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Introduction

 Maths Coursework

Syllabus 1385

TUBES

The basis of my investigation is, when I fold in various ways a piece of card like this, what will occur.

32cm    24cm These are the prisms I will be using:       The first shape I will try to investigate will be rectangles. I will try all the various possibilities. I will start on the smaller side i.e. the 24cm. A substantial rule to remember to find the volume, is Length x Height x Width.

A = 11 x 1 = 11

V = 11 x 32 = 352cm

A = 10 x 2 = 20

V = 20 x 32 = 640cm

A = 9 x 3 = 27

V = 27 x 32 = 864cm

A = 8 x 4 = 32

V = 32 x 32 = 1024cm

A = 7 x 5 = 35

V = 35 x 32 = 1120cm

A = 6 x 6 = 36

V = 36 x 32 = 1152cm

A = 5 x 7 = 35

V = 35 x 32 = 1120cm

I realised in this review that the more cubic the piece of card gets the more possibilities in maximising its volume space.  To prove this notion I even tried to work out the rectangle after it, which resulted in a smaller volume.

After folding along the 24cm side, I moved on to the 32cm edge, the results are worked out in the same way except that the depth is now 24cm instead of 32cm.

A = 15 x 1 = 15

V = 15 x 24 = 360cm

A = 14 x 2 = 28

V = 28 x 24 = 672cm

A = 13 x 3 = 39

V = 39 x 24 = 936cm

A = 12 x 4 = 48

V = 48 x 24 = 1152cm

A = 11 x 5 = 55

V = 55 x 24 = 1320cm

A = 10 x 6 = 60

V = 60 x 24 = 1440cm

A = 9 x 7 = 63

V = 63 x 24 = 1512cm

A = 8 x 8 = 64

V = 64 x 24 = 1536cm

A = 7 x 9 = 63

V = 63 x 24 = 1512cm

Middle I would like to move on to triangles, and determine what their biggest volume would be. To work out the area of a triangle the formulae needed is half base times height. In this particular case I am not given the height, so foremost I have to work out the height, this is done using what’s known as Pythagerous’s Theorem. This method used is, A² + B² = C ².

A is known as the base, B the height, and C the length. But in this specific case the formulae is rearranged to C² + A² = B² . As a matter of interest another way to work out the area of a triangle is AB Sin C, but this only works when the triangle is when you know 2 sides and the angle between them.

The triangles I will be investigating are only isosceles, as any other kind would take too long, and it makes the calculations easier.

Here again, once the area of the triangle is found like last time you have to times the area by the length of the prism to find the volume.  11            11

B = 11 – 1 =  121 – 1  =  120

B =  120 = 10.95

Area = 10.95 x 1 = 10.95cm

Volume = 10.95 x 32 = 350.54cm

2 B = 10 – 2 = 100 – 4              10                                                                 B =  96 = 9.8

Area = 9.8 x 2 = 19.6cm

Volume = 19.6 x 32 = 672.07cm

B = 9 – 3= 81 - 9

B =  72 = 8.49

9                9                                                   Area = 8.49 x 3 = 25.46cm

Volume = 25.46 x 32 = 814.5cm

6

60°

8               8                                                       ½  x A x B x Sin C

= ½ x 8 x 8 x Sin C

Area = 27.71 cm

Volume = 27.71 x 32 = 886.81 cm

8

B = 72 – 52 = 49 – 25

7                7                                                    B =  24 = 4.9

Area = 4.

Conclusion

L/5 ÷ 2 = L/10.

Therefore the height of the right-angled triangle would be

Tan 36˚ x L/10 = H.

Now this is established I have to find the height.

L x L = L²                        =                    L²   x   Tan 36˚   x   H

10 x 10 = 100                                       100

###### FORMULA FOR THE CIRCLE PRISM

I will now try finding a formula for an open-ended cylinder.

First I will investigate a tube with a height of 24cm and a circumference of 32cm.

24cm

2Л

Л

Volume of cylinder:                Лr²h

Volume of cylinder:                Л x 5.0929² x 24

Volume of cylinder:               = 1960cm³ (to 3 s.f.)

Now I will do the same as I have done except that I will use a cylinder with a height 32cm and a perimeter of 24cm.

2Л

Л

Volume of cylinder:                Лr²h

Volume of cylinder:                Л x 3.8197² x 32

Volume of cylinder:                = 1470cm³ (to 3 s.f.)

###### VOLUME FOR ANY CYLINDER

I will now work out a theory to find out the volume of the largest shape, the cylinder.

Acm²

L

If the area of this piece of card is Acm² and the Length is Lcm then it is easy to determine the height will be A

L

If I then turn this card into a cylinder

The height of this cylinder will be A

L

And the circumference will be L.

So:                                       L = 2Лr

Which is the same as:         r = L

2Л

Area of the circle:               Лr²

So:                                       Л  L  ²

2Л Which is the same as:          Л x    L  x  L  2Л    2Л

Which equals:                      L²

4Л Volume of cylinder:             L²   x  height

4Л

Volume of cylinder:             L²  x  A

4Л      L

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