12. VOLUME
12 - 1 Sections
2
12 - 2 Finding bases of prisms
3
12 - 3 Constructing prisms
4
12 - 4 Drawing prisms
5
12 - 5 Making cross-sections of prisms
6
12 - 6 Building prisms
7
12 - 7 Volume of a prism
9
12 - 8 Volume of a cylinder
10
12 - 9 Converting between units of volume
11
12 - 10 Volumes of cash
13
12 - 11 Finding the volume of a playing card
14
12 - 12 Volume of an oblique prism
15
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Activity 12 – 1
Sections
SOLID
BOTTOM
SLICE
MIDDLE
SLICE
TOP
SLICE
A pyramid
A prism
A cone
A cylinder
The prism and the cylinder have parallel sections that are uniform (ie.
all the same shape and size).
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Activity 12 – 2
Finding bases
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Activity 12 – 3
Constructing prisms
The side faces of oblique prisms are parallelograms.
an octagonal prism
a hexagonal prism
an oblique octagonal prism
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Activity 12 – 4
Drawing prisms
120mm
50mm
30mm
140mm
80mm
140mm
140mm
140mm
120mm
30mm
50mm
80mm
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30mm
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Activity 12 – 5
Making cross-sections of prisms
SECTION
SLICE
SLICE
SLICE
SLICE
SLICE
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Activity 12 – 6
Building prisms
Building A
Building B
Building C
Number of blocks Number of layers Number of blocks
in the base
high
in the building
Building A
12
1
12
Building B
12
2
24
Building C
12
3
36
Building D
12
9
48
The sides of each block they used are 1cm.
If the base is the bottom of the building, the perpendicular height of
Building D is 3cm.
Each cross-section has 12 squares.
Each square has an area of 1 cm2.
The cross-sectional area is 12 cm2 .
Building A occupies the space of 12 blocks. This building therefore has a
volume of 12 cm3.
The volume of Building D is 12 x 4 = 48 cm3.
V=Axh
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The area of the top of this cuboid is 15 cm2.
There are 2 layers.
Then the volume of the building = 15 x 2 = 30 cm3.
The area of the face on the left hand side is 10 cm2.
There are 3 layers.
Then the volume of the building = 10 x 3 = 30 cm3.
The area of the face on the right hand side is 6 cm2.
There are 5 layers.
Then the volume of the building = 6 x 5 = 30 cm3.
The 3 dimensions: Length: 5 cm
Width: 3 cm
Height: 2 cm
You can find the volume of a cuboid without counting a whole layer of
blocks by multiplying its 3 dimensions.
A cuboid of blocks with a volume of 36cm3 can have any of the following
sets of dimensions (each in cm): 12 x 3 x 1, 6 x 2 x 3, 4 x 3 x 3,
36 x 1 x 1, 6 x 6 x 1, 4 x 9 x 1, 9 x 2 x 2, 18 x 2 x 1.
The cross-sectional area of this building is 9 cm2 .
If it was 12 stories high,
• its perpendicular height would be 12 cm
• Its volume would be 108 cm2.
Yes, the formula V = A x h could be used to calculate the volume of a
building with a triangular base. If a cuboid is cut into 2 equal triangular
prisms, they each have half the volume of the cuboid and half its base
area, so the formula still works.
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Activity 12 – 7
Volume of a prism
2cm
2.5cm
3cm
2cm
4cm
3cm
Cross-sectional area
= ½ x 4 x 2 cm2
= 4 cm2
Cross-sectional area
= 3 x 3 cm2
= 9 cm2
Volume
= 4 x 2.5 cm3
= 10 cm3
Volume
= 9 x 2 cm3
= 18 cm3
2cm
1cm
4cm
3cm
2cm
2cm
6cm
3cm
Cross-sectional area
= (½ x 6 x 3) + (½ x 2 x 3) cm2
= 12 cm2
Cross-sectional area
= (2 x 1) + (½ x 2 x 2) cm2
= 4 cm2
Volume
= 12 x 2 cm3
= 24 cm3
Volume
= 4 x 4 cm3
= 16 cm3
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Activity 12 – 8
Volume of a cylinder
CROSS-SECTION
SLICE
Yes. The volume of a cylinder can be calculated using the same formula
because a cylinder, like a prism, has a uniform cross-section.
For a cylinder that has a radius of 3cm and a height of 8cm:
Cross-sectional area
= π x 32 cm2
= 28.27 cm2
Volume
= 28.27 x 8 cm3
= 226 cm3
The volume of the piece of hollow pipe that has an outer diameter of
8cm, an inner diameter of 6cm and a length of 8cm:
8cm
Cross-sectional area
= π (82 - 62) cm2
= 28 π cm2
= 88 cm2
6cm 8cm
Volume
= 88 x 8 cm3
= 704 cm3
If a cylinder has a radius of r units, A = π r cm2
If its perpendicular height is h units, its volume in terms of r and h is:
V = π r h cm3
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Activity 12 – 9
Converting between units of volume
A cube with a volume of one cubic decimetre has edges that are each 10
cm long.
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There are 10 centimetres (cm) in a decimetre (dm).
There are 100 square centimetres (cm2) are in a square decimetre (dm2
There are 1 000 cubic centimetres (cm3) are in a cubic decimetre (dm3)
There are 10 millimetres (mm) are in a centimetre (cm).
There are 100 square millimetres (mm2) are in a square centimetre
(cm2).
There are 1 000 cubic millimetres (mm3) are in a cubic centimetre (cm3)
To work out the number of cubic millimetres (mm3) that are in a cubic
decimetre (dm3), find the number of millimetres in a decimetre then cube
that number.
Answer: 1003 mm3 = 1 000 000 mm3
To work out the number of cubic millimetres (mm3) that are in a cubic
metre (m3), find the number of millimetres in a metre then cube that
number.
Answer: 1 0003 mm3 = 1 000 000 000 mm3
When calculating a volume, the measurements of all three dimensions
must first be converted into the same units, This is because a volume is
measured in cubes and a cube has all 3 dimensions the same length.
Volume of a cuboid 5m long, 20cm wide and 6mm high:
Volume = 5 000 x 200 x 6 mm3
= 6 000 000 mm3
= 6 000 cm3
= 6 dm3
1) Space occupied by textbooks = 675 x 250 x 190 x 32 mm3
= 1 026 000 000 mm3
= 1.026 m3
2) Volume of topsoil needed = 8.1 x 3.7 x 0.1 m3
= 2.997 m3
3 m3
3) Volume of asphalt needed = 600 x 5.46 x .025 m3
= 81.9 m3
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Activity 12 – 10
Volumes of cash
Method
Calculate the volume of one dollar worth of each type of coin.
Because there are only 10 five cent coins (worth 50 cents), the volume
will need to be doubled to make a dollar worth of these coins.
To calculate the volume of a pile of coins, measure the diameter of one
coin (eg. by using set squares as calipers as in Activity 9 – 1) and the
height of the pile of coins (using a ruler).
Calculations
Diameter of a twenty cent coin = 28.5 mm
Volume of 5 twenty cent coins = 28.5 x π x 12 mm3
= 1074.4 mm3
Diameter of a ten cent coin = 23.6 mm
Volume of 10 ten cent coins = 23.6 x π x 18 mm3
= 1334.5 mm3
Diameter of a five cent coin = 19.4 mm
Volume of 10 five cent coins = 19.4 x π x 13 mm3
= 792.3 mm3
Volume of 20 five cent coins = 1584.6 mm3
Note: Your calculations could differ slightly from those above because
with a ruler, you would measure diameter to the nearest mm and you
could expect measurement errors of ± 0.5 mm.
Conclusion:
You would spend five cent coins to spend the greatest volume.
Note: This is the conclusion you are likely to arrive at by measuring the
height and diameter of piles of coins but in fact, the coins were designed
so that a dollar worth of each coin (whether it be the five cent, ten cent or
twenty cent coin) has the same mass and volume. There is
measurement error caused by the inaccuracy of rulers and also by the
fact that there is air space within each pile of coins. A more accurate
means of calculating their volume might be to use displacement. This
alternative method is presented in Activity 13 – 7.
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Activity 12 – 11
Finding the volume of a playing card
The number of cards is 52 plus any jokers (or other special cards).
The measurements will differ slightly for different packs of cards.
Calculate the volume of one card by dividing the total volume by the
number of cards.
When you slant the deck to one side:
No. The volume of the deck has not changed because there are still the
same number of cards, each with the same volume.
No. The perpendicular height of the deck has not changed.
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Activity 12 – 12
Volume of an oblique prism
The two parts of your oblique prism make a right prism because the two
sides of the prism are parallel (ie. they have the same slope) and so they
fit together.
The oblique prism and the right prism you made from it have the same
volume because nothing has been taken from or added to the original
volume.
The general formula (V = Ah) still works for an oblique prism because
there is no change in the area of the cross-section of a prism or its
perpendicular height when it is made oblique.
Volume of the triangular prism:
Volume = cross-sectional area x height
= ½ (4 x 3) x 6 cm2
= 36 cm2
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