Geometry: similarity and mensuration -...

VCEcoverageArea of studyUnits 3 & 4 • Geometry and

trigonometry

In thisIn this chachapterpter8A Properties of angles,

triangles and polygons

8B Area and perimeter

8C Total surface area

8D Volume of prisms, pyramids and spheres

8E Maps and similar figures

8F Similar triangles

8G Area and volume scale factors

8

Geometry:similarity andmensuration

Ch 08 FM YR 12 Page 347 Friday, November 10, 2000 11:30 AM

348

F u r t h e r M a t h e m a t i c s

Geometry

Geometry is an important area of study. Many professions and tasks require and use geometrical concepts and its associated techniques. Besides architects, surveyors and navigators, all of us use it in our daily lives — for example, to describe shapes of objects, directions on a car trip and space or position of a house. Much of this area of study is assumed knowledge gained from previous years of study.

Properties of angles, triangles and polygons

In this module, we will often encounter problems where some of the information we need is not clearly given. To solve the problems, some missing information will need to be deduced using the many common rules, definitions and laws of geometry. Some of the more important rules are presented in this chapter.

Interior angles of polygons

For a regular polygon (all sides and angles are equal) of

n

sides, the interior angle is given by 180

° −

.

For example, for a square the interior angle is:

180

° −

=

180

° −

90

°=

90

°

The exterior angle is given by .

Stairways

Bed 4 Bed 3

Bed 2

Bed 1

UPPERLEVEL

Exterior angle

Interior angle

360°n

-----------

360°4

-----------

360°n

-----------

Find the interior and exterior angle of the regular polygon shown.

THINK WRITE

This shape is a regular pentagon, a 5-sided figure.Substitute n = 5 into the interior angle formula.

Interior angle = 180° −

= 180° − 72°= 108°

Substitute n = 5 into the exterior angle formula.

Exterior angle =

= 72°Write your answer. A regular pentagon has an interior angle of

108° and an exterior angle of 72°.

1360°

5-----------

2360°

5-----------

3

1WORKEDExample


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n

349

Geometry rules, definitions and notation rules

The following geometry rules and notation will be most valuable in establishingunknown values in the topics covered and revised in this module.

Definitions of common terms

Some common notations and rules

Less than 90°

A

B C

∠ABC

Acute angleMQ F M t fi 11 05(b)

90°

Right angle

MQ FurMat fig 11.05(c)

180°

Straight angle

MQ F rMat fig 11 05(d)

Obtuse angle

Between 90°and 180°

Reflex angle

Between 180°and 360°

Parallel lines Perpendicular lines MQ FurMat fig 11.05(h)

Line segment

Ray

LineA B

AB

AB

AB

A B

A B

MQ FurMat fig 11.06(a)

Scalene triangle

b

a c

No equal sidesa + b + c = 180°

MQ FurMat fig 11.06(b)

Isosceles triangle

Two equalsides and

angles

Equilateral triangle

All equalsides and

angles

60° 60°

60°

Right-angledisoceles triangle

45°

45°

MQ FurMat fig 11.06(f)

Supplementary angles

a + b = 180°

a b

MQ FurMat fig 11.06(g)

Vertically oppositeangles

a = ba

b

MQ FurMat fig 11.06(e)

Complementary angles

a + b = 90°

ab

CD is a perpendicular bisector of AB

A

C

D

B

MQ F M fi 11 07( )

Alternate angles

a c

a = bc = d

bd

Corresponding angles

ac

a = bc = d

bd

Co-interior angles

a c

a + d = 180°b + c = 180°

bd

MQ FurMat fig 11.07(d)

ab

cd

a + b + c + d = 360°MQ FurMat fig 11 07(e)

a

b

B

A C Dc d

a + b = d

∠BCD is an exterior angle

Right angle at thecircumference in

a semicircle

Ch 08 FM YR 12 49 Friday, November 10, 2000 11:30 AM

350 F u r t h e r M a t h e m a t i c s

Find the values of the pronumerals in the polygon at right.

THINK WRITE

This shape is a regular hexagon. The angles at the centre are all equal.

a =

= 60°The other two angles in the triangle are equal. a + b + c = 180°

b = cSo:60 + 2b = 180°b = 60°c = 60°

The 6 triangles are equilateral triangles, therefore all sides are equal. d cm = 6 cm

c

d cm

6 cm

b

a

1360°

6-----------

2

60°

3

2WORKEDExample

Find the missing pronumerals in the diagram of railings for a set of stairs shown at right.

THINK WRITE

Recognise that the top and bottom of the stair rails are parallel lines.

To find the unknown angle a, use the alternate angle law and the given angle.

Given angle 35°.a = 35°

The unknown angle c is a right angle, using the given right angle and corresponding angle law.

c = 90°Use the straight angle rule to find the unknown angle b.

a + b + c = 180°35° + b + 90° = 180°b = 180° − 125°b = 55°

35°

a bc

1

35°

35°

a b

c

2

3c

4

3WORKEDExample


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 351

Properties of angles, triangles and polygons

1 Find the interior and exterior angles for each of the following regular polygons.a Equilateral triangle b Regular quadrilateralc Hexagon de Heptagonf Nonagong

2 Find the value of the pronumerals in the following figures.

a b c

d e f

rememberProperties of angles, triangles and polygons

1. Draw careful diagrams.

2. Carefully interpret geometric notations, for example from the diagram below.

3. Carefully consider geometric rules, such as isosceles triangles have 2 equal sides and angles. (Refer to the figures in the preceding section on definitions of common terms and common notations and rules.)

Equal sides

remember

8AWORKEDExample

1

WORKEDExample

2

27°

52° a

130°

yx 63°

c

15°c ab

c 50°

b 8 cm

32°m



3 Find the value of the pronumerals in the following figures.

4 Name the regular polygon that has the given angle(s).

a Interior angle of 108°, exterior angle of 72°

b Interior angle of 150°, exterior angle of 30°

c Interior angle of 135°, exterior angle of 45°

d Interior angle of 120°

e Exterior angle of 120°

5 Find the unknown pronumerals.

6The value of a is closest to:

A 30°

B 75°

C 90°

D 120°

E 150°

7An isosceles triangle has a known angle of 50°. The largest possible angle for thistriangle is:

a b c

d e f

a b c

d e

A 80° B 130° C 90° D 65° E 50°

WORKEDExample

3

35°x y

30°

z0°

62°

t

70° b cad

27° m

n140°a

81°

4.2 cm

3.6

cm

h

r

35°x zy

110° db c

a8 cm29° 122°

ab

cd

86° a

b

40°

mmultiple choiceultiple choice

a

150°




Area and perimeterMuch of our world is described by area (the amount of space enclosed by a closedfigure) and perimeter (the distance around a closed figure).

Some examples are the area of a house block, the fencing of a block of land, the size of a bedroom and the amount of paint required to cover an object. In this section we will review the more common shapes.

PerimeterPerimeter is the distance around a closed figure.

Some common rules are:1. For squares, the perimter = 4l 2. For rectangles, the perimter = 2(l + w)

3. Circumference (C) is the perimeter of a circle.C = 2 × π × radius

= 2πr

Lot 658

761m2

13.05

23.55 5.86

32.1

8

32.7

5

Corner block with expansive 23.55 m frontage

$51,000

aº

Lot 603645m2

17

14.07

4.05

36.5

6

37.9

2

Corner blockwith wide

17 m frontage

$47,000

l

l

l l

Squarel

l

w w

Rectangle

r

Cir

cum

ference

of a cirle

Find the perimeter of the closed figure given at right (to the nearest mm).

THINK WRITE

The shape is composed of a semicircle and three sides of a rectangle.

Perimeter = 300 + 2 × 600 + circumference where

of circumference = × 2πr

= π × 150= 471.24

Add together the three components of the perimeter.

Perimeter = 300 + 2 × 600 + 471.24= 1971.24

Write your answer. Perimeter of the closed figure is 1971 mm.

300

mm

600 mm

112---

12--- 1

2---

2

3

4WORKEDExample



Area of common shapesThe areas of shapes commonly encountered are:1. Area of a square: A = length2 = l2

2. Area of a rectangle: A = length × width = l × w

3. Area of a parallelogram: A = base × height = b × h

4. Area of a trapezium: A = (a + b) × h

5. Area of a circle: A = π × radius2 = π × r2

6. Area of a triangle: A = × b × h (see the next chapter)

Area is measured in mm2, cm2, m2, km2 and hectares.1 hectare = 100 m × 100 m = 10 000 m2

l

l

Square

l

w

Rectangle

h

b

Parallelogram

12---

h

a

b

Trapezium

r

O

Circle

12---

h

b

Triangle

Find the area of the garden bed given in the diagram (to the nearest square metre).THINK WRITE

The shape of garden is a trapezium. Use the formula for area of a trapezium. Remember that the lengths of the two parallel sides are a and b and h is the perpendicular distance between the two parallel sides.

a = 7.5 b = 5.7 h = 2.4Area of garden = Area of a trapezium

= (a + b) × h

Substitute and evaluate. = (7.5 + 5.7) × 2.4

= × 13.2 × 2.4

= 15.84 m2

Write your answer. Area of the garden bed is approximately 16 square metres.

5.7 m

2.4

m

7.5 m

1

12---

212---

12---

3

5WORKEDExample



Composite areasOften a closed figure can be identified as comprising two or more different commonfigures. Such figures are called composite figures. The area of a composite figure is thesum of the areas of the individual common figures.

Area of composite figure = sum of the individual common figuresAcomposite = A1 + A2 + A3 + A4 + . . .

Find the area of the hotel foyer from the plans given below (to the nearest square metre).

THINK WRITE

The shape is composite and needs to be separated into two or more common shapes: in this case, a rectangle, a triangle and half of a circle.

Area of foyer = A1 + A2 + A3

Substitute and evaluate each of the shapes. The width of the rectangle and the base of the triangle is twice the radius of the circle; that is, 16 metres.

A1 = area of triangle

= × b × h

= × 16 × 20

= 160 m2

A2 = Area of rectangle= l × w= 25 × 16= 400 m2

A3 = Area of half of a circle

= × π × r2

= × π × 82

= 100.53 m2

Add together all three areas for the composite shape.

Area of foyer = A1 + A2 + A3

= 160 + 400 + 100.53= 660.53 m2

Write your answer. Area of the hotel foyer is 661 m2.

25 m

8 m

20 m

1 25 m

16 m

16 m

8 m

A2A1 A3

20 m

12---

12---

12---

12---

2

3

6WORKEDExample



Conversion of units of areaOften the units of area need to be converted, for example from cm2 to m2 and viceversa.1. To convert to smaller units, for example m2 to cm2, multiply (×).2. To convert to larger units, for example, mm2 to cm2, divide (÷).Some examples are:(a) 1 cm2 = 10 mm × 10 mm = 100 mm2

(b) 1 m2 = 100 cm × 100 cm = 10 000 cm2

(c) 1 km2 = 1000 m × 1000 m = 1 000 000 m2

(d) 1 hectare = 10 000 m2

Area÷102 ÷1002 ÷10002

× 102 × 1002 × 10002

mm2 cm2 m2 km2

Convert 1.12 m2 to square centimetres (cm2).

THINK WRITE

Conversion factor for metres to centimetres is multiply by 100. That is, 1 metre = 100 centimetres.

1.12 m2= 1.12 × 1 metre × 1 metre

Conversion factor for metres2 to centimetres2 is multiply by 1002 or 10 000.

= 1.12 × 100 cm × 100 cm= 11 200 cm2

Write your answer. 1.12 m2 is equal to 11 200 square centimetres (cm2).

1

2

3

7WORKEDExample

Convert 156 000 metres2 to a kilometres2 b hectares.

THINK WRITE

a Conversion factor for metres to kilometres is divide by 1000; that is, 1 metre = kilometre.

a 156 000 m2 = 156 000 × km × km

Conversion factor for metres2 to kilometres2 is divide by 10002 or 1 000 000.

=

= 0.156 km2

Write the answer in correct units. 156 000 m2 = 0.156 square kilometres (km2)

b Conversion factor is10 000 m2 = 1 hectare; that is,1 m2 = hectare

b 156 000 m2 = 156 000 × hectares

= hectares

= 15.6 hectares

Write the answer. 156 000 m2 = 15.6 hectares

1

11000------------

11000------------ 1

1000------------

2156 000

1000 1000×------------------------------

3

1

110 000----------------

110 000----------------

156 00010 000-------------------

2

8WORKEDExample



Area and perimeter

1 Find the areas of the following figures (to the nearest whole units).a b c

d e f

rememberArea and perimeter

1. Perimeter is the distance around a closed figure.(a) For squares, the perimeter = 4l

(b) For rectangles, the perimeter = 2(l + w)

(c) Circumference (C) is the perimeter of a circle.C = 2 × π × radius

2. Area is measured in mm2, cm2, m2, km2 and hectares.3. (a) 1 cm2 = 10 mm × 10 mm = 100 mm2

(b) 1 m2 = 100 cm × 100 cm = 10 000 cm2

(c) 1 km2 = 1000 m × 1000 m = 1 000 000 m2

(d) 1 hectare = 10 000 m2

4. Area of shapes commonly encountered are:(a) area of a square: A = l2

(b) Area of a rectangle: A = l × w(c) Area of a parallelogram: A = b × h(d) Area of a trapezium: A = (a + b) × h(e) Area of a circle: A = πr2

(f) Area of a triangle: A = × b × h5. Area of composite figure = sum of the individual common figures

Acomposite = A1 + A2 + A3 + A4 + . . .

l

l

ll

Square

l

l

w w

Rectangle

r

Circum

feren

ce of a circle

12---

12---

remember

8BWORKEDExample

5 Mathcad

Areaand

perimeter

7 m

12 m

5 m4 m

23.7 cm

17.8

cm

15.4

cm

27.5 cm

13.5 mm

7.5 m

11.5 m

5 m

4 m

120

m

210

m

90 m

70 m

SkillSH

EET 8.1



2 Find the perimeters of the closed figures in question 1.

3 Find the areas of the following figures (to 1 decimal place).a b c

d e f

4 Find the perimeters of the closed figures in question 3.

5 Convert the following areas to the units given in brackets.

6 A kite has the dimensions in the figure at right. Find the area of the kite (to the nearest cm2).

7 Find the area of the regular hexagon as shown in the diagram at left(to 2 decimal places, in m2).

8 A cutting blade for a craft knife has the dimensions shown in the diagram.What is the area of steel in the blade?

a 20 000 mm2 (cm2) b 320 000 cm2 (m2) c 0.035 m2 (cm2)d 0.035 m2 (mm2) e 2 500 000 m2 (km2) f 357 000 m2 (hectares)g 2 750 000 000 mm2 (m2) h 0.000 06 km2 (m2)

WORKEDExample

4

WORKEDExample

612 m

25 m

10 m14 m

20 m2 m

3.5

m

12 m

17 m

125 mm

24 mm90 mm48 mm

21 cm

16 cm

8 cm

10 c

m

12 cm

10 cm

20 m

22 m

7 m

13 m

11 m

34 m

44 m

SkillSH

EET 8.2 WORKEDExample

7, 8

70 cm

180

cm

2.08 m

1.20 m

30 mm

20 m

m

40 mm

5 mm



The area in m2 of the stacked objects shown at right is closest to:A 1.44B 1.68C 1.92D 3.84E 11.52

10

The perimeter of the figure shown in centimetres is:A 34B 24 + 5πC 24 + 2.5πD 29 + 5πE 29 + 2.5π

11

The perimeter of the enclosed figure shown is156.6 metres. The unknown length, x, is closest to:A 20.5 mB 35.2 mC 40.2 mD 80.4 mE Cannot be determined

12 A 3-ring dartboard has dimensions as shown below left. (Give all answers to 1decimal place.)

a What is the total areaof the dartboard?

b What is the area of thebullseye (inner circle)?

c What is the area of the2-point middle ring?

d Express each area of the threerings as a percentage of the totalarea (to 2 decimal places).


0.8

m0.

8 m

0.8

m

1.2 m

1.0 m

0.8 m

0.6 m

0.8

m


12 cm

7 cm

2 cm

3 cm


35.2

m

20.5 mx

MQ FurMat fig 11.59

1

2

3

2

1

6 cm

40 cm20 cm

SkillSH

EET 8.3



13 On a western movie set, a horse is tied to a railing outside a saloon bar. Therailing is 2 metres long; the lead on the horse is also 2 metres long and tied at oneof the ends of the railing.a Draw a diagram of this situation.b To how much area does the horse now have access (to 1 decimal place)?

The lead is now tied to the centre of the railing.c Draw a diagram of this situation.d To how much area does the horse have access (to 1 decimal place)?

14 The rectangular rear window of a car has an area of 1.28 m2.a Find the height of the rear window if its length is 160 centimetres (to the nearest

centimetre).

A wiper blade is 50 cm long and the end just reaches the top of the window as itmakes a semicircular sweep. The base of the wiper is situated at the bottom centre ofthe rear window.b Draw a diagram of the situation.c Find the area of the window that is swept by the wiper (to the nearest cm2).d Find the percentage of the window’s area that is not swept by the wiper.

The manufacturer decides to increase the wiper length by 10 cm.e Find the new area of the rear window that is swept.f Find the percentage of the window’s area that is not

swept by the wiper.

15 A signwriter charges his clients by the width and height of the sign to be painted. A client advises the signwriter to paint 12 words with 10 cm high characters and a 20 cm length for each word.a What is the area of each word?b What are all the different ways of arranging the words

in a rectangular pattern?c If the charge is $2 per 10 cm in height and $1.50 per

10 cm in length, find the minimum cost for the sign and its dimensions.



Total surface areaThe total surface area (TSA) of a solid object is the sum of the areas of the surfaces.

In some cases, we can use established formulas of very common everyday objects. Inother situations we will need to derive a formula by using the net of an object.

Total surface area formulas of common objects

Cubes: Cuboids:

TSA = 6l2 TSA = 2(lw + lh + wh)

Cylinders: TSA = 2πr(r + h)

Cones: Spheres:

TSA = πr(r + s) where TSA = 4πr2

s is the slant height

Cube

l

Cuboid

h l

w

Cylinder

h

r

r

Cone

Slantheighth s

r

Sphere

r

Find the total surface area of a poster tube with a length of 1.13 metres and a radius of 5 cm. Give your answer to the nearest 100 cm2.

THINK WRITE

A poster tube is a cylinder.Express all dimensions in centimetres.Remember 1 metre = 100 centimetres.

Radius, r = 5 cmHeight, h = 1.13 m

= 113 cmSubstitute and evaluate.Remember BODMAS.

TSA of a tube = 2πr(r + h)= 2 × π × 5(5 + 113)= 2 × π × 5 × 118= 3707.08

Write your answer. The total surface area of a poster tube is approximately 3700 cm2.

1.13 m

5 cm

1

2

3

9WORKEDExample



Total surface area using a netIf the object is not a common object or a variation of one, such as an open cylinder, thenit is easier to generate the formula from first principles by constructing a net of the object.

A net of an object is a plane figure that represents the surface of a 3-dimensional object.

Find the total surface area of a size 7 basketball with a diameter of 25 cm. Give your answer to the nearest 10 cm2.THINK WRITE

Use the formula for the total surface area of a sphere. Use the diameter to find the radius of the basketball and substitute into the formula.

Diameter = 25 cmRadius = 12.5 cmTSA of sphere = 4πr2

= 4 × π × 12.52

= 1963.495Write your answer. Total surface area of the ball is approximately 1960 cm2.

1

2

10WORKEDExample

A die used in a board game has a total surface area of 1350 mm2. Find the linear dimensions of the die (to the nearest millimetre).THINK WRITE

A die is a cube. We can substitute into the total surface area of a cube to determine the dimension of the cube. Divide both sides by 6.

TSA = 6 × l2

TSA = 1350 mm2

1350 = 6 × l2

l2 =

= 225Take the square root of both sides to find l.

l = = 15 mm

Write your answer. The dimensions of the die are:15 mm × 15 mm × 15 mm

1

13506

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

2 225

3

11WORKEDExample

MQ FurMat fig 11.68

Slantheight

Net

Square pyramid

Net

NetMQ FurMat fig 11.69



Find the total surface area of the triangular prism shown in the diagram.

THINK WRITE

Form a net of the triangular prism, transferring all the dimensions to each of the sides of the surfaces.

Identify the different-sized common figures and set up a sum of the surface areas. The two triangles are the same.

TSA = A1 + A2 + A3 + 2 × A4

A1 = l1 × w1

= 20 × 10

= 200 cm2

A2 = l2 × w2

= 20 × 8

= 160 cm2

A3 = l3 × w3

= 20 × 6

= 120 cm2

A4 = × b2 × h2

= × 8 × 6

= 24 cm2

Sum the areas. TSA = A1 + A2 + A3 + 2 × A4

= 200 + 160 + 120 + 2 × 24

= 528 cm2

Write your answer. The total surface area of the triangular prism is

528 cm2.

20 cm

10 cm

8 cm

6 cm

1

A1

A4

A4

8 cm

8 cm10 cm

20 c

m10 cm 10 cm

10 cm

20 c

m

20 c

m

6 cm

6 cm

6 cm

A2 A3

2

12---

12---

3

4

12WORKEDExample



Find the surface area of an open cylindrical can that is 12 cm high and 8 cm in diameter (to 1 decimal place).

THINK WRITE

Form a net of the open cylinder, transferring all the dimensions to each of the surfaces.

Identify the different-sized common figures and set up a sum of the surface areas. The length of the rectangle is the circumference of the circle.

TSA = A1 + A2

A1 = 2πr × w= 2 × π × 4 × 12= 301.59 cm2

A2 = π × r2

= π × 42

= 50.27 cm2

Sum the areas. TSA = A1 + A2

= 301.59 + 50.27= 351.86 cm2

Write your answer. The total surface area of the open cylindrical can is 351.9 cm2.

fi

8 cm

12 c

m

1 2 rπ

A1

12 c

m

A24 cm

2

3

4

13WORKEDExample

rememberTotal surface area

1. Total surface area (TSA) is measured in mm2, cm2, m2 and km2.2. The TSAs of some common objects are as follows:

(a) Cubes: TSA = 6l2

(b) Cuboids: TSA = 2(lw + lh + wh)(c) Cylinders: TSA = 2πr(r + h)(d) Cones: TSA = πr(r + s) where s is the slant height(e) Spheres: TSA = 4πr2

3. For all other shapes, form their nets and establish the total surface area formulas.

remember



Total surface area

1 Find the total surface area for each of the solids a to f from the following information.Give answers to 1 decimal place.a A cube with side lengths of 110 cmb A cuboid with dimensions of 12 m × 5 m × 8 m (l × w × h)c A sphere with a radius of 0.8 metresd A closed cylinder with a radius of 1.2 cm and a height of 6 cme A closed cone with a radius of 7 cm and a slant height of 11 cmf An opened cylinder with a diameter of 100 mm and height of 30 mm

2 Find the total surface area of the objects given in the diagrams. Give answers to1 decimal place.

3 Find the unknown dimensions, given the total surface area of the objects. Giveanswers to 1 decimal place.a Length of a cube with a total surface area of 24 m2

b The radius of a sphere with a total surface area of 633.5 cm2

c Length of a cuboid with width of 12 mm, height of 6 cm and a total surface areaof 468 cm2

d Diameter of a playing ball with a total surface area of 157 630 cm2

4 Find the total surface areas for the objects given in the diagrams. Give answers to1 decimal place.a b

a b c

d e f

8CWORKEDExample

9Mathcad

Totalsurface

area

WORKEDExample

10

410 mm

Length = 1.5 m

14 c

m

4 cm7 cm

Diameter = 43 cm

6 cm

28 cm

90 c

m

2 cm

4 cm

8 cm

WORKEDExample

11

WORKEDExample

12

5 cm

10 cm15 cm

4 cm

7 cm30 cm6.

06 c

m



c d

e f

5 Find the total surface area of each of the objects in the diagrams below. Give answersto 1 decimal place.a b

c d

6 A concrete swimming pool is a cuboid with the following dimensions: length of6 metres, width of 4 metres and depth of 1.3 metres. What surface area of tiles isneeded to line the inside of the pool? (Give answer in m2 and cm2 to 1 decimalplace.)

13 cm

8 cm

Area = 22 cm2

80 mm

40 mm105 mm

22 m

m

30 mm

9 mm25 mm

15 mm

12 m

m

7 m

30 m

4 m5 m

6 m

4 m

WORKEDExample

13

250 mm

250

mm

Rubbish bin

20 cm

10 cm

10.5

cm

15 cm

13.5 cm

1.5 m

1.2 m0.9 m

2.5 m2

cm

4.5 cm7 cm

3 cm


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 3677 What is the total area of canvas needed for the tent

(including the base) shown in the diagram at right?Give the answer to 2 decimal places.

8The total surface area of a 48 mm-diameter ball used in a game of pool is closest to:

A 1810 mm2

B 2300 mm2

C 7240 mm2

D 28 950 mm2

E 115 800 mm2

9The total surface of a golf ball of radius 21 mm is closest to:

10The formula for the total surface area for the object shown is:

A abh

B 2 × bh + ab + 2 × ah

C 3( bh + ab)

D bh + 3ab

E bh + 3ab

11The total surface area of a poster tube that is 115 cm long and 8 cm in diameter isclosest to:

12 A baker is investigating the best shape for a loaf of bread. The shape with the smallestsurface area stays freshest. The baker has come up with two shapes: a rectangularprism with a 12 cm-square base and a cylinder with a round end that has a 14 cmdiameter.a Which shape stays fresher if they have the same overall length of 32 cm?b What is the difference between the total surface areas of the two loaves of bread?

A 550 mm2 B 55 cm2 C 55 000 mm2 D 0.055 m2 E 5.5 cm2

A 3000 cm2 B 2900 cm2 C 1500 mm2 D 6200 m2 E 23 000 cm2

4.5 m6.5 m

2.5 m

1.5

m

1.0

m




h

b

a

12---

12---

12---

12---


WorkS

HEET 8.1



Volume of prisms, pyramids and spheresThe most common volumes considered in the real world are the volumes of prisms, pyramids, spheres and objects which are a combination of these. For example, country people who rely on tank water need to know the capacity (volume) of water that the tank is holding.

Volume is the amount of space occupied by a 3-dimensional object.

The units of volume are mm3 (cubic millimetres), cm3 (cubic centimetres or cc), and m3 (cubic metres).

1000 mm3 = 1 cm3

1 000 000 cm3 = 1 m3

Another measure of volume is the litre which is used primarily for quantities of liquids but also for capacity, like the capacity of a refrigerator, or the size of motor car engines.

1 litre = 1000 cm3

1000 litres = 1 m3

Conversion of units of volumeOften the units of volume need to be converted, for example from cm3 to m3 and vice versa.

Volume

÷103 ÷1003

× 103 × 1003

mm3 cm3 m3

Convert 1.12 cm3 to mm3.

THINK WRITE

The conversion from centimetres to millimetres is 1 cm = 10 mm.

1.12 cm3 = 1.12 × 1 cm × 1 cm × 1 cm= 1.12 × 10 mm × 10 mm × 10 mm

The conversion factor for cm3 to mm3 is to multiply by 103 or 1000; that is, 1cm3 = 1000 mm3.

= 1.12 × 1000 mm3

= 1120 mm3

Write the answer in correct units. 1.12 cm3 is equal to 1120 mm3.

1

2

3

14WORKEDExample



Volume of prismsA prism is a 3-dimensional object that has a uniform cross-section.

A prism is named in accordance with its uniform cross-sectional area. Note: Circular prisms are called cylinders.

To find the volume of a prism we need to determine the area of the uniform cross-section (or base) and multiply by the height. This is the same for all prisms.

Volume of a prism, Vprism, can be generalised by the formula:Vprism = area of uniform cross-section × height

V = A × H

Convert 156 000 cm3 to: a m3 b litres.

THINK WRITE

a The conversion factor for centimetres to metres is divide by 100; that is, 1 cm = m.

a 156 000 cm3 = 156 000 × 1 cm × 1 cm × 1 cm

= 156 000 × m × m × m

The conversion factor for cm3 to m3 is divide by 1003 or 1 000 000; that is, 1 000 000 cm3 = 1 m3.

= m3

= 0.156 m3

Write the answer in correct units.

156 000 cm3 = 0.156 cubic metres (m3)

b Conversion factor is1000 cm3 = 1 litre; that is,1 cm3 = litre.

b 156 000 cm3 = 156 000 × litres

= litres

= 156 litres

Write the answer. 156 000 cm3 = 156 litres

1

1100---------

1100--------- 1

100--------- 1

100---------

2156 000

100 100 100××---------------------------------------

3

1

11000------------

11000------------

156 0001000

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

2

15WORKEDExample

Triangular prism

CylinderSquareprism

Uniformcross-section

Height



Given the volume of an object, we can use the volume formula to find an unknowndimension of the object by transposing the formula.

Find the volume of the object (to the nearest cm3).

THINK WRITE

The object has a circle as a uniform cross-section. It is a cylinder. The area of the base is: area of a circle = πr2.Volume is cross-sectional area times height.

Vcylinder = A × Hwhere Acircle = πr2

Vcylinder = π × r2 × H= π × 152 × 20= 4500 π= 14 137.1669 cm3

Write your answer. The volume of the cylinder is 14 137 cm3.

20 c

m

15 cm

1

2

16WORKEDExample

Find (to the nearest mm3) the volume of the slice of bread with a uniform cross-sectional area of 250 mm2 and a thickness of 17 mm.

THINK WRITE

The slice of bread has a uniform cross-section. The area of the cross-section is not a common figure but its area has been given.

V = A × Hwhere A = 250 mm2

V = 250 mm2 × 17 mm= 4250 mm3

Write your answer. The volume of the slice of bread is 4250 mm3.

1

2

17WORKEDExample

17 mm

Area 250 mm2



Volume of pyramidsA pyramid is a 3-dimensional object that has a similar cross-section but the sizereduces as it approaches the vertex.

The name of the pyramid is related to its similar cross-sectional area (or base).Note: Circular pyramids are commonly called cones.

To find the volume of the pyramids above, we take a similar approach to prisms butthe volume of a pyramid is always one-third of a prism with the same initial base andsame height, H. This is the same for all pyramids.

Volume of a pyramid, Vpyramids, can be generalised by the formula:

Vpyramids = × area of cross-section at the base × height

V = × A × H

The height of a pyramid, H, is sometimes called the altitude.

Find the height of the triangular prism from the information provided in the diagram at right (to 1 decimal place).

THINK WRITE

The volume of the object is given, along with the width of the triangular cross-section and the height of the prism.

V = 6.6 m3, H = 1.1 m, b = 2 mV = A × Hwhere A = × b × h

V = × b × h × H

Substitute the values, transpose and evaluate.

6.6 = × 2 × h × 1.1

= 1.1 h

h =

Write your answer. The height of the triangle in the given prism is 6.0 metres.

h

2 m 1.1

m

Volume of prism = 6.6 m3

1

12---

12---

212---

6.61.1-------

3

18WORKEDExample

Cone

Vertex

Triangular pyramid

H

A

13---

13---



Volume of spheres and composite objectsVolume of a sphere

Spheres are unique but common objects that deserve special attention.

The formula for the volume of spheres is:

Vsphere = πr3

where r is the radius of the sphere.

Find the volume of the pyramid at right (to the nearest m3).

THINK WRITE

The pyramid has a square base. It is a square pyramid. The area of the base is:Area of a square = l2.

Vpyramid = × A × Hwhere Asquare = l2

Vpyramid = × l2 × H

= × 302 × 40

= 12 000 m3

Write your answer. The volume of the square pyramid is 12 000 m3.

30 m 30 m

Height of pyramid = 40 m

113---

13---

13---

2

19WORKEDExample

r

43---


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 373Volume of composite objects

Often the object can be identified as comprising two or more different common prisms,pyramids or spheres. Such figures are called composite objects. The volume of a com-posite object is found by adding the volumes of the individual common figures ordeducting volumes. The grain silo can be modelled as the sum of a cylinder and a largecone, less the tip of the large cone.

Volume of composite object = sum of the individual common prisms, pyramids orspheres.

Vcomposite = V1 + V2 + V3 + . . . (or Vcomposite = V1 − V2)

Find the volume of the object shown at right (to the nearest litre).

THINK WRITE

The object is a composite of a cylinder and a square prism.

The volume of the composite object is the sum of volumes of the cylinder plus the prism.

Vcomposite = volume of cylinder + volume ofsquare prism

= Acircle × Hcircle + Asquare × Hsquare

= (πr2 × Hc) + (l2 × Hs)

= (π × 62 × 20) + (182 × 25)= 2261.946 711 + 8100= 10 361.946 711 cm3

Convert to litres using the conversion of 1000 cm2 = 1 litre.

10 362 cm2 = 10.362 litres

Write your answer. The volume of the object is 10 litres.

25 c

m

12 cm

18 cm

20 c

m

1 r = 6 cm

18 cm 25 c

m

18 cm

H =

20

cm

2

3

20WORKEDExample



Volume of prisms, pyramids and spheres

1 Convert the volumes to the units specified.

2 Find the volume of the following prisms to the nearest whole unit.

a 0.35 cm3 to mm3 b 4800 cm3 to m3 c 56 000 cm3 to litres

d 15 litres to cm3 e 1.6 m3 to litres f 0.0023 cm3 to mm3

g 0.000 57 m3 to cm3 h 140 000 mm3 to litres i 250 000 mm3 to cm3

a b c

d e f

rememberVolume of prisms, pyramids and spheres

1. Volume is the amount of space occupied by a 3-dimensional object.2. (a) The units of volume are mm3, cm3 (or cc), m3.

(b) 1000 mm3 = 1 cm3

(c) 1 000 000 cm3 = 1 m3

(d) 1 litre = 1000 cm3

(e) 1000 litres = 1 m3

3. The volume of a prism is Vprism = area of uniform cross-section × heightV = A × H

4. (a) The volume of a pyramid is Vpyramid = × area of cross-section at the base × height

V = × A × H

(b) The height of a pyramid, H, is sometimes called the altitude.5. The volume of a sphere is Vsphere = πr3.6. The volume of a composite object = sum of the individual common prisms,

pyramids or spheres.Vcomposite = V1 + V2 + V3 + . . .

(or Vcomposite = V1 − V2 . . . )

13---

13---

43---

remember

8D

Mathca

d

Volumeformulas

WORKEDExample

14, 15

SkillSH

EET 8.4WORKEDExample

16

4000 mm

75 mm

51.2

cm

104.8 cm

7 cm

23 c

m

15 cm

4 cm

6.4 m4.8 m

2.1 m

20 m

m

30mm

34 mm

22 mm57 mm14

mm


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 3753 Find the volume of the following prisms (to 2 decimal places).

4 Find the measurement of the unknown dimension (to 1 decimal place).

5 Find the volume of these pyramids (to the nearest whole unit).

a b

c d

a b c d

a b c

d e f

WORKEDExample

17

2.9 m

Area = 4.2 m2

0.5 m

Area = 1000 cm2

14.5

mm

Area = 120 mm2

8.5 cm

Area = 15 cm2

Area =32 cm2

WORKEDExample

18

Volume of cube= 1.728 m3

x

15.0 cm

21.4 cmx

Volume of triangular prism =1316.1 cm3

Volume of cylinder = 150 796.4 mm3

120

mm

x

3x

x

Volume of prism = 10 litres1–8

WORKEDExample

19

12 cm 11 cm

VO = 10 cm

V

O

11 cm

35 c

m

8 m 12 m

V

O

VO = 17m

VO = 8 cm4 cm

4 cm VO

Altitude of squarepyramid = 18 mm

12 mm

10 cm

6 cm

6 cm

Base ofpyramid

VO = 15 cm

O

V



6 Find the volume of these objects (to the nearest whole unit).

7 a Find the volume of a cube with sides 4.5 cm long.

b Find the volume of a room, 3.5 m by 3 m by 2.1 m high.

c Find the radius of a baseball that has a volume of 125 cm3.

d Find the volume of a square pyramid, 12 cm square and 10 cm high.

e Find the height of a cylinder that is 20 cm in diameter with a volume of 2.5 litres(to the nearest unit).

f Find the height of a triangular prism with a base area of 128 mm2 and volume of1024 mm3.

g Find the depth of water in a swimming pool which has a capacity of 56 000 litres.The pool has rectangular dimensions of 8 metres by 5.25 metres.

h Find the radius of an ice-cream cone with a height of 12 cm and a volume of9.425 cm3.

8 The medicine cup below has the shape of a cone with a diameter of 4 cm and a heightof 5 cm (not including the cup’s base). Find the volume to the nearest millilitre, where1 cm3 = 1 mL.

a b c

d e f

g h

WORKEDExample

20

r = 8 cm

4 cm

7 cm8 cm

10 c

m

20 cm

3 m

5 m

4 m

4 m6 m

2 m

3 m

10 cm

10 cm

15 cm2.5 m

2.1 m

1 m

6 m

42 m

19 m

42 m

60 m10

0 m

m

25 mm

5 cm

4 cm


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 3779 Tennis balls have a diameter of 6.5 cm and are packaged in a cylinder

that can hold four tennis balls.Assuming the balls just fit inside a cylinder, find:a the height of the cylindrical canb the volume of the can (to 1 decimal place)c the volume of the four tennis balls (to 1 decimal place)d the volume of the can occupied by aire the fraction of the can’s volume occupied by the balls.

10

The volume 200 000 mm3 is equivalent to:

11

The ratio of the volume of a sphere to that of a cylinder of similar dimensions, asshown in the diagram, is best expressed as:

A

B

C

D

E

12

If the volume of the square pyramid shown is 6000 m3, then the perimeter of the baseis closest to:A 900 mB 20 mC 30 mD 80 mE 120 m

13

A tin of fruit is 13 cm high and 10 cm in diameter. Its volume, to 1 decimal place, is:

14 A model aeroplane is controlled by a tethered string of 10 metres length. The operatorstands in the middle of an oval. (Give all answers to the nearest whole unit.)a What is the maximum area of the oval occupied by the plane in flight?b If the plane can be manoeuvred in a hemispherical zone, find:

i the surface area of the airspace that the plane can occupyii the volume of airspace that is needed by the operator for controlling the plane.

c Repeat part b with a new control string length of 15 metres.

A 2 litres B 2 cm3 C 20 cm3 D 200 cm3 E 2000 cm3

A 1021.0 cm3 B 510.5 cm3 C 1021.4 cm3

D 1020.1 cm3 E 4084.1 cm3



43---

23---

43---r

h-----

34---

32---

r

r


VO = 20 m

O

V



378


Maps and similar figures

Maps and scales

We often need to refer to maps for specifying locations or for establishing distancesbetween two locations. Maps are a reduction of lengths in real life; that is, they havethe same shape as the original but are much smaller in size. A measure of the amountof reduction is the

map scale

.There are two types of map scales.

1. A ratio scale where, for example, 1:100 means that 1 unit on the map represents100 units in real life. In the map below one unit on the map represents 50 000 units.

2. A simple conversion scale where, for example, 1 cm

=

100 m means 1 cm on themap represents 100 metres in real life. In the map below 1 cm on the map represents1 km.

Converting from one type of map scale to another is shown in the followingexample.

SCALE 1:50 000METRES 1000 0 1 2 3 KILOMETRES

Kilometres 0 1 2 3 4 5 6 7 8 Kilometres


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n

379

To find the distance represented on a map, use the simple conversion scale andproportion to the desired value as shown in the next two examples.

Converting map distances to real-life distances

Convert the following map ratio scales:a 1:50 000 to a simple conversion scale with units of centimetresb 2:25 model scale to simple scale with units of millimetresc 1:250 000 to simple scale with units of centimetres.

THINK WRITE

a Rewrite the map scale including the unit centimetres.

a 1:50 0001 cm: 50 000 cm

Convert 50 000 cm to a more appropriate unit of length, for example 100 cm = 1 m.

1 cm: m

1 cm = 500 mb Rewrite the map scale including the unit

millimetres. Divide by 2 to reduce to a unit.

b 2:252 mm = 25 mm1 mm = 12.5 mm

c Rewrite the map scale including the unit centimetres.Convert 250 000 cm to a more appropriate unit of length.Remember 100 cm = 1 m

1000 m = 1 km

c 1:250 0001 cm = 250 000 cm

1 cm = m

1 cm = km

1 cm = 2.5 km

1

2 50 000100

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

1

2 250 000100

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

25001000------------

21WORKEDExample

Find the distance in real life represented by:a 7 mm on a map with 1:100 000 scaleb 11.5 cm on a map with a scale 1 cm = 50 km.

THINK WRITE

a Convert map scale ratio to a conversion scale.

a 1:100 0001 mm:100 000 mm1 mm:100 m

A map distance of 7 mm corresponds to an actual distance of 7 times 100 m.

7 × 1 mm = 7 × 100 m7 mm = 700 m

Write your answer. 7 mm on the map represents 700 m in real life.

b Proportion the scale by multiplying both sides by 11.5.

b 1 cm = 50 km11.5 × 1 cm = 11.5 × 50 km

11.5 cm = 575 kmWrite your answer. On a map with a scale of 1 cm = 50 km,

11.5 cm represents 575 km.

1

2

3

1

2

22WORKEDExample


380


Converting real life distances to map distances

Similar figures

Two objects that have the same shape but different size are said to be

similar

.

For two figures to be similar, they must have the following properties:

1. The ratios of the corresponding sides must be equal.

=

common ratio

2. The corresponding angles must be equal.

∠

A

= ∠

A

′ ∠

B

= ∠

B

′ ∠

C

= ∠

C

′ ∠

D

= ∠

D

′

On a map with a map ratio scale of 1:200 000, find the distance that would represent a distance of:a 5 km b 500 m. THINK WRITE

a Convert ratio scale to a simple conversion scale using an appropriate unit of measure.

a 1:200 0001 cm:200 000 cm1 cm = 2000 m1 cm = 2 km

Multiply by 2.5 to go from 2 km to 5 km and do it with both sides.

Write your answer. On a 1:200 000 map, 5 km is represented as 2.5 cm.

b Use 1 cm = 2000 m and divide both sides by 4 to go from 2000 m to 500 m.

Convert or 0.25 cm to mm.

b

0.25 cm = 500 m2.5 mm = 500 m

Write your answer. On a 1:200 000 map, 500 m is represented by 2.5 mm.

1

2 1 cm = 2 km

x cm = 5 km

2.5 cm = 5 km

× 2.5 × 2.5

3

1

214---

1 cm = 2000 m

x cm = 500 m

4 4÷ ÷

3

23WORKEDExample

A′B′AB

------------ B′C′BC

------------ C′D′CD

------------ A′D′AD

------------= = =

B'C'

D'A'

4

2

6

2

2

1

1

3

BC

DA

125°60°

85°

BC

DA

B'C'

D'A'

125°60°

85°



Scale factor, kA measure of the relative size of the two similar figures is the scale factor. The scale factoris the common ratio of the corresponding sides and quantifies the amount of enlargementor reduction one figure undergoes to transform into the other figure. The starting shape iscommonly referred to as the original and the transformed shape as the image.

1. Scale factor, k, is the amount of enlargement or reduction and is expressed as integers, fraction or map scale ratios.

For example, k = 2, k = or 1:10 000.

2. Scale factor, k = =

where for enlargements, k is greater than 1 and for reductions, k is between 0 and 1.

3. For k = 1, the figures are exactly the same shape and size and are referred to as congruent.

Enlargements and reductions are important in many aspects of photography, mapmaking and modelling. Often, photographs are doubled in size (enlarged), while houseplans are an example of a reduction to a scale, for example 1:25.

112------

length of imagelength of original-------------------------------------------- A′B′

AB------------ B′C′

BC------------ C′A′

CA------------= =

B'

A' C'

9 9

3

B

A C

3

1

3

For the similar shapes shown at right:a find the scale factor for the reduction of the shapeb find the unknown length in the small shape.

THINK WRITE

a As it is a reduction, the larger shape is the original and the smaller shape is the image.

a

Original

20 cm

45 c

m

10 cmImage

x

1

24WORKEDExample

Continued over page



THINK WRITE

The two shapes have been stated as being similar, so set up the scale factor ratio, k.

Scale factor, k =

=

=

=

b Use the scale factor to determine the unknown length as all corresponding lengths are in the same ratio.

b Scale factor, k =

k =

=

x = × 45 cm

x = 22.5 cmWrite your answers. The scale factor of reduction is and the

unknown length on the smaller shape is 22.5 cm.

2length of image

length of original-----------------------------------------

A′B′AB

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

10 cm20 cm---------------

12---

112---

length of imagelength of original-----------------------------------------

12---

x45 cm--------------

12---

212---

a Prove that the figures given below are similar.b Given that the scale factor is 2, find the lengths of the two unknown sides s and t.

THINK WRITE

a Firstly, orientate the figures to identify corresponding sides and angles easily. Calculate the missing angles and compare each pair of corresponding angles.

a

Sum of interior angles = 360°All corresponding angles are equal.

40°

30°

20°70 m

100m

s

40°

30°

20°

t

50 m

30 m

40°

30°

20°70 m

100m

s

Image Original

270°

270°40°

30°

20°t

50 m30

m25WORKEDExample



THINK WRITE

b As the scale factor given is for enlargements, the original is the smaller figure.

b Scale factor, k =

For s 2 =

Set up the scale factor ratio for each of the two sides.

s = 2 × 30 m = 60 m

For t 2 =

t =

= 35 m

1length of image

length of original-----------------------------------------

s30 m------------

2

70 mt

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

70 m2

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

rememberMaps and scales

Map scales can be stated as:1. A ratio scale. For example, 1:100 means that 1 unit on the map represents

100 units in real life.

2. A simple conversion scale. For example, 1 cm = 100 m means 1 cm on the map represents 100 metres in real life.

Similar figures

For two figures to be similar, they must have the following properties:1. The ratios of the corresponding sides must be equal.

= common ratio

2. The corresponding angles must be equal.∠A = ∠A′ ∠B = ∠B′ ∠C = ∠C′ ∠D = ∠D′

Scale factor, k

1. Scale factor, k, is the amount of enlargement or reduction and is expressed as integers, fractions or map scale ratios, for example k = 2, k = or 1:10 000.


where for enlargements, k is greater than 1 and for reductions, k is between 0 and 1.

3. For k = 1, the figures are exactly the same shape and size and are referred to as congruent.

B'C'

D'A'

4

2

6

2

2

1

1

3

BC

DA

125°60°

85°

BC

DA

B'C'

D'A'

125°60°

85°

A′B′AB

------------ B′C′BC

------------ C′D′CD

------------ A′D′AD

------------= = =

112------

length of imagelength of original----------------------------------------- A′B′

AB------------ B′C′

BC------------ C′A′

CA------------= = B'

A' C'

9 9

B

A C

3

1

3

3

remember



Maps and similar figures

1 Convert the following map ratio scales to simple conversion scales with cm as the unitof measure.

2 State the real-life distance represented on a map for each of the following:

3 State the distance on a map for each of the following:

4 For each of these pairs of similar shapes, find:

i the scale factor ii the value of x and y.

a 1:500 000 b 1:1000 c 1:125 000

d 2:40 000 e 1:1 750 000 f 1:500

a 22 cm on a 1 cm = 1.5 km map b 8.5 cm on a 1 cm = 200 m map

c 8 mm on a 1 mm = 100 m map d 13 cm on a 1:750 000 map

e 17 cm on a 1:20 000 map f 25 mm on a 1:200 000 map.

a 4 km on a 1:100 000 map b 750 m on a 1:25 000 map

c 100 km on a 1:200 000 map d 25 m on a 1:500 map

e 300 m on a 1:150 000 map f 12 km on a 1:750 000 map.

a b

c d

8EWORKEDExample

21

Mathca

d

Scalefactor

WORKEDExample

22

WORKEDExample

23

WORKEDExample

24

50

70

200 cm

x cm

y cm

50

x cm

20 cm

y cm

4 m

1 m

25 m

etre

s

2 cm

8 cm

8 cm

4 cmx cm

y cm

y mm 42 mm

63 m

m

21 mm

x

7 mm

Ch 08 FM YR 12 4 Friday, November 10, 2000 11:36 AM

C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 3855 Prove that the following pairs are similar figures and find the value of a.

6 A photo has the dimensions 10 cm by 12 cm. The photo is enlarged by a factor of 2.5.Find the new dimensions of the photo.

7 Most scale model cars are in the ratio 1:12. Find:a the length of a real car if the model is 20 cm long (in metres to 1 decimal place)b the height of a real car if the model is 3 cm high (to the nearest centimetre)c the length of a model if the real car is 3 metres long.

8 The dimensions of a student’s room are 4300 mm by 3560 mm. An appropriate scaleto draw a scale diagram on an A4 sheet is 1:20. Find the dimensions of the scale drawing of the room and state whether the drawing should be landscape or portrait on the A4 sheet.

9 The map at right uses a line scale.a Convert the line scale to

a simple conversion scale.b State the map scale ratio.c Find the straight-line

distances between:i McLeod and Thomasii McLeod and Clowesiii Sharpe and Thomas.

a b

c d

WORKEDExample

25

17 mm

17 mm

a

8 mm

1–2

30°

41 c

m

38 cm 37 cm

48 cm

45°15°

60°

22 cm

36 cm

15°

60°45°

30°

24 cm

a

Height of person = 186 cm

a

Photo

40 m

m

62 m

m 24

33°

7.5

12

30

32

40

10

16a°

DAVIS LAND

Blazing

Gold mine

Coal mine

Copper mine

Silver mine

Temple

River

Rome

River

KILOMETRES0 10 20 30 40

N

Temple

Thomas Badger

Parry

Clowes

Sharpe

McLeod

Danby

WestSea



10 Find the distance between the following pairs of locations in the map (to the nearestkilometre).

a From Brambletown to Ross in a straight lineb From Charleston to Markham in a straight linec From Shelly Beach to Baletta in a straight lined From Charleston to Ross in a straight linee From Charleston to Ross via the roads.

(Hint: Use a length of string to measure the distance.)

11 Using the map from question 10, state which town(s) is/are within 15 kilometres ofBrambletown.

12The perimeter of the real object shown in the scale diagram of 1:25 is:A 464 cm B 514 cm C 357 cm D 14.28 cm E 150 cm

13A 1:27 scale model of a truck is made from clay. What is the length of the tray on theoriginal truck, if it is 27 cm length on the model?

14A scale factor of 0.2 is:A a reduction with a scale of 1 cm = 2 cmB an enlargement with a scale of 1 cm = 0.2 cmC an enlargement with a scale of 1 cm = 5 cmD a reduction with a scale of 1 cm = 5 cmE a reduction with a scale of 1 cm = 20 cm

A 1 cm B 100 cm C 270 cm D 540 cm E 729 cm

N

SCALE 1:1 000 000

Kantar

Reneton

Paxton

BRAMBLETOWN

MartinaRive

rStuckley

Sea

Shelly Beach

FosterPlains

Bolivia

1253 mGoldernSea

Baletta

Markham

Charleston

Jewel

3014 m

Ross

Swing

RiverS

nake

Riv

erNewbury

2750 m


4 cm

2 cmmmultiple choiceultiple choice




Similar trianglesSimilar triangles can be used to find the height of trees and buildings or widths of riversand mountains. One extra rule can be used to identify similar triangles to those men-tioned for similar shapes in the previous section. Two triangles are similar if one of thefollowing conditions is identified:1. All three corresponding angles

are equal (AAA).

2. All three corresponding pairs of sides are in the same ratio (linear scale factor) (SSS).

3. Two corresponding pairs of sides are in the same ratio and the included angles are equal (SAS).

As in the previous section, we use the known values of a pair of corresponding sidesto determine the scale factor for the similar triangles.

Scale factor, k = =

6

3

21

2 4sf = = = = 0.51

2--- 2

4--- 3

6---

2

3

6

4

sf = = = 263--- 4

2---

OA′OA---------- length of side of image

length of corresponding side of original----------------------------------------------------------------------------------------------------

For the similar triangles in the diagram, finda the scale factorb the value of the pronumeral, x.

THINK WRITE

a Identify that the two triangles are similar because they have equal angles (AAA). The third angle is not given but use the rule that all angles in a triangle sum to 180°.

a

B'

C'

6

A' x

100°

30°

B

C

4

6A

100°30°

1

B'

C'

6

A' x

100°

30° 50°

B

C

4

6A

100°30° 50°

Original

26WORKEDExample

Continued over page

Image



THINK WRITE

Always select the triangle with the unknown length, x, as the image. Evaluate the scale factor by selecting a pair of corresponding sides from the two triangles with known lengths.

Scale factor,

k =

=

=

= 1.5

b Use the scale factor to find the unknown length, x.Transpose and evaluate.

b Scale factor, k = 1.5

1.5 =

1.5 =

x = 1.5 × 6x = 9

Write answer in the correct units and level of accuracy.

The scale factor is 1.5 and the unknown length, x, is 9 units.

2

length of side of imagelength of corresponding side of original-----------------------------------------------------------------------------------------------

A′B′AB

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

64---

1

A′C′AC

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

x6---

2

For the given similar triangles, find the value of the pronumeral, x.

THINK WRITE

Confirm that the two triangles are similar because they have equal angles (AAA). This conclusion is supported by the parallel lines shown and using corresponding law and common angle, ∠A.

For clear analysis separate the two triangles. Note that the lengths of the sides and are the sum of the given values.

= 4.0 + 3.5= 7.5 m

= (7 + x) m

4.0

7

3.5

CA

All measurements in m

E

B

D

x

1

(7 + x) m

7.5 m

A

E

D

4.0 m

7 mC

A

B Original

Image

2

AE AD

AD

AE

27WORKEDExample



There are many practical applications of similar triangles in the real world. It is particu-larly useful for determining the lengths of inaccessible features such as the height oftall trees or the widths of rivers. This problem is overcome by setting up a trianglesimilar to the feature to be examined, as shown in the next example.

THINK WRITESelect as the image the triangle with the unknown length. Evaluate the scale factor by selecting a pair of corresponding sides from the two triangles with known lengths.

Scale factor,

k =

=

=

k = 1.875Use the scale factor to find the unknown length.Transpose and evaluate.

Scale factor,

k =

1.875 =

1.875 =

1.875 × 7 = 7 + x13.125 = 7 + x

x = 13.125 − 7x = 6.125

Write answer in the correct units and level of accuracy.

The value of x is 6 metres.

3


AD

AB--------

7.54.0-------

4


AE

AC--------

7 x+7

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

518---

Find the height of the tree shown in the diagram at right.Give the answer to 1 decimal place.

THINK WRITE

Confirm that the two triangles are similar because they have equal angles (AAA). This conclusion is supported by parallel lines, assuming the tree and the girl are perpendicular to the ground and using corresponding law and common angle, ∠A.

Sun's ray

s

Girl(168 cm)

Shadow(140 cm)

14 metres

1

14 m

x m140 cm

168 cm

Original

28WORKEDExample

Continued over page

Image



THINK WRITE

For clear analysis separate the two triangles.

Select the triangle with the unknown length as the image. Evaluate the scale factor by selecting a pair of corresponding sides from the two triangles with known lengths.Note: All measurements should be in the same units, preferably in metres.

Scale factor, k =

k =

10 =

Transpose and evaluate. x = 10 × 1.68 = 16.8 m

Write answer in the correct units. Height of the tree is 16.8 metres.

2

3height of tree (image)

height of girl (original)-------------------------------------------------------

141.4------- x

1.68----------=

x1.68----------

4

5

rememberSimilar triangles


2. Two triangles are similar if one of the following conditions is identified:(a) All three corresponding angles are equal (AAA).

(b) All three corresponding pairs of sides are in the same ratio (linear scale factor) (SSS).

(c) Two corresponding pairs of sides are in the same ratio and the included angles are equal (SAS).

OA′OA ---------- length of side of image

length of corresponding side of original-----------------------------------------------------------------------------------------------

6

3

21

2 4

sf = = = = 0.51—2

2—4

3—6

6

8

3

4

remember



Similar triangles

1 State the rule (SSS or AAA or SAS) that proves the pair of triangles are similar anddetermine the scale factor (expressed as an enlargement k > 1).

2 For the given similar triangles, find the value of the pronumeral, a.

3 For the given similar triangles, find the value of the pronumeral, a.

a b c

d e f

a b c

d e f

a b c

8FWORKEDExample

26a

640 m

m

320 mm25°

480 m

m

240 mm25°

5.6

4.4

4.6

8.8

9.2

11.2

4.5

5

9

10

7.010.5

14 3

42

10.5

7

3.5

10.5

WORKEDExample

26bCabri Geometry

Similartriangles

22.5 mm

a m

m

62° 62°

56°

20 m

m

15 mm62° 62°

56°

12 m 14.4 m

a m

15 m

59 cm

45 cm

75 cm

38°

25 cm38°

a

67°

71°7

68a

14

1612

12

4

x9

a x

3.2

9.6

a

13

xx

6

7.8

WORKEDExample

27

12

3

6

a

a

8

12

2

a

10.5

7.5

4.5



4 Find the height of the flagpole shown in the diagram at right (to the nearest centimetre).

5 Find the length of the bridge, , needed to span the river, using similar triangles as shown (to the nearest decimetre).

6 The shadow of a tree is 4 metres and at the same time the shadow of a 1 metre stick is25 cm. Assuming both the tree and stick are perpendicular to the horizontal ground,what is the height of the tree?

7 Find the width of the lake (to the nearest metre) using these surveyor’s notes at right.

8In the given diagram, the length of side b is closest to:A 24B 22C 16D 15E 9.6

Questions 9 and 10 refer to the following information. A young tennis player’s serve is shown in the diagram. Assume the ball travels in a straight line.

9The height of the ball just as it is hit, x, is closest to:A 3.6 m B 2.7 m C 2.5 m D 1.8 m E 1.6 m

10The height of the player, y, as shown is closest to:A 190 cm B 180 cm C 170 cm D 160 cm E 150 cm

d e f

43

17.2

15.2

a

17 m

a m

68 m

18 m

142 mm

108 mm

324 mm

a80°

80°

WORKEDExample

28

0.9 m

9 m1 m

Guy wire

AB

4.3 m12.5 m2.5 m

A

B

(Not to scaleAll measurementare in metres)

1.2 m

2 m

25 m

A

BLake


20

16 12

b

Not to scalex

y1.

1 m

5 m 10 m

0.9 mmmultiple choiceultiple choice


WorkS

HEET 8.2



Area and volume scale factorsAn unknown area or volume of a figure can be found without the need to use knownformulas such as in exercises 8B and 8D. We have seen that two figures that are similarhave all corresponding lengths in the same ratio or (linear) scale factor, k. The same canbe shown for the area and volume of two similar figures.

Area of similar figuresIf the lengths of similar figures are in the ratio a:b or k, then the areas of the similarshapes are in the ratio a2:b2 or k2. Following are investigations to support this relationship.

Different length ratios (or scale factors) of a square

Different length ratios (or scale factors) of a circle

Area = πr 2 = 1π cm2

Area = πr2 = 4π cm2

Area = πr2 = 9π cm2

From above, as long as two figures are similar then the area ratio or scale factor isthe square of the linear scale factor, k. The same applies for the total surface area.

Area scale ratio or factor (asf) =

= square of linear scale factor (lsf)

= (lsf)2

= k2

1 cmArea

=1 cm2

1 cm

3 cm Area = 9 cm2

3 cm

2 cm Area = 4 cm2

2 cm

length of blue squarelength of red square-------------------------------------------------- 2 cm

1 cm----------- 2 k= = =

area of blue squarearea of red square--------------------------------------------- 4 cm2

1 cm2-------------- 4 22 k2= = = =

length of green squarelength of red square

----------------------------------------------------- 3 cm1 cm----------- 3 k= = =

area of green squarearea of red square

------------------------------------------------ 9 cm2

1 cm2-------------- 9 32 k2= = = =

radius length of blue circleradius length of red circle

------------------------------------------------------------------ 2 cm1 cm----------- 2 k= = =

1 cm

area of blue circlearea of red circle------------------------------------------- 4π cm2

1π cm2------------------ 4 22 k2= = = =

2 cm

radius length of green circleradius length of red circle

--------------------------------------------------------------------- 3 cm1 cm----------- 3 k= = =

3 cm

area of green circlearea of red circle

---------------------------------------------- 9π cm2

1π cm2------------------ 9 32 k2= = = =

area of imagearea of original------------------------------------



The steps required to solve for length, area or volume (investigated later) usingsimilarity are:1. Clearly identify the known corresponding measurements (length, area or volume) of

the similar shape.2. Establish a scale factor (linear, area or volume) using known measurements.3. Convert to an appropriate scale factor to determine the unknown measurement.4. Use the scale factor and ratio to evaluate the unknown.

For the 2 similar triangles shown, find the area, x cm2, of the small triangle.

THINK WRITE

Determine a scale factor, in this instance the linear scale factor, from the two corresponding lengths given. It is preferred that the unknown triangle is the image.

Linear scale factor =

k =

=

Determine the area scale factor. Area scale factor = k2

=

=

Use the area scale factor to find the unknown area.

Area scale factor =

=

Transpose the equation to get unknown by itself.

x = × 100x = 25

Write your answer. The area of the small triangle is 25 cm2.

Area = x

2.4 cm

Area = 100 cm2

4.8 cm

1length of small triangle (image)length of large triangle (original)-------------------------------------------------------------------------------

2.4 cm4.8 cm----------------

12---

212---

2

14---

3area of small triangle (image)area of large triangle (original)--------------------------------------------------------------------------

14---

x cm2

100 cm2--------------------

414---

5

29WORKEDExample

For the two similar shapes shown, find the unknown length, x cm.

THINK WRITE

Determine a scale factor, in this instance the area scale factor, as both areas are known. It is preferred that the triangle with the unknown is stated as the image.

Area scale factor =

k2 =

= 25

10 cm2

2 cm

250 cm2

x

1area of image (large trapezium)

area of original (small trapezium)--------------------------------------------------------------------------------

250 cm2

10 cm2--------------------

30WORKEDExample



Volume of similar figuresIf the lengths of similar figures are in the ratio a:b or k, then the volume of the similarshapes are in the ratio a3:b3 or k3. The following is an investigation of two differentobjects, cubes and rectangular prisms.

A cube

A rectangular prism

From above, as long as two figures are similar then the volume ratio or scale factor is the cube of the linear scale factor, k.

Volume scale factor (vsf) =

= cube of linear scale factor (lsf)

= (lsf)3

= k3

THINK WRITE

Determine the linear scale factor.


k = k = 5

Use the linear scale factor to find the unknown length.


5 =


x = 5 × 2x = 10

Write your answer. The length, x, is 10 cm.

2 k2

25

3length of image (large trapezium)

length of original (small trapezium)-------------------------------------------------------------------------------------

x cm2 cm------------

4

5

Volume = 1 × 1 × 1 = 1 cm 3

1 cm1 cm

1 cm

Volume =2 × 2 × 2= 8 cm3

2 cm

2 cm2 cm

length of large (blue) cubelength of small (red) cube--------------------------------------------------------------- 2 cm

1 cm----------- 2 k= = =

volume of large cubevolume of small cube--------------------------------------------------- 8 cm2

1 cm2-------------- 8 23 k3= = = =

Volume =1 × 1 × 3= 3 cm3

3 cm1 cm

1 cm

Volume =2 × 2 × 6= 24 cm3

6 cm2 cm

2 cm

length of small prismlength of large prism--------------------------------------------------- 3 cm

6 cm----------- 1

2--- k= = =

volume of small prismvolume of large prism------------------------------------------------------ 3 cm3

24 cm3----------------- 1

8--- 1

2---

3k3= = = =

volume of imagevolume of original--------------------------------------------



We can use the relationship between linear, area and volume scale factors to find anyunknown in any pair of similar figures as long as a scale factor can be established.

1. Given linear scale factor (lsf) = k area scale factor = k2 volume scale factor = k2

For example: = 2 = 22 = 4 = 23 = 8

2. Given area scale factor (asf) = k2 linear scale factor = volume scale factor = k3

For example: = 4 k = = 2 = 23 = 8

3. Given volume scale factor (vsf) = k3 linear scale factor = area scale factor = k2

For example: = 8 k = = 2 = 22 = 4

For the two similar figures shown, find the volume of the smaller cone.

THINK WRITE

Separate the two figures to clarify the details of the similar figures.

Determine a scale factor, in this instance the linear scale factor, from the two corresponding lengths given. It is preferred that the unknown triangle is the image.


k =

=

Determine the volume scale factor.

Volume scale factor = k3

k =

k =

Use the volume scale factor to find the unknown length.

Volume scale factor =

=

Transpose the equation to get the unknown by itself.

x = × 540x = 160

Write your answer. The volume of the smaller cone is 160 cm3.

Volume oflarge cone= 540 cm3 9 cm6 cm

1 Volume =540 cm3

Volume =x cm3

9 cm6 cm

2length of small triangle (image)length of large triangle (original)-------------------------------------------------------------------------------

6 cm9 cm-----------

23---

323---

3

827------

4volume of small cone (image)volume of large cone (original)---------------------------------------------------------------------------

827------

x cm3

540 cm3--------------------

5827------

6

31WORKEDExample

k2

4

k33

83



For two similar triangular prisms with volumes of 64 m3 and 8 m3, find the total surface area of the larger triangular prism, if the smaller prism has a total surface area of 2.5 m2.

THINK WRITE

Determine a scale factor, in this instance the volume scale factor, from the two known volumes. It is preferred that the larger unknown triangular prism is stated as the image.

Volume scale factor =

k3 =

k3 = 8

Determine the area scale factor. For ease of calculation, change volume scale factor to linear and then to area scale factor.


k = = 2

Area scale factor = k2

= 22

= 4

Use the area scale factor to find the total surface area.

Area scale factor =

4 =


x = 4 × 2.5x = 10

Write your answer. The total surface area of the larger triangular prism is 10 m2.

1volume of larger prism (image)

volume of smaller prism (original)-----------------------------------------------------------------------------------

64 m3

8 m3--------------

2 k33 k=

83

3area of larger prism (image)

area of smaller prism (original)---------------------------------------------------------------------------

x m2

2.5 m2----------------

4

5

32WORKEDExample

rememberArea and volume scale factors

The steps required to solve for length, area or volume using similarity are:1. Clearly identify the known corresponding measurements (length, area or

volume) of the similar shape.2. Establish a scale factor (linear, area or volume) using known pairs of

measurements.3. Convert to an appropriate scale factor to determine the unknown measurement.4. Use the scale factor and ratio to evaluate the unknown.

Area scale factors

Area scale ratio or factor (asf) =

= square of linear scale factor (lsf)= k2

Volume scale factor

Volume scale ratio or factor (vsf) =

= cube of linear scale factor (lsf)= k3



remember



Area and volume scale factors

1 Complete the following table of values.

2 Find the unknown area of the following pairs of similar figures.

3 a Find the unknown length of the following pairs of similar figures.

Linear scale factorsk

Area scale factorsk2

Volume scale factorsk3

2 8

16

3

125

100

64

0.027

36

0.1

100

0.16

400

a b

c d

i ii

8G

Mathca

d

Area andvolumescalefactors

WORKEDExample

29

15 mm 22.5 mm

540 mm2

x mm2

48 c

m

8 cm

12 cm2

x cm2

2 m7 m

122.5 mx m2

214 mm

Surface area = x mm2

Surface area = 100 mm2

21 mm

WORKEDExample

30

Area =1.0 m2

x m1.7 m

Area =6.25 m2

Area =750 cm2

25 cm

Area = 3000 cm2

x


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 399b Two similar trapezium-shaped strips of land have an area of 0.5 hectares and

2 hectares. The larger block has a distance of 50 metres between the parallel sides.Find the same length in the smaller block.

c Two photographs have areas of 48 cm2 and 80 cm2. The smaller photo has a widthof 6 cm. Find the width of the larger photo.

4 Find the unknown volume in the following pairs of similar objects.

5 a For the 2 similar triangular pyramids with volumes of 27 m3 and 3 m3, find thetotal surface area of the larger triangular prism if the smaller prism has a total sur-face area of 1.5 m2.

b For a baseball with diameter of 10 cm and a basketball with a diameter of 25 cm,find the total surface area of the baseball if the basketball has a total surface areaof 1963.5 cm2.

c For a 14 inch car tyre and 20 inch truck tyre that are similar, find the volume (tothe nearest litre) of the truck tyre if the car tyre has a volume of 70 litres.

d For 2 similar kitchen mixing bowls with total surface areas of 1500 cm2 and3375 cm2, find the capacity of the larger bowl if the smaller bowl has a capacity of1.25 litres (to the nearest quarter of a litre).

6 a Find the volume of the b Find the volume of the larger small cone. triangular pyramid

a b

c d

WORKEDExample

31

2400 cm3

x cm3

14 cm

7 cm

Volume of small pyramid= 40 cm

12 cm2 cm

3

Volume of large sphere= 8 litres

Volume= 1200 cm3

30 cm

45 cm

WORKEDExample

32

Volume oflarge cone= 270 cm3

Area = 45 cm2

Area =5 cm2

TSA of small pyramid= 200 cm2

Volume of small pyramid= 1000 cm3

TSA of large pyramid= 288 cm2



c Find total surface area of d Find the diameter of the the small prism small cylinder.

7 A plan of a holiday bungalow has a scale of 1 cm = 50 cm.a What is the area of the plan?b Express the drawing scale as a linear scale factor.c Using similarity, find the actual area of the bungalow

(in m2 to 2 decimal places).d What is the area scale factor (k2)?

Area = 12 cm2

TSA =78 cm2

TSA =x cm2

Area =6 cm2

Volume= 1280 cm3

Volume= 20 cm3

12 cm

x cm

12 cm

10 cm5 cm

8 cm


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 4018 What is the area ratio of:

a two similar squares with side lengths of 3 cm and 12 cm?b two similar circles with diameters of 9 m and 12 m?c two similar regular pentagons with sides of 16 cm and 20 cm?d two similar right-angled triangles with bases of 7.2 mm and 4.8 mm?

9 Find the volume ratios from the similar shapes given in question 8.

10 Find the total surface area of the small cone as given in the diagram.

11 A 1:12 scale model of a car is created from plaster and painted.a If the actual car has a volume of 3.5 m3, find the amount of

plaster needed for the model to the nearest litre.b The model needed 25 millilitres of paint. How much paint would be needed for

the actual car (in litres to 1 decimal place)?

12 Find the ratios of the volume of 2 cubes whose sides are in the ratio of 3:4.

13 An island in the Pacific Ocean has an area of 500 km2. What is the area of itsrepresentation on a map drawn to scale of 1 cm = 5 km?

14 Two statutes of a famous person used 500 cm3 and 1.5 litres of clay. The smallerstatue stood 15 cm tall. What is the height of the other statue (to the nearestcentimetre)?

15 The ratio of the volume of two cubes is 27:8. What is the ratio of:a the lengths of their edges?b the total surface area?

16 The radius of one sphere is equal to the diameter of another sphere. Find the ratio ofthe small sphere to the large sphere:a for total surface areab for volume.

17 A cone is half-filled with ice-cream. What is the ratio of ice cream to empty space?

18A 1:27 scale model of a truck is made from clay. The ratio of volume of the model tothe real truck is:

19The ratio of the volume of the blue portion to the volume of the red portion is:A 1:3B 1:8C 1:9D 1:26E 1:27

20A 1:100 scale model of a building is a cube with sides of 100 cm. The volume of thereal building is:

A 1:3 B 3:1 C 1:9 D 1:729 E 1:19 683

A 10 000 000 m3 B 1 000 000 m3 C 100 000 m3

D 10 000 m3 E 1000 m3

TSA oflarge cone= 840 cm2



h

3h




Properties of angles, triangles and polygons• Draw careful diagrams.• Carefully interpret geometric notations, such as

the diagram at right.• Carefully consider geometric rules, such as

isosceles triangles have 2 equal sides and angles.

Area and perimeter• Perimeter is the distance around a closed figure.• Circumference is the perimeter of a circle.

C = 2 × π × radius= 2πr

• Area is measured in mm2, cm2, m2, km2 and hectares.• 1 cm2 = 10 mm × 10 mm = 100 mm2

1 m2 = 100 cm × 100 cm = 10 000 cm2

1 km2 = 1000 m × 1000 m = 1 000 000 m2

1 hectare = 10 000 m2

• Area of shapes commonly encountered are:1. Area of a square: A = l2

2. Area of a rectangle: A = l × w3. Area of a parallelogram: A = b × h4. Area of trapezium: A = (a + b) × h

5. Area of a circle: A = πr2

6. Area of a triangle: A = × b × h• Area of composite figure = sum of the individual common figures

Acomposite = A1 + A2 + A3 + A4 + . . .

Total surface area (TSA)• Total surface area (TSA) is measured in mm2, cm2, m2 and km2.• The TSAs of some common objects are as follows:

1. Cubes: TSA = 6l2

2. Cuboids: TSA = 2(lw + lh + wh)3. Cylinders: TSA = 2πr(r + h)4. Cones: TSA = πr(r + s) where s is the slant height5. Spheres: TSA = 4πr2

• For all other objects, form their nets and establish the total surface area formulas.

Volume of prisms, pyramids and spheres• Volume is the amount of space occupied by a 3-dimensional object.• The units of volume are mm3, cm3 (or cc) and m3.

1. 1000 mm3 = 1 cm3

2. 1 000 000 cm3 = 1 m3

3. 1 litre = 1000 cm3

4. 1000 litres = 1 m3

summaryEqual sides

12---

12---


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 403• Volume of a prism, Vprism = area of uniform cross-section × height

V = A × H

• Volume of a pyramid, Vpyramid = × area of cross-section at the base × height

V = × A × H

• The height of a pyramid, H, is sometimes call the altitude.

• Volume of a sphere is Vsphere = πr3

• Volume of a composite object = sum of the individual common prisms, pyramids or spheres.

Vcomposite = V1 + V2 + V3 + . . .or Vcomposite = V1 − V2 . . .

Maps and scales

• Ratio scale, for example 1:100, means that 1 unit on the map represents 100 units in real life.

• A simple conversion scale, for example 1 cm = 100 m, means 1 cm on the map represents 100 metres in real life.

13---

13---

43---

SCALE 1:50 000METRES 1000 0 1 2 3 KILOMETRES

Kilometres 0 1 2 3 4 5 6 7 8 Kilometres



Similar figures• Two objects that have the same shape but different size

are said to be similar.• For 2 figures to be similar, they must have the following

properties:(a) The ratios of the corresponding sides must be equal.

= common ratio

(b) The corresponding angles are equal.∠A = ∠A′ ∠B = ∠B′ ∠C = ∠C′ ∠D = ∠D′

Scale factor, k

• Scale factor, k =

where for enlargements, k is greater than 1 andfor reductions, k is between 0 and 1.

• For k = 1, the figures are exactly the same shape and size and are referred to as congruent.

Similar triangles• Two triangles are similar if one of the following conditions is identified:

1. All 3 corresponding angles are equal (AAA).2. All 3 corresponding pairs of sides are in the same ratio (linear scale factor)

(SSS).3. Two corresponding pairs of sides are in the same ratio and the included angles

are equal (SAS).

Area and volume scale factors• The steps required to solve for length, area or volume using similarity are:

1. Clearly identify the known corresponding measurements (length, area or volume) of the similar shapes.

2. Establish a scale factor (linear, area or volume) using known pairs of measurements.

3. Convert to an appropriate scale factor to determine the unknown measurement.4. Use the scale factor and ratio to evaluate the unknown.

Area scale factor

• Area scale ratio or factor (asf) =

= square of linear scale factor (lsf)= k2

Volume scale factor

• Volume scale ratio or factor (vsf) =

= cube of linear scale factor (lsf)= k3

B'C'

D'A'

4

2

6

2

2

1

1

3

BC

DA

125°60°

85°

BC

DA

B'C'

D'A'

125°60°

85°

A′B′AB

------------ B′C′BC

------------ C′D′CD

------------ A′D′AD

------------= = =

length of imagelength of original----------------------------------------- A′B′

AB------------ A′B′

AB------------ B′C′

BC------------ C′A′

CA------------= = = =

B'

A' C'

9 9

B

A C

3

1

3

3





Multiple choice

1 For the triangle shown in a semicircle, x is:A 32°B 58°C 68°D 90°E none of the above

2 A triangle LABC has the following values given. = 10 cm, = 12 cm where and are perpendicular. The area of the triangle is

3 The area of the kitchen bench shown in the plan is closest to:A 1250π + 19 600 cm2

B 1250π + 37 600 cm2

C 1250π + 29 600 cm2

D 2500π + 29 600 cm2

E 30 100 cm2

4 The total surface area of a closed cylinder with a radius of 40 cm and a height of 20 cm is given by:

5 The net of an object is shown in the diagram. An appropriate name for the object is:A rectangular prismB rectangular pyramidC triangular prismD triangular pyramidE trapezium prism

6 The volume of a sphere with a diameter of 15 cm is closest to:

7 The volume of the composite object, given that = 10 cm is closest to:A 1000 cm3

B 1300 cm3

C 1500 cm3

D 2000 cm3

E 10 000 cm3

A 120 cm2 B 30 cm2 C 240 cm2 D 121 cm2 E 60 cm2

A 2 × π × 20 × (40) B 2 × π × 40 × (40) C 2 × π × 40 × (100)D 2 × π × 40 × (60) E 2 × π × 20 × (60)

A 560π cm3 B 900π cm3 C 4500π cm3

D 4500π cm3 E 36 000π cm3

CHAPTERreview

8A

x°32°

8BAB AC ABAC

8B

200

220

50

80

Allmeasurements

in cm

8C

8C

8D

8DVO

O

V



8 A map ratio scale of 1:150 000 expressed as a simple conversion scale is:

9 In the triangle shown, the value of c is:A 3B 6C 9D 12E 4

10 The circumference of the larger cone is closest to:

A 113 mmB 151 mmC 226 mmD 302 mmE 459 mm

11 The diagonal distance on the television screen is used to specify the different sizes available. If the height on a 51 cm television is 45 cm, then a similar 34 cm television has a height, h, which is closest to:

A 67 cm B 45 cm C 34 cm D 30 cm E 26 cm

12 The diagram at right shows the path of a pool ball into the middle pocket of a 12 by 6 billiard table. To achieve this, the expression for the value of x is:

A

B

C

D

E

A 1 cm = 15 m B 1 cm = 150 m C 1 cm = 1500 mD 1 mm = 1.5 km E 1 cm = 15 km

8E

8E 3

7.8

c2.6

8E189 mm

63 mm

24 mm

8F

h cm

45 c

m

34 cm 51 cm

8F

64--- 6 x–

x-----------=

46--- 6 x–

x-----------=

64--- x 6–

x-----------=

126

------ 6 x–x

-----------=

64--- 2 x+

x------------=

6 -

x6

124

6

x


C h a p t e r 8 G e o m e t r y : s i m i l a r i t y a n d m e n s u r a t i o n 40713 Jennifer is standing 2 metres directly in front of her bedroom

window which is 1 metre wide. The width (w) of her view of a mountain range 1 kilometre from her window is (to the nearest metre):

A 1002 metresB 1000 metresC 499 metresD 501 metresE 500 metres

14 The large cone is filled to one-third of its height with water as shown. The ratio of the volume of water to air is:

A 1:27B 1:26C 27:1D 1:9E 1:3

Short answer1 For each of the figures, find the value of the pronumeral.

2 Find the outer perimeter and area of the flower.

3 For the triangular prism:a Sketch an appropriate net for the given solid prism.b Transfer the units appropriately to the net from part a.c Calculate the total surface area of the object.

4 a What is the volume contained by the solid and framed sections (to 1 decimal place)?

b What is the volume of the solid part only?

a b

8F

1 m

w

2 m

1000

m

8G

8A

40°

b

a

c

ba

c

8B

r = 22 mm

r = 11 mm

8C

6 m5 m

3 m4 m

6 m

10 m



5 The dimensions of a rectangular prism tub are 30 cm by 20 cm by 15 cm. The tub is filled completely with water and then transferred into a cylinder tank that is 10 cm in radius and 40 cm tall. How high is the water level in the cylinder?

6 A plan of a region is to the scale 1:200 000.a If the distance on the map between 2 towns is 27 mm, find the actual distances between

the towns.b The distance between the fire station and the local airport is 2.4 km. Find the distance

represented on the plan.

7 Two ladders are placed against the wall at the same angle. The ladders are 2 metres and 3 metres long. If the taller ladder reaches 2.1 metres up the wall, how far up will the second ladder reach (to 1 decimal place)?

8 A yacht is an unknown distance from the shore. A family on the beach make the measurements as shown in the diagram at right. How far is it to the yacht (to the nearest metre)?

9 A plan is drawn to scale of 1:50 000. Find:a the length in centimetres on the plan that represents 1 kmb the area in hectares of a region represented by 4 cm2 on the planc the area on the plan of a region of 25 hectares.

AnalysisA rectangular block of modelling clay has dimensions of 30 cm by 20 cm by 10 cm.1 a What is the volume of the block of clay?

b Express in litres your answer from question 1 a.c What is the total surface area of the clay?

2 The entire block of clay is remoulded to a shape of a cylinder with a height of 30 cm.a Find the diameter of the cylindrical block of clay (to 2 decimal places).b Find the new total surface area of the clay when moulded as a cylinder (to nearest cm2).c What fraction of the volume needs to be removed to turn the cylindrical block into a cone

with the same diameter and height?

3 Clay is moulded to the shape at right to represent a 1:100 scale model of a grain silo.a Find the volume of clay needed to make a scale model grain silo

(to 1 decimal place).b Find the actual volume of the grain silo. Express your answer to

the nearest cubic metre.c What is the ratio of the volume of model to the volume of the

actual grain silo?d If the scale model has a total surface area of 143.14 cm2, find the

total surface area of the actual silo.

4 It is decided that another silo, half the size of the silo in question 3, is to be built. What fraction will this smaller silo be in volume compared to the larger silo?

8D

8E

8F

8F

10 m 1 m

6 m

8G

5 cm

6 cm

6 cm

4.5 cm

testtest

CHAPTERyyourselfourself

testyyourselfourself

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Geometry: similarity and mensuration -...

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