.
EN =ND = 166 2/3 / 2 = 250/3
= 83 1/3m
Angle EMD = 360 / 6
= 60°
Therefore angle NMD = 60 / 2
= 30°
Since angle MND = 90°
Therefore angle MDN = 180 – (90 + 30)
= 60°
tan = opp / adj
tan 30 = 83 1/3 / adj
adjacent = 83 1/3 / tan 30
= 144.3375673
MN = 144.3375673meters
Area of triangle MDN = ½ x base x height
= ½ x 83 1/3 x 144.3375673
= 6014.065304 m2
Therefore area of triangle EMD = area of triangle MDN x 2
= 6014.065304 x 2
= 12028.13061 m2
Area of ABCDEF = area of triangle EMD x 6
= 12028.13061 x 6
= 72168.78365m2
= 72169 (nearest d.p.)
After calculating the area of a regular hexagon I can say that the area of a regular 6-sided shape with a perimeter of 1000 meters is greater than the area of a 5-sided shape with the same perimeter.
72169 > 68819
The next step in my investigation is to get the area of an eight-sided shape (Octagon) with a perimeter of 1000 meters, and compare with the area of the pervious shapes.
An Octagon: 8 sides
A B
H C
G D
F E
N
AB = BC = CD = DE = EF = GH = HA = 125 meters
By dividing the octagon ABCDEFGH into eight equal triangles I can get its area.
In triangle FME:
MN is the perpendicular bisector of FE.
FN = EN = 125 / 2
= 62.5 m
Angle EMD = 360 / 8
= 45
Therefore angle NME = 45 / 2
= 22.5°
Since angle MNE = 90°
Therefore angle MEN = 180 – (90 + 22.5)
= 67.5°
tan = opp / adj
tan 22.5 = 62.5 / adj
adjacent = 62.5 / tan 22.5
= 150.8883476
MN = 150.8883476 meters
Area of triangle MEN = ½ x base x height
= ½ x 62.5 x 150.8883476
= 4715.260864 m2
Therefore area of triangle FME = area of triangle MEN x 2
= 4715.260864 x 2
= 9430.521728 m2
Area of ABCDEFGH = area of triangle FME x 8
= 9430.521728 x 8
= 75444.17382 m2
= 75444 (nearest d.p.)
After calculating the area of a regular octagon I can say that the area of a regular 8-sided shape with a perimeter of 1000 meters is greater than the area of a 6-sided shape with the same perimeter.
75444 > 72169
Instead of having to do the same steps every time I am working out the area of a polygon with a perimeter of 1000 meters, I am going to try and find a formula for all regular polygons.
Formula of a polygon:
I am going to use a polygon to demonstrate what is happening.
A
E B
D N C
n = number of sides of shape
x = angle NMC
h = height of one of the triangles
b= base of big triangle (DMC)
b= 1000/n
2
x = 360 / 2n (because x is the angle in half one of the n triangles)
tan x = (1000/n divided 2) / h
h = (1000/n / 2) / tan x
area of small triangle (MNC) = ½ X base X height
= ½ X 1000/n X ( 1000/n / tan x)
- 2
Area of big triangle (DMC) = 2 (½ X 1000/n X ( 1000/n / tan x) )
2 2
Area of the whole shape = 2n (½ X 1000/n X ( 1000/n / tan x) )
2 2
By simplifying I got:
n ( 1000/n X ( 1000/n / tan (360) )
- 2 2
n ((500 / tan(180) X (500))
n n n
n ( 500 X 1 X n X 500 )
n tan 180 n
n ( 250000 )
n2 tan (180)
n
Formula for any regular polygon equals:
250000
n tan (180)
n
Using this formula I can get the area of different regular polygons, but first I have to try it with a pentagon and see if the results match the ones before or not.
250000
5 tan (180)
5
250000
5 tan 36
= 68819.09602
This is exactly the answer I got when using the other method.
I can now make a table of the area of different shapes with different number of sides.
Looking at this table. It is very obvious that the area increases with the increase of the number of sides. A graph will make this observation more clear.
From the graph I can see that as the number of sides increase the area increases. In the previous shapes, as the number of sides increased the shape was looking more like the shape of a circle. Since the number of sides in a circle cannot be counted, I think that the circle will have the greatest area compared to all the previous shapes.
A Circle:
B
AB is the radius of the circle with perimeter 1000 meters.
Circumference = d
Therefore d = Circumference
= 1000
= 318.3098862 m
Therefore radius = d / 2
= 159.1549431 m
Area of a circle = r2
= X (159.1549431) 2
= 79577.47155
= 79577 (nearest d.p)
The area of the circle with perimeter 1000 meters is greater than the area of all the other shapes that I have tried. Even when there is 100000 sides, still the area of the circle is greater.
Using the formula to get the area of a shape with 100000 sides:
250000
100000tan (180)
100000
= 79577.47152
The area of a circle is equal to the area of a shape with 1000000 sides.
Using the formula to get the area of a shape with 1000000 sides:
250000
1000000tan (180)
1000000
= 79577.47155
A shape with sides more than 1000000 will still have the same area as a circle. It is not possible to establish a shape with this number of sides.
Using the formula to get the area of a shape with 10000000 sides:
250000
10000000tan (180)
10000000
= 79577.47155
This means that no other shape will have a greater area than a circle if they both have the same perimeter.
In conclusion, according to my investigation I can say that no other shape will have an area greater than that of a circle if they both have the same perimeter. Therefore the circle is the best choice for fencing the maximum area with a perimeter of 1000 meters.
