Chapter 6

Chapter Measurement6What you will learnReview of lengthPythagoras’ theoremAreaSurface area—prisms and cylindersSurface area—pyramids and cones 10A

Volume—prisms and cylindersVolume—pyramids and cones 10A

Spheres 10A

6.16.26.36.46.56.66.76.8

ISBN 978-0-521-17866-2 Photocopying is restricted under law and this material must not be transferred to another party.

© David Greenwood, Sara Woolley, Jenny Goodman, Jennifer Vaughan, GT Installations, Georgia Sotiriou, Voula Sotiriou 2011 Cambridge University Press

363

Monolithic domes

Monolithic domes are round one-piece structures with a smooth spherical-like surface. They offer excellent protection from earthquakes, bushfires and cyclones because of their shape and durability. They are extremely energy efficient because of the minimal surface area for the volume contained within the structure. The first type of monolithic domes used were igloos, which are very strong and provide good insulation in freezing conditions. The minimal surface area of the dome means that there is less surface for heat to be transferred to the outside air. A cube, for example, containing the same volume of air as the dome has about 30% more surface area exposed to the outside air. Volumes and surface areas of spheres and other solids can be calculated using special formulas.

Australian CurriculumM e a s u r e M e n t a n d g e o M e t r y

u s i n g u n i t s o f m e a s u r e m e n t

Solve problems involving surface area and volume for a range of prisms, cylindersand composite solids.

10A Solve problems involving surface area and volume of right pyramids, rightcones, spheres and related composite solids.

P y t h a g o r a s a n d

t r i g o n o M e t r y

Solve right-angled triangle problems.10A Apply Pythagoras’ theorem to solve three-dimensional problems in right-angled triangles.

•



Chapter 6 Measurement364

Pre-

test 1 Evaluate the following.

a 2 × 100 b 5 ÷ 102 c 230 ÷ 102

d 0.043 × 1002 e 62 900 ÷ 1000 f 1.38 × 10002

2 Evaluate the following.

a A = l × w if l = 3 and w = 7

b A = b × h if b = 10 and h = 3

c A = 1

2b × h if b = 2 and h = 3.8

d A = 1

2(a + b)h if a = 2, b = 3 and h = 4

3 Find the perimeter of these shapes.

a

3 cm

b

3 m

1 m c

5 cm

3 cm

4 Use these rules to find the circumference (C) and area (A) of the circles correct to one

decimal place: C = 2pr and A = pr2 where r is the radius.

a

2 m

b

12 cm

c

3.9 km

5 Use Pythagoras’ theorem a2 + b2 = c2 to find the value of the pronumeral in these triangles.

Round to one decimal place where necessary.

a

3

4

c

b

1

2

c

c

13

12

a

6 Find the surface area and volume of these solids. Round to one decimal place

where necessary.

a

2 cm5 cm

4 cm

b

2 m

c

4 cm

6 cm



365 Measurement and Geometry

Review of lengthLength measurements are common in

many areas of mathematics, science and

engineering, and are clearly associated

with the basic measures of perimeter and

circumference, which will be studied here.

Let’s start: The simple sectorThis sector looks simple enough but can you describe

how to find its perimeter? Discuss these points to help.

• Recall the rule for the circumference of a circle.

• What is a definition of perimeter?

• What fraction of a circle is this sector?

• Find the perimeter using both exact and rounded numbers.

6.1

100°

10 cm

■■ Converting between metric units of length

×1000

÷1000

km m

×100

÷100

cm

×10

÷10

mm

■■ Perimeter is the distance around the outside

of a closed shape.

■■ The circumference of a circle is the distance

around the circle.

• C = 2pr = pd where d = 2r

■■ Perimeter of a sector

• P = 2r + θ

360 × 2pr

radi

us(r

)

diameter(d)

Key

idea

s

θ r

Engineers at a refinery checking measurements.




Example 1 Finding the perimeter of polygons

Consider the given two-dimensional shape.

a Find the perimeter of the shape if x = 2.6.

b Find x if the perimeter is 11.9 m.

c Write an expression for x in terms of the perimeter P.

Solution Explanation

a Perimeter = 4.5 + 2.1 + 3.4 + 2.6

= 12.6 m

Simply add the lengths of all four sides

using x = 2.6.

b 11.9 = 4.5 + 2.1 + 3.4 + x

= 10 + x

∴ x = 1.9

Add all four sides and set equal to the perimeter 11.9.

Simplify and solve for x by subtracting 10 from both

sides.

c P = 4.5 + 2.1 + 3.4 + x

= 10 + x

∴ x = P - 10

Use P for the perimeter and add all four sides. Simplify

and rearrange to make x the subject.

3.4 m

2.1 m

4.5 m

x m

Example 2 Using the formula for the circumference of a circle

If a circle has radius r cm, find the following, rounding the answer to two decimal places where

necessary.

a The circumference of a circle if r = 2.5

b A rule for r in terms of the circumference C

c The radius of a circle with a circumference of 10 cm

SolUtion Explanation

a Circumference = 2pr

= 2p(2.5)

= 15.71 cm

Write the rule for circumference and substitute r = 2.5,

then evaluate and round as required.

b C = 2pr

∴ =rC

2π

Write the rule for circumference, then divide both sides

by 2p to make r the subject.

c rC=

=

210

2

π

π = 1.59 cm

Substitute C = 10 into the rule from part b and evaluate.

r cm



Measurement and Geometry 367

Example 3 Finding perimeters of sectors

This sector has a radius of 3 cm.

a Find the sector’s exact perimeter.

b Find the perimeter correct to one decimal place.


a P = 2r + θ

360 × 2pr

= 2 × 3 + 240

360 × 2 × p × 3

= 6 + 2

3 × 2 × p × 3

= 6 + 4p cm

The perimeter of a sector consists of two radii and a

fraction 240

360

2

3=

of the circumference of a circle.

6 + 4p is the exact value.

b 6 + 4p ≈ 18.6 cm Round to the required one decimal place using

a calculator.

3 cm240°

Exercise 6A

Unde

rsta

ndin

g

1 Convert the following length measurements into the units given in brackets.

a 4 cm (mm) b 0.096 m (cm) c 0.001 km (m)

d 300 m (km) e 800 cm (m) f 297 m (km)

g 0.0127 m (cm) h 5102 mm (cm) i 0.0032 km (m)

2 What fraction of a circle (in simplest form) is shown in these sectors?

a b c

d

40°

e

105°

f 30°

3 Find the perimeter of these shapes.

a

185 m

220 m

b 0.5 km

2.6 km

c 12 cm

32 cm34 cm




Flue

ncy

4 Consider the given two-dimensional shape.

a Find the perimeter of the shape if x = 8.

b Find x if the perimeter is 33.7 cm.


5 Consider the given two-dimensional shape.

a Find the perimeter of the shape if x = 5.

b Find x if the perimeter is 20 m.


6 Find the circumference of these circles, correct to two decimal places.

a

7 cm

b

1.2 m

c

28.4 mm

d

1.1 km

7 If a circle has radius r cm, find the following, rounding to two decimal places,

where necessary.

a The circumference of a circle if r = 12

b A rule for r in terms of the circumference C

c The radius of a circle with a circumference of 35 m

8 Find the perimeter of these sectors, by:

i using exact values

ii rounding the answer to one decimal place.

a

4 m

b

2 m

c

1 km60°

d

6 cm60°

e

5 mm

80°

f

3 cm

205°

Example 1

6.2 cm

10.4 cm

12 cm

x cm

8.4 m

x m

Example 2

r m

Example 3




Prob

lem

-sol

ving

9 Find the value of x for these shapes with the given perimeters.

a

5

2

7

Perimeter = 17

x

b

5.3Perimeter = 22.9

x

c

2x

Perimeter = 0.072

d 2.8 2x

Perimeter = 16.2

e 11.61

3.72

7.893x

Perimeter = 46.44

f 1.5x

2xx

Perimeter = 10.8

10 A rectangular rose garden of length 15 m and width 9 m is surrounded by a path of

width 1.2 m. Find the distance around the outside of the path.

11 Find the perimeter of these composite shapes correct to two decimal places.

a

1.5

3.6

b 7.9 c

0.3

d

10

e

2.2

3.6 f

6 12

12 Find the perimeter of these shapes, giving answers as exact values.

a 3 cm

b

5 m

2.5 m

c

1 km




Reas

onin

g

13 A bicycle has wheels with diameter 64 cm.

a Find how far, correct to the nearest cm, the bicycle moves if the wheels turn:

i 1 rotation ii 5 rotations

b How many rotations are required for the bike to travel 10 km? Round to the nearest whole number.

c Find an expression for the number of rotations required to cover 10 km if the wheel has

a diameter of d cm.

14 A square of side length n has the same perimeter as a circle. What is the radius of the circle?

Give an expression in terms of n.

15 A square of side length x just fits inside a circle. Find the exact

circumference of the circle in terms of x.

16 Consider a rectangle with perimeter P, length l and width w.

a Express l in terms of w and P.

b Express l in terms of w if P = 10.

c If P = 10 state the range of all possible values of w.

d If P = 10 state the range of all possible values of l.

x

l

w

Enrichment: Rotating circles

17 When a circle rolls around the outside of another circle it will rotate by a certain angle.

For these problems the fixed circle will have radius r. Given the following conditions, by how

many degrees will the moving circle rotate if it rolls

around the fixed circle once?

a Assume the rotating circle has radius r (shown).

b Assume the rotating circle has radius 1

2r.

c Assume the rotating circle has radius 2r.

d Assume the rotating circle has radius 1

3r. r

rRotatingcircle

Fixedcircle




Pythagoras’ theoremYou will recall that for any right-angled triangle we can link the

length of the three sides using Pythagoras’ theorem. Given two of

the sides, we can work out the length of the remaining side. This has

applications in all sorts of two- and three-dimensional problems.

Let’s start: President Garfield’s proofFive years before he became president of the United States of America

in 1881, James Garfield discovered a proof of Pythagoras’ theorem. It

involves arranging two identical right-angled triangles ( 1 and 2 ) to

form a trapezium as shown.

b

a

b

c

c1

a

3

2

• Use the formula for the area of a trapezium 1

2( )a b h+

to find an expression for the area of

the entire shape.

• Explain why the third triangle 3 is right-angled.

• Find an expression for the sum of the areas of the three triangles.

• Hence prove c2 = a2 + b2.

6.2

US President Garfield discovered a proof of Pythagoras’ theorem.

■■ Pythagoras’ theorem states that:

The sum of the squares of the two shorter sides of a right-angled triangle equals

the square of the hypotenuse.

■■ To write an answer using an exact value, use a square root sign

where possible (e.g. 3).

Key

idea

s

a

b

ca2 + b2 = c2




Example 4 Finding side lengths using Pythagoras’ theorem

Find the length of the unknown side in these right-angled triangles, correct to two decimal places.

a

9 cm

5 cmx cm

b

1.5 m

1.1 m

y m

Solution ExPlanation

a c2 = a2 + b2

∴ x2 = 52 + 92

= 106

∴ x = 106

= 10.30

The length of the unknown side is

10.30 cm.

x cm is the length of the hypotenuse.

Substitute the two shorter sides

a = 5 and b = 9 (or a = 9 and b = 5).

Find the square root of both sides and round as

required.

b a2 + b2 = c2

y2 + 1.12 = 1.52

y2 = 1.52 - 1.12

= 2.25 - 1.21

= 1.04

∴ y = 1 04.

= 1.02

The length of the unknown side is

1.02 m.

Substitute the shorter side b = 1.1 and the hypotenuse

c = 1.5.

Subtract 1.12 from both sides.

Find the square root of both sides and evaluate.

For centuries builders, carpenters and landscapers have used Pythagoras’ theorem to construct right-angles for their foundations and plots. The ancient Egyptians used three stakes joined by a rope in a triangular shape with side lengths of three, four and five units, which form a right angle when the rope is taut.





a

A

E

B

4

7

c2=a2+b2

∴ BE2= 42+72

=65

∴ BE = 65

Drawtheappropriatetriangle.

Substitutea =4andb=7.

SolveforBEexactly.

Leaveintermediateanswersinsurdformtoreducethe

chanceofaccumulatingerrorsinfurthercalculations.

b

E B

H

2

√65

BH 2=HE 2+EB 2

=22+ 652( )

=4+65

=69

∴ BE = 69

≈8.31

Drawtheappropriatetriangle.

SubstituteHE=2andEB= 65.

Note 65 65 65 652( ) × == .

Unde

rsta

ndin

g

Exercise 6B

1 Solve for a in these equations leaving your answer in exact form using a square root sign.

Assume a > 0.

a a2 + 32 = 82 b a2 + 52 = 62 c 22 + a2 = 92

d a2 + a2 = 22 e a2 + a2 = 42 f a2 + a2 = 102

2 Write an equation connecting the pronumerals in these right-angled triangles.

a

zx

y b

ba

c

x

c

Example 5 Using Pythagoras’ theorem in 3D

Consider a rectangular prism ABCDEFGH with the side lengths

AB = 7, AE = 4 and EH = 2. Find:

a BE, leaving your answer in exact form

b BH, correct to two decimal places

10A

D E F

GH

C

A B




Flue

ncy

3 Use Pythagoras’ theorem to find the length of the hypotenuse for these right-angled triangles.

Round your answers to two decimal places where necessary.

a

4 cm

3 cm

b

5 m

10 m c

15 km

7 km

d 0.2 mm

1.8 mm

e 0.21 km0.37 km f

72.1 cm

27.3 cm

4 Find the length of the unknown side in these right-angled triangles, correct to two

decimal places.

a

2 m 5 m

b

9 m

12 m

c 0.3 m

0.7 m

d 0.71 cm

1.32 cm

e

19.3 cm24.2 cm

f

0.11 cm

0.14 cm

5 For each of the cuboids ABCDEFGH, find:

i BE, leaving your answer in exact form

ii BH, correct to two decimal places.

a

D

A

C

G

B

F2

35

H

E

b DH

GC

BF

EA

84

14

Example 4a

Example 4b

Example 5

10A




Flue

ncyc

C

GH

E FD

BA 1

7

3

d

F

B8

2

9

A

D

H

E

G

C

6 Use Pythagoras’ theorem to help decide whether these triangles are right-angled. They may

not be drawn to scale.

a

2

3

4

b

5

12 13

c

611

8

d

1210

6

e

1215

9 f

12

√5

Prob

lem

-sol

ving

7 A 20 cm drinking straw sits diagonally in a glass of radius 3 cm and height

10 cm. What length of straw protrudes from the glass? Round to one decimal

place.

8 Use Pythagoras’ theorem to find the distance between points A and B in these

diagrams, correct to two decimal places.

a

A

B

1.2 m

2.6 m

b

A

B

1.7 cm

3 cm

3.5 cm

1.9 cm

c

BA

26 m

35 m

49 m

10A




Prob

lem

-sol

ving

d

5.3 cm

5.3 cm

A

B

e BA

2.1 km 1.8 km

0.5 km f

AB

17.2 km 19.7 km

14.3 km

9 Find the value of x, correct to two decimal places, in these three-dimensional diagrams.

a x mm

11.4 mm

9.3

mm

b x m

15.5 m

3.7 m

c

5 cm

3 cm

x cm

d

3 m

2 m

1.5 m

x m e

6.2 km

8.2 km

6.7 km

x km f 2.93 cm

4.04 cm5.31 cm

x cm

10 Find the exact distance between these pairs of points

on a number plane.

a (0, 0) and (4, 6)

b (-2, 3) and (2, -1)

c (-5, -3) and (4, 7)

11 a Find the length of the longest rod that will fit inside these objects. Give your answer

correct to one decimal place.

i A cylinder with diameter 10 cm and height 20 cm

ii A rectangular prism with side lengths 10 cm, 20 cm and 10 cm

b Investigate the length of the longest rod that will fit in other solids, such as triangular

prisms, pentagonal prisms, hexagonal prisms and truncated rectangular pyramids. Include

some three-dimensional diagrams.

10A

4

6

2

-4

-6

-2-4-6 -2-8

-8

4 62 80

8

y

x

10A




Reas

onin

g

12 Two joining chords in a semicircle have lengths 1 cm and 2 cm as

shown. Find the exact radius, r cm, of the semicircle. Give reasons.

13 The diagonals of a rectangle are 10 cm long. Find the exact

dimensions of the rectangle if:

a the length is twice the width

b the length is three times the width

c the length is 10 times the width.

14Streamers are used to decorate the interior of a room that is

4.5 m long, 3.5 m wide and 3 m high, as shown.

a Find the length of streamer, correct to two decimal

places, required to connect from:

i A to H ii E to B

iii A to C iv A to G via C

v E to C via D vi E to C directly

b Find the shortest length of streamer required, correct to

two decimal places, to reach from A to G if the streamer

is not allowed to reach across open space.

4.5 m

3 m

E

AB

C

GH

F

D

3.5 m

10A

1 cm

2 cm

r cm

Enrichment: How many proofs?

15 There are hundreds of proofs of Pythagoras’ theorem.

a Research some of these proofs using the internet and pick one you understand clearly.

b Write up the proof giving full reasons.

c Present your proof to a friend or the class. Show all diagrams, algebra and reasons.

Mathematical proofs are vital to the subject and are considered to be amongst its greatest achievements. Most mathematicians consider a good proof to be a thing of beauty.




AreaArea is a measure of surface and is expressed

as a number of square units.

By the inspection of a simple diagram like

the one shown, a rectangle with side lengths

2 m and 3 m has an area of 6 square metres

or 6 m2.

Area = 6 m2

2 m

3 m

For rectangles and other basic shapes, we can use area formulas to help us calculate the number of

square units.

Some common metric units for area include square kilometres (km2), square metres (m2), square

centimetres (cm2) and square millimetres (mm2).

Let’s start: Pegs in holesDiscuss with reasons relating to the area of the shapes which is the better fit:

• a square peg in a round hole

• a round peg in a square hole

6.3

■■ Conversion of units of area

×10002

÷10002

km2 m2

×1002

÷1002

cm2

×102

÷102

mm2Key

idea

s

The appropriate unit for the area of a desert is the square kilometre.




Example 6 Converting between units of area

Convert these areas to the units shown in the brackets.

a 2.5 cm2 (mm2) b 2 000 000 cm2 (km2)


a 2.5 cm2 = 2.5 × 100 mm2

= 250 mm2

10 mm

10 mm

1 cm2

1 cm2 = 10 × 10 mm2 = 100 mm2

b 2 000 000 cm2 = 2 000 000 ÷ 1002 m2

= 200 m2

= 200 ÷ 10002 km2

= 0.0002 km2

km m cm2 2 2

1000÷ 2 ÷1002

10002 = 1 000 000 and 1002 = 10 000

■■ The area of a two-dimensional shape can be defined as the number of square units contained

within its boundaries. Some common area formulas are given.

Square

Area = l2

l

Rectangle

Area = lw

w

l

Triangle

Area = 1

2bh

h

b

Rhombus

Area = 1

2xy

yx

Parallelogram

Area = bh

h

b

Trapezium

Area = 1

2(a + b)h

h

b

a

■■ The rule for the area of a circle is:

Area = pr2, where r is the radius

■■ The rule for the area of a sector is A r= θ360

2πr

θ

Key

idea

s

x

y

Area = 1

2xy

Kite




Example 7 Finding the area of basic shapes

Find the area of these basic shapes, correct to two decimal places where necessary.

a

2 cm

3 cm

5 cm

b

5.8 m

3.3 m

c

1.06 km


a A = 1

2(a + b)h

= 1

2(3 + 5)2

= 8 cm2

The shape is a trapezium, so use this formula.

Substitute a = 3, b = 5 and h = 2.

Simplify and include the correct units.

b A = 1

2bh

= 1

2(5.8)(3.3)

= 9.57 m2

The shape is a triangle.

Substitute b = 5.8 and h = 3.3.

Simplify and include the correct units.

c A = pr2

= p (0.53)2

= 0.88 km2

The shape is a circle.

The radius r is half the diameter.

Round to the required number of decimal places.

Example 8 Using area to find unknown lengths

Find the value of the pronumeral for these basic shapes, rounding to two decimal places where

necessary.

a

Area = 11 mm2

l mm2.3 mm

b

Area = 0.5 m2

1.3 m

0.4 ma m




Example 9 Finding areas of sectors and composite shapes

Find the area of this sector and composite shape. Write your answer as an exact value and as a decimal

correct to two places.

a

80°3 cm

b

5 m


a A r= ×

= × ×

=

≈

θ π

π

π

360280

3603

7

21 99

2

2

2. cm

Write the formula for the area of a sector.

Sector angle = 360° − 80° = 280°.

Simplify to express as an exact value (7π) then round

as required.

b A = × − × ×

= −

≈

2 51

45

5025

4

30 37

2 2

2.

π

π

m

The area consists of two squares minus a quarter circle

with radius 5 m.

50 − 25

4

π is the exact value.

SoLuTion ExPLAnATion

a A = lw

11 = l × 2.3

∴ =.

l11

2 3 = 4.78

Use the rectangle area formula.

Substitute A = 11 and w = 2.3.

Divide both sides by 2.3 to solve for l.

b A = 1

2(a + b)h

0.5 = 1

2(a + 1.3) × 0.4

0.5 = 0.2(a + 1.3)

2.5 = a + 1.3

∴ a = 1.2

Use the trapezium area formula.

Substitute A = 0.5, b = 1.3 and h = 0.4.

Simplify then divide both sides by 0.2 and solve for a.




Exercise 6C

Unde

rsta

ndin

g

1 Write the formula for the area of these shapes.

a circle b sector c square d rectangle

e kite f trapezium g triangle h rhombus

i parallelogram j semicircle k quadrant (quarter circle)

2 Convert the following area measurements into the units given in brackets.

a 3000 mm2 (cm2) b 29 800 cm2 (m2) c 205 000 m2 (km2)

d 0.5 m2 (cm2) e 5 km2 (m2) f 0.0001 km2 (m2)

g 0.023 m2 (cm2) h 537 cm2 (mm2) i 0.0027 km2 (m2)

j 10 m2 (mm2) k 0.00022 km2 (cm2) l 145 000 000 mm2 (km2)

Example 6

Flue

ncy

3 Find the area of these basic shapes, rounding to two decimal places where necessary.

a

5 cm

b 5.2 m

10.5 m

c

2.8 km

1.3 km

d

0.2 mm

0.1 mm

0.3 mm e

7 m

f

15 cm10 cm

g 64 m 23 m h 0.4 mm

0.25 mm

i km23

km14

j

√3 km

k 0.82 m

1.37 m

l

√6 cm√2 cm

√10 cm

Example 7




Flue

ncy4 Find the value of the pronumeral for these basic shapes, rounding to two decimal places where

necessary.

a

Area = 15 cm2

w cm

5.2 cm b

l m

Area = 206 m2

c

Area = 1.3 km2

1.8 km

h km

d

1.3 m

2.8 m

a mArea = 2.5 m2

e 5.34 mm

h mm

Area = 10.21 mm2

f

18 m

x mArea = 80 m2

g r cm

Area = 5 cm2

h d m

Area = 0.21 m2

i d m

Area = 26 km2

5 Find the area of each sector. Write your answer as an exact value and as a decimal rounded to

two places.

a

6 cm

b 10 m c

120°

7 m

d

100°2 m

e

150°

6 km

f

3 mm

325°

Example 8

Example 9a

6 A lawn area is made up of a semicircular region with diameter

6.5 metres and a triangular region of length 8.2 metres, as

shown. Find the total area of lawn, to one decimal place.

Prob

lem

-sol

ving

6.5 m

8.2 m




7 Find the area of these composite shapes. Write your answers as exact values and as decimals

correct to two places.

a

5 cm

b

4 m

7 m

10 m

c

1.7 m1.8 m

1.6 m

d

4.2 mm

e 28 km

18 km

26 km

f 0.25 m 0.3 m

0.7 m

8 An L-shaped concrete slab being prepared for the foundation of a new house is made up of

two rectangles with dimensions 3 m by 2 m and 10 m by 6 m.

a Find the total area of the concrete slab.

b If two bags of cement are required for every 5 m2 of concrete, how many whole bags of

cement will need to be purchased for the job?

Example 9b

Prob

lem

-sol

ving

9 1 hectare (1 ha) is 10 000 m2 and an acre is 1

640 square miles (1 mile ≈1.61 km).

Find how many:

a hectares in 1 km2

b square metres in 20 hectares

c hectares in 1 acre (round to one decimal place)

d acres in 1 hectare (round to one decimal place)

Reas

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10 Consider a trapezium with area A, parallel side lengths a and b and height h.

b

a

h A = 1

2 (a + b)h

a Rearrange the area formula to express a in terms of A, b and h.

b Hence, find the value of a for these given values of A, b and h.

i A = 10, b = 10, h = 1.5 ii A = 0.6, b = 1.3, h = 0.2 iii A = 10, b = 5, h = 4

c Sketch the trapezium with the dimensions found in part b iii above. What shape have

you drawn?

11 Provide a proof of the following area formulas, using only the area formulas for rectangles

and triangles.

a parallelogram b kite c trapezium

Reas

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g

Enrichment: Percentage areas

12 Find correct to one decimal place the percentage areas for these situations.

a The largest square inside a circle. b The largest circle inside a square.

c The largest square inside a right d The largest circle inside a right isosceles

isosceles triangle. triangle.

x

x

x

x




Surface area—prisms and cylinders Knowing how to find the area of simple shapes combined with some knowledge about

three-dimensional objects helps us to find the total surface area of a range of solids.

A cylindrical can, for example, has two circular ends and a curved surface that

could be rolled out to form a rectangle. Finding the sum of the two circles and the

rectangle will give the total surface area of the cylinder.

You will recall the following information about prisms and cylinders.

■■ A prism is a polyhedron with a uniform cross-section and two congruent ends.

• A prism is named by the shape of the cross-section.

• The remaining sides are parallelograms.

■■ A cylinder has a circular cross-section.

• A cylinder is similar to a prism in that it has a

uniform cross-section and two congruent ends.

Let’s start: Drawing netsDrawing or visualising a net can help to find total surface

area of a solid. Try drawing a net for these solids.

Square prismCylinder

6.4

Right triangular

prism

Oblique pentagonal

prism

■■ The total surface area (TSA) of a three-dimensional object can be found by finding the sum

of the areas of each of the shapes that make up the surface of the object.

■■ A net is a two-dimensional illustration of all the surfaces of a solid object.

■■ The net and surface area of a cylinder

h

rDiagram

h

r

r

2πr

NetTSA = 2 circles + 1 rectangle

= 2πr2 + 2πrh

= 2πr (r + h)

Key

idea

s




Example 10 Finding the surface area of prisms and cylinders

Find the total surface area of these objects. Round to two decimal places where necessary.

a 3 cm

5 cm8 cm

b

5.3 m

1.7 m


a TSA = 2 × (8 × 3) + 2 × (5 × 3) + 2 × (8 × 5)

= 158 cm2

Draw the net of the solid.

3 cm

5 cm

8 cm

5 cm

3 cm

b TSA = 2pr2 + 2prh

= 2p(1.7)2 + 2p(1.7) × 5.3

= 74.77 m2

5.3 m2πr

1.7 m

1.7 m

■■ Composite solids are solids made up of two or more basic solids.

• To find a total surface area do not include any common faces.

Common face Key

idea

s




Example 11 Finding the surface area of composite solids

A composite object consists of a square-based prism and a cylinder as shown. Find the total surface

area correct to one decimal place.

8 cm

10 cm

20 cm

6 cm


TSA = 4 × (20 × 10) + 2 × (20 × 20) + 2 × p × 6 × 8

= 1600 + 96p= 1901.6 cm2

The common circular area, which should

not be included, is added back on with the

top of the cylinder. So the total surface area of

the prism is added to the curved area only of

the cylinder.

Unde

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Exercise 6D

1 Draw an example of these solids.

a cylinder b rectangular prism c triangular prism

2 Draw a net for each of these solids.

a

10 cm

4 cm b

2 cm

c

2 cm

1 cm3 cm

Flue

ncy

3 Find the total surface area of these solids. Round your answers to two decimal places where

necessary.

a 3 cm

3 cm6 cm

b

2.7 mm

1.3 mm5.1 mm

c 2.1 cm

4.5 cm7 cm

Example 10




Flue

ncyd 8 m

17 m

15 m

20 m

e 12.8 m

9.2 m

f 5.2 m

6.4 m

4 Find the total surface area of these solids.

a

1.2 cm

b 5 mm

3 mm

4 mm

7 mm

c 25 m

10 m

8.66 m

d

16 mm

9 mm

11.5 mm

12 mm

e 1.2 cm

1.5 cm

0.8 cm0.76 cm

0.5 cm f

2.3 m

4.8 m

2.1 m

1.4 m

1.9 m

5 Find the total surface area in square metres of the outside of an open pipe of radius 85 cm and length

4.5 m, correct to two decimal places.

6 What is the minimum area of paper required to wrap a box with dimensions 25 cm wide, 32 cm long

and 20 cm high?

Prob

lem

-sol

ving

7 The cross-sections of these solids are sectors. Find the total surface area, rounding to

one decimal place.

a

3 cm10 cm

b

2 m

1.6 m

c

6 m

4 m d

20 cm

40 cm

40°




Prob

lem

-sol

ving

8 Use Pythagoras’ theorem to determine any unknown side lengths and find the total surface

area of these solids, correct to one decimal place.

a

7

3

2

b

1.4

1.8 3.2 c

8

7

10

d 8

10

e

33

26

f

1.85.2

9 Find the total surface area of these composite solids. Answer correct to one decimal place.

a

2 cm

6 cm

4 cm

4 cm b

10 m

8 m

c 10 cm

20 cm

d

15 cm

40 m20 m

10 Find the total surface area of this triangular prism correct to one decimal place.

5 m2 m

Example 11




11 Find a formula for the total surface area of these solids using the given pronumerals.

a

x

b

a b

c

c

h

d

d

rh

12 Find the exact total surface area for a cylinder with the given dimensions. Your exact answer will

be in terms of p.

a r = 1 and h = 2 b r = 12

and h = 5

13 If the total surface area of a cylinder is given by the rule Total surface area = 2pr(r + h), find the

height, to two decimal places, of a cylinder that has a radius of 2 m and a total surface area of:

a 35 m2 b 122 m2

14 Can you find the exact radius of the base of a cylinder if its total surface area is 8p cm2 and

its height is 3 cm?

Reas

onin

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Enrichment: inventing formulas for special solids

15 Derive the formulas for the total surface area of the following solids.

a A cylinder with its height b A square-based prism with square side

equal to its radius r length x and height y

r

r

y

x

c A half cylinder with radius r and d A solid with a sector cross-section, radius

height h r, sector angle q and height h

r

h

θh

r




Surface area—pyramids and cones 10A Pyramids and cones are solids for which we can also

calculate the total surface area by finding the sum of the

areas of all the outside surfaces.

The total surface area of a pyramid involves finding the

sum of the areas of the base and its triangular faces. The

rule for the surface area of a cone can be developed after

drawing a net including a circle (base) and sector (curved

surface), as you will see in the following examples.

Let’s start: The cone formulaUse compasses to construct a large sector. Use any sector

angle q you like. Cut out the sector and join the points A and

B to form a cone of radius r.

• Give the rule for the area of the base of the cone.

• Give the rule for the circumference of the base of the cone.

• Give the rule for the circumference of a circle with radius s.

• Use the above to find an expression for the area of the base of the cone as a fraction of the area ps2.

• Hence explain why the rule for the surface area of a cone is given by Total surface area = pr2 + prs.

6.5

s

rθ

sB

A

■■ A cone is a solid with a circular base and a curved surface that reaches from

the base to a point called the apex.

• A right cone has its apex directly above the centre of the base.

• The pronumeral s is used for the slant height and r is the radius of the base.

• Cone total surface area.

r

s rs

2πr

Area (base) = πr 2

and

Area = × =2

2

2π

ππ π

r

ss rs

∴ TSA (cone) = πr2 + πrs = πr(r + s)

■■ A pyramid has a base that is a polygon and remaining sides that are

triangles that meet at the apex.

• A pyramid is named by the shape of its base.

• A right pyramid has its apex directly above the centre of the base.

Apex

Right square-based pyramid

Key

idea

s

s

Apex

Right coner




Example 12 Finding the total surface area of a cone and pyramid

Find the total surface area of these solids, using two decimal places for part a.

4.5 m

2 m

a Cone with radius 2 m and slant height 4.5 m

b Square-based pyramid with square-base length 25 mm and triangular

face height 22 mm


a TSA = pr 2 + prs

= p(2)2 + p(2) × (4.5)

= 40.84 m2

The cone includes the circular base plus the curved

part. Substitute r = 2 and s = 4.5.

b TSA = l2 + 4 × 1

2bh

= 252 + 4 × 1

2 × 25 × 22

= 1725 mm2 25 mm

22 mm

Example 13 Finding the slant height and vertical height of a cone

A cone with radius 3 cm has a curved surface area of 100 cm2.

a Find the slant height of the cone correct to one decimal place.

b Find the height of the cone correct to one decimal place.


a Surface area = prs

100 = p × 3 × s

s = 100

3π = 10.6 cm

Substitute the given information into the rule for the

curved surface area of a cone and solve for s.

b h2 + r2 = s2

h2 + 32 = 100

3

2

π

h2 = 100

3

2

π

− 9

h = 100

39

2

π

−

= 10.2 cm

sh

r

Identify the right-angled triangle within the cone

and use Pythagoras’ theorem to find the height

h. Use the exact value of s from part a to avoid

accumulating errors.




Exercise 6E

Unde

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g

1 Write the rule for the following.

a Area of a triangle

b Surface area of the base of a cone with radius r

c Surface area of the curved part of a cone with slant height s and radius r

2 Find the exact slant height for these cones, using Pythagoras’ theorem. Answer exactly, using

a square root sign.

a 5 cm

2 cm

b

14 m

5 m c

6 cm

10 cm

3 Draw a net for each of these solids.

a

2 cm

b

4 cm2 cm

c

3 cm

4 Find the total surface area of these cones, correct to two decimal places.

a 9 mm

12 mm

b

0.5 m

0.8 m

c 11 km

15 km

5 Find the total surface area of these pyramids.

a

6 m

4 m

b

8 cm

5 cm

c

0.3 m

0.2 m

6 Find the area of the curved surface only of each cone, correct to two decimal places.

a

10 m

2 m

b 1.1 cm

1.5 cm

c

26 mm

48 mm

Example 12a

Example 12b

Flue

ncy




Flue

ncy 7 A cone has height 10 cm and radius 3 cm.

a Use Pythagoras’ theorem to find the slant height of the cone, rounding your answer to

two decimal places.

b Find the total surface area of the cone, correct to one decimal place.

Prob

lem

-sol

ving

8 A cone with radius 5 cm has a curved surface area of 400 cm2.



9 A cone with radius 6.4 cm has a curved surface area of 380 cm2.



10 This right square-based pyramid has base side length 4 m and vertical

height 6 m.

a Find the height of the triangular faces correct to one decimal place.

b Find the total surface area correct to one decimal place.

11 Party hats A and B are in the shape of open cones with no base. Hat A has

radius 7 cm and slant height 25 cm and hat B has radius 9 cm and slant height 22 cm. Which hat has the

greater surface area?

12 Find the total surface area of these composite solids correct to one decimal place as necessary.

a

4 cm

5 cm

2 cm

b 3 cm

3 cm

c 4.33 cm

8 cm

5 cm

4.33 cm

Example 13

6 m

4 m




Prob

lem

-sol

ving

d

10 m

3 m

8 m

e

10 mm

6 mm

9 mm

f

5 m

7 m2 m

Enrichment: Carving pyramids from cones

16 A woodworker uses a rotating lathe to produce a cone with radius

4 cm and height 20 cm. From that cone the woodworker then cut

slices off the sides of the cone to produce a square-based pyramid

of the same height.

a Find the exact slant height of the cone.

b Find the total surface area of the cone correct to two decimal

places.

c Find the exact side length of the base of the square-based pyramid.

d Find the height of the triangular faces of the pyramid correct to three decimal places.

e Find the total surface area of the pyramid correct to two decimal places.

f Express the total surface area of the pyramid as a percentage of the total surface area of the

cone. Give the answer correct to the nearest whole percentage.

20 cm

8 cm

13 Explain why the surface area of a cone with radius r and height h is given by the expression

πr r r h+ +( )2 2 .

14 A cone has a height equal to its radius (h = r). Show that its total surface area is given by the

expression pr2(1 + 2).

15 There is enough information in this diagram to find the total surface

area but the side length of the base and the height of the triangular

faces are not given. Find the total surface area correct to one decimal

place.

Reas

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10 cm

10 cm




Volume—prisms and cylindersVolume is the amount of space contained within the outside surfaces of a three-dimensional object and is

measured in cubic units.

The common groups of objects considered in this section are the prism and the cylinder.

Let’s start: Right and oblique prismsRecall that to find the volume of a right prism, you would first find the area of the base and multiply by the

height. Here is a right rectangular prism and two oblique rectangular prisms with the same side lengths.

10

63

10

63

10

63

• What is the volume of the right rectangular prism?

• Do you think the volume of the two oblique prisms would be calculated in the same way as for the

right rectangular prism and using exactly the same lengths?

• Would the volume of the oblique prisms be equal to or less than that of the rectangular prism?

• Discuss what extra information is required to find the volume of the oblique prisms.

• How does finding the volume of oblique prisms (instead of a right prism) compare with finding the

area of a parallelogram (instead of a rectangle)?

6.6




Example 14 Finding the volume of right prisms and cylinders

Find the volume of these solids, rounding to two decimal places for part b.

a

4 m

3 m2 m

b

6 cm

2 cm


a V = lwh

= 2 × 3 × 4

= 24 m3

Write the volume formula for a rectangular prism.

Substitute l = 2, w = 3 and h = 4.

b V = pr2h

= p(2)2 × 6

= 75.40 cm3

The prism is a cylinder with base area pr2.

Substitute r = 2 and h = 6.

Evaluate and round.

■■ Metric units for volume include cubic kilometres

(km3), cubic metres (m3), cubic centimetres (cm3)

and cubic millimetres (mm3).

■■ Units for capacity include megalitres (ML),

kilolitres (kL), litres (L) and millilitres (mL).

• 1 cm3 = 1 mL

■■ For right and oblique prisms and cylinders the volume is given by:

V = Ah where

• A is the area of the base

• h is the perpendicular height.

w

h

l

h

xx r

h

Right rectangular prism Oblique square prism Right cylinder

V = Ah V = Ah V = Ah

= lwh = x2h = pr2h

×10003

÷10003

km3 m3

×1003

÷1003

cm3

×103

÷103

mm3

×1000

÷1000

ML kL

×1000

÷1000

L

×1000

÷1000

mL

Key

idea

s




Example 15 Finding the volume of an oblique prism

Find the volume of this oblique prism.


V = 1

2bh × 8

= 1

2 × 5 × 3 × 8

= 60 cm3

The base is a triangle so multiply the area of the base

triangle by the perpendicular height.

3 cm 8 cm

5 cm

Example 16 Finding the volume of a composite solid

Find the volume of this composite solid correct to one

decimal place.


Radius of cylinder = 6

2 = 3 cm

V = lwh + pr2h

= 6 × 6 × 2 + p × 32 × 2

= 72 + 18p = 128.5 cm3

First find the radius length, which is half the side length

of the square base.

Add the volume of the square-based prism and the

volume of the cylinder.

6 cm2 cm

2 cm

Exercise 6F

Unde

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g

1 Find the volume of these solids with the given base areas.

a

A = 8 cm2

10 cm

b 2 m

A = 16 m2

c

12 mm

A = 9 mm2




Flue

ncy

3 Find the volume of each rectangular prism.

a

2 cm

4 cm5 cm

b

30 m10 m

35 m

c

7 mm

3.5 mm10.6 mm

4 Find the volume of each cylinder, correct to two decimal places.

a

5 m

10 m

b

1.5 cm

2 cm c

12 m

14 m

5 Find the volume of these oblique solids. Round to one decimal place for part b.

a

3 cm

2 cm

b

20 m

5 m c 1.2 mm

2.4 mm

1.7 mm

6 Find the volume of these solids, rounding your answers to three decimal places where necessary.

a

2 km10 km

3 km

b

3.5 cm

8 cm

7 cm

c

7 m3 m

3 m

2 m

Example 14a

Example 14b

Example 15

2 Convert these volume measurements into the units given in brackets.

a 2 cm3 (mm3) b 0.2 m3 (cm3) c 0.015 km3 (m3)

d 5700 mm3 (cm3) e 28 300 000 m3 (km3) f 762 000 cm3 (m3)

g 0.13 m3 (cm3) h 0.000 001 km3 (m3) i 2.094 cm3 (mm3)

j 2.7 L (mL) k 342 kL (ML) l 35 L (kL)

m 5.72 ML (kL) n 74 250 mL (L) o 18.44 kL (L)

Unde

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ndin

g




Flue

ncyd

5 cm2 2 cm

e

0.12 m

0.38 m

f 21.2 cm

18.3 cm

24.5 cm

g 0.5 m

0.8 m

h 12 cm

12 cm

i

2 m

3 m

2 m

6 m

Prob

lem

-sol

ving7 How many containers holding 1000 cm3 (1 L) of water will you need to fill 1 m3?

8 How many litres of water are required to fill a rectangular fish tank that is 1.2 m long, 80 cm wide

and 50 cm high?

9 Find the volume of these composite objects, rounding to two decimal places where necessary.

a

8 cm

2 cm

5 cm

b

5 m

5 m

c

2.3 m

1.6 m

1.1 m 2.6 m

d

5 mm

8 mm11 mm

e

9 cm

3 cm

4 cm

5 cm

f 4 cm

6 cm

10 Find the exact volume of a cube if its surface area is:

a 54 cm2 b 18 m2

Example 16




11 Use the rule V = pr2h to find the height of a cylinder, to one decimal place, with radius 6 cm and

volume 62 cm3.

12 A fellow student says the volume of this prism is given by V = 4 × 3 × 5.

Explain his error.

13 Find a rule for the volume of a cylindrical portion with angle

q, radius r and height h as shown.

14 Decide whether there is enough information in this diagram

of a triangular prism to find its volume. If so, find the volume


5

34

h

r θ

10 m4 m

Reas

onin

g

Enrichment: Concrete poles

15 A concrete support structure for a building is made up of a cylindrical base and a square-based

prism as the main column. The cylindrical base is 1 m in diameter and 1 m high and the square

prism is 10 m long and sits on the cylindrical base as shown.

10 m

1 m

1 m

a Find the exact side length of the square base of the prism.

b Find the volume of the entire support structure correct to one decimal place.




Volume—pyramids and cones 10A

The volume of a cone or pyramid is a certain fraction of the volume of a prism with the same base area.

This particular fraction is the same for both cones and pyramids and will be explored in this section.

The volume of sand needed to construct these garden features can be calculated from the formula for the volume of a cone.

Let’s start: is a pyramid half the volume of a prism?Here is a cube and a square pyramid with equal base side lengths and equal heights.

• Discuss whether or not you think the pyramid is half the volume of the cube?

• Now consider this diagram of the cube with the pyramid inside.

The cube is black.

The pyramid is green.

The triangular prism is blue.

• Compared to the cube, what is the volume of the triangular prism (blue)? Give reasons?

• Is the volume of the pyramid (green) more or less than the volume of the triangular prism (blue)?

• Do you know what the volume of the pyramid is as a fraction of the volume of the cube?

6.7




Example 17 Finding the volume of pyramids and cones

Find the volume of this pyramid and cone. Give the answer for part b correct to two decimal places.

a

1.4 m

1.2 m

1.3 m

b

29 mm

23 mm


a V Ah

l w h

=

= × ×

= × ×

=

1

31

31

31 4 1 2 1 3

0 728 3

( )

( . . ) .

. m

The pyramid has a rectangular base with area l × w.

Substitute l = 1.4, w = 1.2 and h = 1.3.

Evaluate the answer.

b V Ah

r h

=

=

= ×

=

1

31

31

311 5 29

4016 26

2

2

3

π

π( . )

. mm

The cone has a circular base of area pr2.

Substitute r = 23

2 = 11.5 and h = 29.

Evaluate and round the answer.

■■ For pyramids and cones the volume is given by: V = 1

3Ah

where A is the area of the base and h is the perpendicular height.

Right square pyramid Right cone Oblique cone

xx

h

h

r

h

r

V = 13

Ah V = 1

3Ah V =

1

3Ah

= 1

3x2h =

1

3pr2h =

1

3pr2h

Key

idea

s




Exercise 6G

Unde

rsta

ndin

g

1 A cylinder has volume 12 cm3. What will be the volume of a cone with the same base area and

perpendicular height?

2 A pyramid has volume 5 m3. What will be the volume of a prism with the same base area and

perpendicular height?

3 Find the volume of these solids with the given base areas.

a

A = 5 m2

6 m

b

A = 2 cm2

4 cm

c

7 mm

A = 25 mm2

Flue

ncy

4 Find the volume of the following pyramids. For the oblique pyramids (parts d, e, f) use the given

perpendicular height.

a

2 cm

3 cm b

18 m

15 m

13 m

c

12 km

5 km

5 km

d 2 cm

e 4 cm

3 cm

f 1.4 mm

0.6 mm

Example 17a

Perpendicular

height = 2 cm

Perpendicular

height = 4 cmPerpendicular

height = 1.2 mm




Prob

lem

-sol

ving

5 Find the volume of the following cones, correct to two decimal places.

a

2.6 m

1.1 m

b

3.5 mm

1.6 mm c

60 m

20 m

d

6 m

2 m

e

1.3 m

0.6 m f

5 cm

20 cm

6 A medicine cup is in the shape of a cone with base radius 3 cm and height 5 cm. Find its capacity in

mL correct to the nearest mL.

7 Find the volume of these composite objects, rounding to two decimal places where necessary.

a

8 m

6 m

3 m

b

5 cm

6 cm

c

4 m

3 m6 m

2 m

d

4.8 mm

1.5 mm

5.2 mm

e

3.96 m

2.8 m

f

7.1 m

10 m

2.6 m

Example 17b




9 A wooden cylinder is carved to form a cone that has the same base area and same height as the

original cylinder. What fraction of the wooden cylinder is wasted? Give a reason.

10 A square-based pyramid and a cone are such that the diameter of the cone is equal to the length

of the side of the square base of the pyramid. They also have the same height.

a Using x as the side length of the pyramid and h as its height, write a rule for:

i the volume of the pyramid in terms of x and h

ii the volume of the cone in terms of x and h

b Express the volume of the cone as a fraction of the volume of the pyramid. Give an exact

answer.

11 a Use the rule V = 1

3pr2h to find the base radius of a cone, to one decimal place, with height

23 cm and volume 336 cm3.

b Rearrange the rule V = 1

3pr2h to write:

i h in terms of V and r ii r in terms of V and h

Reas

onin

gPr

oble

m-s

olvi

ng

8 The volume of ice-cream in the cone is half the volume of

the cone. The cone has a 3 cm radius and 14 cm height. What

is the depth of the ice-cream correct to two decimal places?

14 cmIce-cream

depth ofice-cream

3 cm

Enrichment: Truncated cones

12 A truncated cone is a cone that has its apex cut off by an intersecting plane.

In this example the top has radius r2, the base has radius r1 and the two

circular ends are parallel.

a Give reasons why rr

h

h1

2

1

2

= .

b Find a rule for the volume of a truncated cone.

c Find the volume to one decimal place of a truncated cone if r1 = 2 cm, h1 = 5 cm and

h2 equals:

i 1

2 1h ii 2

3 1h

r1

h1

r2

h2




Spheres 10A

Planets are spherical in shape due to the effects of

gravity. This means that we can describe a planet’s

size using only one measurement: its diameter or

radius. Mars, for example, has a diameter of about

half that of the Earth, which is about 12 756 km. The

Earth’s volume is about 9 times that of Mars and this

is because the volume of a sphere varies with the

cube of the radius. The surface area of the Earth is

about 3.5 times that of Mars because the surface of a

sphere varies with the square of the radius.

Let’s start: What percentage of a cube is a sphere?A sphere of radius 1 unit just fits inside a cube.

• First guess the percentage of space occupied by the sphere.

• Draw a diagram showing the sphere inside the cube.

• Calculate the volume of the cube and the sphere. For the sphere use V = 4

3pr3.

• Now calculate the percentage of space occupied by the sphere. How close was your guess?

6.8

Mars is approximately spherical in shape.

Example 18 Finding the surface area and volume of a sphere

Find the surface area and volume of a sphere of radius 7 cm, correct to two decimal places.


TSA = 4pr 2

= 4p(7)2

= 615.75 cm2

V = 4

3pr3

= 4

3p(7)3

= 1436.76 cm3

Write the rule for the surface area of a sphere and

substitute r = 7.


Write the rule for the volume of a sphere and

substitute r = 7.


■■ The surface area of a sphere depends on its radius r and is given by:

Surface area = 4pr 2

■■ The volume of a sphere depends on its radius r and is given by:

Volume = 4

3pr 3

r

Key

idea

s




Example 19 Finding the radius of a sphere

Find the radius of a sphere with volume 10 m3 correct to two decimal places.

V = 10 m3


V = 4

3pr3

10 = 4

3pr3

30 = 4pr3

15

2π = r3

∴ r=15

23

π

= 1.34

∴ the radius is 1.34 m.

Substitute V = 10 into the formula for the volume of a

sphere.

Solve for r3 by multiplying both sides by 3 and then

dividing both sides by 4p. Simplify 30

4

15

2π π= .

Take the cube root of both sides to make r the subject

and evaluate.

r m

Example 20 Finding the surface area and volume of composite solids with sphere portions

This composite object includes a hemisphere and cone as shown.

a Find the surface area, rounding to two decimal places.

b Find the volume, rounding to two decimal places.


a Radius r = 7 - 5 = 2

s2 = 52 + 22

s2 = 29

s = 29

TSA = 1

2 × 4pr 2 + prs

= 1

2 × 4p(2)2 + p(2) 29( )

= 8p + 2 29p = 58.97 cm2

First find the radius, r cm, of the

hemisphere.

Calculate the slant height, s, of the

cone using Pythagoras’ theorem.

Write the rules for the surface area of

each component and note that the top

shape is a hemisphere (half sphere). Only the curved

surface of the cone is required.


Simplify and then evaluate, rounding as required.

5 cm 7 cm

5 cm

2 cm

√29




b V r r h= × +

= × +

=

1

2

4

3

1

31

2

4

32

1

32 5

16

3 2

3 2

π π

π π

π

( ) ( ) ( )

33

20

336

312

37 70 3

+

=

=

=

π

π

π

. cm

Volume (object) = 1

2Volume(sphere) + Volume(cone)


Simplify and then evaluate, rounding as required.

Exercise 6H

Unde

rsta

ndin

g

1 Evaluate and round to two decimal places.

a 4 × p × 52 b 4 × p × 2.22 c 4 × p × 1

2

2

d 4

3 × p × 23 e

4

3 × p × 2.83 f

4 7

3

3π( )

2 Rearrange S = 4pr 2 to write r in terms of S.

3 Rearrange V = 4

3pr 3 to write r in terms of V.

4 What fraction of a sphere is shown in these diagrams?

a b c

5 Find the surface area and volume of the following spheres, correct to two decimal places.

a

2 cm

b 0.5 m

c

38 mm

d

18 cm

e 0.92 km

f 1.36 m

Example 18

Flue

ncy




Flue

ncy

6 Find the total surface area and volume of a sphere with the given dimensions. Give the answer

correct to two decimal places.

a radius 3 cm b radius 4 m c radius 7.4 m

d diameter 5 mm e diameter 7 m f diameter 2.2 km

7 a Find the radius of these spheres with the given volumes, correct to two decimal places.

i

V = 15 cm3

r cm

ii

V = 180 cm3

rcm

iii

V = 0.52 km3

rkm

b Find the radius of these spheres with the given surface area, correct to two decimal places.

i

S = 10 m2

rm

ii

S = 120 cm2

rcm

iii

S = 0.43 mm2

rmm

Example 19

Prob

lem

-sol

ving

8 A box with dimensions 30 cm long, 30 cm wide and 30 cm high holds 50 tennis balls of radius

3 cm. Find:

a the volume of 1 tennis ball, correct to two decimal places

b the volume of 50 tennis balls, correct to one decimal place

c the volume of the box not taken up by the tennis balls, correct to one decimal place.

9 An expanding spherical storage bag has 800 cm3 of water pumped into it. Find the diameter of the

bag correct to one decimal place, after all the water has been pumped in.

10 A sphere just fits inside a cube. What is the surface area of the sphere as a percentage of the

surface area of the cube? Round to the nearest whole percentage.

11 Find the volume of these portions of a sphere correct to two decimal places.

a 5 cm b 4 cm c 1 m

12 Two sports balls have radii 10 cm and 15 cm. Find the difference in their total surface areas,





Prob

lem

-sol

ving

13 A monolithic structure has a cylindrical base of radius 4 m

and height 2 m and a hemispherical top.

a What is the radius of the hemispherical top?

b Find the total volume of the entire monolithic structure


14 Find the total surface area for these solids, correct to two

decimal places.

a 5 m b

1.3 cm

0.7 cm c

4 mm

8 mm

d

1 m

5 m3 m

e

2 m

f 3 cm

2 cm

√2 cm

15Find the volume of the following composite objects correct to two decimal places.

a

3 m

1 m

b

13 m

19 m

c

2 cm

d 28 cm

20 cm

e

2 m

2 m

2 m

f

1 cm

Hollow

2 m4 m

Example 20a

Example 20b




Prob

lem

-sol

ving

16 A spherical party balloon is blown up to help decorate a room.

a Find the volume of air, correct to two decimal places, needed for the balloon to be:

i 10 cm wide ii 20 cm wide iii 30 cm wide

b If the balloon pops when the volume of air reaches 120 000 cm3, find the diameter of the

balloon at that point, correct to one decimal place.

17 A hemisphere sits on a cone and two height measurements are

given as shown. Find:

a the radius of the hemisphere

b the exact slant height of the cone in surd form

c the total surface area of the solid, correct to one decimal place.

15 cm10 cm

18 a Find a rule for the radius of a sphere with surface area S.

b Find a rule for the radius of a sphere with volume V.

19 A ball’s radius is doubled.

a By how much does its surface area change? b By how much does its volume change?

20 Show that the volume of a sphere is given by V = 1

6 pd3 where d is the diameter.

21 A cylinder and a sphere have the same radius r and volume V. Find a rule for the height of the

cylinder in terms of r.

Reas

onin

g

Enrichment: Comparing surface areas

22 Imagine a cube and a sphere that have the

same volume.

a If the sphere has volume 1 unit3, find:

i the exact radius of the sphere

ii the exact surface area of the sphere

iii the value of x (the side length of the cube)

iv the surface area of the cube

v the surface area of the sphere as a percentage of the surface area of the cube correct to one

decimal place.

b Now take the radius of the sphere to be r units. Write:

i the rule for the surface area of the sphere

ii the rule for x in terms of r, given the volumes are equal

iii the surface area of the cube in terms of r.

c Now write the surface area of the sphere as a fraction of the surface area of the cube using your

results from part b.

Simplify to show that the result is π6

3 .

d Compare your answers from part a v and part c.

xx

x

r




Similar objects

Enlarging a rectangular prism

The larger rectangular prism shown here is an enlargement of the smaller prism by a factor of 1.5.

4

26

6

3 9

a Write down the ratios of the side lengths.

b Find the surface areas of the prisms and write these as a ratio in simplest form.

c Find the volumes of the prisms and write these as a ratio in simplest form.

d What do you notice about the ratios for length, area and volume? Explain.

e Can you write all three ratios using indices with the same base?

Using similarity

From the previous section, we note that if two similar objects have a length ratio a : b the following

apply.

Length ratio = a : b Scale factor = b

a

Area ratio = a2 : b2 Scale factor = b

a

2

2

Volume ratio = a3 : b3 Scale factor = b

a

3

3

a A sphere of radius 1 cm is enlarged by a length scale factor of 3.

i Find the ratio of the surface areas of the two spheres.

ii Find the ratio of the volumes of the two spheres.

iii Use your ratio to find the surface area of the larger sphere.

iv Use your ratio to find the volume of the larger sphere.

b This rectangular prism is enlarged by a factor of 1

2, which means that the prism is reduced in size.

10 m

2 m12 m

i Find the ratio large : small of the surface areas of the prisms.

ii Find the ratio large : small of their volumes.

iii Use your ratios to find the surface area of the smaller prism.

iv Use your ratios to find the volume of the smaller prism.

inve

stig

atio

n




DensityDensity is defined as the mass or weight of a substance per cubic unit of volume.

Densitymass

volume= or Mass = density × volume

Finding mass

Find the total mass of these objects with the given densities, correct to one decimal place where

necessary.

a

1 m2 m

3 m

Density = 50 kg per m3

b

Density = 100 kg per m3

2 m

1 m c

Density = 0.05 kg per m3

10 cm

5 cm

6 cm

Finding density

a Find the density of a compound with the given mass and volume measurements, rounding to two

decimal places where necessary.

i Mass 30 kg, volume 0.4 m3 ii Mass 10 g, volume 2 cm3

iii Mass 550 kg, volume 1.8 m3

b The density of a solid depends somewhat on how its molecules are packed together. Molecules

represented as spheres are tightly packed if they are arranged in a triangular form. The following

relates to this packing arrangement.

1 cmA

C

i Find the length AC for three circles, each of radius 1 cm, as shown. Use exact values.

ii Find the total height of four spheres, each of radius 1 cm, if they are packed to form a

triangular-based pyramid. Use exact values.

First note that AB = 2BC for an equilateral triangle (shown). Pythagoras’

theorem can be used to prove this, but this is trickier.

C

A

B




ConcentrationConcentration is associated with the purity of dissolved substances and will be considered here

using percentages.

Concentration (%)volume of substance

total v=

oolume× 100

1

Finding concentration

Find the concentration of acid as a percentage, if 10 cm3 of pure acid is mixed into the given containers

that are half full (half the volume) of water. Give your answers correct to two decimal places.

a

10 cm

5 cm

b

2 cm

12 cm

c 6 cm

8 cm

Finding volumes

a Find the volume of pure acid required to give the following concentrations and total volumes.

i Concentration 25%, total volume 100 cm3

ii Concentration 10%, total volume 90 m3

iii Concentration 32%, total volume 1.8 L

b A farmer mixes 5 litres of a chemical herbicide into a three-quarter full tank of water. The tank is

cylindrical with diameter 1 metre and height 1.5 metres. Find:

i the number of litres of water in the tank, correct to three decimal places

ii the concentration of the herbicide after it has been added to the water, correct to two decimal

places

iii the possible diameter and height measurements of a tank that would make the concentration

of herbicide 1%.




1 A cube has a surface area that has the same value as its volume. What is the side length of this

cube?

2 The wheels of a truck travelling at 60 km/h make 4 revolutions per second. What is the diameter

of each wheel in metres to one decimal place?

3 A sphere fits exactly inside a cylinder, and just touches the top, bottom and curved surface. Show

that the surface area of the sphere equals the curved surface area of the cylinder.

4 A sphere and cone with the same radius r have the same volume. Find the height of the cone in

terms of r.

5 Four of the same circular coins of radius r are placed such that they are just

touching as shown. What is the area of the shaded region enclosed by the

coins in terms of r?

6 Find the exact ratio of the equator to the distance around the Earth at latitude 45° north. (Assume

that the Earth is a perfect sphere.)

45°

7 A ball just fits inside a cylinder. What percentage of the volume of the cylinder is taken up by the

ball? Round to the nearest whole percentage.

Chal

leng

es




Chap

ter s

umm

ary

Measurement

Pythagoras’ theorem

For prisms and pyramids draw thenet and add the areas of all the faces.

e.g.

Surface areaVolume

r

r

h

rh

h2πr

s

r

TSA = 4 × triangles + square base

TSA = 2πrh + 2πr2

curved base surface ends

TSA = 4πr2

TSA = πrs + πr2

curved + base

Cylinder

Cone

h

r

Cone:

Sphere:

Sphere

For composite solids considerwhich surfaces are exposed.

×10003

÷10003

km3 m3

×1003

÷1003

cm3

×103

÷103

mm3

×1000

÷1000

megalitres(ML)

kilolitres(kL)

millilitres(mL)

Litres(L)

×1000

÷1000

×1000

÷1000

h

wl

V = πr2hV = lwh

V = Ah13

V = πr2h13

V = πr343

LengthUnits of length

Perimeter is the distance aroundthe outside of a closed shape.Circumference of a circle,C = 2πr = πdPerimeter of a sector

×1000

÷1000

km m

×100

÷100

cm

×10

÷10

mm

P = 2r + × 2πrθ360

θ

r

cb

a

c2 = a2 + b2

Can occur in 3D shapes

AreaUnits of area

Formulas

×10002

÷10002

km2 m2

×1002

÷1002

cm2

×102

÷102

mm2

r

Circle

Sector Rhombus Parallelogram Trapezium Kite

A = πr2

l

Square

A = l 2

w

l

Rectangle

A = lw

h

b

Triangle

A = bh12

A = xy12A = xy1

2A = × πr2θ360

A = h (a + b)A = bh 12

θ h

b

h

b

ax

yxy

+

where A is the area of the base

Units of volume

Rectangular prism

For pyramids and cones:

Cylinder

Units of capacity

1 cm3 = 1 mL

For right and oblique prisms and cylindersV = Ah, A is the area of the base h is the perpendicular height




Multiple-choice questions 1 If the perimeter of this shape is 30.3 cm the value of x is:

A 12.8

B 6.5

C 5.7

D 6.4

E 3.6

2 The perimeter of the sector shown, rounded to one decimal place, is:

A 8.7 m

B 22.7 m

C 18.7 m

D 56.7 m

E 32.7 m

3 The value of x in this triangle is closest to:

A 4.9 B 3.5 C 5.0

D 4.2 E 3.9

4 The exact shaded area in square units is:

A 32 − 72p B 120 − 36p C 32 + 6pD 120 − 18p E 48 + 18p

5 0.128 m2 is equivalent to:

A 12.8 cm2 B 128 mm2 C 1280 cm2 D 0.00128 cm2 E 1280 mm2

6 A cube has a total surface area of 1350 cm2. The side length of the cube is:

A 15 cm B 11 cm C 18 cm D 12 cm E 21 cm

7 A cylindrical tin of canned food has a paper label glued around its curved surface. If the

can is 14 cm high and has a radius of 4 cm, the area of the label is closest to:

A 452 cm2 B 352 cm2 C 126 cm2 D 704 cm2 E 235 cm2

8 The exact surface area of a cone of diameter 24 cm and slant height 16 cm is:

A 216p cm2 B 960p cm2 C 528p cm2 D 336p cm2 E 384p cm2

9 A cone has a radius of 7 m and a slant height of 12 m. The exact height

in metres of the cone is:

A 52 B 193 C 85

D 137 E 95

10 The volume of the hemisphere shown correct to the nearest cubic

centimetre is:

A 1257 cm3 B 628 cm3 C 2094 cm3

D 1571 cm3 E 4189 cm3

10A

10A

10A

x cm

7.1 cm10.4 cm

100°5 m

x km7 km

12

10

7 m

12 m

10 cm




11 The volume of this oblique square-based prism is:

A 72 cm3 B 48 cm3 C 176 cm3

D 144 cm3 E 120 cm3

12 The volume of air in a sphere is 100 cm3. The radius of the sphere

correct to two decimal places is:

A 1.67 cm B 10.00 cm C 2.82 cm D 23.87 cm E 2.88 cm

Short-answer questions1 Convert the given measurements to the units in the brackets.

a 0.23 m (cm) b 270 mm2 (cm2) c 2.6 m3 (cm3)

d 8.372 litres (mL) e 638 250 mm2 (m2) f 0.0003 km2 (cm2)

2 Find the perimeter of these shapes correct to one decimal place where necessary. Note

Pythagoras’ theorem may be required in part b.

a

8 m

8 m

b

9 m

7 m

4 m

c

80°6 cm

3 A floral clock at the Botanic Gardens is in the shape of a circle and has a circumference of 14 m.

a Find the radius of the clock in exact form.

b Hence, find the area occupied by the clock. Answer to two decimal places.

4 For the rectangular prism with dimensions as shown, use

Pythagoras’ theorem to find:

a AF, leaving your answer in exact form

b AG to two decimal places.

9 cm

4 cm

10A

D

H G

C

B

FE

A

2

47

10A




5 Find the area of these shapes. Round to two decimal places where necessary.

a

4.2 m

6.2 m

b

12 m

18 m

13 m

c

7 m11 m

d

10 cm

e

6.4 m

9.8 m

f 13 m

7 m4.2 m

8.1 m

6 A backyard deck as shown has an area of 34.8 m2.

w m

9.2 m

5.3 m

deck

a Find the width w metres of the deck.

b Calculate the perimeter of the deck correct to two decimal places. Pythagoras’ theorem will

be required to calculate the missing length.

7 For each of these solids find, correct to two decimal places where necessary:

i the total surface area ii the volume

a

6 m

5 m8 m

b

6 cm10 cm

5 cm4 cm

c 8 cm

20 cm

d 13 cm

10 cm

10 cm

12 cm

e

4 m

f

12 cm

10 cm

13 cm

8 A cone has a radius of 6 cm and a curved surface area of 350 cm2.

a Find the slant height of the cone in exact form.


10A

10A




9 A papier mache model of a square-based pyramid with base length 30 cm has volume 5400 cm3.

a What is the height of the pyramid?

b Use Pythagoras’ theorem to find the exact height of the triangular faces.

c Hence, find the total surface area of the model correct to one decimal place.

10 A cylinder and a cone each have a base radius of 1 m. The cylinder has a height of 4 m.

Determine the height of the cone if the cone and cylinder have the same volume.

11 For each of the following composite solids find, correct to two decimal places where necessary:


a

10 cm

3 cm

3 cm

b

7 m

4 m

5 m c

2.5 mm 1.6 mm

0.9 mm

12 For each of the following composite solids find, correct to two decimal places where necessary:


a

4 cm

5 cm

b 3.9 cm

5 cm

3 cm

2 cm

c

3.2 cm

1.8 cm

13 A water buoy is in the shape shown.

a Find the volume of air inside the buoy in exact form.

b Find the surface area of the buoy in exact form.

10A

10A

10A

10A

39 cm

51 cm

36 cm




Extended-response questions1 A water ski ramp consists of a rectangular floatation container and a triangular angled section as

shown.

6 m

2 m 8 m

1 m

c m

a What volume of air is contained within the entire ramp structure?

b Find the length of the angled ramp (c metres) in exact surd form.

The entire structure is to be painted with a waterproof paint costing $20 per litre. One litre of

paint covers 25 square metres.

c Find the total surface area of the ramp, correct to one decimal place.

d Find the number of litres and the cost of paint required for the job. Assume you can only

purchase 1 litre tins of paint.

2 A circular school oval of radius 50 metres is marked with spray paint to

form a square pitch as shown.

a State the diagonal length of the square.

b Use Pythagoras’ theorem to find the side length of the square in exact

surd form.

c Find the area of the square pitch.

d Find the percentage area of the oval that is not part of the square pitch. Round to the nearest

whole percentage.

Two athletes challenge each other to a one-lap race around the oval. Athlete A runs around the

outside of the oval at an average rate of 10 metres per second. Athlete B runs around the outside

of the square at an average rate of 9 metres per second. Athlete B’s average running speed is less

because of the need to slow down at each corner.

e Find who comes first and the difference in times, correct to the nearest hundredth of a second.

