To prove that this formula works I will input values for the equilateral triangle and confirm that this formula will give me the volume:
A = 8cm, B = 8cm, C = 8cm, Y = 32cm.
82 - 8 2 x 8 x 32 1773.62
2 → 2 → 886.61cm3
2
The fact that the formula has given me the correct answer, which I worked out previously, proves that this formula gives us accurate volumes for equilateral and isosceles triangular prisms.
Cuboids
I will now move onto quadrilaterals; I will investigate both a square and 4 rectangles.
32cm To work out the volume of any quadrilateral, you must work out the
area of the cross section by multiplying
4cm. the sides. To get the volume you must multiply the area of the cross section by
8cm. the length of the prism.
area of cross section = 8 x 4 → 32cm2
volume of prism = 32 x 32 = 1024cm3
Area of cross section = 7 x 5 = 35cm2
Volume of prism = 35 x 32= 1120cm3
5cm
32cm
7cm
The following formula must be used to work out the length of each side of a square
Length of perimeter (24)
No. of sides (4) = 6cm
32cm
To work out the volume of this prism I will use the
Same technique I used for the rectangular prism.
6cm are of cross section = 8 x 8 = 36cm2
Volume = 36 x 32 → 1152cm3
6cm
Area of cross section = 3 x 9 = 27cm2
Volume of prism = 27 x 32 = 864cm3
3cm
32cm
9cm
Area of cross section = 2 x 10 = 20cm2
32cm Volume of prism = 20 x 32 = 640cm3
2cm
10cm
I have also worked out a formula to work out the volumes of quadrilateral:
c V= length of prism. Multiplies
with area to gives us the volume.
a
This gives us the area of the cross section
b
To show that this formula will give me the volume of a quadrilateral, I must put the figures for one of my quadrilaterals in and check to see whether or not the formula gives me the correct volume.
A = 6cm, B= 6cm, C = 32cm
6 x 6 x 32 = 1152cm3
1152cm3 is the same volume as I previously worked out, and therefore proves that this formula is correct.
I have realised that the square is the quadrilateral give the largest volume. Taking into account both the results I have obtained from the triangular and quadrilateral prisms, I conclude that regular shapes give larger volumes than irregular shapes do, therefore I will now only work out the volumes for various regular shapes.
To prove that infact the square will give us the largest volume; I will consider values close to the dimensions of the square.
32cm
A
B
The third shape I will be looking into is the regular pentagonal prism; a pentagonal prism has 5 sides. Because it is a regular shape, to find the length of each shape we simply do the following→ perimeter (24cm) = 4.8cm
No. of sides (5)
To work out the area of the cross section, we must divide the pentagon into 5 triangles:
Angle → 360 = 720
4.8cm no. of sides (5)
4.8cm
72/2= 360
H
2.4cm
Using trigonometry we must find the height:
Tan = opposite (opp) adjacent (adj)
So, adj (H) = opp (2.4)
Tan360
H= 3.30cm
We then work out the area of the triangle using: b x h
2
4.8 (b) x 3.30 (h) → 15.84 → 7.92cm2.
- 2
We will then multiply this by the length of the prism:
7.92 x 32 = 253.44cm3
Because in total there are 5 triangles in this prism, we must multiply the volume of the triangle by 5.
253.44 x 5 = 1267.2cm3
My next shape, is a hexagonal prism, which has 6 sides:
Again we have split it up into 6 triangles.
360/6= 600
h
Base = 24/6 = 4cm
60/2 = 300
h
4/2 = 2cm
Adj = opp (2)
Tan300 = 3.64cm
Area of triangles = 3.464 x 4 → 6.93cm2
2
Volume = 6.928 x 32 → 221.696 x 6 = 1330.176cm3
Using the formulas I have used to find out the volumes for both the pentagonal and hexagonal prism, I have derived a formula to work out the volume of an ‘n’ sided shape.
L= length of prism
P = perimeter of shape
N= number of sides
We divide the perimeter by the number of sides to give
Volume us the length of the base. We multiply this by the height
and divide it by 2 to get the area of the triangle.
It gives us the height Our last step is to
The triangle multiply the whole
This formula is the thing by the length
equivalent to of the prism to get
Adj = opp volume.
Tan We multiply the area of one triangle by the number of
Sides to get the area of the whole cross-section.
Because in the pervious formula we divided by n then multiplied by n, they cancel each other out.
Here, I have written a simplified formula:
To make sure that this formula I have derived is accurate I will input values for the pentagon
P = 24cm, N = 5cm, L = 32cm
24/10
Tan180 x 24 x 32 → 2536.947 = 1268.473578cm3
5 2
2
The answer that the formula has given me is correct, so I will continue using this formula.
I will now use the above formula to work out the volumes for the rest of the prisms.
Cylinder
The last shape I will be investigating will be a cylinder:
Working out the volume of a cylinder is fairly easy,
The method is similar to the one used to find the volume of any prism.
We know the circumference is 24cm. 32cm
π x diameter (D) = circumference (C)
Therefore, C = D
π
D = 7.639cm
D
2 =Radius (R). R= 3.82cm
πR2 = area of cross section. R2 = 14.59cm x π = 45.84cm2
Now to find the volume, we must multiply the cross section by the length of the cylinder.
45.84 x 32 = 1466.77cm3
R2 x π gives us the area of the
cross section.
Formula→
C = circumference
L = length of prism.
π = Pie (3.14)
This gives us the multiply the cross section by the
value for R. length of prism, to get volume.
The only way for me to prove that this formula works is to put the values for the cylinder into the formula to check whether it will give us the correct volume.
C = 24cm, L = 32cm
24/π 2 x π x 32 → 1466.77cm
2
This is the correct volume for the cylinder; I will now be able to use this formula later in this piece of coursework.
Out of all these shapes, the cylinder has given me the largest volume, I will now swap the sides around, making 24cm the length of my prism (L) and 32cm the perimeter (P). I will use the formulas I have worked out previously to help me work out the volumes of the prisms.
32cm→ P
24→ L
Again, I will begin with working out the volume of isosceles and equilateral triangles.
C
A
B 24cm
Now moving onto equilateral prisms:
24cm
b
a a
b
To prove that the square has the largest volume, I will again consider values close to the dimensions of the square.
Clearly we can see that from the quadrilaterals the square has the largest volume.
The third shape I will be investigating the volume of is a regular pentagonal prism.
Volume= P/2N _ 32/2x 5
Tan180 x PL Tan180 x 32 x 24
N → 5
- 2
= 1691.298104cm3
Cylinder:
C = 32cm
24cm
Using the previously formed formula, I will work out the volume:
C/π 2 32/π 2
2 x π x L → 2 x π x 24
25.93822301 x T = 81.48733086 x 24
= 1955.695941 cm3
After this investigation, I can now say that the cylinder has the largest volume, I have also realised that as the number of sides in the prism increases, so does the volume.
Part 2
For part 2 of this investigation, we have a piece of card with an area of 1200cm2:
B (perimeter)
A (length of prism)
I will need to find the prism that I can make from this piece of card, which will give me the largest volume; I will use the knowledge I have gained during part 1 to help me work this out. After swapping the lengths and perimeters in part 1, I could clearly see that the prism with the shorter length, and the larger perimeter gave the largest volume, this is why I will keep the perimeter as large as possible and the length as small as possible.
Because I know that the cylinder is most likely to give me the largest volume, the first shape I will work the volumes out for is cylinders.
Circumference = 1200cm
1cm
I will use the formula I worked out in part 1 to work out the volume of this cylinder:
C/π 2 1200/π 2
2 x π x L → 2 x π x 1
= 114591.559cm3
As you can see, the cylinder with the largest circumference and the smallest length gave me the largest volume, as I predicted.
Because we know that regular shapes give us the largest volume I will only work out the volumes for regular prisms.
Squares
I will work out the volumes for the various squared, by using the formula I worked out in part 1:
AB x C
c
a
b
Equilateral triangle.
I will also use the formula I previously formed to work out the volume of the triangular:
y
a c
b
c2 - b 2 x b x y
2 _
2
Icosagon
Out of the polygons I investigated, the 20 sided regular icosagon had the largest volume, therefore I will now work out the volume for an icosagon using the formula for an ‘n’ sided shape:
P/2N
Tan180 x PL
N
2
From this whole investigation, I have found out that a cylinder will give us the largest volume, this may be due to the fact that it is has a smooth curve rather than many sides. Now, I understand that as the Perimeter of a prism increases and the length decreases the volume increases. So, if I were to make a container that would be able to carry the largest amount of water, I would take into account the fact that cylinders with a large perimeter yet small length will give the largest volume.