We can see from this table that again a ninety degrees angle gives the greatest volume and that a scalene triangle gives a smaller volume than an isosceles triangle. This means that the triangle, which gives the greatest volume, is the isosceles triangle of side twelve centimetres and with angle of ninety degrees. And although this is not a regular triangle it is still a regular quadrilateral, i.e. a square bisected diagonally. Which means my hypothesis is still valid.
I will now investigate rectangles, as they should have the next least volume. I shall use investigate it in the same way as I investigated the various triangles. I expect that it shall be the square that has the greatest volume and that it shall be identical to that of the largest triangle.
As I predicted the maximum volume for the rectangular shape’s is the same as that for triangular shapes. Also the graphs are very similar. I can tell that 28800 centimetres cubed is the maximum volume since the points on the graph are symmetrical and it is the same distance from each. So far my hypothesis is sound and on both occasions regular shapes have had the greatest volume; as such I shall now investigate only regular shapes for greatest volume. I will now place regular polygons into a table and graph the volumes. Since the polygons will be halved I shall only investigate even number sided shapes since the odd numbered ones would only be irregular variants of these.
To calculate the area of each polygon I first find the area of a single triangle inside the shape, ie in a hexagon:
I use trigonometry to do this – A B SinC 0.5. Then I multiply by half the number of sides to find the cross-sectional area. This is process I used in my table (The number of sides is the number in the whole shape, whereas only half the shape shall be formed by the net and as such the volume only reflects that of half the shape)
The graph shows that the area increases with each additional side; and although the increase keeps getting less it will never stop increasing. I predict that this shall never be greater than or equal the volume of the semi-circle, as the increase is probably due to the shapes becoming rounder.
I shall now investigate the semi-circle, as it is the last shape I am considering. There is only one possibility for a semi-circle, which is half a circle of circumference forty-eight.
This is the greatest possible volume for the net given. I now feel I have enough information to draw up my conclusion.
Conclusion
In conclusion I can say that my hypothesis was, beyond reasonable doubt, true. The semi-circle had the greatest volume and the greater the number of sides the greater the volume in regular shapes. I may also be able to workout an equation by combing the steps I used to calculate the area of the polygons. S shall represent the number of sides. I will first calculate an equation in terms of S for the base and angle in the triangle:
Angle – 360/2S
Base – 24 0.5S
I can now use these two formulas to calculate the formula for the cross-sectional area of a polygon:
24/S/Sin(360/2S)
(24/S/Sin(360/2S)) ² Sin(360/S) 0.5
S/2((24/S/Sin(360/2S)) ² Sin(360/S) 0.5)
I shall test this now to find the area of an eight-sided shape:
=8/2((24/8/SIN(360/16))² SIN(360/8) 0.5)
=4((3/SIN22.5)² SIN45 0.5)
=86.91168825
Since the solution of the equation is the same as the area displayed in the table it proves my formula. I will now use this formula to help me prove my hypothesis by calculating the area of a thirty-one-sided shape:
=31/2((24/31/SIN(360/62))² SIN(360/31) 0.5)
=15.5((0.774193548/SIN5.806451613))² SIN11.61290323 0.5)
=91.35919972cm²
This shows that a polygon’s area cannot exceed that of a circle made from the same net. As such I shall choose the semi-circular guttering as it has the greatest volume and thus the greatest capacity for fluid, and also since it is being used in guttering it shall require less cleaning since dirt and leaves are more likely to gather in corners
