# 04_Volume_Practice

Volume Practice
Volumes
This unit explains how the volumes of
various solids are calculated.
It includes simple applications of
formula and clear examples.
It also contains a variety of challenges
and problems.
Volumes
What 3-D solid is this?
Volumes
What 3-D solid is this?
Volumes - Contents
List of Formulae
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Formulae Summary
Cuboid
=
width x length x height
Prism
=
area of end x height
Cylinder =

r2 x

height
Cone
1
= 3
Pyramid
1
= 3 base area x height
r2 x
height
Volumes - Contents
List of Formulae
A. Problems / Challenges
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
PACKETS IN A BOX PROBLEMS
In each problem, find
the number of smaller
packets that will fit
neatly into the box.
1
8
7
6
5
4
3
2
PACKETS IN A BOX EXAMPLE 1
In this problem, find the number of smaller
packets that will fit neatly into the box.
Packet
2cm
4cm
Box
Dimensions
2cm
12 x 4 x 6
4 x 2 x 2
3 x 2 x 3 = 18
12cm x 4cm x 6cm
18
PACKETS IN A BOX EXAMPLE 2
In this problem, find the number of smaller
packets that will fit neatly into the box.
Packet
Box
Dimensions
3cm
5cm
2cm
20 x 8 x 6
5 x 2 x 3
4 x 4 x 2 = 32
20cm x 8cm x 6cm
32
PACKETS IN A BOX EXAMPLE 3
Be careful with this one!
There is an extra step!
Packet
2cm
3cm
Box
Dimensions
9cm x 8cm x 25cm
5cm
9 x 8 x 25
3 x 2 x 5
3 x 4 x 5 = 60
Note change
of order !!
60
PACKETS IN A BOX PROBLEMS 1
In each problem, find the number of smaller
packets that will fit neatly into the box.
1
Box
10 x 8 x 6
Packet 5 x 2 x 3
2 x 4 x 2 = 16
2
Box
12 x 15 x 8
Packet 2 x 3 x 2
6 x 5 x 4 = 120
3
Box
30 x 8 x 10
Packet 6 x 4 x 5
5 x 2 x 2 = 20
4
Box
50 x 20 x 15
Packet 5 x 4 x 3
10 x 5 x 5 = 250
PACKETS IN A BOX PROBLEMS 2
In each problem, find the number of smaller
packets that will fit neatly into the box.
1
Box
9 x 10 x 10
Packet 3 x 2 x 5
3 x 5 x 2 = 30
2
Box
24 x 25 x 30
Packet 6 x 5 x 3
4 x 5 x 10 = 200
3
Box
40 x 20 x 20
Packet 5 x 4 x 2
8 x 5 x 10 = 400
4
Box
18 x 21 x 15
Packet 3 x 3 x 3
6 x 7 x 5 = 210
PACKETS IN A BOX PROBLEMS 3
In some of these problems you may have to
re-order the numbers!
1
Box
50 x 24 x 16
Packet 5 x 4 x 4
5 x 4 x 4 = 80
2
Box
15 x 6 x 8
Packet 52 x 5
2 x 2
3 x 3 x 4 = 36
3
Box
20 x 18 x 14
Packet 65 x 7
6 xx 75
4 x 3 x 2 = 24
4
Box
28 x 40 x 15
Packet 75 x 7
4 x 4
5
7 x 10 x 3 = 210
PACKETS IN A BOX PROBLEMS 4
In each problem, find the number of smaller
packets that will fit neatly into the box.
1
Box
24 x 20 x 90
5 x 9
Packet 8
5 x 89
3 x 4 x 10 = 120
2
Box
30 x 80 x 25
5 x 3
Packet 3
8 x 85
10 x 10 x 5 = 500
3
Box
75 x 100 x 60
Packet 25
6 x 25
20 x 20
6
3 x 5 x 10 = 150
4
Box
70 x 40 x 30
Packet 15
7 x 7
8 x 15
8
10 x 5 x 2 = 100
Volumes - Contents
List of Formulae
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
GOING FOR GOLD PROBLEM
This problem requires knowledge of
how to calculate the volume of a
“trapezoidal prism” and a “cylinder”.
Look up the relevant sections if some
background work is necessary.
GOING FOR GOLD PROBLEM
In a bullion robbery, a gang
of thieves seize 100 gold
ingots with the dimensions
shown on the right. Find the
volume of one ingot.
6cm
5cm
25cm
10cm
The 100 ingots are melted
3cm
0.5cm souvenir medals.
How many medals could
be produced?
GOING FOR GOLD PROBLEM
STEP 1
Find volume of one ingot.
6cm
Area of End
1
= (10 + 6) x 5 = 40 cm2
2
Vol = 40 x 25 = 1000 cm3
STEP 2
Volume of 100 ingots =
100 x 1000 =100000 cm3
5cm
25cm
10cm
Volume of Prism
= Area of End x
Length
GOING FOR GOLD PROBLEM
STEP 3
Find volume of
one disc using the formula
for a “cylinder”.
Vol =  x 3 x 3 x 0.5
= 14.13 cm3
STEP 4
Vol. of gold = 100000 cm3
Number of “medals” ?
= 100000  14.13
3cm
0.5cm
Volume of a
Cylinder ?
=
 r2 x
height
= 7077 (approx)
GOING FOR GOLD PROBLEM 2
Similar problem ... Different numbers!!!
In a bullion robbery, a gang
of thieves seize 500 gold
ingots with the dimensions
shown on the right. Find the
volume of one ingot.
4cm
3cm
10cm
6cm
The 500 ingots are melted
2cm
0.5cm souvenir medals.
How many medals could
be produced?
Volumes - Contents
List of Formulae
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Half Full Problem
There is a famous saying concerning the way
different people look at situations.
“The glass is half full v the glass is half empty”
BUT ...
What do we mean by
HALF FULL???
Half Full Problem
4cm
8cm
See next slide for hints.
1
This ice cream cone
has a height of 8cm
and circular face of
When full it contains
134 cm3. (Check)
What height will the
ice cream be at when
it is half full?
Half Full Problem
Half the height.
2 cm
4cm
Try other measurements
2
Half the volume
= half of 134 cm3
= 67 cm3
BUT what do you
notice when you
find the volume of
this “half sized”
cone?
Only ... 16.6 cm3
Half Full Problem
is half the height.
3cm
6cm
3
TARGET VOLUME
= half of 134cm3
67 cm3
This one on the left
gives a volume of ...
56.5 cm3
Is this any closer?
Try others!!
Volumes - Contents
List of Formulae
A. Problems / Challenges
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Cuboid Example 1
Volume of a cuboid
= width x length x height
Vol of cuboid
4 cm
= 5 x 3 x 4
Vol = 60 cm3
3 cm
5 cm
Cuboid Example 2
Volume of a cuboid
= width x length x height
Vol of cuboid
= 10 x 3 x 4
Vol = 120 cm3
4 cm
10 cm
3 cm
Cuboid Example 3
Volume of a cuboid
= width x length x height
Vol of cuboid
6 cm
= 6 x 6 x 6
Vol = 216 cm3
6 cm
6 cm
Cuboid Example 4
Volume of a cuboid
= width x length x height
Vol of cuboid
4 cm
= 8 x 5 x 4
Vol = 160
5 cm
cm3
8 cm
Cuboids
Basic Exercise A
1
2
36cm3
6cm
3cm
32cm3
2cm
3
8cm
2cm
2cm
4
36cm3
4cm
3cm
3cm
56cm3
7cm
4 cm
2 cm
Cuboids
Basic Exercise B
1
2
60cm3
4cm
48cm3
3cm
8cm
5cm
3
3cm
2cm
4
120cm3
6cm
5cm
4cm
21cm3
3.5cm
3 cm
2 cm
Cuboids
Gaps Exercise A
Fill in the missing values.
Width
Length
Height
Volume
1
2
3
5
30 cm3
2
4
5
10
200 cm3
3
3
3
5
45 cm3
4
5
2
6
60 cm3
5
4
8
10
320 cm3
Cuboids
Gaps Exercise B
Fill in the missing values.
Width
Length
Height
Volume
1
4
5
8
160 cm3
2
8
2
3
48 cm3
3
5
10
20
1000 cm3
4
4
6
10
240 cm3
5
6
5
9
270 cm3
Cuboids
Gaps Exercise C
Fill in the missing values.
Width
Length
Height
Volume
1
2.5
5
10
125 cm3
2
4
6
4
96 cm3
3
6
6
6
216 cm3
4
5
4
8
160 cm3
5
6
8
20
960 cm3
Volumes - Contents
List of Formulae
A. Problems / Challenges
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Triangular Prism Example 1
Volume of a prism
= area of end x length
Area of End
1
=
(5 x 4) = 10 cm2
2
4cm
Vol = 10 x 8 = 80 cm3
8cm
5cm
Triangular Prism Example 2
Volume of a prism
= area of end x length
Area of End
1
=
(2 x 6) = 6 cm2
2
Vol = 6 x 7 = 42 cm3
2cm
7cm
6cm
Triangular Prism Example 3
Volume of a prism
= area of end x length
Area of End
= 1 (4 x 3) = 6 cm2
2
Vol = 6 x 9 = 54 cm3
3cm
9cm
4cm
Triangular Prism Example 4
Volume of a prism
= area of end x length
Area of End
= 1 (5 x 6) =15 cm2
2
Vol = 15 x 12 = 180cm3
12cm
6cm
5cm
Triangular Prisms
1
6cm3
4cm
1.5cm
Basic Ex A
2
4cm
45cm3
3cm
6cm
4cm
4cm
2cm
3
32cm3
5cm
4
24cm3
2cm
6cm
4cm
Triangular Prisms
1
35cm3
7cm
2cm
Basic Ex B
2
30cm3
5cm
4cm
3cm
5cm
3
105cm3
5cm
6cm
7cm
4
105cm3
3cm
10cm
7cm
Right Angle Triangular Prisms
Ex A
Fill in the missing values.
Base
Height
Length
Volume
1
2
5
6
30 cm3
2
2
4
4
16 cm3
3
2
5
10
50 cm3
4
3
4
5
30 cm3
5
5
5
8
100 cm3
Right Angle Triangular Prisms
Ex B
Fill in the missing values.
Base
Height
Length
Volume
1
4
5
6
60 cm3
2
2
6
3
18 cm3
3
6
10
10
300 cm3
4
5
5
10
125 cm3
5
3
5
10
75 cm3
Right Angle Triangular Prisms
Ex C
Fill in the missing values.
Base
Height
Length
Volume
1
3
5
8
60 cm3
2
3
3
5
22.5 cm3
3
8
6
20
480 cm3
4
4
6
10
120 cm3
5
6
5
9
135 cm3
Volumes - Contents
List of Formulae
A. Problems / Challenges
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Trapezoidal Prism Example 1
Volume of a prism
= area of end x length
Area of End
= 1 (4 + 2) x 3 = 9 cm2
2
Vol = 9 x 5 = 45 cm3
2cm
3cm
5cm
4cm
Trapezoidal Prism Example 2
Volume of a prism
= area of end x length
Area of End
= 1 (8 + 4) x 6 = 36 cm2
2
4cm
6cm 10cm
Vol = 36 x 10 = 360 cm3
8cm
Trapezoidal Prism Example 3
Volume of a prism
= area of end x length
Area of End
= 1 (3 + 2) x 5 =12.5 cm2
2
Vol = 12.5 x 8 = 100 cm3
2cm
8cm
5cm
4cm
3cm
Trapezoidal Prisms
1
48cm3
Basic Ex A
2
2cm
3cm
2cm
6cm
4cm
5cm
4cm
5cm
3
60cm3
160cm3
4cm 4cm
6cm
8cm
4
10cm3
4cm
2cm
1cm
3cm
Trapezoidal Prisms
1
180cm3
Basic Ex B
2
3cm
4cm
3cm
10cm
10cm
5cm
5cm
8cm
3
200cm3
144cm3
4cm
3cm
6cm
8cm
4
10cm
55cm3
4cm
1cm
7cm
Volumes - Contents
List of Formulae
A. Problems / Challenges
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Cylinders
Cylinders occur a lot in everyday life.
Many containers are this shape.
Volume of a cylinder ???
=
r2
x height
Cylinder Example 1
Volume of a cylinder
=

r2 x
height
Volume of cylinder
=

x3x3x6
Vol = 169.65 cm3
6cm
3cm
Cylinder Example 2
Volume of a cylinder
=

r2 x
height
Volume of cylinder
=

x2x2x8
Vol = 100.53
cm3
8cm
2cm
Cylinder Example 3
Volume of a cylinder
=

r2 x
height
Volume of cylinder
=

x5x5x4
4cm
5cm
Vol = 314.16 cm3
Cylinders
1
Basic Exercise A
226.2cm3
2
307.9cm3
7cm
3cm
3
2cm
3c
8cm
197.9cm3
4
8cm
37.7cm3
7cm
3cm
1cm
12cm
Cylinders
1
Basic Exercise B
88.0cm3
2
3cm 141.4cm3
5cm
2cm
3
7cm
3c
42.4cm3
6cm
1.5cm
4
8cm
235.6cm3
5cm
3cm
Volumes - Contents
List of Formulae
A. Problems / Challenges
Problems and challenges involving the formulae for the volumes of a variety of shapes.
Packets in a Box
Challenge (Cuboids)
Ingots Problem
(Prisms and Cylinders)
Half Full Problem
(Cones)
Each of the below sections is a a mixture of explanations and
basic consolidation activities
B. Cuboids
C. Triangular Prisms
D. Trapezoidal Prisms
E. Cylinders
F. Cones and Pyramids
Cone Example 1
Volume of a cone
1
=
r 2 x height
3

Volume of cone
1
=
3

x3x3x6
Vol = 56.55 cm3
6cm
3cm
Cone Example 2
Volume of a cone
1
=
r 2 x height
3

Volume of cone
1
=
3

x4x4x5
Vol = 83.78 cm3
5cm
4cm
Cone Example 3
Volume of a cone
1
=
r 2 x height
3

Volume of cone
1
=
3

x2x2x8
8cm
Vol = 33.51 cm3
2cm
Pyramid Example 1
Volume of a pyramid
1
=
x base area x height
3
Vol of pyramid
1
=
x 6 x 6 x 10
3
10cm
Vol = 120cm3
6cm
Pyramid Example 2
Volume of a pyramid
1
=
x base area x height
3
Vol of pyramid
1
=
x8x8x6
3
Vol = 128 cm3
6cm
8cm
Pyramid Example 3
Volume of a pyramid
1
=
x base area x height
3
Vol of pyramid
1
=
x 10 x 10 x 4
3
Vol = 133.33
cm3
4cm
10cm
Cones and Pyramids
1
47.1cm3
Basic Ex A
2
9 cm3
3cm
5cm
3cm
3cm
3
150.8cm3
4
8cm
3c
130.7cm3
8cm
4cm
6cm
7cm
Cones and Pyramids
1
402.1cm3
6cm
Basic Ex B
2
64 cm3
3cm
8cm
8cm
3
54 cm3
2cm
4
8cm
3c
183.3cm3
7cm
5cm
9cm