= 3 x 5
= 15m2
Now we can work out the area of the triangles. I will now work out the different areas starting with 10m as the base and 450m for each side.
By using Pythagoras’s theorem
(52)+ b2 = (4502)
25 + b2 = 202500
b2 = 202475
b = 449.972m
½ 10 x 449.972
A = 2249.861m2
(102) + b2 = (4902)
100 + b2 = 240100
b2 = 240000
b = 489.89m
Area
½ 20 x 489.89
A = 4898.979m2
(152) + b2 = (4852)
225 + b2 = 235225
b2 = 235000
b = 484.768
Area
½ 30 x 484.768
A = 7271.520m2
As we can see by increasing the base the area of the triangle also increases. Eventually the area of the triangle will peak; this is when it reaches its maximum area.
Below is a table showing the area of triangles as the base increases.
As you can see from the table above the maximum area occurs when all the sides are the same length. The length is 333 1/3m Hence the triangle with the largest area is an equilateral triangle.
Four sided shape
Again for the four sided shapes I will start with a base of a 10m and increase keep on increasing it. I predict that the largest area again will occur when all the sides will be equal. I will first start with a rectangle. Below will show a quadratic curve which will show what lengths give the maximum area for a four sided shape.
Perimeter = 2(500–x) + 2(x)
Area of rectangle = x(500-x)
X(500 – x) is a quadratic curve and looks like this.
y = x(500-x)
The formula for working out the area of a rectangle is.
Area = Length x Width
E.g.
Area = Length x Width
A = 100 x 400
A = 4000m2
A = 10 x 490
A = 4900m2
A = 20 x 480
A = 9600m2
A = 30 x 470
A= 14100m2
A = 40 x 460
A = 18400m2
A = 50 x 450
A = 22500m2
After showing 5 rectangles we can see a pattern which increase, this was the same with the triangle. The width of the rectangle will carry on increases until it reaches its peak point again. Below is a table showing as the width increases so does the area.
Again like the triangle the maximum area was achieved when the all sides in the rectangle were equal, a rectangle with equal sides is a square. A square is a regular four sided shape. We can now say that the maximum area occurs for any sided shape when it is in the form of a regular polygon.
Five sided shape
From working out the areas of triangles and four sided shapes, we can see a pattern emerging. This is patter is for achieving
Prediction
I predict that the pentagon with all equal sides will have the maximum area. This is the one which lengths are 200m.
For the five sided shape I will be using the same process. To work out the area of a 5 sided shape you have you divide the shape into equal isosceles triangles.
The pentagon will look like this. To work out the total area we need to take out 1 triangle and work out its area.
To work out the area for this triangle we have to again divide it into 2 to make it into a right angled triangle.
At this point we do not no the height but by using trigonometry we can work it out.
Tan = opposite
adjacent
Tan 36 = 5
h
h = 5
Tan 36
h = 6.882m
Now we have the height we can work out the area of the right angle triangle.
A = ½ base x height
A = ½ 5 x 6.882
A = 17.205m2
To work out the area of the isosceles triangle we multiply the area of the right angle triangle by 2
Area of isosceles triangle = 17.205 x 2
A = 34.410 m2
To work out the total area of the pentagon you multiply the area of the isosceles triangle by 5
Area of pentagon = 34.410 x 5
A = 172.048m2 (2dp)
The pentagon that has the largest area is a regular pentagon.
Area of isosceles triangle
A = ½ 200 x 137.638 Tan 36 = 100
h
A = 13763.819m2
h = 100
Area of pentagon Tan 36
A = 13763.819 x 5 h = 137.638m
A = 68819.10m2 (2dp)
Six sided shape
As we have seen in the previous shapes the maximum area was achieved by all the sides of the shape being equal. So I will work out the hexagon with equal sides. This is called a regular hexagon.
Above is how to work out the area of a hexagon. From the original hexagon you take out one of the isosceles triangles. Then you divide that triangle into 2 to form 2 right angle triangles. To work you the area of the right angle triangle you use trigonometry.
Tan = opposite
adjacent
Tan 30 = 83 1/3
h
h = 83 1/3
Tan 30
h = 144.338m
Area of right angle triangle = ½ Base x height
A = ½ (83 1/3) x 144.338
A = 6014.065
Area of isosceles triangle = area of right angle triangle x 2
A = 6014.065 x 2
Area of isosceles triangle = 12028.131m2
Area of hexagon = area of isosceles triangle x 6
A = 12028.131 x 6
A = 72168.78m2 (2dp)
Below is a table that shows the maximum area for a certain number of sides.
From looking at the table we can say that the area increases at the number of sides do. We can say that the numbers of side are inversely proportional to the area.
A
When ‘n’ tends to infinity
n ∞ n = number of sides
The length of each side tends to 0
l 0 l = length
and the area tend a fixed value
A a fixed value A = area
As the number of sides increase the shape becomes more circular, and a circle can be considered to be a regular polygon with infinite sides.
Below is a graph showing as the maximum area for a shape with a certain number of sides.
Circle
Area of circle = ∏r2
Since we do not know the radius we have to work it out by using the perimeter formula which is
Circumference = 2∏r
1000 = 2 x ∏ x r
500 = ∏ x r
500 = r
∏
r = 159.155m (2dp)
Area = ∏r2
A = ∏ (159.1552)
A = 79577.472m2 (3dp)
Conclusion
In conclusion we can say that as the number of sides increase so does the area of a regular polygon. This is shown in the table above. We can also say that there is a general formula to find out the maximum area for a regular polygon and that is;
Area for any regular 1000 2
polygon = n
4Tan 360
where n =number of sides 2n
The diagram below shows how a circle achieves the largest area.
As the number of sides increase it takes up a larger area, this eventually leads to the circle which has the greatest area.
