7a and polygons 7b 7c 7d geometry: 7e 7f 7g similar ...mathsbooks.net/jacplus books/12 further...

310

7a Properties of angles, triangles and polygons

7b Area and perimeter 7c Total surface area 7d Volume of prisms, pyramids

and spheres 7e Similar fi gures 7F Similar triangles 7G Area and volume scale factors

Pythagoras’ theorem in two and three dimensions • and the use and applications of similarityCalculation of surface area and volume of • regular and composite solids

Application of the effect of changing linear • dimensions (that is, if the linear scale factor is k, then the area scale factor is k2 and the volume scale factor is k3).

areas oF sTUdY

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7Geometry: similarity and mensuration

GeometryGeometry is an important area of study. Many professions and tasks require and use geometrical concepts and 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 polygonsIn this module, we will often encounter problems in which 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, defi nitions and laws of geometry. Some of the more important rules are presented in this chapter.

interior angles of polygons

Stairways

Bed 4 Bed 3

Bed 2

Bed 1

UPPERLEVEL

7a

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Interactivityint-0259

The sum of external angles of a

polygon

maths Quest 12 Further mathematics for the Casio Classpad

311Chapter 7 Geometry: similarity and mensuration

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

the interior angle is given by 180° - 360°

n . For example, for a

square the interior angle is:

180° - 360

4° = 180° - 90°

= 90°The exterior angle is given by

360°n

.

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

Think WriTe

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

Interior angle = 180° - 360

5°

= 180° - 72° = 108°

2 Substitute n = 5 into the exterior angle formula. Exterior angle = 360

5°

= 72°3 Write your answer. A regular pentagon has an interior angle of 108°

and an exterior angle of 72°.

Geometry rules, defi nitions and notation rulesThe following geometry rules and notation will be most valuable in establishing unknown values in the topics covered and revised in this module.

defi nitions of common terms

Less than 90

A

B C

ABC

Acute angle

90

Right angle

180

Straight angle Obtuse angle

Between 90and 180

Reflex angle

Between 180and 360

Parallel lines Perpendicular lines Line segment

Ray

LineA B

AB

AB

AB

A B

A B

Exterior angle

Interior angle

Worked example 1

312

some common notations and rules

Scalene triangle

b

a c

No equal sidesa b c 180

Isosceles triangle

Two equalsides and

angles

Equilateral triangle

All equalsides and

angles

60 60

60

Right-angledisoceles triangle

45

45

Complementary angles

a b 90

ab

Supplementary angles

a b 180

a b Vertically oppositeangles

a ba

b

CD is a perpendicular bisector of AB

A

C

D

B

Alternate angles

a c

a = bc = d

bd

Corresponding angles

ac

a bc d

bd

Co-interior angles

a c

a d 180b c 180

bd

ab

cd

a b c d 360

a

b

B

A C Dc d

a b d

BCD is an exterior angle.

Right angle at thecircumference in

a semicircle

Calculate the values of the pronumerals in the polygon shown.

c

d cm

6 cm

b

a

Think WriTe

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

a = 360

6°

= 60°

Worked example 2



2 The other two angles in the triangle are equal.

60

a + b + c = 180°b = cSo:60 + 2b = 180°b = 60°c = 60°

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

Calculate the missing pronumerals in the diagram of railings for a set of stairs shown below.

35

a bc

Think WriTe

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

35

35

a b

c

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

Given angle 35°.a = 35°

3 Using the corresponding angle law and the given right angle, recognise that the unknown angle c is a right angle.

c

c = 90°4 Use the straight angle rule to find the

unknown angle b.a + b + c = 180°35° + b + 90° = 180°b = 180° - 125°b = 55°

Worked example 3

314

Draw diagrams carefully.1. Carefully interpret geometric notation; for example, the triangle below has two sides 2. equal in length.

Equal sides

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

rememBer

properties of angles, triangles and polygons 1 We 1 Calculate the interior and exterior angles for each of the following regular polygons.

a Equilateral trianglec Hexagone Heptagonf Nonagon

b Regular quadrilaterald

g

2 We2 Calculate the value of the pronumerals in the following figures.

a 27

52 a

b

130

yx

c

63c

d

15c ab

e

c 50

b 8 cm

f

32m

3 We3 Calculate the value of the pronumerals in the following figures.

a

35x y

b 30

z0

exerCise

7a



c 62

t

d 70 b ca

d

e 27 m

n

f

140a

81

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 Calculate the unknown pronumerals.

a

4.2 cm

3.6

cm

h

r b

35x zy

110c

ab

cd

86

d

a

b

40

6 mC In the figure at right, the value of a is closest to:a 30° b 75° c 90°d 120° e 150°

7 mC An isosceles triangle has a known angle of 50°. The largest possible angle for this triangle is:a 80° b 130° c 90° d 65° e 50°

area and perimeterMuch of our world is described by area (the amount of space enclosed by a closed figure) 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.

exam Tip Many students provided answers without suitable working, usually in the Geometry module. Where the answer is correct, full marks are awarded, but without working, incorrect answers or consequential answers do not receive any marks.

[Assessment report 2 2005]

a

150

7B

Lot 658

761m2

13.05

23.55 5.86

32.1

8

32.7

5

Corner block with expansive 23.55 m frontage

$251000Lot 603645 m2

17

14.07

4.05

36.5

6

37.9

2

Corner blockwith wide

17 m frontage

$147000

316

perimeterPerimeter is the distance around a closed fi gure.

Some common rules are:1. For squares, the perimeter = 4l

l

l

l l

Square

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

l

l

w w

Rectangle

3. Circumference (C ) is the perimeter of a circle.

C = 2 × p × radius = 2pr

r

C

ircu

mfe

renc

e of a circle

Calculate the perimeter of the closed fi gure given at right (to the nearest mm).

Think WriTe

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

Perimeter = 300 + 2 × 600 + 12 circumference where

12 of circumference =

12 × 2pr

= p × 150 = 471.24

2 Add together the three components of the perimeter.

Perimeter = 300 + 2 × 600 + 471.24 = 1971.24

3 Write your answer. The perimeter of the closed fi gure is 1971 mm, correct to the nearest millimetre.

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

l

l

Square

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

l

w

Rectangle

Worked example 4

300

mm

600 mm



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

h

b

Parallelogram

4. Area of a trapezium: A = 12 (a + b) × h

h

a

b

Trapezium

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

r

Circle

6. Area of a triangle: A = 12 × b × h

(see the next chapter)

h

b

Triangle

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

Calculate the area of the garden bed given in the diagram (to the nearest square metre).

5.7 m

2.4

m

7.5 m

Think WriTe

1 The shape of the 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 =

1 Area of a trapezium1

Area of a trapezium

2(a + b) × h

2 Substitute and evaluate. = 12(7.5 + 5.7) × 2.4

= 12 × 13.2 × 2.4

= 15.84 m2

3 Write your answer. The area of the garden bed is approximately 16 square metres.

Composite areasOften a closed fi gure can be identifi ed as comprising two or more different common fi gures. Such fi gures are called composite fi gures. The area of a composite fi gure is the sum of the areas of the individual common fi gures.

Area of a composite fi gure = sum of the areas of the individual common fi gures Acomposite = A1 + A2 + A3 + A4 + . . .

Worked example 5

318

Calculate the area of the hotel foyer from the plans given at right (to the nearest square metre).

Think WriTe

1 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.

16 m

16 m

8 m

A2A1 A3

20 m

25 m

Area of foyer = A1 + A2 + A3

2 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 = 1

2 × b × h

= 12 × 16 × 20

= 160 m2

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

A3 = Area of half of a circle

= 12 × p × r2

= 12 × p × 82

= 100.53 m2

3 Add together all three areas for the composite shape.

Area of foyer = A1 + A2 + A3

= 160 + 400 + 100.53 = 660.53 m2

4 Write your answer. The area of the hotel foyer is approximately 661 m2.

Conversion of units of areaOften the units of area need to be converted, for example, from cm2 to m2 and vice versa.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

Convert 1.12 m2 to square centimetres (cm2).

Think WriTe

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

1.12 m2 = 1.12 × 1 metre × 1 metre

Worked example 6

25 m

8 m

20 m

102 1002 10002

102 1002 10002

mm2

Area

cm2 m2 km2

Worked example 7



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

= 1.12 × 100 cm × 100 cm = 11 200 cm2

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

Convert 156 000 metres2 to:a kilometres2 b hectares.

Think WriTe

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

1000 kilometre.

a 156 000 m2 = 156 000 × 1

1000 km × 1

1000 km

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

= 156 000

1000 1000×

= 0.156 km2

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

b 1 Conversion factor is 10 000 m2 equals 1 hectare; that is, 1 m2 equals 1

10 000 hectare.b 156 000 m2 = 156 000 × 1

10 000 hectares

= 15600010 000

hectares

= 15.6 hectares156 000 m2 = 15.6 hectares

2 Write the answer.

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

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

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

l

l

l l

Square

l

l

l l

l

l

w w

Rectangle

r

Cir

cum

fere

nce of a circle

rememBer

Worked example 8

320

Area is measured in mm2. 2, cm2, m2, km2 and hectares.(a) 1 cm3. 2 = 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 of shapes commonly encountered are:4. (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 = 1

2(a + b) × h

(e) Area of a circle: A = p r2

(f) Area of a triangle: A = 12 × b × h

Area of a composite fi gure 5. = sum of the areas of the individual common fi gures Acomposite = A1 + A2 + A3 + A4 + . . .

area and perimeter 1 We4 Calculate the perimeters of the following figures

(to the nearest whole units).

a

7 m

b 12 m

5 m4 m

c 23.7 cm

17.8

cm

15.4

cm

27.5 cm

d

13.5 mm

e

120

m

210

m90 m

70 m

2 We5 Calculate the areas of the closed figures in question 1 .

3 We6 Calculate the areas of the following figures (to 1 decimal place).

a

12 m

25 m

10 m14 m

20 m

b

2 m

3.5

m

exerCise

7B

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Digital docSkillSHEET 7.1

Substitution into a formula



c

12 m17

m

d

125 mm

24 mm90 mm48 mm

e

21 cm

16 cm

8 cm

10 c

m

12 cm

10 cm

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

5 We 7, 8 Convert the following areas to the units given in brackets.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)

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

7 A cutting blade for a craft knife has the dimensions shown in the diagram.

What is the area of steel in the blade (to the nearest mm2)?

8 mC The perimeter of the figure shown, in centimetres, is:a 34b 24 + 5pc 24 + 2.5pd 29 + 5pe 29 + 5p

9 mC The perimeter of the enclosed figure shown is 156.6 metres.The unknown length, x, is closest to:

a 20.5 mb 35.2 mc 40.2 md 80.4 me 90.6 m

exam Tip Many students relied heavily on using formulas they did not understand.


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Conversion of units of length

and area

2.08 m

1.20 m

30 mm

20 m

m

40 mm5 mm

12 cm

7 cm

2 cm

3 cm

35.2

m

20.5 mx

322

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

a What is the total area of the dartboard?b What is the area of the bullseye (inner circle)?c What is the area of the 2-point middle ring?d Express each area of the three rings as a percentage of the total area (to 2 decimal places).

11 On the set of a western movie, a horse is tied to a railing outside a saloon bar. The railing is 2 metres long; the reins are also 2 metres long once tied at one of 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 reins are 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)?

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. In other situations we will need to derive a formula by using the net of an object.

Total surface area formulas of common objectsCube

l

Cuboid

h l

w

Cylinder

h

r

r

Cylinders:TSA = 2pr (r + h)

Cubes:TSA = 6l2

Cuboids:TSA = 2(lw + lh + wh)

Cone

Slantheighth s

r

Sphere

r

Cones:TSA = pr (r + s), where s is the slant height

Spheres:TSA = 4pr2

1

2

3

2

1

6 cm

40 cm20 cmeBookpluseBookplus


Expressing one number as a

percentage of another

7C



Calculate 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

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

Radius, r = 5 cmHeight, h = 1.13 m = 113 cm

2 Substitute and evaluate.Remember BODMAS.

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

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

Calculate 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

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

Diameter = 25 cmRadius = 12.5 cmTSA of sphere = 4pr2

= 4 × p × 12.52

= 1963.495

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

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

Think WriTe/displaY

Method 1: Using the rule

1 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 = 1350

6 = 225

Worked example 9eBookpluseBookplus

Tutorialint-0458

Worked example 9

1.13 m

5 cm

Worked example 10

25 cm

Worked example 11

324

2 Take the square root of both sides to fi nd l. l = 225 = 15 mm

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

Method 2: Using a CAS calculator

1

2 Write your answer. Given the length, l, is 15 mm, the dimensions of the cube are 15 mm × 15 mm × 15 mm.

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

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

Slantheight

Net

Square pyramid

Trapezoidal prism

Net

Net

Cylinder


On the NumSolve screen, use the )/V soft keyboard to complete the entry line as:t = 6 × l2Then press E.The list of variables will appear. Enter the known values and select the variable to solve by checking the adjacent bullet. Then tap:• 1 (on the tool bar)• OK


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

Think WriTe

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

A1

A4

A4

8 cm 6 cm

8 cm10 cm

20 c

m

10 cm 10 cm

10 cm

20 c

m

20 c

m

20 c

m

6 cm

6 cm

6 cm

A2 A3

2 Identify the different-sized common fi gures 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 = 12 × b × h

= 12 × 8 × 6

= 24 cm2

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

= 200 + 160 + 120 + 2 × 24 = 528 cm2

4 Write your answer. The total surface area of the triangular prism is 528 cm2.

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

8 cm

12 c

m


Tutorialint-0459

Worked example 12

20 cm

10 cm

8 cm

6 cm

Worked example 13

326

Think WriTe/draW

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

2 r

A1

12 c

m

A24 cm

2 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 = 2pr × w = 2 × p × 4 × 12 = 301.59 cm2

A2 = p × r2

= p × 42

= 50.27 cm2

3 Sum the areas. TSA = A1 + A2

= 301.59 + 50.27 = 351.86 cm2

4 Write your answer. The total surface area of the open cylindrical can is 351.9 cm2, correct to 1 decimal place.

Total surface area (TSA) is measured in mm1. 2, cm2, m2 and km2.

The TSAs of some common objects are as follows:2.

(a) Cubes: TSA = 6l 2

(b) Cuboids: TSA = 2(lw + lh + wh)(c) Cylinders: TSA = 2p r (r + h)(d) Cones: TSA = p r (r + s) where s is the slant height(e) Spheres: TSA = 4p r2

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

rememBer

Total surface area 1 We9 Calculate 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

exerCise

7C



2 We 10 Calculate the total surface area of the objects given in the diagrams. Give answers to 1 decimal place.a

410 mm

Length = 1.5 m b

14 c

m

4 cm7 cm

c

Diameter = 43 cm

3 We 11 Calculate the unknown dimensions, given the total surface area of the objects. Give answers 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 area of 468 cm2

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

4 We 12 Calculate the total surface areas for the objects given in the diagrams. Give answers to 1 decimal place.

a

5 cm

10 cm15 cm

4 cm

b

7 cm30 cm6.

06 c

m

c

13 cm

8 cm

Area = 22 cm2 d

9 mm25 mm

15 mm

12 m

m

5 We 13 Calculate the total surface area of each of the objects in the diagrams below. Give answers to 1 decimal place.a

250 mm

250

mm

Rubbish bin b

20 cm

10 cm

10.5

cm

15 cm

13.5 cmc

2 cm

4.5 cm7 cm

3 cm

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

7

4.5 m6.5 m

2.5 m

1.5

m

1.0

m

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.

328

8 mC The 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

9 mC The total surface of a golf ball of radius 21 mm is closest to:a 550 mm2 b 55 cm2 c 55 000 mm2 d 0.055 m2 e 5.5 cm2

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

a 12 abh b 2 × 1

2 bh + ab + 2 × ah c 3(1

2 bh + ab)

d 12 bh + 3ab e bh + 3ab

11 mC The total surface area of a poster tube that is 115 cm long and 8 cm in diameter is closest to:a 3000 cm2 b 2900 cm2 c 1500 mm2 d 6200 m2 e 23 000 cm2

12 A baker is investigating the best shape for a loaf of bread. The shape with the smallest surface area stays freshest. The baker has come up with two shapes: a rectangular prism with a 12 cm square base and a cylinder with a round end that has a 14 cm diameter.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?

Volume of prisms, pyramids and spheresThe most common volumes considered in the real world are the volumes of prisms, pyramids, spheres and objects that are a combination of these. For example, 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, such as 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.

h

b

a

eBookpluseBookplus

Digital docWorkSHEET 7.1

7d

Volume

103 1003

103 1003

mm3 cm3 m3



Convert 1.12 cm3 to mm3.

Think WriTe

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

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

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

= 1.12 × 1000 mm3

= 1120 mm3

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

Convert 156 000 cm3 to: a m3 b litres.

Think WriTe

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

100 m.

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

= 156 000 × 1100

m × 1100

m × 1100

m

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

= 156000

100 100 100× ×100× ×100 m3

= 0.156 m3

3 Write the answer in correct units. 156 000 cm3 = 0.156 cubic metres (m3)

b 1 Conversion factor is 1000 cm3 equals

1 litre; that is, 1 cm3 equals 1100

litre.

b 156 000 cm3 = 156 000 × 11000

litres

= 156000

1000 litres

= 156 litres156 000 cm3 = 156 litres2 Write the answer.

Volume of prismsA prism is a 3-dimensional object that has a uniform cross-section. It is named in accordance with its uniform cross-sectional area. Note: Circular prisms are called cylinders.

To fi nd 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

Worked example 14

Worked example 15

Triangular prism

CylinderUniformcross-section

Height

Squareprism

330

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

Think WriTe

1 The object has a circle as a uniform cross-section. It is a cylinder. The area of the base is: area of a circle = pr2.Volume is cross-sectional area times height.

Vcylinder = A × H, where Acircle = pr2

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

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

Calculate (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

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

V = A × Hwhere A = 250 mm2

V = 250 mm2 × 17 mm = 4250 mm3

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

Given the volume of an object, we can use the volume formula to fi nd an unknown dimension of the object by transposing the formula.

Worked example 16

20 c

m

15 cm

Worked example 17

uniform cross-sectional area of 250 mm2

Area 250 mm2

17 mm



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

Think WriTe/displaY


1 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 × H,where A = 1

2 × b × h

V = 12 × b × h × H

2 Substitute the values, transpose and evaluate. 6.6 = 12

× 2 × h × 1.1

6.6 = 1.1 h

h = 6 61 16 6.6 6.1 1.1 1

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


1

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

Volume of pyramidsA pyramid is a 3-dimensional object that has a similar cross-section but the size reduces 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.

Worked example 18 eBookpluseBookplus

Tutorialint-0460

Worked example 18

h

2 m 1.1

m

Volume of prism = 6.6 m3

Triangular pyramid Cone

Vertex

On the NumSolve screen, use the )/V soft keyboard to complete the entry line as:V = 0.5 × b × h × d Then press E.The list of variables will appear. Enter the known values and select the variable to solve by checking the adjacent bullet. Then tap:• 1 (on the tool bar)• OK

Note: An alternative such as d is used in place of H.

332

To fi nd the volume of pyramids, we take a similar approach to prisms but the volume of a pyramid is always one-third of a prism with the same initial base and same height, H. This is the same for all pyramids.

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

Vpyramid = 13 × area of cross-section at the base × height

V = 13 × A × H

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

Calculate the volume of the pyramid below (to the nearest m3).

Think WriTe/displaY


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

Vpyramid = 13 × A × H, where Asquare = l2

Vpyramid = 13 × l2 × H

= 13 × 302 × 40

= 12 000 m3

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


1

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

H

A

Worked example 19

30 m 30 m

Height of pyramid = 40 m


On the NumSolve screen, use the )/V soft keyboard to complete the entry line as:

V = 1

3 × l2 × h

Then press E.The list of variables will appear. Enter the known values and select the variable to solve by checking the adjacent bullet. Then tap:• 1 (on the tool bar)• OK


Volume of spheres and composite objectsVolume of a sphereSpheres are unique (but common) objects that deserve special attention.

The formula for the volume of a sphere is:Vsphere = 4

3π r 3

where r is the radius of the sphere.

Volume of composite objectsOften the object can be identifi ed as comprising two or more different common prisms, pyramids or spheres. Such fi gures are called composite objects. The volume of a composite object is found by adding the volumes of the individual common fi gures or deducting volumes. Each of the structures at right can be roughly modelled as the sum of a cylinder and a cone.

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

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

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

THINK WRITE

1 The object is a composite of a cylinder and a square-based prism.

r = 6 cm

18 cm

25 c

m

18 cm

H =

20

cm

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

Vcomposite = volume of cylinder + volume of square-based prism = Acircle × Hcylinder + Asquare × Hsquare

= (πr2 × Hc) + (l2 × Hs) = (π × 62 × 20) + (182 × 25) = 2261.946 711 + 8100 = 10 361.946 711 cm3

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

10 362 cm2 = 10.362 litres

3 Write your answer. The volume of the object is 10 litres, calculated to the nearest litre.

r

WORKED EXAMPLE 20eBookpluseBookplus

Tutorialint-0461

Worked example 20

25 c

m

12 cm

18 cm

20 c

m

334

Volume is the amount of space occupied by a 3-dimensional object.1. (a) The units of volume are mm2. 3, 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

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

(a) The volume of a pyramid is 4. Vpyramid = 13 × area of cross-section at the base × height

V = 13 × A × H

(b) The height of a pyramid, H, is sometimes called the altitude.The volume of a sphere is 5. Vsphere = 4

3p r 3.

The volume of a composite object equals the sum of the individual common prisms, 6. pyramids or spheres.

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

rememBer

Volume of prisms, pyramids and spheres 1 We 14, 15 Convert the volumes to the units specified.

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 litresi 250 000 mm3 to cm3

2 We 16 Calculate the volume of the following prisms to the nearest whole unit.a

4000 mm

75 mm

b

51.2

cm

104.8 cm

c 7 cm

23 c

m

15 cm

4 cm

d

6.4 m4.8 m

2.1 m

e

20 m

m

30mm

34 mm

f

22 mm57 mm14

mm

exerCise

7deBookpluseBookplus


Conversion ofunits of volume

and capacity



3 We 17 Calculate the volume of the following prisms (to 2 decimal places).a

2.9 m

Area 4.2 m2 b

14.5

mm

Area 120 mm2 c

8.5 cm

Area 15 cm2

Area32 cm2

4 We 18 Calculate the measurement of the unknown dimension (to 1 decimal place).a Volume of cube

= 1.728 m3

x

b 15.0 cm

21.4 cmx

Volume of triangular prism1316.1 cm3

c

Volume of cylinder 150 796.4 mm3

120

mm

x d

3x

x

Volume of prism = 10 litres1–8

5 We 19 Calculate the volume of these pyramids (to the nearest whole unit).

a

12 cm 11 cm

VO = 10 cm

V

O

b

11 cm

35 c

m

c

8 m 12 m

V

O

VO 17 m

d

Altitude of squarepyramid 18 mm

12 mm e

10 cm

6 cm

6 cm

Base ofpyramid

VO = 15 cm

O

V

exam Tip Some students appeared to have diffi culty in dealing with . . . fraction(s) on their calculator when using the formula V = 1

3 × area of

base × height when fi nding the volume of a pyramid.[Assessment report 2 2007]

336

6 We 20 Calculate the volume of these objects (to the nearest whole unit).a

r = 8 cm

b 4 cm

7 cm8 cm

10 c

m

20 cm

c 3 m

5 m

4 m

d

10 cm

10 cm

15 cm

e

2.5 m

2.1 m

1 m

6 m

f

42 m

19 m

42 m

60 m

7 a Calculate the volume of a cube with sides 4.5 cm long.b Calculate the volume of a room, 3.5 m by 3 m by 2.1 m high.c Calculate the radius of a baseball that has a volume of 125 cm3.d Calculate the height of a cylinder that is 20 cm in diameter with a volume of 2.5 litres (to

the nearest unit).e Calculate the height of a triangular prism with a base area of 128 mm2 and volume of

1024 mm3.f Calculate the depth of water in a swimming pool that has a capacity of 56 000 litres. The

pool has rectangular dimensions of 8 metres by 5.25 metres.

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

5 cm

4 cm

9 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, calculate: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 mC The volume 200 000 mm3 is equivalent to:a 2 litres b 2 cm3

c 20 cm3 d 200 cm3

e 2000 cm3

exam Tip Numbers (for example, the length of an object) found during working out within a question must not be rounded when used to determine the answer to that question.




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

a 43

b 23

c 43r

h d 34

e 32

12 mC If the volume of the square pyramid shown is 6000 m3, then the perimeter of the base is closest to:a 900 mb 20 mc 30 md 80 me 120 m

13 mC A tin of fruit is 13 cm high and 10 cm in diameter. Its volume, to 1 decimal place, is:a 1021.0 cm3

b 510.5 cm3

c 1021.4 cm3

d 1020.1 cm3

e 4084.1 cm3

14 A model aeroplane is controlled by a tethered string of 10 metres length. The operator stands 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 fl ight?b If the plane can be manoeuvred in a hemispherical zone, fi nd:

i the surface area of the airspace that the plane can occupy ii 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.

similar fi guresObjects that have the same shape but different size are said to be similar.

r

r

VO = 20 m

O

V

7e

338

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

be equal.

′ ′ = ′ ′ = ′ ′ = ′ ′A B′ ′A B′ ′AB

B C′ ′B C′ ′BC

C D′ ′C D′ ′CD

A D′ ′A D′ ′AD

= common ratio

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

scale factor, kA measure of the relative size of the two similar fi gures is the scale factor. The scale factor is the common ratio of the corresponding sides and quantifi es the amount of enlargement or reduction one fi gure undergoes to transform into the other fi gure. The starting shape is commonly 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 scale ratios.

For example, k = 2, k = 112

or 1 : 10 000.

2. Scale factor, k = length of image

length of original =

′ ′ = ′ ′ = ′ ′A B′ ′A B′ ′AB


C A′ ′C A′ ′CA

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

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

Enlargements and reductions are important in many aspects of photography, map making and modelling. Often, photographs are increased in size (enlarged) to examine fi ne detail without distortion, while house plans are an example of a reduction to a scale; for example, 1 : 25.

B'C'

D'A'

4

2

6

2

2

1

1

3

BC

DA

B'C'

D'A'

12560

85

12560

85

BC

DA

B

A C

3

1

3

B'

A' C'

9 9

3



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

Think WriTe

a 1 As it is a reduction, the larger shape is the original and the smaller shape is the image.

a

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

Scale factor, k = length of image

length of original

= ′ ′A B′ ′A B′ ′

AB

= 10 cm20 cm

= 12

b 1 Use the scale factor to determine the unknown length as all corresponding lengths are in the same ratio.

b Scale factor, k = 12

k = length of image

length of original

12 =

x45 cm

x = 12 × 45 cm

x = 22.5 cm

2 Write your answers. The scale factor of reduction is 12 and the

unknown length on the smaller shape is 22.5 cm.

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

Worked example 21

Original

20 cm

45 c

m

10 cm

Image

x

Worked example 22

40

30

2070 m

100 m

s

40

30

20

t

50 m

30 m

340

Think WriTe

a First, orient the fi gures to identify corresponding sides and angles easily. Calculate the missing angles and compare each pair of corresponding angles.

a

40

30

2070 m

100 m

s

Image

270

Original

27040

30

20

t

50 m30

m

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

b 1 As the scale factor given is for enlargements, the original is the smaller fi gure.

b Scale factor, k = length ongth ongt f image

length ongth ongt f originalnalna

For s 2 = s30 m

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

s = 2 × 30 m = 60 m

For t 2 = 70 mt

t = 70 m2

= 35 m

rememBer

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

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

′ ′ ′ ′ ′ ′ ′ ′A B′ ′A B′ ′AB




= == = = = common ratio

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

Scale factor, 3. k, is the amount of enlargement or reduction and is expressed as integers, fractions or scale ratios; for example, k = 2, k = 1

12 or 1 : 10 000.

B'C'

D'A'

4

2

6

2

2

1

1

3

BC

DA

B'C'

D'A'

12560

85

12560

85

BC

DA



Scale factor,4. k = length of image

length of original = ′ ′ ′ ′ ′ ′A B′ ′A B′ ′

AB



= == =

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

5. For k = 1, the fi gures are exactly the same shape and size and are referred to as congruent.

B

A C

3

1

3

B'

A' C'

9 9

3

similar fi gures 1 We21 For each of these pairs of similar shapes, calculate:

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

a 200 cm

x cm

y cm

50 cm

70 cm 50 c

m

b

4 m

1 m25

m

x cm

20 cm y cm

2 cm

8 cm

8 cm

4 cmx cm

y cm

c

2 We22 Prove that the following pairs are similar figures and calculate the value of a.

a

17 mm

17 mm

a

8 mm

1–2

b

Photo

40 m

m

62 m

m

a

Height of person = 186 cm

exerCise

7e

342

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

10 cm

12 cm

4 Most scale model cars are in the ratio 1 : 12. Calculate: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.

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

6 mC The perimeter of the real object shown in the scale diagram of 1 : 25 at right is: a 464 cm b 514 cmc 357 cm d 14.28 cme 150 cm

7 mC A 1 : 27 scale model of a truck is made from clay. What is the length of the tray on the original truck, if its length is 27 cm on the model? a 1 cm b 100 cmc 270 cm d 540 cme 729 cm

8 mC A 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

4 cm

2 cm



similar trianglesSimilar triangles can be used to fi nd the height of trees and buildings or the width of rivers and mountains. One extra rule can be used to identify similar triangles to those mentioned for similar shapes in the previous section. Two triangles are similar if one of the following conditions is identifi ed: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 sides to determine the scale factor (sf ) for the similar triangles.

Scale factor, k = OA

OA

length of side of imagelength of corresp

′=

onoono ding side of original

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

Think WriTe

a 1 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°.

aB

C

4

6A

10030 50

Original

B'

C'

6

A' x

100

30 50

Image

2 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 = length of side of image

length of corresponding side of original

= ′ ′A B′ ′A B′ ′AB

= 64

= 1.5

7F

section. Two triangles are similar if one of the following conditions is identifi ed:

eBookpluseBookplus

Interactivityint-0188

Scale factors

3

21

6

2 4sf = 1

2 = 2

4 = 3

6 = 0.5 = k

2

3sf = 6

3 = 4

2 = 2 = k6

4

Worked example 23

B

C

4

6A

10030

B'

C'

6

A' x

100

30

344

b 1 Use the scale factor to fi nd the unknown length, x.Transpose and evaluate.

b Scale factor, k = 1.5

1.5 = ′ ′A C′ ′A C′ ′AC

1.5 = x6

x = 1.5 × 6 x = 9

2 Write the answer in the correct units and level of accuracy.

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

For the given triangles, fi nd the value of the pronumeral, x.

4.0

7

3.5

CA

All measurements in metres

E

B

D

x

Think WriTe

1 Confi rm 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.

4.0 m

7 mC

A

BOriginal

exam Tip Labelling and/or drawing diagrams may help students to gain marks for applying the correct method.


2 For clear analysis, separate the two triangles. Note: The lengths of the sides AE and AD are the sum of the given values.

AD = 4.0 + 3.5 = 7.5 m

AE = (7 + x) m

3 Select 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,


length of corresponding side of original

= AD

AB

= 7.54.0

k = 1.875


Tutorialint-0462

Worked example 24

(7 + x) m

7.5 m

A

E

DImage



4 Use the scale factor to fi nd the unknown length. Transpose and evaluate.

Scale factor,


length of corresponding side of original Alternative method

1.875 = AE

AC AE

AC = AD

AB

= 7

7+ x

77+ x =

7 54

7 5.7 5

1.875 × 7 = 7 + x 4(7 + x) = 7 × 7.5 13.125 = 7 + x 28 + 4x = 52.5 x = 13.125 - 7 4x = 24.5 x = 6.125 x = 6.125

5 Write the answer in the correct units and level of accuracy.

The value of x is 6 18 metres.

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

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

Think WriTe

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

2 For clear analysis, separate the two triangles.

3 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 = height of tree (image)

height of girl (original)

k = 141.4

= x1 681 6.1 6

10 = x

1 681 6.1 6

Worked example 25

Sun’s ray

s

Girl(168 cm)

Shadow(140 cm)

14 metres

140 cm

Original

168 cm

14 m

x m

Image

346

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

5 Write the answer in the correct units.

The height of the tree is 16.8 metres.

Scale factor, 1. k = OA

OA

′ = length of side of imagelength of corresponding side of original

Two triangles are similar if one of the following 2. conditions is identifi ed:(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).

3

21

6

2 4

scale factor 0.5 k1—2

2—4

3—6

3

4

6

8

rememBer

similar triangles 1 We23a State the rule (SSS or AAA or SAS) that proves the pair of triangles are similar and

determine the scale factor (expressed as an enlargement k > 1).a

640 m

m

320 mm25

b5.6

4.4

4.6

8.8

9.2

11.2

c

4.5

5

9

10

d

7.010.5

e

14

10.5

7

f

3.5

10.5

exerCise

7F

480 m

m

240 mm25

3

42



2 We23b For the given similar triangles, find the value of the pronumeral a.

a b

12 m 14.4 m

a m

15 m

c

25 cm38

a

d

a14

1612

e

12

4

9

a

3.2

9.6

f

6

7.8

a

13

3 We24 For the given similar triangles, find the value of the pronumeral a.

a

12

3

6

a

ba

8

12

2

c

a

10.5

7.5

4.5

d

43

17.2

15.2

a

e

17 m

a m

68 m

18 m

f

142 mm

108 mm

324 mm

a80

80

4 We25 Find the height of the flagpole shown in the diagram below (to the nearest centimetre).

20 m

m

15 mm62 62

56

22.5 mm

a m

m

62 62

56

59 cm

45 cm

75 cm38

67

717

68

0.9 m

9 m1 m

Guy wire

348

5 Find the length of the bridge, AB , 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 is 25 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 the surveyor’s notes at right.

8 mC In 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 at right. Assume the ball travels in a straight line.

9 mC The 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

10 mC The 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

area and volume scale factorsAn unknown area or volume of a fi gure can be found without the need to use known formulas such as in exercises 7B and 7D. We have seen that two fi gures that are similar have all corresponding lengths in the same ratio or (linear) scale factor, k. The same can be shown for the area and volume of two similar fi gures.

area of similar fi guresIf the lengths of similar fi gures are in the ratio a : b or k, then the areas of the similar shapes are in the ratio a2 : b2 or k2. Following are investigations to support this relationship.

4.3 m12.5 m

2.5 m

A

B

(Not to scale.)

1.2 m

2 m

25 m

ABLake

20

16 12

b

Not to scale

x

y1.

1 m

5 m 10 m

0.9 m

eBookpluseBookplus

Digital docWorkSHEET 7.2

7G



different length ratios (or scale factors) of a square

length of blue squarelength of red square

= 2 cm1 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

different length ratios (or scale factors) of a circleradius length of blue circleradius length of red circle

= 2 cm1 cm

= 2 = k

Area = pr2 = 1p cm2

area of blue circlearea of red circle

= 4p cm2

1p cm2 = 4 = 22 = k2

Area = pr2 = 4p cm2

radius length of green circleradius length of red circle

= 3 cm1 cm

= 3 = k

Area = pr2 = 9p cm2

area of green circlearea of red circle

= 9p cm2

1p cm2 = 9 = 32 = k2

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

Area scale ratio or factor (asf ) = area of image

area of original = square of linear scale factor (lsf ) = (lsf )2

= k2

The steps required to solve for length, area or volume (investigated later) 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 measurements.3. Convert to an appropriate scale factor to determine the unknown measurement.4. Use the scale factor and ratio to evaluate the unknown.

1 cm Area 1 cm2

1 cm

2 cm Area = 4 cm2

2 cm

3 cm Area = 9 cm2

3 cm

1 cm

2 cm

3 cm

350

For the two similar triangles shown, fi nd the area, x cm2, of the small triangle.

Think WriTe

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

k = 2.4 cm4.8 cm

= 1

2

2 Determine the area scale factor. Area scale factor = k2

= 12

2

= 1

4

3 Use the area scale factor to fi nd the unknown area.

Area scale factor = area of small triangle (image)area of large triangle (original)

14 =

x cm

100 cm

2

2

4 Transpose the equation to get the unknown by itself.

x = 14 × 100

= 255 Write your answer. The area of the small triangle is 25 cm2.

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

Think WriTe

1 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 = area of image (large trapezium)

area of original (small trapezium)

k2 = 250 cm

10 cm

2

2

= 25

2 Determine the linear scale factor. Linear scale factor = k2

k = 25 = 5

Worked example 26

Area x

2.4 cm

Area 100 cm2

4.8 cm

Worked example 27

10 cm2

2 cm

250 cm2

x



3 Use the linear scale factor to fi nd the unknown length.

Linear scale factor = length of image (large trapezium)

length of original (small trapezium)

5 = x cm

cm2


x = 5 × 2 = 10

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

Volume of similar fi guresIf the lengths of similar fi gures are in the ratio a : b or k, then the volume of the similar shapes are in the ratio a3 : b3 or k3. The following is an investigation of two different objects, cubes and rectangular prisms.

a cubelength of large (blue) cubelength of small (red) cube

= 2 cm1 cm

= 2 = k

volume of large cubevolume of small cube

= 8 cm2

1 cm2 = 8 = 23 = k3

a rectangular prismlength ongth ongt f small prisprispr mismislength ongth ongt f largeargear prisprispr mismis

3 cm6 cm

= = == === == 12

k

volume of small prisprispr mismisvolume of largeargear prisprispr mismis

3 cm=3

24 cmccmc

1 13

23

8 2= == = 1 11 1

8 28 28 28 28 28 2

= k

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

Volume scale factor (vsf ) = volume of image

volume of originalnalna = cube of linear scale factor (lsf ) = (lsf )3

= k3

Volume 1 1 1 1 cm 3

1 cm1 cm

1 cm

Volume2 2 2 8 cm3

2 cm

2 cm2 cm

Volume1 1 3 3 cm3

3 cm1 cm

1 cm

Volume2 2 6 24 cm3

6 cm2 cm

2 cm

352

For the two similar fi gures shown, fi nd the volume of the smaller cone.

Think WriTe

1 Separate the two fi gures to clarify the details of the similar fi gures.

Volumex cm3

6 cm

Volume540 cm3

9 cm

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

k = 6 cm9 cm

= 23

3 Determine the volume scale factor. Volume scale factor = k3

= 23

3

= 827

4 Use the volume scale factor to fi nd the unknown length.

Volume scale factor = volume of small cone (image)volume of large cone (original)

827

= x cm

540 cm

3

3


x = 827

× 540

x = 160

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

We can use the relationship between linear, area and volume scale factors to fi nd any unknown in any pair of similar fi gures as long as a scale factor can be established.

Given Then

Linear scale factor (lsf ) = kExample (k = 2) = 2

Area scale factor = k2

= 22

= 4

Volume scale factor = k3

= 23 = 8

Area scale factor (asf ) = k2

Example = 4Linear scale factor = k2

= 4 = 2

Volume scale factor = k3

= 23 = 8

Volume scale factor (vsf ) = k3

Example = 8Linear scale factor = k33

= 83

= 2

Area scale factor = k2

= 22 = 4


Tutorialint-0463

Worked example 28

Volume oflarge cone

540 cm39 cm6 cm



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

Think WriTe

1 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 = volume of larger prism (image)

volume of smaller prism (original)

k3 = 64 m

8 m

3

3

k3 = 8

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

Linear scale factor = k33 = k

k = 83 = 2 Area scale factor = k2

= 22

= 4

3 Use the area scale factor to fi nd the total surface area.

Area scale factor = area of larger prism (image)

area of smaller prism (original)

4 = x m

m

2

2 5 22 5.2 5


x = 4 × 2.5 = 10

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

The steps required to solve for length, area or volume using similarity:1.

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

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

(c) Convert to an appropriate scale factor to determine the unknown measurement.(d) Use the scale factor and ratio to evaluate the unknown.

Area scale ratio or factor (asf ) 2. = area of image

area of original = square of linear scale factor (lsf )

= k2

Volume scale ratio or factor (vsf ) 3. = volume of image

volume of original

= cube of linear scale factor (lsf )

= k3

rememBer

Worked example 29

354

area and volume scale factors 1 Complete the following table of values.

Linear scale factors k

Area scale factors k2

Volume scale factors k3

2 816

3125

10064

0.02736

0.1100

0.16400

2 We26 Find the unknown area of the following pairs of similar figures.

a

15 mm 22.5 mm

540 mm2

x mm2

b

48 c

m

8 cm

12 cm2

x cm2

c

Surface area x mm2

21 mm

14 mm

Surface area 100 mm2

3 a We27 Find the unknown length of the following pairs of similar figures. i

x mArea 6.25 m2

Area 1.0 m2

1.7 m

ii

Area 750 cm2

x

25 cm

Area 3000 cm2

b 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 width of 6 cm. Find the width of the larger photo.

4 We28 Find the unknown volume in the following pairs of similar objects.

a

x cm3

7 cm

2400 cm3

14 cm

b Volume of small pyramid 40 cm3

12 cm2 cm

exerCise

7G



c

Volume of large sphere 8 litres

d

45 cm

Volume 1200 cm3

30 cm

5 a We29 For the two similar triangular pyramids with volumes of 27 m3 and 3 m3, calculatethe total surface area of the larger triangular prism if the smaller prism has a total surface area of 1.5 m2.

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

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

d For two similar kitchen mixing bowls with total surface areas of 1500 cm2 and 3375 cm2, calculate the capacity of the larger bowl if the smaller bowl has a capacity of 1.25 litres (to the nearest quarter of a litre).

6 a Calculate the volume of the small cone.

Volume oflarge cone

270 cm3

Area 45 cm2

Area5 cm2

b Calculate the volume of the larger triangular pyramid.

TSA of small pyramid 200 cm2

Volume of small pyramid 1000 cm3

TSA of large pyramid 288 cm2

c Calculate the total surface area of the small prism.

Area 12 cm2

TSA 78 cm2

TSA x cm2

Area 6 cm2

d Calculate the diameter of the small cylinder.

Volume 1280 cm3

12 cm

Volume 20 cm3

x cm

exam Tip A continuing area of student difficulty, which needs particular attention from teachers, is scale factors. Many students fail to take into account the dimensions of the quantities being scaled.


356

7 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?

8 Calculate the volume ratios from the similar shapes given in question 7 .

9 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, calculate 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)?

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

11 An island in the Pacific Ocean has an area of 500 km2. What is the area of its representation on a map drawn to scale of 1 cm ⇔ 5 km?

12 Two statues of a famous person used 500 cm3 and 1.5 litres of clay. The smaller statue stood 15 cm tall. What is the height of the other statue (to the nearest centimetre)?

13 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?

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

15 mC A 1 : 27 scale model of a truck is made from clay. The ratio of volume of the model to the real truck is:a 1 : 3 b 3 : 1 c 1 : 9 d 1 : 729 e 1 : 19 683

16 mC The ratio of the volume of the blue portion to the volume of the red portion is:a 1 : 3 b 1 : 8 c 1 : 9 d 1 : 26 e 1 : 27

17 mC A 1 : 100 scale model of a building is a cube with sides of 100 cm. The volume of the building is:a 10 000 000 m3 b 1 000 000 m3 c 100 000 m3

d 10 000 m3 e 1000 m3

Area 500 km2

h

3h



sUmmarY

Properties of angles, triangles and polygons

When solving geometry problems:draw diagrams carefully•

• carefully interpret geometric notation; for example, the triangle at right has two sides equal in length.carefully consider geometric rules; for example, isosceles triangles have • two equal sides and angles.

area and perimeter

Perimeter is the distance around a closed fi gure.• Circumference is the perimeter of a circle.• C = 2 × p × radius

= 2prArea is measured in mm• 2, cm2, m2, km2 and hectares.1 cm• 2 = 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:Area of a square: 1. A = l2

Area of a rectangle: 2. A = l × wArea of a parallelogram: 3. A = b × hArea of a trapezium: 4. A = 1

2 (a + b) × h

Area of a circle: 5. A = pr2

Area of a triangle: 6. A = 12 × b × h

Area of composite fi gure • = sum of the areas of the individual common fi guresAcomposite = A1 + A2 + A3 + A4 + . . .

Total surface area (TSa)

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

Cubes: TSA 1. = 6l2

Cuboids: TSA 2. = 2(lw + lh + wh)Cylinders: TSA 3. = 2pr(r + h)Cones: TSA 4. = pr(r + s), where s is the slant heightSpheres: TSA 5. = 4pr 2

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 mm• 3, cm3 (or cc) and m3.

1000 mm1. 3 = 1 cm3

1 000 000 cm2. 3 = 1 m3

1 litre 3. = 1000 cm3

1000 litres 4. = 1 m3

Volume of a prism, • Vprism = area of uniform cross-section × heightV = A × H

Equal sides

358

Volume of a pyramid, • Vpyramid = 1

3 × area of cross-section at the base × height

V = 1

3 × A × H

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

Volume of a sphere is • Vsphere = 4

3 pr3.

Volume of a composite object • = sum of the volumes of the individual common prisms, pyramids or spheres. Vcomposite = V1 + V2 + V3 + . . .or Vcomposite = V1 - V2 . . .

Similar figures

• Two objects that have the same shape but different size are said to be similar.For two fi gures to be similar, they must have the following properties:• (a) The ratios of the corresponding sides must be equal.

′ ′ = ′ ′ = ′ ′ = ′ ′A B′ ′A B′ ′

AB




= common ratio

(b) The corresponding angles are equal.

∠A = ∠A′ ∠B = ∠B′ ∠C = ∠C′ ∠D = ∠D′

Scale factor, k

Scale factor, • k = length of imagelength of original

= ′ ′ = ′ ′ = ′ ′A B′ ′A B′ ′AB



where for enlargements, k is greater than 1 and for reductions, k is between 0 and 1.For • k = 1, the fi gures 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 identifi ed:• All 3 corresponding angles are equal (AAA).1. All 3 corresponding pairs of sides are in the same ratio (linear scale factor) (SSS).2. Two corresponding pairs of sides are in the same ratio and the included angles are equal (SAS).3.

area and volume scale factors

The steps required to solve for length, area or volume using similarity are:• Clearly identify the known corresponding measurements (length, area or volume) of the similar shapes.1. Establish a scale factor (linear, area or volume) using known pairs of measurements.2. Convert to an appropriate scale factor to determine the unknown measurement.3. Use the scale factor and ratio to evaluate the unknown.4.

Area scale ratio or factor (asf ) • = area of image

area of original

= square of linear scale factor (lsf ) = k 2

Volume scale ratio or factor (vsf ) • = volume of image

volume of original = cube of linear scale factor (lsf ) = k 3

B'C'

D'A'

4

2

6

2

2

1

1

3

BC

DA

B'C'

D'A'

12560

85

12560

85

BC

DA

B'

A' C'

9 9B

A C

3

1

3

3



ChapTer reVieW

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 ∆ABC has the following values: AB = 10 cm, AC = 12 cm where AB and AC are perpendicular. The area of the triangle is:a 120 cm2

b 30 cm2

c 240 cm2

d 121 cm2

e 60 cm2

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

b 1250p + 37 600 cm2

c 1250p + 29 600 cm2

d 2500p + 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:a 2 × p × 20 × (40)b 2 × p × 40 × (40)c 2 × p × 40 × (100)d 2 × p × 40 × (60)e 2 × p × 20 × (60)

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:a 560p cm3

b 900p cm3

c 3600p cm3

d 4500p cm3

e 36 000p cm3

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

b 1300 cm3

c 1500 cm3

d 2000 cm3

e 10 000 cm3

8 In the triangle shown, the value of c is:a 3b 6c 9d 12e 4

9 The circumference of the larger cone is closest to:a 113 mmb 151 mmc 226 mmd 302 mme 459 mm

10 The diagonal distance on the plasma screen is used to specify the different sizes available. If the height on a 51 cm plasma is 45 cm, then a similar 34 cm television has a height, h, which is closest to:a 67 cmb 45 cmc 34 cmd 30 cme 26 cm

x32

200

220

50

80

Allmeasurements

in cm

O

V

3

7.8

c2.6

189 mm

63 mm

24 mm

34 cm

h cm

51 cm

45 c

m

360

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

a 64

6= − xx

b 46

6= − xx

c 64

6= −xx

d 126

6= − xx

e 64

2= + xx

6 –

x

6

124

6

x

12 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

13 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

14 A solid cylinder has a height of 30 cm and a diameter of 40 cm. A hemisphere is cut out of the top of the cylinder, as shown.In square centimetres, the total surface area

of the remaining solid (including its base) is closest to:a 1260 b 2510 c 6280d 7540 e 10 050

[Vcaa 2007]

exam Tip [This question] involved fi nding the total surface area of a solid. While only 28% of students successfully answered the question, most students had started correctly but either failed to include the base of the solid (the 27% who chose option C) or focussed solely on the hemispherical surface (the 25% who chose option B). To ensure that they include all of the surfaces involved, students might fi nd it helpful to begin the solution to questions such as this by writing down a statement like: Total surface area = surface area of the hemispherical bowl + surface area of the side of the cylinder + surface area of the base of the cylinder.


15 A block of land has an area of 4000 m2. When represented on a map, this block of land has an area of 10 cm2. On the map, 1 cm would represent an actual distance of:a 10 m b 20 m c 40 md 400 m e 4000 m

[Vcaa 2007]

exam Tip Only 18% of students correctly answered [this question]. In this question, students were given the actual area of a block of land. They were also given its area on a map. From this information a scale factor for area (k2 = 400) can be determined. The majority of students (58%) apparently obtained this area scale factor but then incorrectly applied it directly to scaling the given length rather than fi rst converting it into the

corresponding linear scale factor, k = 400 = 20.[Assessment report 1 2007]

shorT ansWer

1 For each of the figures, calculate the values of the pronumerals.a

40

b

a

c

bb

ac

1 m

w

2 m

1000

m

40 cm

30 cm



2 Calculate the outer perimeter and area of the flower.

r 22 mm

r 11 mm

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.

6 m5 m

3 m4 m

exam Tip All students are encouraged to show their working out in all questions that require some calculation. This may allow a method mark to be awarded if a question is worth more than one mark. Also, if a question requires a previous answer for its calculation and that previous answer was incorrect, a consequential mark is only available if the working shows the correct application of that previous incorrect answer. However, many students continue to give answers only, without showing any relevant working. Where the answer was correct, full marks were awarded, but if working was absent or difficult to follow, method and consequential marks could not be awarded where the answer was incorrect.


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?c What is the total surface area of the solid part

only (to 1 decimal place)?

6 m

10 m

exam Tip With questions involving finding the total surface area of a solid, while some students successfully answer the question, most students tend to start correctly but either fail to include the base of the solid or focus solely on the hemispherical surface. To ensure that they include all of the surfaces involved, students might find it helpful to begin the solution to questions such as this by writing down a statement like: Total surface area = surface area of the hemispherical bowl + surface area of the side of the cylinder + surface area of the base of the cylinder.


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 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)?

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

10 m 1 m

6 m

8 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.

362

exTended response

Task 1A 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 the 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 the 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 the 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 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?

Task 2The 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).

b A wiper blade is 50 centimetres long and just reaches the top of the window as it makes a semicircular sweep. The base of the wiper is situated at the bottom centre of the rear window. i Draw a diagram of the situation. ii Find the area of the window swept by the wiper (to the nearest cm2). iii Find the percentage of the window’s area not swept by the wiper.

c The manufacturer decides to increase the wiper length by 10 centimetres i Find the new area of the window that is swept by the wiper (to the nearest cm2). ii Find the percentage of the window’s area that is not swept by the wiper.

Task 3 1 The scale used on a contour map is 1 : 40 000.

Determine the length of a line (in centimetres) on the contour map that represents a distance of 2 km.

[Vcaa 2005]

5 cm

6 cm

6 cm

4.5 cm

exam Tip Surprisingly, this question proved difficult for most students. Common incorrect answers were 0.5 cm or 0.05 cm, while others did not see any problem with unreasonable answers of up to 8 000 000 000 cm on a map.




2 The cross-section, ABC, of an A-frame hut near the camping ground is an isosceles triangle as shown below.Angle ACB is 50°.AB is 5 metres in length and is the fl oor line.MN is 2 metres in length and is the ceiling line inside the hut.AB and MN are parallel.

50

ceiling

floor

2 m

5 m

N

C

M

A B

a Calculate the vertical height of the ceiling, MN, above the fl oor, AB. Write your answer in metres, correct to 1 decimal place.

b The hut has the shape of a triangular prism. The space inside the hut above the ceiling is used for storage.

The total space inside the hut (including storage) is V m3. What fraction of V is used for storage?[Vcaa 2005]

50

ceiling storage

2 m

5 mfloor

N

C

M

A B

exam Tip This question was poorly answered by most students. The common misconception was that the volume fraction was the cube of a length fraction and many students obtained the incorrect answer of

25

3 8125

= . This would apply if the lengths in all three

dimensions had been enlarged; however, only the height and width of the two end triangles were scaled up. The lengths of the triangular prisms were equal and so the fraction for volume is the same as the

fraction for end area: 25

2 425

= .[Assessment report 2 2005]

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364

eBookpluseBookplus aCTiViTies

chapter openerDigital doc

10 Quick Questions: Warm up with a quick quiz on • similarity and mensuration. (page 310)

7a Properties of angles, trianglesand polygons

Interactivity int-0259The sum of external angles of a polygon: Use • the interactivity to investigate external angles of polygons. (page 310)

7b area and perimeterDigital docs

SkillSHEET 7.1: Practise substitution into a formula. • (page 320)SkillSHEET 7.2: Practise conversion of units of • length. (page 321)SkillSHEET 7.3: Practise expressing one number as • a percentage of another. (page 322)

7c Total surface area Digital doc

WorkSHEET 7.1: Convert units, calculate perimeter, • area, total surface area and volume (page 328)

Tutorials

We9 • int-0458: Watch a tutorial on how to calculate the total surface area of a poster tube. (page 323) We 12 • int-0459: Watch a tutorial on how to calculate the total surface area of a triangular prism. (page 325)

7d Volume of prisms, pyramids and spheresDigital doc

SkillSHEET 7.4: Practise conversion to units of • volume and capacity. (page 334)

Tutorials

We 18 • int-0460: Watch a worked example on how to calculate the volume of a prism. (page 331) We20 • int-0461: Watch a worked example on how to calculate the volume of a composite 3-dimensional object. (page 333)

7F Similar trianglesDigital doc

WorkSHEET 7.2: Interior and exterior angles, • perimeter, area, total surface area and volume, convert units and calculate unknown quantities in similar triangles (page 348)

Tutorial

We24 • int-0462: Watch a worked example on how to use similar fi gures to calculate unknown dimensions. (page 344)

Interactivity int-0188Scale factors: Use the interactivity to consolidate • your understanding of 1-, 2- and 3-dimensional scale factors. (page 343)

7G area and volume scale factorsTutorial

We28 • int-0463: Watch a worked example on how to use scale factors to calculate volume. (page 352)

chapter reviewDigital doc

Test Yourself:• Take the end-of-chapter test to test your progress. (page 363)

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7a and polygons 7b 7c 7d geometry: 7e 7f 7g similar ...mathsbooks.net/jacplus books/12 further...

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