Height (a) Hypotenuse (b)
Half of base (c)
In this case we wont to find (a) the formula for this is b2-c2=a2. Square rooting (a) would then give us the height.
To do this on the table it was slightly different as the symbols on the computer had to be substituted for mathematical symbols e.g. SQRT((POWER(C2,2)-POWER((B2/2),2))) was the formula typed into the table to produce the results.
Once the height column was complete I had sufficient data to find the area of the triangle. The formula for the area of a triangle is half base times height.
Therefore all I had to do for the final column was type in this formula so the computer could understand it. In the first case it would be (B2*0.5)*E2. This now gave me the area of a triangle with the lengths given in the columns B, C and D.
Now that all the data is filled in I can find the maximum area for this polygon, which is highlighted on the table as, 48578.45 meters squared. Therefore this is the maximum area of containment the farmer can have in his field if he uses a triangular shaped fence with a perimeter of 1000 meters.
After I had investigated a three-sided pentagon I moved on to a four sided which is a rectangle. For this I predicted that the largest area would be found when the rectangle turns into a square where all the sides are equal
Quadrilateral:
After the Triangle I investigated the next polygon. Which was four sided. I went about it in the same way as the triangle and assembled a table.
I predicted that a four-sided shape would have a larger area than a triangle as it has more sides.
The table was composed in much the same way and a number of combinations of side length to equal 1000 were laid out. This time however it was much more simple as there was no need to work out the height. The formulas for the side lengths were derived in much the same way as the triangle, but there is a different formula for the area of a regular quadrilateral. Which is Width multiplied by the length. The area for a quadrilateral was found to be considerably larger than the area for a triangle.
Area of Quadrilateral
Area in Triangle
As shown by this diagram the area for the Quadrilateral is larger than the triangle. This is the first sign of a trend that will be apparent as I proceed with the investigation. This is the fact that the more sides the polygon has the larger the area will be.
I found the maximum area of the quadrilateral to be a square, with lengths of 250 on each side creating an area of 62500 meters squared as shown on the table.
Pentagon:
Once I found out that there appeared to be a pattern the next test would be to do the same experiment on the next polygon. A pentagon. A regular pentagon however cannot have a series of different combinations for the side lengths, as there is only one combination where all sides (the perimeter) add up to 1000 meters or else the pentagon would be irregular. This is 1000 divided by 5, which is 200.
200m 200m Perimeter = 1000
200m 200m
200m
There for the area found is the maximum area for a regular pentagon with a perimeter of 1000.
To find this, the pentagon has to be divided up into 5 triangles each exactly the same. The area of one of these triangles is found and then multiplied by 5 to get the overall area of the pentagon. To find the area of the triangles the rule half base times height comes into play again. This time however, the triangle has no known lengths other than the base. But the angle is known. To find the angle 5 must divide 360. So the angle of the triangle is 72 decrees. To find the height, as before, a perpendicular must be drawn so the new triangle is a right-angled triangle with two unknown lengths and one known angle.
36
Height Hypotenuse
100m (half base)
The angle is 36 because it is half the previous angle of 72. To find the height a formula must be made using trigonometry. Considering we know the side opposite the angle and want to find out the side adjacent to the angle and also knowing the angle itself, Tan must be used to solve the height. The formula is Height=100/Tan36 so height = 137.638192.
Now the height has been found the area can be found. By multiplying this number with 100, half the base. Now the area of the triangle is found it is just a question of multiplying this area by 5, which equals 68819.09602 meters squared. There fore this is the maximum area for the regular pentagon. Which is again larger than the previous polygon. This strongly supports my theory of the more sides the shape has the larger the maximum area will be. To prove this I need to experiment on more time with the next polygon. A hexagon.
Hexagon:
This can be investigated in exactly the same way as the pentagon, but the numbers are different. So the sides are now 1000 divided by 6 and the angles are now 360 divide by 6.
30
166.6666666666667 (1000/6) 83.333(half base
So Tan is used again.
Height = 83.333333/Tan30
=144.3375673
Height x base = Area
144.3375673 x83.3333 = 12028.1306
X 5 = 72168.74971
This is the maximum area for the hexagon and is again larger than the previous polygon.
Polygons:
Now the pattern has been proven there must be an answer to the initial question. What shape of the plot of land contains the maximum area? Following the trend the shape with the largest area must be a shape with infinite sides.
A Circle.
Now I had found the maximum area for a few polygons I wanted to know all the maximum areas and how they differed. To do this I had to find the formula for an n sided polygon.
To do this I set out a table quite different from the others, which included everything needed to find out the area of a polygon.
The columns needed to find theses maximum areas were firstly the n’th column which determined the amount of sides, the angle in decrees which was determined by the formula 360/n’th term in the first cell, the angle in radians which is the units in which the computer reads the angles which is found by adding the formula
Base, which is 1000/the n’th term and height of the triangles within the polygons, which is found using the same method as before. Dropping the perpendicular and calculating the height using Tan. All this information is enough to find the area for every triangle within the polygons using the same methods as before. Finally to find the maximum areas in the last column a simple formula is needed which is the triangle area multiplied by the n’th term this will give you the maximum area for every polygon up to infinity.
I conclude that this experiment agrees with firstly my prediction, which was that the maximum area would increase as the polygon sides increased. The results table and graphs show this clearly. Also I conclude that the best shape to use to contain this plot of land is a circle, as it will give the largest area with a perimeter of 1000. I think that this investigation went well and I managed to prove my theory and discover that although perimeters in a shape can stay the same increasing the shapes sides alters the area dramatically.
I could expand on this experiment by finding the formula in a shortened version, as one fraction so it would be faster and more practical to discover maximum areas.
