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Think!

In a movie theatre, the snack bar sells ice cream cones and buckets of ice cream for the same price. The base diameter of both the cone and the bucket is 5 cm. The cone is 10 cm deep and the bucket is 5 cm deep. Which contains more ice cream (if the ice cream cone is filled to capacity including the hemisphere of ice cream on top of the cone)?

02-NCM10EX2SB_TXT.fm Page 36 Monday, September 12, 2005 3:06 PM

Wordbank

Z

pyramid

A solid with a polygon for a base and triangular faces that meet at a point or vertex called the apex.

Z

cone

A solid figure with a curved face and a circular base.

Z

sphere

A ball shape. The three-dimensional shape for which all points lie the same distance from its centre.

Z

hemisphere

Half a sphere. The shape formed by making a plane cut through the centre of a sphere.

Z

slant height

For a right pyramid, this is the distance from the apex to the base along a triangular face. For a cone, the slant height is the distance from the apex to the circumference of the circular base.

Z

curved surface area

The area of the curved surface of a solid such as a cylinder, cone or hemisphere. The curved surface of a cylinder is a rectangle when flattened. The curved surface of a cone is a sector when flattened.

Z

similar solids

Solids which have the same shape (one being an enlargement of the other) and lengths of their matching sides in the same ratio.

In this chapter you will:

Z

calculate the surface area and volume of right prisms and right cylinders

Z

use Pythagoras’ theorem to find slant height, base length or perpendicular height of pyramids and right cones

Z

devise and use methods to find the surface area of pyramids

Z

develop and use the formula to calculate the surface area of cones

Z

use the formula to calculate the surface area of spheres

Z

calculate the surface area of composite solids

Z

find the dimensions of solids, given their surface area, by substitution into a formula to generate an equation

Z

use formulas to find the volumes of pyramids and cones

Z

use a formula to find the volume of a sphere

Z

calculate the volume of composite solids

Z

find the dimensions of solids, given their volume, by substitution into a formula to generate an equation

Z

solve practical problems related to surface area, volume and capacity

Z

establish and apply the fact that in two similar figures with similarity ratio 1 :

k

, matching areas are in the ratio 1 :

k

2

and matching volumes are in the ratio 1 :

k

3

.

surface area and volumesurface area and volume

MEASUREMENT


38

N EW CENTURY MATHS 10 : S TAGES 5 .2/5 .3

1 Evaluate each of the following, correct to three significant figures:

a 6.3 × 4.7 × 3.9 b × 8.5 × 3.7 c π × 12

d 2 × π × 5.3 × 2.1 e π × 82 f × 8.7 × 17.5

g + 3.6) × 17 h × π × 63 i × 15.8 × 9.2

2 Calculate the area of each of the following shapes. (All measurements are in millimetres.)a b c

d

g i

3 Use Pythagoras’ theorem to find the value of the unknown in each of the following shapes (correct to the nearest millimetre):

a b c

4 Find the circumference and area of each of the following circles, correct to the nearest whole unit. (C = πd or C = 2πr, and A = πr2)a b c d

12---

12---

12---(8.5 4

3--- 1

3---

41

25

66

48

52

95

50

47

35

85

e48

75

f

1426

20

28

35

h 14

2818

30

25 mm

14 mm

y mm

20 mm

35 mmk mm

32 mm

25 mmd mm

5.2 cm

28 cm63 cm 185 cm

Start up

Worksheet 2-01

Brainstarters 2

Worksheet 2-02

Area ID

Skillsheet 2-01

Pythagoras’ theorem


SURFACE AREA AND VOLUME

39

CHAPTER 2

Surface area of prisms and cylinders

A prism is a solid with congruent ends and a uniform cross section.

In a

right prism

, the edges of the sides are perpendicular to the congruent ends. If the sides are not perpendicular to the ends, the prism is an

oblique prism

.

5 Calculate the perimeter and area of each of the following sectors (correct to three decimal places).

a b c

6 Calculate the volume of each of these solids. (All measurements are in metres.)

a b c

1.2 m120°

15 cm

2 m

110°

3

7

5

516

12

8

1610

Skillsheet 2-02

What is volume?

Skillsheet 2-03Three-

dimensional shapes (solids)

A right prism

cross section

An oblique prism

50°


40

N EW CENTURY MATHS 10 : S TAGES 5 .2/5 .3

You may find it useful to draw a net when finding surface areas. A net may be used to form an open solid or a closed solid. A sealed cardboard carton is an example of a

closed solid

. The cardboard carton with the lid removed is an example of an

open solid

.

The surface area of a solid is the sum of the areas of all its faces. To calculate the surface area of a solid, find the area of each face and then add the areas together.

Example 1

Calculate the surface area of this prism. (All measurements are in centimetres.)

SolutionSurface area = 2 trapeziums + 4 rectangles

= 2 × × (10 + 24) × 12 + (10 × 40) + (15 × 40) + (24 × 40) + (13 × 40)= 408 + 400 + 600 + 960 + 520

∴ A = 2888 cm2

12

24

10

13 4015

24

12

10

1315

40

Trapezoidal prism

Net of trapezoidal prism

12---

The surface area of a closed cylinder is given by:Surface area = 2πr2 + 2πrh

where: area of two circular ends = 2πr2and: area of curved surface = 2πrh

r

h

Example 2

Find the surface area of a cylinder with radius 15 mm and height 40 mm (correct to the nearest mm2).

40 mm

15 mm


SURFACE AREA AND VOLUME 41 CHAPTER 2

Surface area of composite solids made of prisms and cylindersMany solids are made by combining right prisms and cylinders. When calculating surface areas of these solids, remember not to include the areas common to both solids.

SolutionSurface area = 2 circular ends + curved surface

= 2πr2 + 2πrh= 2 × π × 152 + 2 × π × 15 × 40= 5183.627…

∴ A = 5184 mm2

40 mm

15 mm

end

circumference

curved surface

height

15 mm

Worksheet2-03

A page of prisms and cylinders

Example 3

Find the surface area of this solid, correct to the nearest centimetre.

SolutionSurface area = 3 ‘faces’ of half cylinder + 5 faces of rectangular prism

= ( curved surface + 2 semi-circular ends) + (5 rectangular faces)

curved surface = × 2 × π × 28 × 40

= 3518.58… cm22 semi-circular ends = area of circle

= π × 282= 2463.00… cm2

5 rectangular faces = (40 × 56) + 2 × (40 × 25) + 2 × (56 × 25)= 7040 cm2

Surface area = 3518.59… + 2463.00… + 7040= 13 021.59…

∴ A ≈ 13 022 cm2Note that the partial answers have not been rounded or truncated because this would make the final answer for the surface area inaccurate.

40 cm

56 cm

25 cm

12---

12--- 1

2---


42 NEW CENTURY MATHS 10 : S TAGES 5 .2/5 .3

1 Find the surface area of each of the following prisms:

a b c

2 Calculate the surface area of each of the following right prisms:

a b c

d e f

3 Calculate (to the nearest m2) the surface area of each of the following cylinders. (All lengths shown are in metres.)a closed cylinder b cylinder with one open end

c cylinder open both ends d closed half cylinder

3 m

12 m7 m

3 m

8 m

5 m

10 m

6 m

2.5 m

10 m

10 cm

8.4 cm

20 cm

15 cm

8 cm

13 mm

15 mm

24 mm10 mm

6 m

3 m 2 m 10 m

10 cm

9 cm5 cm

12 cm

18 cm 12 cm

9 cm8 cm

x 14 mm48 mm

50 mm

x

2

2.5

1.83.4

8.4

10.2

10

3.6

Exercise 2-01Example 1

Example 2

SkillBuilders19-01 to 19-04

Area

SkillBuilders20-04 to 20-05Surface area of a rectangular

prism

CAS 2-01

Surface area of a cylinder

SkillBuilders20-08 to 20-09Surface area of

a cylinder



e half cylinder with open top, f half cylinder with open topone end open

4 Find (correct to one decimal place) the total surface area of each solid below. Do not include common areas. (All lengths shown are in centimetres.)

a b c

d e

5 An above-ground swimming pool is constructed as shown.a Find (correct to two decimal places) the total

external area of the pool wall (not including the floor).

b The pool cover covers the top with a 0.5 overhang. Calculate its area to the nearest m2.

6 This wooden table has been sanded ready for staining. What is the area that is to be stained, correct to the nearest cm2? (The bottoms of the legs do not need to be stained.)

*The height of each table leg is 40cm.

1.4

7.62.4

4.6

12

8

9

16

10

10

3020

20

20

10

10

21.2

15

35

1.21.4

9

1.5

m

4 m

3 m

40 cm*

6 cm

150 cm100 cm

5 cm

Example 3

Geometry 2-01

Surface area of a cylinder

02-NCM10EX2SB_TXT_SI.fm Page 43 Tuesday, September 5, 2006 10:52 AM


Working mathematicallyApplying strategies and reasoning: Surface area of right prisms1 Consider the right prism below.

Surface area = 2 ‘L-shaped’ ends + 6 rectanglesA = 4500 m2

a The six rectangles can be considered as one rectangle, as shaded in the diagram above. What is the length of this rectangle?(Note: x = 18, y = 20.)

b Find the perimeter of the ‘L-shaped’ end face. Compare your result with your answer to part a. What do you notice?

c Copy and complete:Total length of 6 rectangles = p_________ of the end face.

d Find the surface area of the right prism by copying and completing the following:Surface area = 2 ‘L-shaped’ ends + shaded rectangle

= 2 ‘L-shaped’ ends + (perimeter of end face × distance between the 2 ends)= 2 × (15 × 30 + 20 × 12) + … × 24= …

2 a Copy and complete the following to find the surface area of a right prism (closed).Surface area = … × area of ends + perimeter of end face × distance between …

= 2A + phwhen A = area of end face, p = perimeter of end face, and h = distance between end faces.

b Use the method given above to find the surface area of each of the following solids. (All measurements are in centimetres.)

i ii iii

35

30

15

24

12

yx

15 30

30

35

15

x

x

y

y

12

12

24

18

20

10

8

1410

6

12

5

13

15 17 24

24



Surface area of a pyramidA pyramid is classified by the shape of its base.

The slant height of a right pyramid is the distance from the apex to the base, along a triangular face. For a cone, the slant height is the distance from the apex to a point on the circumference of its circular base.

The perpendicular height of a right pyramid or cone is the perpendicular distance from the apex to the base.

Square pyramid Triangular pyramid Rectangular pyramid

slant height

perpendicular height

apex

The surface area of a pyramid is calculated by adding the area of the base and the area of the triangular faces.

Example 4

Find the surface area of this square pyramid.

SolutionSurface area = area of square base + area of 4 triangular faces

= 14 × 14 + 4 × × 14 × 20= 196 + 560= 756 cm2

1 Find the surface area of a square pyramid with base edge 10 cm and perpendicular height 12 cm.

SolutionFirst find the slant height AP using Pythagoras’ theorem.In ∆AOP, AP2 = AO2 + OP2

= 122 + 52 (OP = × 10)= 169

∴ AP = 13 cmSurface area = area of square base + area of 4 triangular faces

= 10 × 10 + 4 × × 10 × 13= 360 cm2

14 cm

20 cm

14 cm

12---

Example 5

10 cm

12 cm

OP

A

12---

12---

STAGE5.3



2 Find, correct to one decimal place, the surface area of a rectangular prism with base 10 cm by 8 cm and perpendicular height of 15 cm.

SolutionFirst, calculate the slant heights AP and AQ.In ∆AOP, AP2 = AO2 + OP2

= 152 + 42= 241

∴ AP = In ∆AOQ, AQ2 = AO2 + OQ2

= 152 + 52= 250

∴ AQ =

Surface area = area of rectangle base + 2 × (area ∆ABC) + 2 × (area ∆ADC)= 10 × 8 + 2 × × 10 × + 2 × × 8 ×

= 80 + 10 + 8 = 361.732≈ 361.7 cm2, correct to one decimal place.

It is better to leave the lengths of AP and AQ as exact answers ( cm and cm) rather than convert them to decimal approximations. As we discovered in Chapter 1, rounding partial calculations may affect the accuracy of the final result.

8 cm

10 cm

15 cm

E

BPC

OQ

D

A

241 cm

250 cm

12--- 241 1

2--- 250

241 250

241 250

1 Calculate the surface area of each of these pyramids (correct to one decimal place where necessary):

a b c

2 Calculate the surface area (correct to one decimal place where necessary):

a b c

5 m

25 m

18 mm

10 mm

13 mm15 mm

8 cm

4 cm

24 c

m

20 cm

5 m

8 m

16 mm24 mm

60 mm

25 cm

8 cm

16 cm


Example 5

STAGE5.3



3 Calculate the surface area of each of these pyramids, correct to the nearest cm2 where necessary. (All measurements are in centimetres.)

a b c

4 Find the surface area of each of these nets and, therefore, the surface area of the corresponding pyramid. (All measurements are in centimetres.)

a b c

5 Use the given dimensions to calculate the surface area of each square pyramid described below, correct to the nearest m2:a base edge 6 m, height of each triangular face 9 mb base edge 6 m, height 9 mc base edge 6 m, slant edges 9 m

6 The great pyramid of Khufu (or Cheops) has a height of 147 m, and each side of its square base measures 137 m. Find the surface area (excluding the base), correct to the nearest m2.

7 Calculate the surface area of each pyramid below (to one decimal place where necessary):

a b c

8 A square pyramid has a surface area of 4704 m2. If the area of its square base is 1764 m2, find:a the length of the base edgeb the area of each triangular facec the slant height of each triangular faced the perpendicular height of the pyramid

20

16

16

32

24

40

25

37

1212

28.3

24

20

36

10

7 mm36 mm

25 mm24 mm

12 cm 9 cm

5 cm

12 m

20 m

10 m



Surface area of a coneCalculating the curved surface area of a coneThe curved surface area of a cone has the same area as the sector from which it was formed.

The length of the major arc AB is equal to the circumference of the base of the cone.

∴ Arc length AB = 2πrNow the ratio of the area of the sector to the area of the whole circle should be equal to the ratio of the arc length of the sector to the circumference of the whole circle.

This can be written as:

=

=

Area of sector = × πl2

= πrlBut the area of the sector equals the area of the curved surface of the cone.

∴ Area of curved surface of cone = πrl

The Platonic solidsThe cube is a regular polyhedron because it has six identical square faces. The regular polyhedrons are also known as Platonic solids and the format name for a cube is regular hexahedron. The only other Platonic solids are the regular tetrahedron, regular octahedron, regular dodecahedron and regular icosahedron. These regular polyhedrons were proved in the time of Plato to be the only five possible regular polyhedrons.

The tetrahedron, cube and octahedron occur in nature in the form of certain crystals. Briefly describe the types of faces of these polyhedrons.

Regular tetrahedron4 faces

Regular octahedron8 faces

Regular dodecahedron12 faces

Regular icosahedron20 faces

Just for the record

sector

O

A

Blcone

O

AB

l

r

arc AB = 2πr Base radius rcircumference, AB,

of base = 2πr

Area of sectorArea of circle---------------------------------- Length of sector arc

Circumference of circle---------------------------------------------------------

Area of sectorπl2

---------------------------------- 2πr2πl---------

2πr2πl---------

STAGE5.3



Calculating the total surface area of a cone

Surface area = area of curved surface + area of circular base= πrl + πr2

where:• l is the slant height• r is the radius of the base

l l

r r

h

For a closed cone:

Surface area = πrl + πr2

where• l is the slant height• r is the radius

Example 6

For this cone, find correct to one decimal place:

a the curved surface area

b the total surface area

Solution

For the cone, r = 9 cm and l = 18 cm

a Curved surface area = πrl= π × 9 × 18≈ 508.9 cm2

b Total surface area = πrl + πr2= π × 9 × 18 + π × 92= 763.407 …≈ 763.4 cm2

9 cm

18 cm



Example 7

Find the surface area of the closed cone, expressing the answer exactly in terms of π.

SolutionFirst calculate the slant height:

l2 = 82 + 152 (using Pythagoras’ theorem)l2 = 289l = 17

Surface area = πrl + πr2

= π × 8 × 17 + π × 82= 136π + 64π

∴ Total surface area = 200π cm2

A cone has a surface area of 800 cm2 and the radius of the base is 10 cm. Find (correct to three significant figures):a the slant height b the perpendicular height

Solutiona Surface area = πrl + πr2

∴ 800 = π × 10 × l + π × 102800 = 31.415… l + 314.59 … or 800 = 10πl + 100π

∴ 31.415 … l = 485.84 … ∴ 10 πl = 800 − 100π

l = l =

= 15.4647… = 15.4647… ≈ 15.5 cm ≈ 15.5 cm

b l2 = h2 + r2∴ h2 = l2 − r2

= 15.52 − 102= 140.25

∴ h = = 11.8427… ≈ 11.8 cm

16 cm

15 cm

Example 8

485.84 …31.415…------------------------ 800 100π–

10πl---------------------------

140.25

1 Calculate the curved surface area, correct to the nearest cm2, of each cone below. (All measurements are in centimetres.)

a b c

4

8

10

20

44

35


CAS 2-02

Surface area of a cone

STAGE5.3



2 Find the surface area, correct to one decimal place, of each cone below:

a b c

3 Find the total surface area of each closed cone below, giving your answers in terms of π:a b c

4 Use the given dimensions to calculate the total surface area of each cone described below, correct to the nearest cm2:a base radius 10 cm, slant height 20 cmb base diameter 10 cm, slant height 20 cmc base radius 10 cm, perpendicular height 20 cmd base diameter 10 cm, perpendicular height 20 cm

5 Find the amount of sheet metal needed to form a conical funnel of base radius 30 cm and vertical height 50 cm, allowing for a 0.5 cm overlap at the join.

6 A cone is made from a sector of a circle with radius 8 cm which subtends an angle of 216° at the centre. Find:a the fraction of the area of the circle used to form the curved

surfaceb the length of the arc of the circle which forms the

circumference of the basec the radius of the base of the coned the slant height of the conee the total surface area of the cone

7 A cone has a surface area of 350 cm2 and the radius of the base is 5 cm. Find (correct to three significant figures):a the slant heightb the perpendicular height

8 The curved surface area of a cone is 200 cm2 and the slant height is 9 cm. Find (correct to two significant figures):a the base radiusb the perpendicular heightc the total surface area

9 The curved surface area of a cone is 460 cm2 and the base radius is 6 cm. Find (correct to one decimal place):a the slant heightb the total surface area

5 mm

20 mm

8 m

4 m

14 cm

7 cm

12 m

5 m

14 mm24 mm

22 cm

40 cm

8 cm

216°

Example 7

Example 8



Surface area of a sphere

Note that the surface area of a sphere is four times the area of the circle (which slices through the centre of a sphere).

For a hemisphere, the curved surface is 2πr2 and this is double the area of its (circular) base.

10 A cone has a total surface area of 400 cm2 and a base area of 50 cm2. Find (correct to three significant figures):a the radius of the base b the slant heightc the perpendicular height

11 A cone has a total surface area of 2000 cm2. If the area of the base is 150 cm2, find:a the radius of the base b the slant heightc the perpendicular height

12 Find the radius, slant height and perpendicular height (to three significant figures) for each cone described below:a surface area 347 cm2, base area 125 cm2 b surface area 815 cm2, base area 375 cm2

c surface area 1238 cm2, base area 538 cm2

For a sphere:

Surface area = 4πr2

where r is the radius of the sphere

r

r

rr

Example 9

Find the surface area, correct to one decimal place, of a sphere with radius 26 cm.

SolutionSurface area = 4πr2

= 4 × π × 262= 8494.8665…≈ 8494.9 cm2

26 cm

STAGE5.3



Example 10

Find the surface area, correct to one decimal place, of an open hemisphere with radius 5 m.

SolutionSurface area = area of curved hemisphere

= 2πr2

= 2 × π × 52

= 157.0796…≈ 157.1 m2

A sphere has a surface area of 747 cm2. Find its radius correct to three significant figures.

SolutionA = 4πr2

747 = (4 × π) × r2

r2 =

∴ r =

= 7.710 01…≈ 7.71 cm

5 m

Example 11

7474 π×( )

-----------------

7474 π×( )

-----------------

1 Calculate the surface area of each sphere below, correct to two decimal places:

a b c

2 Find the surface area of each hemisphere below in terms of π:a b c

3 Calculate the surface area, correct to the nearest m2, of each shape described below:a an open hemisphere with radius 10 mb a sphere with diameter 10 mc a closed hemisphere with diameter 10 md a sphere with radius 10 m

4 If the radius of the Earth is approximately 6400 km, find its surface area, in scientific notation, correct to two significant figures.

15 mm 11 m 10.8 cm

24 m

8 cm 16 mm


Example 10

CAS 2-03

Surface area of a sphere



Surface area of composite solidsWhen calculating the surface areas of composite solids, remember not to include the area common to both solids.

5 The spherical shell shown has an inner radius of 6 m and an outer radius of 8 m.a Calculate its surface area, correct to one decimal place.b If one litre of paint covers 20 m2 calculate the amount of

paint needed to cover the shell twice, correct to the nearest litre.

6 Find the radius of each solid below if it has a surface area of 600 cm2. Give your answer correct to three significant figures.a a sphereb a closed hemispherec an open hemisphere

7 The radius of a hemisphere is r units.a Show that the surface of the hemisphere is 3πr2.b Use the result from part a to find the surface area (correct to two significant figures) of a

hemisphere with:i radius 20 cm ii diameter 1.9 m

c If the surface area of a hemisphere is 400 cm2, find the radius (correct to one decimal place).

6 m8 m

Example 11

Example 12

Find the surface area, correct to the nearest centimetre, of each of the following solids. (All measurements are in centimetres.)

a b

Solutiona Surface area = curved surface area of cone + area of hemisphere

= πrl + 2πr2 where r = 11, h = 60 and l = ?Find l: l2 = r2 + h2 (using Pythagoras’ theorem)

= 112 + 602= 3721

l = = 61 cm

60

11

50

50

3721

STAGE5.3



So: Surface area = π × 11 × 61 + 2 × π × 112= 2868.274…≈ 2868 cm2

b Surface area = curved surface area of cylinder + area of one end + area of hemisphere= 2πrh + πr2 + 2πr2 where r = 25 and h = 50= 2 × π × 25 × 50 + π × 252 + 2 × π × 252= 13 744.467…≈ 13 744 cm2

1 Find (correct to one decimal place) the total surface area of each solid. (All lengths shown are in centimetres.)

a b c

d e f

g h i

7

24

8

88

6

12

12

12

40

25

15

10

10

6

10

10

5

5

12

6

4

10

19

24

14




j k l

m n o

44

24

12

24

18

60

30

20

4530

24 18

24

5

8

10

6

Divisibility testsHow can you tell if a number is divisible by 2? Look at its last digit. If that digit is 2, 4, 6, 8 or 0, then the number is divisible by 2 (that is, it is even).

How can you tell if a number is divisible by 5? If its last digit is 0 or 5, then the number is divisible by 5.

These are examples of divisibility tests, that is, rules for checking whether a number is divisible by a certain number. The table below shows some common divisibility tests.

1 Examine these examples.a Test whether 748 is divisible by 2, 3 or 4.

• Last digit is 8 (even). ∴ 748 is divisible by 2.• 7 + 4 + 8 = 19, which is not divisible by 3. ∴ 748 is not divisible by 3.• 48 is divisible by 4. ∴ 748 is divisible by 4. (748 ÷ 4 = 187).

b Test whether 261 is divisible by 3, 5 or 9.• 2 + 6 + 1 = 9, which is divisible by 3. ∴ 261 is divisible by 3. (261 ÷ 3 = 87).• Last digit, 1, is not 0 or 5. ∴ 261 is not divisible by 5.• 2 + 6 + 1 = 9, which is divisible by 9. ∴ 261 is divisible by 9. (261 ÷ 9 = 29).

A number is divisible by:

if:

2 its last digit is 2, 4, 6, 8 or 0

3 the sum of its digits is divisible by 3

4 its last two digits form a number divisible by 4

5 its last digit is 0 or 5

6 it is even and the sum of its digits is divisible by 3

9 the sum of its digits is divisible by 9

10 its last digit is 0

Skillbank 2

SkillTest 2-01

Divisibility tests

STAGE5.3



Volume of right prisms and cylindersThe volume of a solid is the amount of space it occupies. Volume is measured in cubic units (mm3, cm3, m3, and so on).

Volume and capacityThere is a very important distinction between volume and capacity. Volume is the space that an object occupies, while capacity measures the maximum amount the object can contain. Capacity is measured in mL and L for fluids, and cm3 and m3 for solids.

A normal cardboard carton (a rectangular prism) measuring 10 cm × 20 cm × 30 cm will have a volume of 6000 cm3. If the cardboard walls of the box had no thickness, its internal measurements would be the same and so its capacity would also be 6000 cm3.

In the real world, of course, the cardboard walls of the box do have a thickness, but in most cases this measurement will be quite small when compared to the dimensions of the carton, and so can usually be ignored.

The relationship between the units of volume and capacity can be summarised as follows:

Volume factsRight prism

In a right prism, the side faces are all perpendicular to the base.

Volume of a right prism = area of base × heightV = Ah

c Test whether 570 is divisible by 4, 6 or 10.• 70 is not divisible by 4. ∴ 570 is not divisible by 4.• 570 is even and 5 + 7 + 0 = 12, which is divisible by 3.

∴ 570 is divisible by 6. (570 ÷ 6 = 95).• Last digit is 0. ∴ 570 is divisible by 10. (570 ÷ 10 = 57).

2 Test whether each of the following numbers is divisible by 2, 3, 5 or 6:a 250 b 189 c 78 d 465e 1024 f 840 g 715 h 627

3 Test whether each of the following numbers is divisible by 4, 9 or 10:a 144 b 280 c 522 d 4170e 936 f 726 g 342 h 5580

Skillsheet 2-02

What is volume?

Skillsheet 2-03Three-

dimensional shapes (solids)

Worksheet 2-03

A page of prisms and cylinders

1 m3 = 1 kL1 cm3 = 1 mL

height

base

cross-section

side face

a rightprism

an obliqueprism



Right cylinder

Volume = area of circular base × heightV = Ah

Since A = πr2, we also have thatV = πr2h where r = radius of the base

r

h

Example 13

Calculate the volume of each of these solids:

a b

Solution

Circular wafer biscuits of diameter 5 cm are packed in a cardboard box of height 21 cm.

a Calculate the surface area of the box.

b How much packaging would be saved (correct to the nearest cm2) if the biscuits were packed into a cylindrical box?

Solutiona Surface area of box = 2 squares + 4 congruent rectangles

= 2 × (5 × 5) + 4 × (5 × 21)= 470 cm2

b Surface area of cylinder = curved surface + 2 ends (with radius 2.5 cm, and height 21 cm)= 2 × π × 2.5 × 21 + 2 × π × 2.52= 369.13… cm

∴ Packaging saved = 470 − 364.13…= 100.86≈ 101 cm2

a Area of cross section = × (9 + 15) × 8

= 96 cm2V = Ah

= 96 × 12= 1152 cm3

b Area of cross section = × π × 252

A = 654.498… mm2V = Ah

= 654.498… × 40= 26 179.938…

∴ V ≈ 26 180 mm3

15 cm

8 cm12 cm

9 cm 40 mm25 mm

120°

12---

120360---------

Example 14

5 cm

WAFER

S

21 cm



1 Calculate (to the nearest 0.1 m3) the volume of each of these. (All lengths are in metres.)

a b c

d e f

2 Calculate the volumes of the cylinders with the following dimensions, giving your answers correct to one decimal place:a radius 7 m, height 10 mb diameter 35 cm, height 15 cmc diameter 6.2 m, height 7.5 md radius 0.8 m, height 2.35 m

3 Find the volume of each of these to the nearest cm3. (All lengths shown are in centimetres.)

a b c

d e f

g h i

1.8

612

9

18

11

13

2.4

3

4.2

10.16.4 12.8

3.5 2.42.8

5.5

11.3

7.7

25

4816

48

8

12

20

40

10 10

radius of circle = 4 cm

11.3

7.2

19.6

12.73.2

36

8 625

15

30

2435

15

12140°

5 14

100°


CAS 2-04

Volumes

SkillBuilders

20-01 to 20-03Volume of a rectangular

prism

SkillBuilders

20-06 to 20-07Volume of a

cylinder



4 A cylindrical loaf of bread 30 cm long with a diameter of 8 cm is cut into slices 15 mm thick. Find:a the number of slices in a loafb the surface area of each slice, correct

to the nearest cm2

c the total surface area of the sliced loaf of bread, correct to the nearest 0.1 m2

5 A manufacturer sells tents in the shape of a triangular prism or a half cylinder with dimensions as shown. Which tent has the greater surface area?

(Note: The floor is included for both tents.)

6 A cylindrical lampshade is to be covered. Find the area of material needed if 10% extra is allowed for seams and overlaps.(Note: The lampshade is a cylinder open at both ends.)

7 A cylindrical loaf of bread 30 cm long with a diameter of 8 cm is cut into slices 15 mm thick as shown in Question 4 above. Calculate (to the nearest 0.1 cm3):a the volume of the loafb the volume of each slice

8 A wedge of cheese is cut from a cylindrical block of height 10 cm and diameter 40 cm, as shown. Find the total surface area of the wedge, correct to two decimal places.

8 cm

30 cm

15 mm

2 m

2 m5 m

2.24 m

2 m5 m

20 cm

30 cm

wedge

40 cm

10 cm60°

60°

Example 14



9 A fish tank 55 cm long, 24 cm wide and 22 cm high can be filled to 4 cm below the top. Calculate:a the volume of the tank in cm2 b the recommended capacity of the tank (to the nearest litre)c the recommended capacity of the tank in kilolitres, correct to three significant figures

10 Robert’s birthday cake had a diameter of 30 cm and a height of 9 cm. Ilhea cut a wedge from the cake as shown in the diagram below. Calculate the volume of cake left.

11 A wedge is cut from a cylindrical piece of wood as shown in the diagram below. The remainder of the wood is to be painted. Calculate, to one decimal place, the area to be painted.

12 A cylinder has a volume of 774.5 cm3. The area of a circular end is 72.38 cm2. Calculate the height of the cylinder.

13 The surface area of the curved surface of a can is 27 143.4 mm2. If the height is 120 mm, find the radius of the can.

14 A right prism has a volume of 4.2 m3. If the height of the prism is 1.3 m, find its cross-sectional area, correct to one decimal place.

15 The external surface area of an open cube is 1125 cm2. Find its volume.

16 The volume of half a cube is 32 m3. Find its surface area to the nearest whole number.

80°

70°1.2 m

0.2 m

Volume = 32 m3



Volume of pyramids, cones and spheresVolume of a pyramid

Working mathematicallyApplying strategies and reasoning: Pyramids make prismsRead through this activity and consider your approach. Then form into groups of two or three to carry it out.

You will need: cardboard, scissors, sticky tape, compasses, a ruler, sand

1 Construct the net of an open square prism (cube) with sides of 10 cm. Allow a 1-cm tab as shown. Cut out the net, fold it and tape it together.

2 Construct the net of a square pyramid with the measurements shown. Allow a 1-cm tab as shown. Cut out the net, fold it and tape it together. (Note: if the height of the square pyramid is 10 cm, then the slant height must be equal to 11.2 cm, as shown by Pythagoras’ theorem.)

3 Carefully remove the base of the pyramid to form an open square pyramid. Check that the pyramid fits snugly inside the prism.

4 a Fill the pyramid with sand and empty it into the prism. Repeat until the prism is full.

b How many times did you do this to fill the prism?

5 Copy and complete:The volume of a square prism = ___________________ × volume of a square pyramid.

6 Rewrite:The volume of a square pyramid = ___________________ × volume of a square prism.

7 Repeat this experiment for other prisms and pyramids with the same base and height.

8 Write the result, relating the volume of a pyramid to the volume of the corresponding prism.

10 cm

Cube

tab

tab

tab

tab

10 cm

Squarepyramid

11.2 cm

tab

tab

tabta

b



Volume of pyramids, by layersWe can think of a pyramid as being made up of layers of prisms as shown below.

The volume of each layer can be easily calculated. Finding the sum of the layers will then give an approximation of the volume of the pyramid.

The layers are of equal thickness (T = The length and base of the pyramid

are decreased by the amounts and from layer to layer.

This spreadsheet approximates the volume of a pyramid with the base length 8 units, base breadth 6 units and perpendicular height 10 units.

Set up your spreadsheet as shown.

1 a Enter 10 in cell D2.b Copy each formula down to row 13.c Explain the results in cells E13 and F13.d How accurate was your result in F13? Explain.e Print out your spreadsheet and paste it into your workbook.

2 a Enter 40 in cell D2 and copy each formula down to row 43.b In one or two sentences compare your result in F43 with the previous result in F13.

3 a Enter each of these values in cell D2 and copy the formulas down to the appropriate row:i 100 (copy down to row 103)

ii 200 (copy down to row 203)iii 400 (copy down to row 403)

b Use the formula V = to calculate the exact volume of the pyramid.c Write a brief report about your results in parts a and b.

A B C D E F

1 Number of layers =

2

3 H L B Thickness of layers Volume of layer Sum of volumes

4 10 8 6 =$A$4/$D$2 =B4*C4*D4 =E45 =B4−$B$4/$D$2 =C4−$C$4/$D$2 =E5+F46

…

13

6

8

10

Hnumber of layers----------------------------------------).

Lnumber of layers---------------------------------------- B

number of layers----------------------------------------

13---Ah

Using technology

Spreadsheet



The volume of a pyramid is given by:

V = where: A is the area of the baseand: h is the perpendicular height

13---Ah

Example 15Calculate the volume of each of the following pyramids (correct to one decimal place where necessary):

a b

Solutiona V = b V =

= × (20 × 25) × 30 = × ( × 10 × 7) × 8

∴ V = 5000 mm3 = ∴ V ≈ 93.3 m3

Find the capacity (to the nearest millilitre) of a square pyramid with base edge 64 mm and slant height 40 mm.

SolutionFirst find h, the perpendicular height of the pyramid.Using Pythagoras’ theorem: h2 = 402 − 322

h2 = 576∴ h = 24 mm

Volume of pyramid =

= × (64 × 64) × 24

= 32 768 mm3= 32.768 cm3 (1 cm3 = 1000 mm3)= 32.768 mL

∴ Capacity = 33 mL (to the nearest mL)

20 mm25 mm

30 mm 8 m

10 m

7 m

13---Ah 1

3---Ah

13--- 1

3--- 1

2---

93.3̇

Example 16

40 mm

64 mm

64 mm

h

13---Ah

13---



Example 17The volume of a square pyramid is 100 cm3. If its height is 12 cm, calculate:a the area of its base b the length of its base

Solutiona Volume of pyramid =

∴ 100 = × A × 12

100 = 4 × A

∴ A =

= 25 cm2b Area of square base = 25

∴ Side length = 5 (5 × 5 = 25)

13---Ah

13---

1004

---------

1 Calculate the volume of each of the following pyramids (correct to one decimal place where necessary):

a b c

d e f

2 Calculate the volume of each pyramid described below (correct to one decimal place where necessary):a base area 248 cm2, perpendicular height 11.5 cmb square base of sides 3.7 m, perpendicular height 9 mc rectangular base 25 mm by 16 mm, perpendicular height 18 mmd triangular base, with base triangle having base 14 cm and height 12 cm, and pyramid

height 30 cme square base of sides 44 mm, perpendicular height 23 mmf rectangular base 4.3 m by 2.5 m, perpendicular height 6.1 mg base area 500 cm2, perpendicular height 47.3 cmh triangular base, with base triangle having base 24 mm and height 32 mm, and pyramid

height 63 mm

8 cm

7 cm

10 cm

10 cm

6 cm

8 cm

10 cm

9 cm

14 m

18 m

8 m

20 cm

12 cm

15 cm

5 m

8 m

8 m


CAS 2-05

Formulas

SkillBuilders

20-15 to 20-16The volumes of pyramids



3 For each of the following, calculate:i the perpendicular height (correct to one decimal place)

ii the volume iii the capacity

a b c

d e f

4 Calculate the volume (in cm3) and capacity (in litres, correct to three decimal places) of each solid shown below (All measurements are in centimetres.):

a b c

d e f

5 A grain hopper is in the shape of an inverted square

20 cm

20 cm

26 cmh

h

96 m

25 m30 m

52 m41 mm 41 mm

18 mm18

mm

68 mm 61 mm

11 mm

11 mm32 mm 32 mm

8.5 m 8.5 m

3.6 m 3.6 m3.6 m 3.6 m

160 cm

126 cm

116 cm

105 cm

4

7

7

9

10

10

6

12

12

12

20

15

12

25

18 24

21

30

20

10

15

4.5 m

5 m

4.5 m

Example 16



Volume of a cone

The formula V = refers to a special case of the ‘volume of a pyramid’ formula.

6 The great pyramid of Khufu (or Cheops) was built on a square base with side lengths approximately 139 m.

a Find the volume in cubic metres if the original height of the pyramid was 147 m.b There are an estimated 2.3 million stone blocks in the pyramid. Calculate the average

volume of each block, correct to three decimal places.

7 The area of the base of a pyramid is 40 m2. If its volume is 360 m3, calculate the perpendicular height of the pyramid.

8 The volume of a pyramid is 200 cm2. If its height is 25 cm, calculate the area of the pyramid’s base.

9 The volume of a square pyramid is 400 mm3. If the length of its base is 8 mm, calculate the height of the pyramid (correct to the nearest millimetre).

10 A square pyramid has a volume of 80 cm3 and a height of 10 cm. Calculate the length of the base of the pyramid (correct to one decimal place).

139 m

147 m

Example 17

The volume of a cone is given by:

V =

where: A is the area of the baseand: h is the perpendicular height.Since A = πr2, we also have:

V = where r is the radius of the base.

r

h

13---Ah

13---πr2h

13---πr2h



Example 18

Calculate the volume of this cone correct to the nearest cm3.

SolutionVolume of cone =

= × π × 10.52 × 30

= 3463.605…∴V ≈ 3464 cm3

A cone has a slant edge of 61 mm and a base radius of 11 mm. Find its volume, correct to one decimal place.

SolutionFirst, find the height.Using Pythagoras’ theorem: h2 = 612 − 112

h2 = 3600∴ h = 60

Volume of cone =

= × π × 112 × 60

= 7602.654…∴ V ≈ 7602.7 mm2

A cone has a volume of 165 cm3 and a height of 12 cm. Find its radius, correct to three significant figures.

SolutionV =

165 = × π × r2 × 12

165 = (4 × π) × r2

r2 = (dividing both sides by 4 × π)

r =

= = 3.6235…

∴ r ≈ 3.62 cm

21 cm

30 cm13---πr2h

13---

Example 19

h11 mm

61 mm

13---πr2h

13---

Example 20

13---πr2h

13---

165(4 π)×----------------

165(4 π)×----------------

13.1302…



Volume of a cone by layersThe volume of a cone can be approximated by calculating the volume of layers as shown in the diagram. The layers that make up the cone

are of equal thickness (T =

and the radius is decreased by

This spreadsheet approximates the volume of a cone with base radius 8 units and perpendicular height 12 units.


1 a Enter 20 (the number of layers) in cell C2.b Copy each formula down to row 23.c What do the values in cells D23 and E23 represent?d Print out your spreadsheet and paste it into your workbook.

2 a Enter the following values in cell C2 and copy the formulas down to the appropriate row:i 100 (copy down to row 103) ii 200 (copy down to row 203)

iii 400 (copy down to row 403)b Record the results for cells E103, E203 and E403.c Compare your results to the result that would be obtained using the formula V =

Write a brief report, explaining how the accuracy of the results may be improved.

A B C D E

1 Number of layers =

2

3 H r T Volume of layer Sum of volumes

4 12 8 =$A$4/$C$2 =PI()*B4^2*C4 =D45 =B4−$B$4/$C$2 =D5+E4

…

23

12

8

r = radius of baseH = height of cone

Hnumber of layers----------------------------------------)

rnumber of layers----------------------------------------.

13---Ah.

Using technology

Spreadsheet

1 Calculate the volume of each of the following cones, correct to the nearest whole number:

a b c

8 m

5 m

14 cm

17 cm

15 mm

18 mm


CAS 2-05

Formulas

SkillBuilders

20-10 to 20-12Volume of a

cone



d e f

2 Find the volume of each of these cones correct to one decimal place:a radius 20 cm, height 30 cm b diameter 20 cm, height 30 cmc radius 30 cm, height 20 cm d diameter 30 cm, height 20 cm

3 Calculate the perpendicular height and the volume of each of these cones, giving your answers correct to one decimal place:a b c

d e f

4 For each net below:i draw the cone

ii calculate the perpendicular height of the coneiii find the volume of the cone correct to one decimal place

a b

c d

5 Find, correct to the nearest centimetre, the radius of each of the following cones:a V = 800 cm3, h = 10 cm b V = 160 cm3, h = 12 cmc V = 15 000 cm3, h = 65 cm d V = 12 cm3, h = 2 cm

12 cm7 cm 10 cm

15 cm

30 mm

18 mm

8 cm

4 cm

44 m

35 m

13.5 cm

10.8 cm

0.8 m

3.6 m

68 m

247 m

83 cm

83 cm

41 mm

9 mm

10 cm

13 cm

3.4 m

1.6 m

75 m

m

42 mm

Example 19

Example 20



Volume of a sphere

6 Find, correct to one decimal place, the height of each of the following cones:a V = 600 m3, r = 10 m b V = 1468 cm3, r = 12 cmc V = 154 cm3, r = 4 cm d V = 20 m3, r = 1.5 m

7 A cone of volume 60 cm3, has a base radius of 5 cm. Calculate the height of the cone, correct to the nearest centimetre.

8 A cone has a volume of 150 mm3. If the height and radius of the cone are equal in length, calculate the radius of the cone. Give your answer to two decimal places.

The volume of a sphere is given by:

V = where r is the radius of the sphere.

r43---πr3

Example 21

Calculate the volume of each of these solids, giving your answers correct to one decimal place.a b c

Solutiona V = b V = (where r = 9 cm)

= × π × 83 = × π × 93

= 2144.660… = 3053.628…∴ V ≈ 2144.7 cm3 ∴ V ≈ 3053.6 cm3

c Volume of hemisphere = × volume of sphere

= × π × 1.33

= 4.601…∴ V ≈ 4.6 m3

Find the radius (correct to the nearest 0.1 millimetre) of a sphere that has a volume of 400 mm3.

SolutionV =

400 = × π × r3

= r3 (dividing by × π to make r3 the subject)

8 cm

18 cm

1.3 m

43---πr3 4

3---πr3

43--- 4

3---

12---

12--- 4

3---×

Example 22

43---πr3

43---

400

(43--- π)×

---------------- 43---



r =

= = 4.5707…

∴ r ≈ 4.6 mm

400

(43--- π)×

----------------3

95.4929…3

1 Calculate the volume of each of these solids, giving your answers correct to the nearest whole number:

a b c

d e f

2 Find the volume, correct to one decimal place of:a a sphere with radius 20 cm b a hemisphere with radius 3.6 mc a sphere with diameter 7 cm d a hemisphere with diameter 1.5 me a sphere with radius 4.4 cm f a hemisphere with diameter 2.5 cm

3 Find, correct to the nearest whole number, the radius of a sphere with:a V = 300 mm3 b V = 1585 cm3 c V = 84 m3

4 Find the volume of each of these solids, correct to three significant figures:

a b c

d e f

5 The Earth has a radius of approximately 6400 km. Assume that the Earth is a sphere and find its volume, correct to two significant figures.

15 mm 11 m10.8 cm

24 m8 cm

16 mm

14 mm27 cm

116 mm

0.8 m

4.2 m 46 mm


Example 22

CAS 2-05

Formulas



Volume of a sphere by shellsThe volume of a sphere can be approximated by calculating and adding the volumes of spherical shells. The volume of a shell can be found by multiplying the surface area of a shell by its thickness. The surface area can be found using SA = 4πr2. The thickness of each shell is

equal to and the radius decreases by

This spreadsheet approximates the volume of a sphere with radius 12 units.


1 a Enter 40 (the number of shells) in cell B2.b Copy each formula down to row 43.c Explain the values in these cells:

i A43ii C43

iii D43

2 a Enter the following values in cell B2 and copy the formulas down to the appropriate row:i 100 (copy down to row 103)

ii 200 (copy down to row 203)iii 400 (copy down to row 403)

b Record the results for cells D103, D203 and D403.

c Print out your spreadsheet and paste it into your workbook.

d Use the result V = (where R = 12) to comment on the accuracy of your results.

A B C D

1 Number of shells =

2

3 R Thickness of shells Volume of shell Sum of volumes of shells

4 12 =$A$4/$B$2 =4*PI()*A4^2*B4 =C45 =A4−$A$4/$B$2 =D4 + C5

…

43

Rnumber of layers---------------------------------------- R

number of layers----------------------------------------.

R

R = radius of sphere

43---πR3

Using technology

Spreadsheet



Summary of surface area and volume formulasPrisms:

V = Ah

Cylinder:

Curved surface area = 2πrhTotal surface area = 2πrh + 2πr2

V = πr2h

Map projectionsAlthough a globe is the most accurate way of representing the Earth, it is often necessary to use a two-dimensional representation of the planet. (Why?)

Map projections are used to provide two-dimensional representations. When making a map, the following aims are important:• The map should give correct shapes to the Earth’s

features.• The map should give correct directions to the Earth’s

features.• The map should keep areas in proportion.

Distortion will occur when projecting from a spherical surface to a flat surface. The choice of map projection will depend on whether shape, direction or area is most important. Map projections can be classified into three groups: cylindrical, conical and azimuthal projections.

Obtain pictures of the three projections above and briefly describe how each projection is constructed.

Cylindrical(For example Mercator’s

projection)

Conical(For example Bonne’s

projection)

Azimuthal(For example Lambert’s equal

area projection)

Just for the record

Worksheet 2-04

Formula matching game

Ah

h

r



Pyramid:

V = Ah

Cone:

Curved surface area = πrlTotal surface area = πrl + πr2

V = πr2h

Sphere:

Surface area = 4πr2

V = πr3

The formulas given above for surface area involve r2 or rh (two dimensions) and are measured in square units, while the formulas for volume involve lbh, r2h or r3 (three dimensions) and are measured in cubic units.

Volumes of composite solids

h

A

13---

lh

r 13---

r43---

Worksheet 2-05

A page of solid shapesExample 23

Calculate:a the volume (to the nearest cm3) of this solidb the capacity (to the nearest litre) of the solid

Solutiona Volume of solid = volume of cylinder + volume of hemisphere (where r = 20 ÷ 2

= π × 102 × 35 + × π × 103 = 10 cm)= 13 089.969…

∴ V ≈ 13 090 cm3

b Capacity ≈ 13 090 mL (1 cm3 = 1 mL)= 13.09 L (1 L = 1000 mL)

∴ Capacity ≈ 13 L

20 cm

35 cm

12--- 4

3---×

Worksheet 2-06

Back-to-front problems 1

Worksheet 2-07

Back-to-front problems 2

1 A storage tank like the one shown in the diagram on the right is completely filled with water.a What solids make up the tank?b Calculate the volume of the tank, to the nearest m3.c What is the capacity of the tank to the nearest kilolitre?

4 m

2 m

4 m




2 For each of these solids, calculate the volume (to the nearest cm3) and the capacity (in litres, correct to three decimal places). All measurements are in centimetres.

a b c

3 For each of these solids, find the volume (to the nearest cm3) and the capacity (in litres). All measurements are in centimetres.

a b c

d e f

4 A conical tank (A) and a hemispherical tank (B) have measurements as shown.a Calculate the volume of each tank,

correct to two decimal places.b How much more does tank B hold

compared to tank A?

5 Spherical balls of diameter 10 cm are stacked inside a box in the shape of a rectangular prism, as shown.a How many balls will fit in the bottom layer?b If the balls are stacked in the same manner as the bottom

layer until the box is full, how many balls will fit in the box?c Calculate the volume of the space occupied by the balls when

the box is full.d What percentage of the box is empty space? Give your

answer to the nearest whole percentage.

40

15

2014

24

5

56

12

7

24

28

14

10

10

6

88

8

30

50

18

36

3 m

3 mA B

3 m

3 m

30 cm40 cm

50 cm



6 An hourglass has measurements as shown.a The total volume of sand in the hourglass is three-quarters the volume

of the bottom cone. Calculate the volume of sand in the hourglass, correct to the nearest cm3.

b When the hourglass is inverted, it takes 1 hour for all the sand to fall to the bottom cone. At what rate is the sand falling? Give your answer in cm3/s, correct to two significant figures.

7 A swimming pool is constructed in the shape shown.a Calculate the volume of the

pool, in m3.b Calculate the capacity of the

pool if it is filled to a depth 20 cm from the top.

c If water costs $0.98/kL, find the cost of filling the pool.

8 Jupiter has a volume more than 1318 times the volume of Earth. If the radius of Earth is 6400 km, approximate the volume of Jupiter, correct to four significant figures.

20 cm

50 cm

10 m

1 m

20 m

10 m

2 m

Working mathematicallyApplying strategies and reasoning: Packing and stackingYou work for a company that produces large tins of canned fruit and you are given the job of designing a storage system to be used for the finished products. The main decisions you must make are:• the dimensions of a cardboard box

to hold a dozen cans• the best arrangement for the boxes

on the wooden pallet• the number of boxes the pallet is

able to support.12 cm

10 cm



Areas of similar figuresSimilar figures have the same shape, their matching angles are equal and the lengths of their matching sides are in the same ratio.

1 The boxesTwo ways of arranging the tins in cardboard boxes are:

a Determine the dimensions required for a box for each type of arrangement.b Draw the net that could be folded to create each of the boxes, including the lid. Show

the measurements of each side (make no allowance for overlapping the sides).c Allow for cardboard 5 mm thick. How does this affect the dimensions of the box?d Determine the unused space in each box when it holds a dozen cans.e Which style of box is the better value for the company?

2 Arrangement on the palletThe filled boxes are stacked onto pallets so that a forklift can be used to move large quantities around the building. The pallets measure 1700 mm by 1500 mm and the boxes are not to overhang the edges. Report on the best way to arrange the boxes so that the maximum number is fitted in one layer.

3 The number of layers on the palletThe packed pallets are stored in a rack system that can take a pallet with overall height of 1000 mm.The pallet is 160 mm high and each pallet has a mass of 25 kg. The maximum lifting mass of the forklift is 1000 kg. Each can of fruit has a mass of 800 g (gross).Calculate and report on the total number of boxes that can be placed on a pallet.

1700 mm

1500 mm

160 mm

Working mathematicallyApplying strategies and reasoning: Sides and areas of similar figures1 a Draw two squares, with sides measuring 3 cm and 6 cm.

i Find their areas.ii What is the ratio of their sides?

iii What is the ratio of their areas?b i The two rectangles A and B are similar. Why?

ii What is the ratio of their matching sides (A to B)?iii What is the ratio of their areas (A to B)?

20 m

8 m

4 m

10 m

A

B

Worksheet 2-08

Investigating paper sizes

Worksheet 2-09

Areas and volumes of

similar figures

STAGE5.3



If the ratio of the matching sides is expressed in the form 1 : k, then the matching areas are in the ratio 1 : k2.

c Draw the two triangles represented by this diagram.

i What is the ratio of their matching sides (X to Y)?

ii Find the areas of both triangles. What is the ratio of their areas (X to Y)?

d Copy and complete:When you double the length in two dimensions, the area is increased 2 × ______ = ______2 = ______ times.

2 a Draw two squares with sides measuring 2 cm and 6 cm.i What is the ratio of their sides? ii What is the ratio of their areas?

b i Write the area of a rectangle with length 5 cm and width 3 cm.ii The length and width of the rectangle are both multiplied by 3. What is the new

area?iii What is the ratio of the matching sides of the two rectangles from parts i and ii?iv What is the ratio of the areas of the two rectangles from parts i and ii?

c Copy and complete:When you multiply the length and width by ______, the area is increased 3 × ______ = ______2 = ______ times.

3 If the sides of a square are multiplied by 4, by how many times is the area increased?

4 a How is the ratio of the areas of similar figures related to the ratio of their matching sides?

b Check your answers to part a by finding the areas of two circles of radii 5 cm and 8 cm, and then comparing the matching ratios of their radii and areas.

4 cm

6 cm

3 cm

2 cmX

Y

If the matching sides of two similar figures are in the ratio m : n, then their areas are in the ratio m2 : n2.

Example 24

1 Two similar triangles have matching sides whose lengths are in the ratio 4 : 1. Find the ratio of their areas.

Solution1 If matching sides are in the ratio m : n, their areas are in the ratio m2 : n2

Ratio of sides = 4 : 1∴ Ratio of areas = 42 : 12

= 16 : 1

2 What is the ratio of the areas of the similar rectangles shown?

20 mm

35 mm14 mm

8 mm

AB



Surface area and volume of similar solids

Solution2 Ratio of matching sides (A to B) = 35 : 14

= 5 : 2∴ Ratio of areas (A to B) = 52 : 22

= 25 : 4

Two similar figures have areas in the ratio 36 : 49. Find the ratio of the lengths of matching sides.

SolutionRatio of areas = m2 : n2 = 36 : 49

∴ Ratio of sides = m : n = = 6 : 7

Two similar figures have sides in the ratio 2 : 3. If the area of the larger figure is 144 cm2, find the area of the smaller figure.

SolutionLet the area of the smaller figure be A cm2.

A : 144 = 22 : 32

=

A = × 144

= 64∴ The area of the smaller figure is 64 cm2.

Example 25

36 : 49

Example 26

A144--------- 4

9---

49---

Working mathematicallyApplying strategies and reasoning: sides, surface areas and volumes of similar solids1 a Calculate the volume of this rectangular prism.

b i If the length is doubled what happens to the volume?

ii If the width is doubled (but length and height stay as shown), what happens to the volume?

iii If the height is doubled (but length and width stay as shown) what happens to the volume?

c Copy and complete:If the length, width and height are all doubled, the volume is increased ______ times.

2 cm

6 cm

8 cm

STAGE5.3



If the ratio of the matching sides is expressed in the form 1 : k (the similarity ratio), then:• matching surface areas are in the ratio 1 : k2

• matching volumes are in the ratio 1 : k3

2 a Explain why these rectangular prisms are similar.b What is the ratio of their matching sides?c What is the ratio of their surface areas?d What is the ratio of their volumes?

3 For the spheres A and B, find:a the ratio of their radiib the ratio of their surface areasc the ratio of their volumes

4 a How is the ratio of surface areas of similar solids related to the ratio of matching sides?b How is the ratio of volumes of similar solids related to the ratio of their matching

sides?c Compare your results with those of other students.

2 cm

1 cm3 cm

2 cm

6 cm

4 cm

9 cm

3 cm

A

B

If the matching sides of two similar figures are in the ratio m : n, then:• the ratio of their surface areas is m2 : n2

• the ratio of their volumes is m3 : n3

Example 27

1 Two similar rectangular prisms have their matching sides in the ratio 3 : 1. Find the ratio of the:a surface areas b volumes

Solution1 a Ratio of sides = 3 : 1 b Ratio of sides = 3 : 1

∴ Ratio of surface areas = 32 : 12 ∴ Ratio of volumes = 33 : 13= 9 : 1 = 27 : 1

2 For these similar triangular prisms, find the ratio of:a the surface areasb the volumes

2.2 cm

3.3 cm

3.6 cm

4.5 cm

2.4 cm3 cm

Y

X



Solution2 Ratio of sides (X to Y) = 3 : 4.5

= 2 : 3a Ratio of surface areas = 22 : 32 b Ratio of volumes = 23 : 33

= 4 : 9 = 8 : 27

Two similar solids have their surface areas in the ratio 25 : 36. If the volume of the smaller solid is 250 cm3, find the volume of the larger solid.

SolutionRatio of surface areas = 25 : 36

∴ Ratio of matching sides = = 5 : 6

∴ Ratio of volumes = 53 : 63= 125 : 216

Let the volume of the larger cube be V cm3.V : 250 = 216 : 125

=

V = × 250

= 432∴ Volume of the larger cube is 432 cm3

Example 28

25 : 36

V250--------- 216

125---------

216125---------

1 Pairs of similar figures have sides in the following ratios. Find the ratio of the areas for each pair.a 1 : 4 b 3 : 4 c 5 : 2 d 2 : 7

2 For each of the following pairs of similar figures, find the ratio of the areas:a b

c d

1 cm3 cm

1.5 m

2.5 m

9 cm 5 cm 4 cm 6 cm


Spreadsheet 2-01

Area of similar shapes

STAGE5.3



3 The ratio of the areas for pairs of similar figures are as follows. Find the ratio of matching sides for each pair:a 9 : 25 b 1 : 100 c 64 : 25 d 16 : 81

4 Find the ratio of the lengths of matching sides in each of the following pairs of similar figures:a b

5 Find the value of x if the two triangles shown are similar.

6 a Two similar circles have radii in the ratio 3 : 5. If the larger area is 150 cm2, find the area of the smaller circle.

b Similar squares have sides in the ratio 7 : 4. If the area of the smaller square is 14.4 cm2, find the area of the larger square.

c Two similar triangles have areas in the ratio 4 : 9. If the length of the base of the smaller triangle is 5 cm, find the length of the base of the larger triangle.

d Two similar rectangles have their areas in the ratio 36 : 121. If the width of the smaller rectangle is 84 cm, find the width of the larger rectangle.

e i If the radius of a circle is doubled, how has its area changed?ii If the area of a square is halved, how have the sides changed?

7 Two similar solids have the lengths of their matching sides in the ratio 1 : 3.4. What is the ratio of:a their surface areas? b their volumes?

8 a Two spheres have their radii in the ratio 2 : 5. Find the ratio of:i their surface areas ii their volumes

b For the similar rectangular prisms shown, find the ratio of:i the surface areas

ii the volumesc Two cubes have sides in the ratio 7 : 4. Find the

ratio of:i their surface areas

ii their volumes

9 a Two similar solids have their volumes in the ratio 27 : 8. Find the ratio of their:i matching sides ii surface areas

b The surface areas of two similar prisms are in the ratio 25 : 9. Find the ratio of their:i matching sides ii volumes

c i If the sides of a cube are doubled, how has the volume changed?ii If the volume of a cube is halved, how have the sides changed?

A = 49 cm2 A = 36 cm2

A = 49 cm2

A = 81 cm2

Area = 12 cm2

7 cmx cm

Area = 3 cm2

5 m

10 m

Example 26

Example 27

Example 25



10 a The shaded areas of these similar triangular prisms are in the ratio 9 : 4. If the volume of the smaller prism is 48 cm3, find the volume of the larger triangular prism.

b Two soup cans are similar. The larger can is 10 cm high and contains 300 mL of soup. If the smaller can is 8 cm high how much soup does it contain (to the nearest millilitre)?

c Two similar pyramids have volumes of 125 cm3 and 343 cm3. What is the ratio of their surface areas?

d Two similar cylindrical containers have capacities of 1.25 L and 900 mL respectively. If the base of the larger container has a radius of 12 cm, find the area of the base of the smaller container. Give your answer to two decimal places.

e A balloon in the shape of a sphere has a radius of 5 cm. By what factor is the surface area increased if the radius increases to 15 cm?

f A spherical balloon has a radius of 8 cm. By what factor is the volume decreased if the radius changes to 6 cm?

11 A rectangular prism has dimensions 3 cm × 4 cm × 6 cm.a What is its volume (V1)?b What is the volume of the new prism (V2) if the original length and

width are doubled? Express your answer in terms of the original volume.

c What is the volume of the new prism (V3) if the original length and width are doubled and the height is halved?

d What is the volume of the new prism (V4) if the original length is doubled, the width is tripled and the height is increased 1.5 times?

e What is the volume of the new prism (V5) if the original length and width are halved and the height is tripled? Express your answer in terms of the original volume.

12 A cylinder of radius 6 cm and height 8 cm has its height doubled and its radius increased 1.5 times. What is the percentage increase in volume?

3

46

Example 28

1 For any square prism and square pyramid with the same surface area and the same base edge, show that the slant height, l, is five halves of the base edge, s; that is, show that l =

2 For any cylinder (for which height equals 2 × radius) and cone with the same surface area and the same base radius as the cylinder, show that the slant height is five times the base radius (l = 5r).

l

ss

ss

s52---s.

r

lh

2r

r

2r

2r

Power plus

STAGE5.3



3 If a cylinder with diameter 2r and height 2r has the same

surface area as a sphere of radius R, show that R =

4 For any sphere and cone with the same volume and radius, show that the height, h, is four times the radius, r (that is, h = 4r).

5 A sphere and a cone fit inside identical cylinders with the same base diameter and height.a Find the ratio of the volume of the cone to the

volume of the sphere to the volume of the cylinder.b Show that the volume of the cone plus the volume of

the sphere equals the volume of the cylinder.

6 A cube is dissected into six identical square pyramids as shown, each with perpendicular height half the length of the base edge.

Show that the volume of each pyramid is one-third the volume of a square prism with the same base edge and perpendicular height.

7 a A sphere has its surface area doubled. By what factor is its radius increased?b The radius of a cylinder of height H is doubled. What is the new height if the volume of the

cylinder does not change?

R

2r

2r

32---r .

r

h

r

2r

2r

2r

2r

2s

2s2s

s

2s

Language of mathsapex base capacity closedcomposite cone cross-section curved surfacecylinder diameter dimension dissectedhemisphere net oblique openperpendicular height prism pyramid Pythagoras’ theorem radius ratio right similar figuressimilar solids similarity ratio slant height spheresurface area volume

1 Which word means a ‘slice’ of a prism?

2 Name three solids that have a curved surface area.

3 What is the formula for the curved surface area of a cone?

4 Explain the difference between volume and capacity?

5 For similar solids what is the relationship between matching sides and volume?

6 Explain the difference between the perpendicular height and the slant height of a pyramid.

Worksheet 2-10

Surface area and volume crossword



Topic overview• Copy and complete the table below.

• Copy and complete the summary of this topic shown below. Ask your teacher to make sure nothing is missing or incorrect.

The best part of this chapter was…

The worst part was…

New work…

I need help with…?

Applications Compositesolids

Rightprisms

Cylinder Cone Sphere Pyramids

NetsUnits

SURFACEAREA

Similar solids• ratio of areas :

Units

VOLUME Similar solids• ratio of volumes :



1 Find the surface area of each of these solids, correct to one decimal place where necessary:

a b c

d e f

2 Calculate the surface area of each of these pyramids:

a b c

3 Find, correct to three significant figures, the surface area of each of these cones. (All measurements are in millimetres.)

a b c

4 Calculate the surface area of each of the following to the nearest m2:

a b c

Chapter 2 Review

Ex 2-01

7 cm

48 cm50 cm

3.6 m

12 m

3 m

8 m

4.8 mCylinder,open atone end

2.7 m

50 cm

50 cm

20 cm5 cm

5 cm

15 cm

30 cm

30 cm30 cm

18 cm 34 cm

25 cm

Ex 2-02

16 cm16 cm

22 cm

54 cm

36 cm

30 cm

14 cm

25 cm

Ex 2-03

8

20

closed

48

40

open

11

60

closed

Ex 2-04

6 m

17 m

25 m

Topic testChapter 2



5 Find, correct to the nearest cm2, the surface area of each solid below. (All measurements are in centimetres.)

a b c

d e f

6 Caclulate the volume of each of these solids. (All lengths are in centimetres. Give your answers to the nearest cm3.)

a b c

7 A greenhouse is constructed with measurements as shown in the diagram.a Find the volume of the greenhouse to the nearest m3.b If the greenhouse costs 0.5c per m3 per hour to heat, how much

is this per day?

8 Calculate the volume of each of these solids, correct to the nearest whole number.

a b c

Ex 2-05

30

16

18

12

25

25

13

1313

9

4848

48

45

20

3024

Ex 2-06

1.6 2.5

5.4

24

42

28

18

25

25 12

12

3 m

4 m 10 m

Ex 2-06

Ex 2-07

11 m

11 m

8 m

15 cm 18 cm

12 cm

8 cm

6 cm6 cm

13 cm



9 a Calculate the volume of each of the following cones. (All units are in metres.) Give your answers correct to one decimal place.

i ii iii

b i Find, correct to one decimal place, the height of a cone with a volume of 115 cm3 and a radius of 3.2 cm.

ii Find, correct to two significant figures, the radius of a cone with a volume of 94.1 m3 and a height of 1.9 m.

10 a Find, correct to three significant figures, the volume and capacity of:i a sphere with radius 35 mm ii a hemisphere with diameter 2.8 m

b Find, correct to the nearest whole number, the radius of a sphere with volume:i 8000 mm3 ii 150 m3

11 Calculate the volume of each of these solids. Give your answers correct to the nearest whole unit.

a b c

d e f

12 a The areas of the bases of similar rectangular prisms are in the ratio of 25 : 64. If the volume of the larger prism is 1024 cm2, find the volume of the smaller prism.

b i The radius of a sphere is tripled. By what factor is the volume increased?ii The radius of a circle is halved. By what factor is the area decreased?

Ex 2-08

8

20

28

50

3.5

Area = 27.2 m2

Ex 2-09

Ex 2-10

80 mm

45 mm

80 mm

45 mm

45 mm

45 mm

6 cm

8 cm

8 cm

8 cm

12 m

12 m

12 m

18 cm

24 cm

12 cm

24 m

44 m

Ex 2-11


Student textImprint pageContentsPrefaceHow to use this bookHow to use the CD-ROMAcknowledgements Syllabus reference gridCh 1 - Working with numberFractions Approximation Accuracy of results when rounding Recurring decimals Expressing fractions in decimal form Expressing decimals in fraction form Solving percentage problems Scientific notation Ratios Scale drawings Rates Topic overview Chapter review

Ch 2 - Surface area and volumeSurface area of prisms and cylindersSurface area of composite solids made of prisms and cylinders Surface area of a pyramid Surface area of a cone Surface area of a sphere Surface area of composite solids Volume of right prisms and cylinders Volume of pyramids, cones and spheres Volume of a sphere Summary of surface area and volume formulas Volumes of composite solidsAreas of similar figures Surface area and volume of similar solidsTopic overview Chapter review

Ch 3 - Surds and indicesRational and irrational numbers Simplifying surds Multiplying and dividing surds Adding and subtracting surds Binomial products and surds Rationalising the denominator Index notation Summary of index laws and propertiesTopic overview Chapter review

Mixed revision 1 Ch 4 - Equations and inequalitiesSolving linear equations Solving word problems Working with formulas Changing the subject of the formula Solving linear inequalities Simultaneous equations Solving problems using simultaneous equationsSolving quadratic equations of the form x2 = cGeneral form of a quadratic equation: ax2 + bx + c = 0Solving quadratic equations by completing the squareSolving quadratic equations by using the quadratic formulaSolving quadratic equations: which method?Solving problems involving quadratic equationsVariable substitutionSimultaneous linear and quadratic equationsTopic overview Chapter review

Ch 5 - Deductive geometryDeduction in geometry Properties of triangles and quadrilaterals Properties of convex polygons Proving congruent triangles Proving properties of triangles and quadrilaterals Tests for quadrilaterals Properties of similar figures Proving geometrical results Pythagoras’ theorem Topic overview Chapter review

Ch 6 - Saving and borrowingInterest Simple interest Compound interest Depreciation Methods of purchasing goods Loan repayments Credit cards Topic overview Chapter review

Mixed revision 2 Ch 7 - Coordinate geometryFormulas for the midpoint, distance and gradient Points lying on a line Parallel and perpendicular lines Graphing linear equations The gradient–intercept form of a linear equation Graphing equations of the form y=mx + bThe general form of a linear equation The point–gradient form of a linear equation Equations of parallel and perpendicular linesCoordinate geometry problems Graphing regions on the number plane 309Graphing non-linear equations Topic overview Chapter review

Ch 8 - TrigonometryRight-angled triangle trigonometry Angles of elevation and depression Bearings Trigonometric relations Trigonometric ratios of obtuse angles The sine rule The cosine rule The area of a triangle Applications of the sine and cosine rules Topic overview Chapter review

Ch 9 - Analysing dataAnalysing data Features of a display of data Comparing the mean, median and modeCumulative frequency tables and graphs Grouped data Measures of spread Standard deviation Comparing sets of data Comparing the range, interquartile range and standard deviation 40Topic overview Chapter review

Mixed revision 3 Ch 10 - ProbabilityProbability Equally likely outcomes Probabilities of simple events Range of probability Complementary events Relative frequency Theoretical probability Independent and dependent events Two-stage events Compound events Topic overview Chapter review

Ch 11 - GraphsDescribing change Distance–time graphs Graphs and other rates of change Interpreting graphs Graphs of equations The axis of symmetry and the vertex of a parabola Summary The cubic curve The hyperbola The exponential curve The circle Identifying graphs Topic overview Chapter review

Mixed revision 4 General revision Ch 12 - Circle geometryThe language of circles Chord properties of circles Angle properties of circles Tangents to a circle Tangent and secant properties of circlesProofs using circle theorems

Ch 13 - Curve sketching and polynomialsThe cubic curve y=a(x -r)(x - s)(x - t)The graph of y = axn The curve y = axn + k The curve y = a(x − r)n The equations of circles Point of intersection of a line with a curve Polynomials Adding, subtracting and multiplying polynomials

Ch 14 - Functions and logarithmsFunctions Inverse functions Logarithms Logarithm laws The graphs of y = ax and y = logax Exponential and logarithmic equations

AnswersIndex

Glossary AB CDE FG HIK L M NO PQ RST UV X Y

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