• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Tube Maths Investigation

Extracts from this document...


For this project we had to investigate the volumes of open-ended tubes made from a rectangular piece of card with the dimensions, 32cm by 24cm. Either of the two sides could be folded to make the perimeter, with the other side being the height. I made a couple of conjectures before I started. They were:

· the more sides, the bigger the volume

· regular polygons are the best

1. For the first question, the volumes have to be investigated from various polygon shapes as the base side.

The first shape I used was a square, with a height of 24cm, and perimeter of 32cm:

To find the area of the base, I used the formula: A = x2, where ‘x’ is the side of the square. Once the area was found it was multiplied by the height.

A = 82

V = 1536cm3

For a square with height 32cm and perimeter 24cm, it is:

A = 62

V = 1152cm3

From the method I used, a formula can be found for an open ended tubes with square bases. If we let ‘h’ be the height, and ‘p’ be the perimeter, it would be:

V = p 2 x h


V = p 2 x h


V = p2h


The next shape was an equilateral triangle based tube:

2 3

10h = 24



...read more.


 2 - p 2 x x 6 x h

12 6 12

V = p x v p2 x h

4 36

The next shape was a cylindrical tube.

h = 24

p = 32

A = pr2

A = 32 2 x p


V = 1956cm3

For the cylindrical tube with a height of 32cm and circumference of 24cm, the volume was:

A = 24 2 x p


V = 1467cm3

From these two results, a formula could be deduced, using the same method used, but turning it into an algebraic equation, by substituting in ‘p’ and ‘h’.

V = p 2 x p x h


V = p2 h


I collected results from other shaped bases, and they are recorded in the table below:

no. of sidesperimeter(cm)height(cm)volume(cm3)cylinder24321467cylinder32241955324328873322411824243211524322415366243213306322417738243213918322418541024321418103224189112243214331232241911All of these bases were constructed from regular polygons, as they have the optimum area. Showing the volume of a tube with height of 32cm, and a base of 4cm by 8cm can prove this.

A = 4 x 8

V = 1024cm3

This value is more then 100cm3 less when compared to the volume of a tube with height 32cm, and base of a 6cm square.

This proves that a regular polygon based tube has the optimum volume.These results show a pattern, that the more sides there are, the greater the area. However, a cylinder has the greatest volume, as it is infinity sides.

...read more.


3. This question involves investigating the volumes of tubes made from card with an area equal to that of 24cm by 32cm, (768cm2).

These figures are for a square based tube made from a card with area 768cm2.

perimeter(cm)height(cm)volume(cm3)17684841921921448672481423041924921676813686476800.136864076,8000.013686400These results show an increase in volume when the perimeter is greater. The perimeter length and the volume are directly proportional,

eg: ‘p’ from 768 to 7680, ‘V’ from 36864 to 368640. The perimeter is 1000% bigger in one, and so is the volume.

The perimeter length and the volume are directly proportionalThis graph shows this:

The graph shows the way in which the volume increases as the perimeter increases. Similarly, the volume can get nearer and nearer to zero with the decrease in value of ‘p’.

The area of the card is equal to ‘ph’, perimeter multiplied by height. This can be substituted into the formulae, thereby removing the necessity of having ‘h’ in the equation. This leaves the final formulae to be:

V = Ap _4n x tan 180/nV = Ap_4pgeneral formula cylindrical formula

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Geography Investigation: Residential Areas

    Extrapolation I will calculate the average area rating from all of the other data I have collected. For each of the hypothesis I have just mentioned the main method I will use throughout the investigation as a breakdown of the last few pages.

  2. GCSE Maths Coursework Growing Shapes

    51 D1 D2 As there are all 5's in the D2 column, the formula contains 2.5n2. Pattern no. (n) 2.5n2 No. of Pentagons - 2.5n2 1 1 1.5 2 3 4 3 5 6.5 4 7 9 5 9 11.5 D1 Pattern no.

  1. Fencing - maths coursework

    and I have the adjacent (180/n) 500 The formula for Tan ...�= Opposite Adjacent We have the angle 180� n x This is the adjacent that we are This is the opposite that we already have. trying to find out the height.

  2. Biological Individual Investigation What Effects Have Management Had On Grasses In Rushey Plain, Epping ...

    When I picked my grassland study area, I had to choose an area, which was 10m x 10m in size, and did not cross any paths. The site that met these two requirements was partially in the shadow of an oak tree.

  1. Geography As Environmental Investigation

    41 7 7 4 46 +11 40 4 4 3 47 +13 43 1 3 0 48 +11 42 1 6 2 49 +10 43 3 5 5 50 +14 40 6 7 3 Spearman's Rank Correlation In order to see whether my different pieces of data can be correlated, I will need to use higher statistical techniques.

  2. Based on the development of the BristolHarbourside the title that I chosen for my ...

    Conclusion and Evaluation Evaluation: The harbourside has been modernised, as happens in any rejuvenation project, but it has also been given a whole new totally different image compared to the one it previously had. Bristol has always been a key city in the UK especially when it comes down to trade.

  1. In this investigation, we have been told to find out the largest possible volume ...

    In addition, I can see that it is better to use the 32cm side as the base and the 24cm one as the height.

  2. Fencing investigation.

    Quadrilaterals Squares 250m 250m 250m 250m Area = length x breadth As we already know both the length and the width of the square, it is the easiest shape to solve the area of. Area = length x breadth Area = 250m x 250m Area = 62,500m2 Quadrilaterals Rectangles 400m

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work