Rectangles – 50Metre pitch
I notice from the diagrams and table that as the length increases, the area increases but after the length is 250m, when the length becomes 300+, the area starts to decrease.
Also in this graph, it shows the length at 250m gives us the rectangle with the largest area. As the pitch in this test is too big, I will repeat the test with a smaller pitch to see if there is any difference in the maximum area. I also notice in the table that the rectangle that has the largest area is a square with all the sides having the same length. It is also a regular shape, which I will take into account.
Rectangles - 10metre pitch
The next thing that I am going to do is to decrease the pitch to 10m. This is so I can see which length give me the largest area with sides 10m apart. The reason I decreased the pitch is because the shape that gives me the maximum area could have a length between the maximum and one of the other two points next to it. So decreasing the pitch would give me more reliable results. As you can still see, when the length is 250m, the area is still the largest.
As you can see in this graph, the points form a curve that is very similar to the curve of the previous graph. This is because the points I have chosen are all between the three highest points on the previous graph.
Rectangles - 1metre pitch
Now I have decreased the pitch even further to 1metre. As you can see from this table and graph, it shows that the length that gives us the biggest area is still 250m, which is a regular four-sided shape or a square. On the other hand, it shows that the two points next to the peak have very close values to the peak value, they are only 1cm^2 away. Although all three tests show me the same result, I will do one more with an even smaller pitch to confirm this result.
Other 4-sided shapes
Parallelogram
Since the rectangle that had the biggest area had all sides with equal length, I predict that the parallelogram with the biggest area would also have the biggest area if all four sides were equal. From just looking at this diagram, I can tell that the area of it would not be larger than that of a square with the same sides. This is because if I were to cut the right side perpendicular from the bottom, I would be able to fit it on top of the left side making it a rectangle. This rectangle would be of length 250m but with the width less than 250m. I know this since the hypotenuse of the cut out triangle if 250m so the height is definitely less. So parallelograms do not have a larger area than a square if their perimeters are the same.
Trapezium
For Trapeziums, the base and height are always different in length. For this shape, I could do the same thing that I did to the parallelogram to find out the area. If I cut one side, and flipped it over, I could fit it on top of the other side to make a rectangle. In addition, it has two hypotenuses that would make the perimeter longer giving us lesser area than the maximum quadrilateral. So this shape does not give us a larger area than the square if they had the perimeter of the same length.
Scalene Triangles
Now I am going to see if a triangle with all three sides having a different would give a greater area than that of an equilateral triangle.
I will test two scalene triangles to see if it will give me a counter example giving me a larger area than the equilateral triangle. The length of the first of the two triangles will be as close to one third of 1000m as possible because I will test if the closer the lengths are to each other, the greater the area would be.
The first triangle will have lengths of 320m, 330m and 350m.
The triangle would look a bit like this.
To find the area of this shape, I would have to use my knowledge of trigonometry. Fist of all, I need to work out one of the angles. I have chosen to find angle x
CosA = (b^2+c^2-a^2)/2bc
Cosx = (320^2+350^2-330^2)/(2x320x350)
Cosx = 116000/224000
x = cos-1(116000/224000)
x = 58.8*
Area = 1/2bc sinA
Area = 1/2(320x350) sin58.8
Area = 56000x0.85536426
Area = 47900.4m^2(1dp)
Now I will test another scalene triangle, which will have lengths that are not close to each other with 325m, 225m and 450m.
The triangle would look a bit like this.
Now I will use my knowledge of trigonometry. First of all, I need to work out one of the angles. I will use the same formula as before
CosA = (b^2+c^2-a^2)/2bc
Cosx = (325^2+225^2-450^2)/(2x325x225)
Cosx = -46250/146250
x = cos-1(-46250/146250)
x = 108.4*
Area = 1/2bc sinA
Area = 1/2(225x325) sin108.4
Area = 36562.5x0.948876011
Area = 34693.3m^2(1dp)
Looking at the area of these two scalene triangles, I can see that they both do not exceed the area of the regular, equilateral triangle. Also, as the first triangle is much closer to that of the regular triangle, I can say that the closer the lengths are to each other, the bigger the area would be since the lengths that the first triangle has are closer to each other. This is also why the equilateral triangle gives us the maximum area since all three sides are equal in length.
Triangle 1 Area = 47900.4m^2
Triangle 2 Area = 34693.3m^2
Regular triangle = 48112.5m^2
Right-angled Triangles
I am going to try to find a counter example with right-angled triangles to see if it produces the maximum area for a triangle. The first one I am going to test is one with sides in ration 3:4:5 since I know that this already produces a right-angled triangle.
So 3+4+5 = 12. The triangle would look a bit like this.
1000/ 12 = 83.33
3 x 83.33 = 250m
4 x 83.33 = 333.33m
5 x 83.33 = 416.67m
The area of right angled triangles is
(Base x height)/ 2
= (333.33 x 250)/ 2
= 41666.6m^2
This area is nowhere that of the equilateral triangle, which was over 48000. I will test another right-angled triangle to see if it gives us the maximum area since testing only one triangle is not enough to give us results.
The next triangle I will be testing is with their sides with a ratio of 5:12:13
So 5+12+13=30 This is a sketch of what the triangle would look like.
1000/30=33.33
5x33.33=166.67m
12x33.33=400m
13x33.33=433.33m
Area of right angled triangles is
(Base x height)/2
= (400 x 166.67)/ 2
= 33333.33m^2
The area of this right-angled triangle is even less than the previous triangle and is even more further away from the area of the equilateral triangle. So I conclude that the area of right angled triangles do not produce the maximum area for all triangles.
Conclusion of Quadrilaterals and Triangles
After the tests that I have done, I have found out that regular shapes give me the largest area for a perimeter of 1000m. I also found out that a regular four-sided shape (62500m^2) gives me a larger area than a regular three-sided shape (48112.52243m^2) with equal perimeters. This may mean that the more sides a shape has, the bigger the area would be with a fixed perimeter. I will test this out now by using regular polygons with five or more sides.
Conclusion
In my investigation, I have found out that whatever number of sides you have in a shape, regular shapes always give the maximum area. I know this because for three and four sided shapes, the regular shapes always gave me the maximum area. I tried to show a counter example but I couldn’t find one. For example, for quadrilaterals, I drew a parallelogram and trapezium to see if they could be a counter example but it didn’t work. Also, in triangles, only the regular equilateral triangle gave the maximum area. So for the rest of the investigation, I used regular shapes because I knew that these gave the maximum area for a shape with any number of sides.
When I tested polygons, I found a reliable method to find the area of the shapes. For every polygon, I used the same method or formula but changing the number of sides for each shape. Now, I can write out the formula to show you what I did.
This is the most simplified version of the general formula.
