Menu © Boardworks Ltd 2001 Volume Practice. Menu © Boardworks Ltd 2001 Volumes This unit explains how the volumes of various solids are calculated. It

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© Boardworks Ltd 2001

Volume Practice

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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.

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Volumes

What 3-D solid is this?

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Volumes

What 3-D solid is this?

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Volumes - Contents

B. Cuboids

C. Triangular Prisms

D. Trapezoidal Prisms

E. Cylinders

List of Formulae

F. Cones and Pyramids

A. Problems / Challenges Menu

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

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Formulae Summary

Cylinder = r 2 x height

Cone = r 2 x height13

Pyramid = base area x height13

Cuboid = width x length x height

Prism = area of end x height

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Volumes - Contents

B. Cuboids



E. Cylinders

List of Formulae


A. Problems / Challenges






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In each problem, find the number of smaller packets that will fit neatly into the box.

PACKETS IN A BOX PROBLEMS

12345678

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In this problem, find the number of smaller packets that will fit neatly into the box.

PACKETS IN A BOX EXAMPLE 1

4cm

2cm2cm

Packet

Dimensions

12cm x 4cm x 6cm

Box

12 x 4 x 6 4 x 2 x 2 3 x 2 x 3 = 18

Answer

18

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In this problem, find the number of smaller packets that will fit neatly into the box.


5cm

3cm2cm

Packet

Dimensions

20cm x 8cm x 6cm

Box

20 x 8 x 6 5 x 2 x 3 4 x 4 x 2 = 32

Answer

32

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Be careful with this one! There is an extra step!


5cm

2cm

3cm

Packet

Dimensions

9cm x 8cm x 25cm

Box

9 x 8 x 25 3 x 2 x 5 3 x 4 x 5 = 60

Note change of order !!

Answer

60

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PACKETS IN A BOX PROBLEMS 1

Box 10 x 8 x 6 Packet 5 x 2 x 3

1 2 x 4 x 2 = 16


2 6 x 5 x 4 = 120


3 5 x 2 x 2 = 20

Box 50 x 20 x 15 Packet 5 x 4 x 3

4 10 x 5 x 5 = 250

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1 3 x 5 x 2 = 30

Box 24 x 25 x 30 Packet 6 x 5 x 3

2 4 x 5 x 10 = 200

Box 40 x 20 x 20 Packet 5 x 4 x 2

3 8 x 5 x 10 = 400

Box 18 x 21 x 15 Packet 3 x 3 x 3

4 6 x 7 x 5 = 210

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In some of these problems you may have to re-order the numbers!


Box 50 x 24 x 16 Packet 5 x 4 x 4

1 5 x 4 x 4 = 80


2 3 x 3 x 4 = 36

Box 20 x 18 x 14 Packet 6 x 7 x 5

3 4 x 3 x 2 = 24

Box 28 x 40 x 15 Packet 5 x 7 x 4

4 7 x 10 x 3 = 210

Packet 5 x 2 x 2

Packet 5 x 6 x 7

Packet 7 x 4 x 5

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Box 24 x 20 x 90 Packet 5 x 9 x 8

1 3 x 4 x 10 = 120

Box 30 x 80 x 25 Packet 5 x 3 x 8

210 x 10 x 5 = 500

Box 75 x 100 x 60 Packet 6 x 25 x 20

3 3 x 5 x 10 = 150

Box 70 x 40 x 30 Packet 15 x 7 x 8

4 10 x 5 x 2 = 100

Packet 8 x 5 x 9

Packet 3 x 8 x 5

Packet 25 x 20 x 6

Packet 7 x 8 x 15

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Volumes - Contents

B. Cuboids



E. Cylinders

List of Formulae








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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.

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The 100 ingots are melted down and made into souvenir medals. How many medals could be produced?

0.5cm3cm

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.

10cm

25cm

6cm

5cm

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STEP 1 Find volume of one ingot.

10cm

25cm

6cm

5cm Area of End

= (10 + 6) x 5 = 40 cm212

Vol = 40 x 25 = 1000 cm3

Volume of Prism

= Area of End x

Length

STEP 2 Volume of 100 ingots =

100 x 1000 =100000 cm3

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STEP 3 Find volume of one disc using the formula for a “cylinder”.

Vol = x 3 x 3 x 0.5

= 14.13 cm3

Volume of a Cylinder ?

STEP 4 Vol. of gold = 100000 cm3

Number of “medals” ?

= 100000 14.13

0.5cm3cm

= 7077 (approx)

= r2 x height

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GOING FOR GOLD PROBLEM 2

The 500 ingots are melted down and made into souvenir medals. How many medals could be produced?

0.5cm2cm

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. 6cm

10cm

4cm

3cm

Similar problem ... Different numbers!!!

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Volumes - Contents List of Formulae







B. Cuboids



E. Cylinders


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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???

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Half Full Problem 1

This ice cream cone has a height of 8cm and circular face of radius 4cm.

When full it contains 134 cm3. (Check)

What height will the ice cream be at when it is half full?

4cm

8cm

See next slide for hints.

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Half Full Problem 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?

2 cm

4cm

Try other measurements

Half the height.Half the circle radius.

Only ... 16.6 cm3

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Half Full Problem 3

TARGET VOLUME = half of 134cm3

This one on the left gives a volume of ...

3cm

6cm

Is this any closer?

The circle radius is half the height.

56.5 cm3

67 cm3

Try others!!

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Volumes - Contents


B. Cuboids



E. Cylinders


List of Formulae






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Volume of a cuboid

= width x length x height

5 cm3 cm

4 cm

Vol of cuboid

= 5 x 3 x 4

Vol = 60 cm3

Cuboid Example 1

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Volume of a cuboid


10 cm3 cm

4 cm

Vol of cuboid

= 10 x 3 x 4

Vol = 120 cm3

Cuboid Example 2

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Volume of a cuboid


6 cm 6 cm

6 cm

Vol of cuboid

= 6 x 6 x 6

Vol = 216 cm3

Cuboid Example 3

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Volume of a cuboid


8 cm

5 cm

4 cm Vol of cuboid

= 8 x 5 x 4

Vol = 160 cm3

Cuboid Example 4

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6cm2cm

3cm

1

8cm2cm

2cm

2

4cm 3cm

3cm

3

7cm2 cm

4 cm

4

36cm3

32cm3

36cm3 56cm3

Cuboids Basic Exercise A

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5cm3cm

4cm

1

8cm2cm

3cm

2

6cm 4cm

5cm

3

3.5cm2 cm

3 cm

4

60cm348cm3

120cm3 21cm3

Cuboids Basic Exercise B

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Cuboids Gaps Exercise A

Fill in the missing values.

Width Length Height Volume

2 3 5 30 cm31

4 5 10 200 cm32

3 3 53

5 2 6 60 cm34

4 8 10 320 cm35

45 cm3

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Cuboids Gaps Exercise B



4 5 8 160 cm31

8 2 3 48 cm32

6 5 9

3

4 6 10 240 cm34

5 10 20 1000 cm3

5 270 cm3

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Cuboids Gaps Exercise C



2.5 5 10 125 cm31

4 6 4 96 cm32

3

5 4 8 160 cm34

6 8 20 960 cm35

6 6 6 216 cm3

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Volumes - Contents


B. Cuboids



E. Cylinders


List of Formulae






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Volume of a prism

= area of end x length

8cm4cm

5cm

Area of End

= (5 x 4) = 10 cm212

Vol = 10 x 8 = 80 cm3

Triangular Prism Example 1

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Volume of a prism


7cm

6cm

Area of End

= (2 x 6) = 6 cm212

Vol = 6 x 7 = 42 cm3

2cm


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Volume of a prism


Area of End

= (4 x 3) = 6 cm212

Vol = 6 x 9 = 54 cm3

9cm3cm

4cm


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Volume of a prism


Area of End

= (5 x 6) =15 cm212

Vol = 15 x 12 = 180cm3 12cm6cm

5cm


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2cm1.5cm

4cm

1

5cm

3

4cm

6cm 2cm

4

6cm3

45cm3 24cm3

Triangular Prisms Basic Ex A

4cm

4cm 4cm

2 32cm3

6cm3cm

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5cm

2cm7cm

1

7cm

3

10cm

4

35cm3

105cm3

Triangular Prisms Basic Ex B

5cm 4cm

2 30cm3

6cm5cm

3cm

3cm7cm

105cm3

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Volume

2 5 6 30 cm31

2 4 4 16 cm32

5 5 8

3

3 4 5 30 cm34

2 5 10 50 cm3

5 100 cm3

Right Angle Triangular Prisms Ex A

Base Height Length

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Volume

4 5 6 60 cm31

2 6 3 18 cm32

3 5 10

3

5 5 10 125 cm34

6 10 10 300 cm3

5 75 cm3

Right Angle Triangular Prisms Ex B

Base Height Length

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Volume

3 5 8 60 cm31

3 3 5 22.5 cm32

6 5 9

3

4 6 10 120 cm34

8 6 20 480 cm3

5 135 cm3

Right Angle Triangular Prisms Ex C

Base Height Length

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Volumes - Contents


B. Cuboids



E. Cylinders


List of Formulae






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Volume of a prism


Area of End

5cm

2cm

4cm

3cm = (4 + 2) x 3 = 9 cm21

2

Vol = 9 x 5 = 45 cm3

Trapezoidal Prism Example 1

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Volume of a prism


Area of End

= (8 + 4) x 6 = 36 cm212

Vol = 36 x 10 = 360 cm3

10cm

4cm

8cm

6cm


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Volume of a prism


Area of End

8cm

2cm

4cm5cm

= (3 + 2) x 5 =12.5 cm212

Vol = 12.5 x 8 = 100 cm3

3cm


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6cm

5cm

1

8cm

3 4

48cm3

160cm3

Trapezoidal Prisms Basic Ex A

5cm

2 60cm3

6cm 4cm

3cm

10cm3

3cm

4cm 2cm

2cm

2cm

1cm

4cm

4cm

4cm

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10cm

8cm

1

6cm

3 4

180cm3

144cm3

Trapezoidal Prisms Basic Ex B

2 200cm3

8cm 10cm

7cm

55cm3

4cm

4cm

3cm

3cm

1cm

5cm

10cm

4cm

5cm

3cm

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Volumes - Contents


B. Cuboids



E. Cylinders


List of Formulae






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Cylinders

Cylinders occur a lot in everyday life. Many containers are this shape.

Volume of a cylinder ???

= r 2 x height

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Volume of a cylinder

= r 2 x height

6cm

3cm

Volume of cylinder

Vol = 169.65 cm3

= x 3 x 3 x 6

Cylinder Example 1

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= r 2 x height

Volume of cylinder

Vol = 100.53 cm3

= x 2 x 2 x 8 2cm

8cm

Cylinder Example 2

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= r 2 x height

4cm 5cm

Volume of cylinder

Vol = 314.16 cm3

= x 5 x 5 x 4

Cylinder Example 3

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8cm 3c3cm

1

8cm

2cm

2

3cm7cm

3

1cm 12cm

4

Cylinders Basic Exercise A

307.9cm3

37.7cm3

7cm

226.2cm3

197.9cm3

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7cm 3c2cm

1

8cm

5cm

2

1.5cm6cm

3

5cm3cm

4

Cylinders Basic Exercise B

88.0cm3

42.4cm3 235.6cm3

3cm 141.4cm3

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Volumes - Contents


B. Cuboids



E. Cylinders


List of Formulae






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Volume of a cone

= r 2 x height13

6cm

3cm

Volume of cone

Vol = 56.55 cm3

= x 3 x 3 x 6 13

Cone Example 1

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Volume of a cone

= r 2 x height13

5cm

4cm

Volume of cone

Vol = 83.78 cm3

= x 4 x 4 x 5 13

Cone Example 2

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Volume of a cone

= r 2 x height13

8cm

2cm

Volume of cone

Vol = 33.51 cm3

= x 2 x 2 x 8 13

Cone Example 3

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Volume of a pyramid

= x base area x height13

6cm

Vol of pyramid

Vol = 120cm3

= x 6 x 6 x 10 13

10cm

Pyramid Example 1

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Volume of a pyramid


Vol of pyramid

Vol = 128 cm3

= x 8 x 8 x 6 13

6cm

8cm

Pyramid Example 2

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Volume of a pyramid


Vol of pyramid

Vol = 133.33 cm3

= x 10 x 10 x 4 13

4cm

10cm

Pyramid Example 3

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3c

1

8cm

3cm

2

3

7cm

4

Cones and Pyramids Basic Ex A

47.1cm3

150.8cm3 130.7cm3

9 cm3

8cm

3cm

4cm6cm

5cm

3cm

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3c

1

8cm

8cm

2

9cm

3 4

Cones and Pyramids Basic Ex B

402.1cm3

54 cm3 183.3cm3

3cm

64 cm3

2cm7cm

5cm

6cm8cm

Documents

Menu © Boardworks Ltd 2001 Volume Practice. Menu © Boardworks Ltd 2001 Volumes This unit explains how the volumes of various solids are calculated. It