Chapter

Review of length

(Consolidating)

Pythagoras’ theorem

Area (Consolidating)

Surface area – prisms and

cylinders

Surface area – pyramids and

cones (10A)

Volume – prisms and

cylinders

Volume – pyramids and

cones (10A)

Spheres (10A)

Limits of accuracy

(Extending)

6A

6B6C6D

6E

6F

6G

6H6I

M E A S U R E M E N T A N D G E O M E T R Y

Using units of measurement

Solve problems involving surface area and volume for a range of prisms, cylinders and composite solids.(10A) Solve problems involving surface area and volume of right pyramids, right cones, spheres and related composite solids.

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

What you will learn

Australian curriculum

6 Measurement

Pythagoras and trigonometry

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.

Chapter pre-testVideos of all worked examplesInteractive widgetsInteractive walkthroughsDownloadable HOTsheetsAccess to HOTmaths Australian Curriculum courses

Online resources

Monolithic domes

••••

100°

10 cm

× 1000

÷ 1000

km m

× 100

÷ 100

cm

× 10

÷ 10

mm

•

radi

us (r

)

diameter (d)

• θ r

3.4 m

2.1 m

4.5 m

x m

∴

∴

r cm

∴

3 cm240°

( )

40° 105°30°

185 m

220 m

0.5 km

2.6 km

12 cm

32 cm34 cm

6.2 cm

10.4 cm

12 cm

x cm

8.4 m

x m

7 cm 1.2 m

28.4 mm

1.1 km

r m

4 m

2 m

1 km60°

6 cm60°

5 mm

80°

3 cm

205°

5

2

7

Perimeter = 17

x

5.3Perimeter = 22.9

x2x

Perimeter = 0.072

2.8 2x

Perimeter = 16.2

11.61

3.72

7.893x

Perimeter = 46.44

1.5x

2xx

Perimeter = 10.8

1.5

3.6

7.9

0.3

10 2.2

3.6

6 12

3 m5 m

2.5 m 1 km

x

l

w

r

rrotatingcircle

fixedcircle

( )

b

a

b

c

c1

a

3

2

•( )

•••

a

b

c

√

9 cm

5 cmx cm

1.5 m

1.1 m

y m

∴

∴√

∴√

D E F

GH

C

A B

A

E

B

4

7

∴

∴√

E B

H

2

√65

√

∴√

√√ √ √

zx

y

ba

d

x

c

4 cm

3 cm5 m

10 m

15 km

7 km

0.2 mm

1.8 mm

0.21 km0.37 km

72.1 cm

27.3 cm

2 m 5 m

9 m

12 m

0.3 m

0.7 m

0.71 cm

1.32 cm

19.3 cm24.2 cm

0.11 cm

0.14 cm

D

A

CG

B

F2

35

H

E

DH

GC

BF

EA

84

14

C

GH

E FD

BA 1

7

3

F

B8

5

9

A

D

H

E

G

C

2

3

4

5

12 13 611

8

1210

6

1215

9

12

√5

A

B

1.2 m

2.6 m

A

B

1.7 cm

3 cm

3.5 cm

1.9 cm

BA

26 m

35 m

49 m

5.3 cm

5.3 cm

A

B

BA

2.1 km 1.8 km

0.5 km

AB

17.2 km 19.7 km

12.3 km

x mm

11.4 mm

9.3

mm

x m

15.5 m

3.7 m 5 cm

3 cm

x cm

3 m

2 m

1.5 m

x m

6.2 km

8.2 km

6.7 km

x km 2.93 cm

4.04 cm5.31 cm

x cm

4

6

2

−4

−6

−2−4−6 −2−8

−8

4 62 8O

8

y

x

2 cm1 cm

r cm

4.5 m

3 m

E

A B

C

GH

F

D

3.5 m

Area = 6 m2

2 m

3 m

••

× 10002

÷ 10002

km2 m2

× 1002

÷ 1002

cm2

× 102

÷ 102

mm2

1 m

= 1

00 c

m

1 m = 100 cm

1 m2

l w

l

h

b

yx h

b

h

b

a x

y

rθ

10 mm

10 mm

1 cm2

km m cm2 2 2

÷ 10002 ÷ 1002

1002 = 10 000 and 10002 = 1 000 000

2 cm

3 cm

5 cm 5.8 m

3.3 m 1.06 km

Area = 11 mm2

l mm2.3 mm

Area = 0.5 m2

1.3 m

0.4 ma m

∴

∴

( )

80°3 cm

5 m

◦ ◦ ◦

5 cm

5.2 m

10.5 m

2.8 km

1.3 km

0.2 mm

0.1 mm

0.3 mm

7 m

15 cm10 cm

64 m 23 m 0.4 mm

0.25 mm

km23

km14

√3 km

0.82 m

1.37 m √6 cm√2 cm

√10 cm

w cm

5.2 cm

l m

1.8 km

h km

1.3 m

2.8 m

a m

5.34 mm

h mm 18 m

Area = 80 m2x m

r cmd m

d m

6 cm

10 m

7 m

120°

100°2 m 150°

6 km3 mm

325°

6.5 m

8.2 m

5 cm

4 m

7 m

10 m

1.7 m1.8 m

1.6 m

4.2 mm

28 km

18 km

26 km

0.25 m 0.3 m

0.7 m

b

a

h

x

x

x

x

�••

�•

h

r

h

r

r

2πr

•

common face

3 cm

5 cm8 cm

5.3 m

1.7 m

3 cm

5 cm

8 cm

5 cm

3 cm

5.3 m2πr

1.7 m

1.7 m

8 cm

10 cm

20 cm

6 cm

10 cm

4 cm

2 cm2 cm

1 cm3 cm

3 cm

3 cm6 cm 2.7 mm

1.3 mm5.1 mm

2.1 cm

4.5 cm7 cm

8 m

17 m

15 m

20 m

12.8 m

9.2 m

5.2 m

6.4 m

1.2 cm

5 mm

3 mm

4 mm

7 mm

25 m10 m

8.66 m16 mm

5 mm

6 mm12 mm

1.2 cm

1.5 cm

0.8 cm0.76 cm

0.5 cm

2.3 m

4.8 m

2.1 m

1.4 m

1.9 m

3 cm10 cm 2 m

1.6 m

6 m

4 m20 cm

40 cm

40°

7

3

2

1.4

1.8 3.2

8

7

10

8

10 33

26

1.85.2

2 cm

6 cm

4 cm

4 cm

10 m

8 m

10 cm

20 cm15 m

40 m20 m

5 m2 m

x

a b

c

h

d rh

r

r

y

x

r

h

θh

r

••

s

rθ

sB

A

•••

•••

r

s rs

2πr

∴

s

apex

right coner

••

apex

right square-based pyramid

4.5 m

2 m

25 mm

22 mm

( )( )√( )

sh

r

5 cm

2 cm

14 m

5 m6 cm

10 cm

2 cm4 cm

2 cm

3 cm

9 mm

12 mm

0.5 m

0.8 m

11 km

15 km

6 m

4 m

8 cm

5 cm0.3 m

0.2 m

10 m

2 m

1.1 cm

1.5 cm

26 mm

48 mm

6 m

4 m

4 cm

5 cm

2 cm

3 cm

3 cm

4.33 cm

8 cm

5 cm

4.33 cm

10 m

3 m

8 m

10 mm

6 mm

9 mm 5 m

7 m2 m

√

√

10 cm

10 cm

20 cm

8 cm

11 cm

13 cm

7 cm5 cm 4 cm

8 cm

3 cm

13 cm

12 cm

x cm9 cm

40 cmx cm

8 cm

5 cmx cm

12 cm

4 mm 17.6 mm

5 cm

80°5 cm

4 cm

210°

4 cm

6 cm7 cm

10 cm

5 cm

8 cm

3 cm

1.1 m

50 cm

12 cm

7.2 cm8 cm

6 cm

◦

120°7.5 cm

O

10

63

10

63

10

63

••

•••

× 10003

÷ 10003

km3 m3

× 1003

÷ 1003

cm3

× 103

÷ 103

mm3

•

× 1000

÷ 1000

ML kL

× 1000

÷ 1000

L

× 1000

÷ 1000

mL

••

w

h

l

h

xxr

h

4 m

3 m2 m

6 cm

2 cm

3 cm8 cm

5 cm

6 cm2 cm

2 cm

10 cm

2 m

12 mm

2 cm

4 cm5 cm

30 m10 m

35 m7 mm

3.5 mm10.6 mm

5 m

10 m1.5 cm

2 cm

12 m

14 m

3 cm

2 cm

20 m

5 m

1.2 mm

2.4 mm

1.7 mm

2 km10 km

3 km

3.5 cm

8 cm

7 cm

7 m3 m

3 m

2 m

5 cm2 2 cm

0.12 m

0.38 m

21.2 cm

18.3 cm

24.5 cm

0.5 m

0.8 m

12 cm

12 cm

2 m

3 m

2 m

6 m

8 cm

2 cm

5 cm

5 m

5 m

2.3 m

1.6 m

1.1 m 2.6 m

5 mm

8 mm11 mm 9 cm

3 cm

4 cm

5 cm

4 cm

6 cm

5

34

h

r θ

10 m4 m

10 m

1 m

1 m

••

•••

xx

h h

r

h

r

1.4 m

1.2 m

1.3 m

29 mm

23 mm

A = 5 m2

6 m4 cm

A = 2 cm2

7 mm

A = 25 mm2

2 cm

3 cm

18 m

15 m

13 m

12 km

5 km

5 km

2 cm

Perpendicularheight = 2 cm

4 cm

3 cm

Perpendicularheight = 4 cm

1.4 mm

0.6 mm

Perpendicularheight = 1.2 mm

2.6 m

1.1 m

3.5 mm

1.6 mm

60 m

20 m

6 m

2 m

1.3 m

0.6 m

5 cm

20 cm

8 m

6 m

3 m

5 cm

6 cm

4 m

3 m6 m

2 m 4.8 mm

1.5 mm

5.2 mm

3.96 m

2.8 m 7.1 m

10 m

2.6 m

6 cmice-cream

depth ofice-cream

3 cm

r1

h1

r2

h2

••

•

•

r

r m

∴√

∴

5 cm 7 cm

√

√√

5 cm

2 cm

s cm

( )

2 cm0.5 m

38 mm

18 cm0.92 km

1.36 m

√ √

r cmr cm

r km

r m r cm r mm

5 cm 4 cm 1 m

2 m4 m

5 m

1.3 cm

0.7 cm

4 mm

8 mm

1 m

5 m3 m

2 m

3 cm

2 cm

√2 cm

3 m

1 m 13 m

19 m

2 cm

28 cm

20 cm

2 m

2 m

2 m

1 cm

hollow

15 cm10 cm

xx

x

r

√

•

•

•

•

85 86.0

86.585.5

87

•

85.8 85.9 86.0 86.1 86.2

85.95 86.05

4

26

6

3 9

6 m

2 m10 m

1 m2 m

4 m2 m

1 m

10 cm

5 cm

6 cm

1 cmA

C

C

A

B

10 cm

5 cm

2 cm

12 cm

6 cm

8 cm

Up for a challenge?If you get stuck on a question,check out the 'Working withunfamiliar problems' poster

at the end of the bookto help you.

◦

45°

Measurement

Pythagoras’ theorem

Volume

rh

h

r

Cone:

Sphere:

× 10003

÷ 10003

km3 m3

× 1003

÷ 1003

cm3

× 103

÷ 103

mm3

× 1000

÷ 1000

megalitres(ML)

kilolitres(kL)

millilitres(mL)

litres(L)

× 1000

÷ 1000

× 1000

÷ 1000

hw

lV = pr2hV = lwh

V = Ah,13

V = pr2h13

V = pr343

Perimeter is the distance aroundthe outside of a closed shape.

Circumference of a circleC = 2pr = pd

Perimeter of a sector

LengthUnits of length

× 1000

÷ 1000

km m

× 100

÷ 100

cm

× 10

÷ 10

mm

P = 2r + × 2prθ360

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

sector rhombus parallelogram trapeziumkite

A = xyh 12A = xy1

2A = × pr2θ360

A = (a + b)A = bh 12

θ h

b

h

b

ax

yxy

where A is the area of the base.

r

circle

A = pr2

l

square

A = l 2

w

l

rectangle

A = lw

triangle

A = bh12

h

b

r

Usually ± 0.5 × the smallestunit of measuremente.g. 72 cm is 71.5 cm to 72.5 cm.

Limits of accuracy

Units of volume

rectangular prism

For pyramids and cones: (10A)

cylinder

Units of capacity

1 cm3 = 1 mL

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

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

e.g.

Surface area

r

r

h h2pr

s

r

TSA = 4 × triangles + square base

TSA = 2prh + 2pr2

curved base surface ends

TSA = 4pr2

TSA = prs + pr2

curved + base

Cylinder

Cone

Sphere

For composite solids considerwhich surfaces are exposed.

+

(10A)

(10A)

(10A)

x cm

7.1 cm10.4 cm

100°5 m

x km7 km

12

10

√ √ √√ √

7 m

12 m

10 cm

9 cm

4 cm

8 m

8 m

9 m

7 m

4 m80°

6 cm

D

H G

C

B

FE

A

2

47

5.2 m

6.2 m 12 m

18 m

13 m7 m

11 m

10 cm6.4 m

9.8 m

13 m

7 m4.2 m

8.1 m

w m

9.2 m

5.3 m

deck

6 m

5 m8 m 6 cm

10 cm

5 cm4 cm

8 cm

20 cm

10A13 cm

10 cm

10 cm

12 cm

10A

4 m

10A

12 cm

10 cm

13 cm

10 cm

3 cm3 cm

7 m

4 m

5 m

2.5 mm 1.6 mm

0.9 mm

4 cm

5 cm

3.9 cm

5 cm

3 cm

2 cm 3.2 cm

1.8 cm

39 cm

51 cm

36 cm

6 m

2 m 8 m

1 m

c m