Ch 08 FM YR 12 Page 347 Friday, November 10, 2000 11:30 AM
Geometry:
similarity and
mensuration
8
VCE coverage
Area of study
Units 3 & 4 • Geometry and
trigonometry
In this cha
chapter
pter
8A 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
Ch 08 FM YR 12 Page 348 Friday, November 10, 2000 11:30 AM
348
Further Mathematics
Geometry
UPPER
LEVEL
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.
Bed 1
Bed 2
Bed 4
Bed 3
Stairways
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
360°
n sides, the interior angle is given by 180° − ----------- .
n
For example, for a square the interior angle is:
360°
180° − ----------- = 180° − 90°
4
= 90°
360°
The exterior angle is given by ----------- .
n
Exterior angle
Interior angle
WORKED Example 1
Find the interior and exterior angle of the regular polygon shown.
THINK
1
This shape is a regular pentagon, a
5-sided figure.
Substitute n = 5 into the interior angle
formula.
2
Substitute n = 5 into the exterior angle
formula.
3
Write your answer.
WRITE
360°
Interior angle = 180° − ----------5
= 180° − 72°
= 108°
360°
Exterior angle = ----------5
= 72°
A regular pentagon has an interior angle of
108° and an exterior angle of 72°.
Ch 08 FM YR 12 Page 349 Friday, November 10, 2000 11:30 AM
349
Chapter 8 Geometry: similarity and mensuration
Geometry rules, definitions and notation rules
The following geometry rules and notation will be most valuable in establishing
unknown values in the topics covered and revised in this module.
Definitions of common terms
A
∠ABC
Straight angle
90°
MQ FurMat fig 11.05(c)
Less than 90°
B
Between 90°
and 180°
180°
Acute angle C
Obtuse angle
Right angle
MQ F rMat fig 11 05(d)
MQ F M t fi 11 05(b)
A
Between 180°
and 360°
B
Line
A
AB
B
AB
A
B
AB
Line segment
Ray
Reflex angle
Parallel lines
Perpendicular lines
MQ FurMat fig 11.05(h)
Some common notations and rules
a
All equal
sides and
60° angles
Two equal
sides and
angles
a + b + c = 180°
No equal sides
b
c
Scalene triangle
MQ FurMat fig 11.06(a)
45°
45°
60°
60°
Equilateral triangle
Isosceles triangle
Right-angled
isoceles triangle
MQ FurMat fig 11.06(b)
a + b = 90°
C
a=b
a + b = 180°
a
a
b
a
b
b
Vertically opposite
angles
Supplementary angles
Complementary angles
MQ FurMat fig 11.06(f)
MQ FurMat fig 11.06(e)
a=b
c=d
a c
d b
Alternate angles
MQ FurMat fig 11.06(g)
a=b
c=d
a
c
b
d
Corresponding angles
A
B
D
CD is a perpendicular
bisector of AB
a + d = 180°
b + c = 180°
a c
d b
Co-interior angles
MQ F M fi 11 07( )
B a+b=d
b
b
a
c
d
a + b + c + d = 360°
MQ FurMat fig 11.07(d)
a
c d
A
C D
∠BCD is an exterior angle
MQ FurMat fig 11 07(e)
Right angle at the
circumference in
a semicircle
Ch 08 FM YR 12 Page 350 Friday, November 10, 2000 11:30 AM
350
Further Mathematics
WORKED Example 2
b
Find the values of the pronumerals in the polygon at right.
THINK
1
2
a
WRITE
This shape is a regular hexagon. The
angles at the centre are all equal.
The other two angles in the triangle are
equal.
60°
3
c
The 6 triangles are equilateral triangles,
therefore all sides are equal.
360°
a = ----------6
= 60°
6 cm
a + b + c = 180°
b=c
So:
60 + 2b = 180°
b = 60°
c = 60°
d cm = 6 cm
WORKED Example 3
Find the missing pronumerals in the diagram of railings
for a set of stairs shown at right.
c
a b
THINK
1
WRITE
Recognise that the top and bottom of
the stair rails are parallel lines.
35°
c
a b
35°
35°
2
To find the unknown angle a, use the
alternate angle law and the given angle.
3
The unknown angle c is a right angle,
using the given right angle and
corresponding angle law.
Given angle 35°.
a = 35°
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°
m
dc
Ch 08 FM YR 12 Page 351 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
351
remember
remember
Properties of angles, triangles and polygons
1. Draw careful diagrams.
2. Carefully interpret geometric notations, for example from the diagram below.
Equal sides
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.)
8A
WORKED
Example
1
WORKED
Example
2
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 quadrilateral
c Hexagon
d
e Heptagon
f Nonagon
g
2 Find the value of the pronumerals in the following figures.
a
b
c
27°
130°
52°
y
a
d
c
x
e
15°
63°
f
b
c a
b
c
8 cm
50°
m
32°
Ch 08 FM YR 12 Page 352 Friday, November 10, 2000 11:30 AM
WORKED
Example
3
Further Mathematics
3 Find the value of the pronumerals in the following figures.
a
b
x
y
35°
d
70°
a
d
c
30°
0°
e
62°
z
t
f
m
27°
b c
81°
n
140°
a
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.
a
b
r
c
8 cm
29°
110°
y z
x
d
a
b 122° c
35°
h
3.6 cm
352
d
e
40°
a
4.2 cm
b
a
86°
d
c
b
6 multiple choice
The value of a is closest to:
A 30°
B 75°
a
C 90°
D 120°
150°
E 150°
7 multiple choice
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°
Ch 08 FM YR 12 Page 353 Friday, November 10, 2000 11:30 AM
353
Chapter 8 Geometry: similarity and mensuration
Area and perimeter
4.0
5
Perimeter
14.07
$47,000
5
32.7
32.18
37.92
36.56
Much 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
aº
Corner block with expansive
17
23.55 m frontage
of a house block, the fencing
Lot 603
13.05
$51,000
of a block of land, the size of
2
6
4
5
m
a bedroom and the amount of
paint required to cover an
Lot 658
object. In this section we will
2
761m
Corner block
review the more common
with wide
shapes.
17 m frontage
6
5.8
23.55
Perimeter 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)
Square
Rectangle
l
l
l
l
w
w
l
3. Circumference (C) is the perimeter of a circle.
C = 2 × π × radius
= 2πr
Circumfer
e
l
a c
e of irle
nc
r
WORKED Example 4
THINK
1
The shape is composed of a semicircle
and three sides of a rectangle.
WRITE
Perimeter = 300 + 2 × 600 +
where
1
--2
2
3
Add together the three components of
the perimeter.
Write your answer.
600 mm
300 mm
Find the perimeter of the closed figure given at right
(to the nearest mm).
of circumference =
1
--2
1
--2
circumference
× 2π r
= π × 150
= 471.24
Perimeter = 300 + 2 × 600 + 471.24
= 1971.24
Perimeter of the closed figure is 1971 mm.
Ch 08 FM YR 12 Page 354 Friday, November 10, 2000 11:30 AM
354
Further Mathematics
Area of common shapes
The areas of shapes commonly encountered are:
1. Area of a square:
A = length2 = l2
Square
l
2. Area of a rectangle:
l
Rectangle
A = length × width = l × w
w
l
Parallelogram
3. Area of a parallelogram: A = base × height = b × h
h
4. Area of a trapezium:
A=
1
--2
b
Trapezium
a
(a + b) × h
h
b
5. Area of a circle:
A = π × radius = π × r
2
2
Circle
r
O
6. Area of a triangle:
A = 1--2- × b × h
(see the next chapter)
Triangle
h
b
Area is measured in mm2, cm2, m2, km2 and hectares.
1 hectare = 100 m × 100 m = 10 000 m2
WORKED Example 5
=
3
Write your answer.
1
--2
2.4 m
5.7 m
Find the area of the garden bed given in the diagram
(to the nearest square metre).
THINK
WRITE
7.5 m
a = 7.5
b = 5.7
h = 2.4
1 The shape of garden is a trapezium.
Use the formula for area of a trapezium.
Area of garden = Area of a trapezium
Remember that the lengths of the two parallel
= 1--2- (a + b) × h
sides are a and b and h is the perpendicular
distance between the two parallel sides.
= 1--2- (7.5 + 5.7) × 2.4
2 Substitute and evaluate.
× 13.2 × 2.4
= 15.84 m2
Area of the garden bed is approximately
16 square metres.
Ch 08 FM YR 12 Page 355 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
355
Composite areas
Often a closed figure can be identified as comprising two or more different common
figures. Such figures are called composite figures. The area of a composite figure is the
sum of the areas of the individual common figures.
Area of composite figure = sum of the individual common figures
Acomposite = A1 + A2 + A3 + A4 + . . .
WORKED Example 6
Find the area of the hotel foyer from the plans given below (to the nearest square metre).
25 m
20 m
8m
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.
25 m
A1
A2
8m
1
WRITE
16 m
16 m
THINK
A3
20 m
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
=
1
--2
×b×h
=
1
--2
× 16 × 20
= 160 m2
A2 = Area of rectangle
=l×w
= 25 × 16
= 400 m2
A3 = Area of half of a circle
=
1
--2
× π × r2
=
1
--2
× π × 82
= 100.53 m2
2
Add together all three areas for the
composite shape.
Area of foyer = A1 + A2 + A3
= 160 + 400 + 100.53
= 660.53 m2
3
Write your answer.
Area of the hotel foyer is 661 m2.
Ch 08 FM YR 12 Page 356 Friday, November 10, 2000 11:30 AM
356
Further Mathematics
Conversion of units of area
Often 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
Area
÷102
÷1002 ÷10002
(b) 1 m2 = 100 cm × 100 cm = 10 000 cm2
2
2
2
mm
cm2
m2
km2
(c) 1 km = 1000 m × 1000 m = 1 000 000 m
(d) 1 hectare = 10 000 m2
× 102 × 1002 × 10002
WORKED Example 7
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 = 100 centimetres.
1.12 m2= 1.12 × 1 metre × 1 metre
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).
WORKED Example 8
Convert 156 000 metres2 to a kilometres2
b hectares.
THINK
WRITE
a
a 156 000 m2 = 156 000 ×
b
1
Conversion factor for metres to
kilometres is divide by 1000;
1
- kilometre.
that is, 1 metre = ----------1000
2
Conversion factor for metres2 to
kilometres2 is divide by 10002 or
1 000 000.
3
Write the answer in correct units.
1
Conversion factor is
10 000 m2 = 1 hectare; that is,
1
- hectare
1 m2 = --------------10 000
2
Write the answer.
1
-----------1000
km ×
1
-----------1000
km
156 000
= -----------------------------1000 × 1000
= 0.156 km2
156 000 m2 = 0.156 square kilometres (km2)
b 156 000 m2 = 156 000 ×
=
156 000
------------------10 000
1
---------------10 000
hectares
= 15.6 hectares
156 000 m = 15.6 hectares
2
hectares
Ch 08 FM YR 12 Page 357 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
357
remember
remember
Area and perimeter
1. Perimeter is the distance around a closed figure.
(a) For squares, the perimeter = 4l
Square
l
l
l
l
Rectangle
l
(b) For rectangles, the perimeter = 2(l + w)
w
w
l
mf
ircu ere
o
nce f a
r
circle
(c) Circumference (C) is the perimeter of a circle.
C = 2 × π × radius
C
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 = 1--2- (a + b) × h
(e) Area of a circle: A = π r 2
(f) Area of a triangle: A = 1--2- × b × h
5. Area of composite figure = sum of the individual common figures
Acomposite = A1 + A2 + A3 + A4 + . . .
8B
15.4 cm
cm
4m
.8
27.5 cm
e
70 m
f
5m
13.5 mm
11.5 m
7.5 m
210 m
d
Area
and
perimeter
90
m
HEET
SkillS
7m
Math
cad
5m
17
5
1 Find the areas of the following figures (to the nearest whole units).
a
b
c
12 m
23.7 cm
120 m
Example
4m
WORKED
Area and perimeter
8.1
Ch 08 FM YR 12 Page 358 Friday, November 10, 2000 11:30 AM
358
WORKED
Example
Further Mathematics
2 Find the perimeters of the closed figures in question 1.
4
WORKED
Example
6
3 Find the areas of the following figures (to 1 decimal place).
a
b
c
10 m
14 m
3.5 m
17 m
12 m
2m
12 m
20 m
m
0c
12
cm
22 m
34 m
1
125 mm
f
16 cm
m
13
e
11 m
90 mm
10 cm
24 mm
8 cm
d 48 mm
44 m
25 m
21 cm
7m
20 m
4 Find the perimeters of the closed figures in question 3.
WORKED
Example
7, 8
5 Convert the following areas to the units given in brackets.
b 320 000 cm2 (m2)
c
a 20 000 mm2 (cm2)
2
2
2
2
e 2 500 000 m (km )
f
d 0.035 m (mm )
g 2 750 000 000 mm2 (m2) h 0.000 06 km2 (m2)
0.035 m2 (cm2)
357 000 m2 (hectares)
6 A kite has the dimensions in the figure at right. Find the
area of the kite (to the nearest cm2).
70 cm
180 cm
8.2
0m
1.2
7
Find the area of the regular hexagon
as shown in the diagram at left
(to 2 decimal places, in m2).
2.08 m
30 mm
8 A cutting blade for a craft knife has the dimensions
shown in the diagram.
What is the area of steel in the blade?
20 mm
SkillS
HEET
5 mm
40 mm
Ch 08 FM YR 12 Page 359 Friday, November 10, 2000 11:30 AM
359
Chapter 8 Geometry: similarity and mensuration
The area in m2 of the stacked objects shown at right
is closest to:
A 1.44
B 1.68
C 1.92
D 3.84
E 11.52
0.8 m 0.8 m 0.8 m 0.8 m
9 multiple choice
0.6 m
0.8 m
1.0 m
1.2 m
10 multiple choice
The perimeter of the figure shown in centimetres is:
A 34
B 24 + 5π
C 24 + 2.5π
D 29 + 5π
E 29 + 2.5π
7 cm
2 cm
3 cm
12 cm
11 multiple choice
20.5 m
35.2 m
The perimeter of the enclosed figure shown is
156.6 metres. The unknown length, x, is closest to:
A 20.5 m
B 35.2 m
C 40.2 m
D 80.4 m
E Cannot be determined
x
12 A 3-ring dartboard has dimensions as shown below left. (Give all answers to 1
decimal place.)
40 cm
20 cm
6 cm
1
2
3
2
1
MQ FurMat fig 11.59
HEET
SkillS
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).
8.3
Ch 08 FM YR 12 Page 360 Friday, November 10, 2000 11:30 AM
360
Further Mathematics
13 On a western movie set, a horse is tied to a railing outside a saloon bar. The
railing is 2 metres long; the lead on the horse is also 2 metres long and 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 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 it
makes a semicircular sweep. The base of the wiper is situated at the bottom centre of
the 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.
Ch 08 FM YR 12 Page 361 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
361
Total surface area
The 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 objects
Cube
Cylinder
Cuboid
r
l
h
l
w
Cubes:
TSA = 6l2
h
Cuboids:
TSA = 2(lw + lh + wh)
Cone
r
Sphere
Cylinders:
TSA = 2π r(r + h)
Slant
s height
h
r
r
Cones:
TSA = π r(r + s) where
s is the slant height
Spheres:
TSA = 4π r 2
WORKED Example 9
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.
3m
5 cm
1.1
THINK
1
2
3
A poster tube is a cylinder.
Express all dimensions in centimetres.
Remember 1 metre = 100 centimetres.
Substitute and evaluate.
Remember BODMAS.
Write your answer.
WRITE
Radius, r = 5 cm
Height, h = 1.13 m
= 113 cm
TSA of a tube = 2π r(r + h)
= 2 × π × 5(5 + 113)
= 2 × π × 5 × 118
= 3707.08
The total surface area of a poster tube is
approximately 3700 cm2.
Ch 08 FM YR 12 Page 362 Friday, November 10, 2000 11:30 AM
362
Further Mathematics
WORKED Example 10
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
1
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.
2
Write your answer.
Diameter = 25 cm
Radius = 12.5 cm
TSA of sphere = 4π r 2
= 4 × π × 12.52
= 1963.495
Total surface area of the ball is approximately 1960 cm2.
WORKED Example 11
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
1
2
3
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.
Take the square root of both
sides to find l.
Write your answer.
TSA = 6 × l2
TSA = 1350 mm2
1350 = 6 × l2
l2 =
1350
-----------6
= 225
l = 225
= 15 mm
The dimensions of the die are:
15 mm × 15 mm × 15 mm
Total surface area using a net
If 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 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.
Square pyramid
Slant
height
Net
MQ FurMat fig 11.68
Net
MQ FurMat fig 11.69
Net
Ch 08 FM YR 12 Page 363 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
363
WORKED Example 12
10 cm
8 cm
Form a net of the triangular prism,
transferring all the dimensions to each
of the sides of the surfaces.
10
10 cm
A1
10 cm
2
Identify the different-sized common
figures and set up a sum of the surface
areas. The two triangles are the same.
cm
A4 6 cm
8 cm
20 cm
1
WRITE
A2
A3
20 cm
THINK
20 cm
20
cm
6 cm
Find the total surface area of the triangular prism shown in the diagram.
8 cm 6 cm
10 A4 6 cm
cm
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 =
1
--2
× b2 × h 2
=
1
--2
×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.
Ch 08 FM YR 12 Page 364 Friday, November 10, 2000 11:30 AM
364
Further Mathematics
WORKED Example 13
12 cm
Find the surface area of an open cylindrical can that is 12 cm high and 8 cm in diameter
(to 1 decimal place).
8 cm
fi
1
WRITE
Form a net of the open cylinder,
transferring all the dimensions to each
of the surfaces.
2π r
A1
12 cm
THINK
A2
4 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.
3
Sum the areas.
4
Write your answer.
TSA = A1 + A2
A1 = 2 π r × w
= 2 × π × 4 × 12
= 301.59 cm2
A2 = π × r 2
= π × 42
= 50.27 cm2
TSA = A1 + A2
= 301.59 + 50.27
= 351.86 cm2
The total surface area of the open cylindrical
can is 351.9 cm2.
remember
remember
Total 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 = 2π r(r + h)
(d) Cones: TSA = π r(r + s) where s is the slant height
(e) Spheres: TSA = 4π r2
3. For all other shapes, form their nets and establish the total surface area formulas.
Ch 08 FM YR 12 Page 365 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
8C
WORKED
Example
10
2 Find the total surface area of the objects given in the diagrams. Give answers to
1 decimal place.
a Length = 1.5 m
b
c
410 mm
7 cm
d
Diameter = 43 cm
4 cm
e
f
90 cm
6 cm
8 cm
28 cm
2 cm
WORKED
Example
12
4 Find the total surface areas for the objects given in the diagrams. Give answers to
1 decimal place.
a
b
10 cm
m
15 c
6.06 cm
11
3 Find 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 cm
Example
5 cm
WORKED
m
30 c
7 cm
Math
cad
9
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 cm
b A cuboid with dimensions of 12 m × 5 m × 8 m (l × w × h)
c A sphere with a radius of 0.8 metres
d A closed cylinder with a radius of 1.2 cm and a height of 6 cm
e A closed cone with a radius of 7 cm and a slant height of 11 cm
f An opened cylinder with a diameter of 100 mm and height of 30 mm
4 cm
Example
Total surface area
14 cm
WORKED
365
Total
surface
area
Ch 08 FM YR 12 Page 366 Friday, November 10, 2000 11:30 AM
Further Mathematics
d
Area = 22 cm2
8 cm
c
22 mm
366
30
mm
mm
105
40 mm
80 mm
13 cm
f
15
m
m
30
m
5 Find the total surface area of each of the objects in the diagrams below. Give answers
to 1 decimal place.
a Rubbish bin
b
13.5 cm
cm
10.5 cm
10
15
cm
20 cm
250 mm
2.5 m
1.2 m
0.9 m
d
3 cm
c
2 cm
13
4m
7m
2
250 mm
Example
6m
5 mm
9 mm
WORKED
5m
4m
12 mm
e
4.5 cm
7 cm
1.5 m
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 answer in m2 and cm2 to 1 decimal
place.)
Ch 08 FM YR 12 Page 367 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
2.5
m
1.5 m
1.0 m
7 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.
367
4.5 m
6.5 m
8 multiple choice
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 multiple choice
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 multiple choice
The formula for the total surface area for the object shown is:
A
1
--- abh
2
B 2×
1
--2
bh + ab + 2 × ah
C 3( 1--2- bh + ab)
D
1
--2
bh + 3ab
E bh + 3ab
h
a
b
11 multiple choice
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
ET
SHE
Work
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?
8.1
Ch 08 FM YR 12 Page 368 Friday, November 10, 2000 11:30 AM
368
Further Mathematics
Volume of prisms, pyramids and spheres
The 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 volume
Often the units of volume need to be converted, for example from cm3 to m3 and vice versa.
Volume
÷103
mm3
× 103
÷1003
cm3
m3
× 1003
WORKED Example 14
Convert 1.12 cm3 to mm3.
THINK
1
The conversion from centimetres to
millimetres is 1 cm = 10 mm.
2
The conversion factor for cm3 to mm3
is to multiply by 103 or 1000; that is,
1cm3 = 1000 mm3.
Write the answer in correct units.
3
WRITE
1.12 cm3 = 1.12 × 1 cm × 1 cm × 1 cm
= 1.12 × 10 mm × 10 mm × 10 mm
= 1.12 × 1000 mm3
= 1120 mm3
1.12 cm3 is equal to 1120 mm3.
Ch 08 FM YR 12 Page 369 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
WORKED Example 15
Convert 156 000 cm3 to:
a m3
369
b litres.
THINK
WRITE
a
a 156 000 cm3 = 156 000 × 1 cm × 1 cm × 1 cm
b
1
The conversion factor for
centimetres to metres is
divide by 100; that is,
1
- m.
1 cm = -------100
= 156 000 ×
2
The conversion factor for cm3 to m3
is divide by 1003 or 1 000 000;
that is, 1 000 000 cm3 = 1 m3.
3
Write the answer in correct
units.
1
Conversion factor is
1000 cm3 = 1 litre; that is,
1
- litre.
1 cm3 = ----------1000
2
1
--------100
m×
1
--------100
m×
1
--------100
m
156 000
= --------------------------------------- m3
100 × 100 × 100
= 0.156 m3
156 000 cm3 = 0.156 cubic metres (m3)
b 156 000 cm3 = 156 000 ×
=
156 000
------------------1000
1
-----------1000
litres
litres
= 156 litres
156 000 cm3 = 156 litres
Write the answer.
Volume of prisms
A prism is a 3-dimensional object that has a
uniform cross-section.
Triangular prism
Cylinder
Square
prism
A prism is named in accordance with its uniform crosssectional area. Note: Circular prisms are called cylinders.
Uniform
cross-section
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
Height
Ch 08 FM YR 12 Page 370 Friday, November 10, 2000 11:30 AM
370
Further Mathematics
WORKED Example 16
THINK
1
2
WRITE
15 cm
20 cm
Find the volume of the object (to the nearest cm3).
Vcylinder = A × H
where Acircle = π r 2
The object has a circle as a uniform
cross-section. It is a cylinder. The area
of the base is: area of a circle = π r 2.
Volume is cross-sectional area times
height.
Vcylinder = π × r 2 × H
= π × 152 × 20
= 4500 π
= 14 137.1669 cm3
The volume of the cylinder is 14 137 cm3.
Write your answer.
WORKED Example 17
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.
Area 250 mm2
17 mm
THINK
1
2
WRITE
The slice of bread has a uniform crosssection. The area of the cross-section is
not a common figure but its area has
been given.
Write your answer.
V=A×H
where A = 250 mm 2
V = 250 mm2 × 17 mm
= 4250 mm3
The volume of the slice of bread is 4250 mm3.
Given the volume of an object, we can use the volume formula to find an unknown
dimension of the object by transposing the formula.
Ch 08 FM YR 12 Page 371 Friday, November 10, 2000 11:30 AM
371
Chapter 8 Geometry: similarity and mensuration
WORKED Example 18
Volume of prism = 6.6 m3
Find the height of the triangular prism from the information
provided in the diagram at right (to 1 decimal place).
2m
THINK
1.1
m
h
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 m
V=A×H
where A = 1--2- × b × h
2
Substitute the values, transpose and
evaluate.
3
Write your answer.
6.6 = 1--2- × 2 × h × 1.1
= 1.1 h
6.6
h = ------1.1
The height of the triangle in the given prism is
6.0 metres.
1
V=
1
--2
×b×h×H
Volume of pyramids
A pyramid is a 3-dimensional object that has a similar cross-section but the size
reduces as it approaches the vertex.
Vertex
Triangular pyramid
Cone
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 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, Vpyramids, can be generalised by
the formula:
Vpyramids =
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.
H
A
Ch 08 FM YR 12 Page 372 Friday, November 10, 2000 11:30 AM
372
Further Mathematics
WORKED Example 19
Find the volume of the pyramid at right (to the nearest m3).
THINK
1
Height of pyramid = 40 m
WRITE
The pyramid has a square base. It is a
square pyramid. The area of the base is:
Area of a square = l 2.
30 m
30 m
Vpyramid = × A × H
where Asquare = l 2
1
--3
Vpyramid = 1--3- × l 2 × H
= 1--3- × 302 × 40
2
= 12 000 m3
The volume of the square pyramid is 12 000 m3.
Write your answer.
Volume of spheres and composite objects
Volume of a sphere
Spheres are unique but common objects that deserve special attention.
The formula for the volume of spheres is:
Vsphere = 4--3- π r 3
where r is the radius of the sphere.
r
Ch 08 FM YR 12 Page 373 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
373
Volume 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 composite object is found by adding the volumes of the individual common figures or
deducting volumes. The grain silo can be modelled as the sum of a cylinder and a large
cone, less the tip of the large cone.
Volume of composite object = sum of the individual common prisms, pyramids or
spheres.
Vcomposite = V1 + V2 + V3 + . . . (or Vcomposite = V1 − V2)
WORKED Example 20
12 cm
25 cm
20 cm
Find the volume of the object shown at right (to the nearest litre).
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.
18 cm
r = 6 cm
18
cm
25 cm
1
WRITE
H = 20 cm
THINK
18 cm
Vcomposite = volume of cylinder + volume of
square prism
= Acircle × Hcircle + Asquare × Hsquare
= (π r 2 × Hc) + (l2 × Hs)
2
3
Convert to litres using the conversion of
1000 cm2 = 1 litre.
Write your answer.
= (π × 62 × 20) + (182 × 25)
= 2261.946 711 + 8100
= 10 361.946 711 cm3
10 362 cm2 = 10.362 litres
The volume of the object is 10 litres.
Ch 08 FM YR 12 Page 374 Friday, November 10, 2000 11:30 AM
374
Further Mathematics
remember
remember
Volume 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 × height
V=A×H
4. (a) The volume of a pyramid is Vpyramid = 1--3- × area of cross-section at the
base × height
V=
1
--3
×A×H
(b) The height of a pyramid, H, is sometimes called the altitude.
5. The volume of a sphere is Vsphere = 4--3- π r 3.
6. The volume of a composite object = sum of the individual common prisms,
pyramids or spheres.
Vcomposite = V1 + V2 + V3 + . . .
(or Vcomposite = V1 − V2 . . . )
8D
Example
16
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
2 Find the volume of the following prisms to the nearest whole unit.
a
b
75
c
7 cm
mm
cm
WORKED
b 4800 cm3 to m3
104.8 cm
4000 mm
4 cm
23
8.4
a 0.35 cm3 to mm3
15 cm
d
e
f
2.1 m
m
4.8
6.4 m
14 mm
SkillS
HEET
14, 15
51.2 cm
Volume
formulas
1 Convert the volumes to the units specified.
m3m0
Mat
Example
20 mm
WORKED
d
hca
Volume of prisms, pyramids
and spheres
22 mm
34 mm
m
57 m
Ch 08 FM YR 12 Page 375 Friday, November 10, 2000 11:30 AM
375
Chapter 8 Geometry: similarity and mensuration
Example
17
3 Find the volume of the following prisms (to 2 decimal places).
a
b
2
Area = 4.2 m
0.5
m
WORKED
2.9 m
c
Area = 1000 cm2
d
Area = 15 cm2
14.5 mm
Area = 120 mm2
Area =
32 cm2
WORKED
Example
8.5
cm
4 Find the measurement of the unknown dimension (to 1 decimal place).
18
b Volume of triangular prism =
1316.1 cm3
= 1.728 m3
x
c
x
d
x
15
.0 c
m
Volume of prism = 10 1–8 litres
120 mm
a Volume of cube
cm
21.4
Volume of cylinder
= 150 796.4 mm3
VO = 10 cm
11 cm
O
8m
12 m
O
12 cm
11 cm
d
e 12 mm
O
4 cm
VO = 8 cm
f
V
V
VO = 15 cm
Altitude of square
pyramid = 18 mm
O
6 cm
Base of
pyramid
6 cm
19
5 Find the volume of these pyramids (to the nearest whole unit).
a
b
c VO = 17m V
V
35 cm
Example
4 cm
WORKED
3x
x
10 cm
Ch 08 FM YR 12 Page 376 Friday, November 10, 2000 11:30 AM
8
7 cm
8 cm
3m
5m
r=
c
4 cm
cm
4m
b
10 cm
a
20 cm
d
e
f
1m
10 cm
2.1 m
6m
15 cm
6m
10 cm
4m
g
2.5 m
h
42 m
100 mm
60 m
42 m
25 mm
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 of
1024 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 of
9.425 cm3.
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). Find the volume to the nearest millilitre, where
1 cm3 = 1 mL.
4 cm
5 cm
20
6 Find the volume of these objects (to the nearest whole unit).
2m
Example
3m
WORKED
Further Mathematics
19 m
376
Ch 08 FM YR 12 Page 377 Friday, November 10, 2000 11:30 AM
Chapter 8 Geometry: similarity and mensuration
377
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, find:
a the height of the cylindrical can
b 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 air
e the fraction of the can’s volume occupied by the balls.
10 multiple choice
The volume 200 000 mm3 is equivalent to:
C 20 cm3
A 2 litres
B 2 cm3
D 200 cm3
E 2000 cm3
11 multiple choice
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
4
--3
B
2
--3
4
--- r
3
---C h
D
3
--4
E
3
--2
r
r
12 multiple choice
If the volume of the square pyramid shown is 6000 m3, then the perimeter of the base
is closest to:
V
A 900 m
VO = 20 m
B 20 m
C 30 m
D 80 m
O
E 120 m
13 multiple choice
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
3
3
D 1020.1 cm
E 4084.1 cm
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 flight?
b If the plane can be manoeuvred in a hemispherical zone, find:
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.
Ch 08 FM YR 12 Page 378 Friday, November 10, 2000 11:36 AM
378
Further Mathematics
Maps and similar figures
Maps and scales
We often need to refer to maps for specifying locations or for establishing distances
between two locations. Maps are a reduction of lengths in real life; that is, they have
the same shape as the original but are much smaller in size. A measure of the amount
of 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 represents
100 units in real life. In the map below one unit on the map represents 50 000 units.
SCALE 1:50 000
METRES 1000
0
1
2
3 KILOMETRES
2. A simple conversion scale where, for example, 1 cm = 100 m means 1 cm on the
map represents 100 metres in real life. In the map below 1 cm on the map represents
1 km.
Kilometres 0
1
2
3
4
5
6
7
8 Kilometres
Converting from one type of map scale to another is shown in the following
example.
Ch 08 FM YR 12 Page 379 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
379
WORKED Example 21
Convert the following map ratio scales:
a 1:50 000 to a simple conversion scale with units of centimetres
b 2:25 model scale to simple scale with units of millimetres
c 1:250 000 to simple scale with units of centimetres.
THINK
WRITE
a 1 Rewrite the map scale including the
a 1:50 000
unit centimetres.
1 cm: 50 000 cm
2 Convert 50 000 cm to a more
50 000
1 cm: ---------------- m
appropriate unit of length, for
100
example 100 cm = 1 m.
1 cm = 500 m
b Rewrite the map scale including the unit
b 2:25
millimetres. Divide by 2 to reduce to a
2 mm = 25 mm
unit.
1 mm = 12.5 mm
c 1:250 000
c 1 Rewrite the map scale including the
1 cm = 250 000 cm
unit centimetres.
2 Convert 250 000 cm to a more
250 000
1 cm = ------------------- m
appropriate unit of length.
100
Remember 100 cm = 1 m
2500
1 cm = ------------ km
1000 m = 1 km
1000
1 cm = 2.5 km
To find the distance represented on a map, use the simple conversion scale and
proportion to the desired value as shown in the next two examples.
Converting map distances to real-life distances
WORKED Example 22
Find the distance in real life represented by:
a 7 mm on a map with 1:100 000 scale
b 11.5 cm on a map with a scale 1 cm = 50 km.
THINK
a
1
2
3
b
1
2
WRITE
a 1:100 000
1 mm:100 000 mm
1 mm:100 m
A map distance of 7 mm corresponds
7 × 1 mm = 7 × 100 m
to an actual distance of 7 times 100 m.
7 mm = 700 m
Write your answer.
7 mm on the map represents 700 m in
real life.
Proportion the scale by multiplying
b
1 cm = 50 km
both sides by 11.5.
11.5 × 1 cm = 11.5 × 50 km
11.5 cm = 575 km
Write your answer.
On a map with a scale of 1 cm = 50 km,
11.5 cm represents 575 km.
Convert map scale ratio to a
conversion scale.
Ch 08 FM YR 12 Page 380 Friday, November 10, 2000 11:36 AM
380
Further Mathematics
Converting real life distances to map distances
WORKED Example 23
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
1
Convert ratio scale to a simple
conversion scale using an
appropriate unit of measure.
2
Multiply by 2.5 to go from 2 km to
5 km and do it with both sides.
a 1:200 000
1 cm:200 000 cm
1 cm = 2000 m
1 cm = 2 km
× 2.5
1 cm = 2 km
× 2.5
x cm = 5 km
2.5 cm = 5 km
3
b
1
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.
÷4
1 cm = 2000 m
÷4
x cm = 500 m
2
3
Convert 1--4- or 0.25 cm to mm.
Write your answer.
0.25 cm = 500 m
2.5 mm = 500 m
On a 1:200 000 map, 500 m is represented
by 2.5 mm.
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:
B'
C'
2
1. The ratios of the corresponding sides must be equal. 4
A′B′
B′C′
C′D′
A′D′
------------ = ------------ = ------------ = ------------ = common ratio
AB
BC
CD
AD
A'
6
2
C
B 1
2
3
A 1
D'
D
C'
B' 125°
2. The corresponding angles must be equal.
∠A = ∠A′ ∠B = ∠B′ ∠C = ∠C′ ∠D = ∠D′
60°
B
125°
A
85°
A'
D'
C
60°
85°
D
Ch 08 FM YR 12 Page 381 Friday, November 10, 2000 11:36 AM
381
Chapter 8 Geometry: similarity and mensuration
Scale factor, k
A measure of the relative size of the two similar figures is the scale factor. The scale factor
is the common ratio of the corresponding sides and quantifies the amount of enlargement
or reduction one figure undergoes to transform into the other figure. 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 map scale ratios.
For example, k = 2, k =
1
-----12
B'
or 1:10 000.
B
length of image
A′B′
B′C′
C′A′
2. Scale factor, k = -------------------------------------------- = ------------ = ------------ = -----------length of original
AB
BC
CA
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.
3
9
3
9
A 1 C
A'
3
C'
Enlargements and reductions are important in many aspects of photography, map
making and modelling. Often, photographs are doubled in size (enlarged), while house
plans are an example of a reduction to a scale, for example 1:25.
WORKED Example 24
THINK
WRITE
a
a
1
As it is a reduction, the larger shape
is the original and the smaller shape
is the image.
20
Original
cm
45 cm
For the similar shapes shown at right:
a find the scale factor for the reduction of the shape
b find the unknown length in the small shape.
Image
cm
10
Continued over page
x
Ch 08 FM YR 12 Page 382 Friday, November 10, 2000 11:36 AM
382
Further Mathematics
THINK
b
WRITE
length of image
Scale factor, k = ----------------------------------------length of original
A′B′
= -----------AB
10 cm
= --------------20 cm
1
= --2-
2
The two shapes have been stated
as being similar, so set up the
scale factor ratio, k.
1
Use the scale factor to determine
the unknown length as all
corresponding lengths are in the
same ratio.
2
b Scale factor, k =
1
--2
length of image
k = ----------------------------------------length of original
x
1
--- = -------------2
45 cm
x = 1--2- × 45 cm
x = 22.5 cm
The scale factor of reduction is 1--2- and the
unknown length on the smaller shape is 22.5 cm.
Write your answers.
WORKED Example 25
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.
40°
s
100m
20°
t
20°
30
m
40°
m
70
WRITE
a
40°
30°
70
20°
m
Image
40°
270°
30°
50 m
270°
30
m
s
100m
THINK
a Firstly, orientate the figures to identify
corresponding sides and angles easily.
Calculate the missing angles and
compare each pair of corresponding
angles.
30°
50 m
30°
t
20°
Original
Sum of interior angles = 360°
All corresponding angles are equal.
Ch 08 FM YR 12 Page 383 Friday, November 10, 2000 11:36 AM
383
Chapter 8 Geometry: similarity and mensuration
THINK
WRITE
b
length of image
b Scale factor, k = ----------------------------------------length of original
s
For s
2 = -----------30 m
s = 2 × 30 m
= 60 m
70 m
For t
2 = -----------t
70 m
t = -----------2
= 35 m
1
As the scale factor given is for
enlargements, the original is the
smaller figure.
2
Set up the scale factor ratio for each
of the two sides.
remember
remember
Maps 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
B'
For two figures to be similar, they must have the
following properties:
4
1. The ratios of the corresponding sides must be equal.
A′B′
B′C′
C′D′
A′D′
------------ = ------------ = ------------ = ------------ = common ratio
AB
BC
CD
AD
2. The corresponding angles must be equal.
∠A = ∠A′ ∠B = ∠B′ ∠C = ∠C′ ∠D = ∠D′
A'
C'
2
2
C
B 1
6
D'
B' 125°
2
3
A 1
D
C'
60°
C
B
125°
A
85°
A'
60°
85°
D
D'
Scale factor, k
1. Scale factor, k, is the amount of enlargement or reduction and is expressed as
1
- or 1:10 000.
integers, fractions or map scale ratios, for example k = 2, k = ----12
length of image
A′B′
B′C′
C′A′
2. Scale factor, k = ----------------------------------------- = ------------ = ------------ = -----------length of original
BC
CA
AB
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'
B
3
3
9
9
A1C
A' 3 C'
Ch 08 FM YR 12 Page 384 Friday, November 10, 2000 11:36 AM
Further Mathematics
8E
21
WORKED
22
c
1:125 000
d 2:40 000
e 1:1 750 000
f
1:500
2 State the real-life distance represented on a map for each of the following:
a 22 cm on a 1 cm = 1.5 km map
c
8 mm on a 1 mm = 100 m map
e 17 cm on a 1:20 000 map
WORKED
Example
23
f
25 mm on a 1:200 000 map.
b 750 m on a 1:25 000 map
c
d 25 m on a 1:500 map
100 km on a 1:200 000 map
f
12 km on a 1:750 000 map.
4 For each of these pairs of similar shapes, find:
24
i the scale factor
a
ii the value of x and y.
b
y cm
200 cm
x cm
x cm
50
50
70
1m
20 cm
4m
c
d
m
Example
d 13 cm on a 1:750 000 map
a 4 km on a 1:100 000 map
e 300 m on a 1:150 000 map
WORKED
b 8.5 cm on a 1 cm = 200 m map
3 State the distance on a map for each of the following:
2c
Scale
factor
b 1:1000
42 mm
y mm
8 cm
m
Mat
Example
a 1:500 000
8c
d
hca
1 Convert the following map ratio scales to simple conversion scales with cm as the unit
of measure.
y cm
Example
x
y cm
x cm
4 cm
63 mm
WORKED
Maps and similar figures
25 metres
384
21 mm
7 mm
Ch 08 FM YR 12 Page 385 Friday, November 10, 2000 11:36 AM
385
Chapter 8 Geometry: similarity and mensuration
m
1– m
82
15°
a
24 cm
45°
60°
30°
22
41 cm
mm
a
60°
15°
38 cm
c
30°
cm
d
48 cm
17
36
c
m
mm
17
45°
37 cm
7.5
12
24
33°
a
25
5 Prove that the following pairs are similar figures and find the value of a.
a
b
40 mm
Example
62 mm
WORKED
30
32
Photo
16
a°
Height of person = 186 cm
10
40
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 scale
to draw a scale diagram
on an A4 sheet is 1:20.
Thomas
N
Badger
Find the dimensions of
the scale drawing of the
Blazing
room and state whether
er
Riv
the drawing should be
DAVIS
LAND
Temple
landscape or portrait on
ple
Tem
the A4 sheet.
Rom
e
Parry
Danby
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 Thomas
ii McLeod and Clowes
iii Sharpe and Thomas.
Clowes
West
Sea
Riv
er
Sharpe
0
KILOMETRES
10
20
30
40
McLeod
Gold mine
Copper mine
Coal mine
Silver mine
Ch 08 FM YR 12 Page 386 Friday, November 10, 2000 11:36 AM
Further Mathematics
10 Find the distance between the following pairs of locations in the map (to the nearest
kilometre).
SCALE 1:1 000 000
N
2750 m
Kantar
Martina
r
ve
Ri
Paxton
Stuckley
Sea
Shelly Beach
Reneton
Foster
Plains
Baletta
River
ing
1253 m
Newbury
Sw
Goldern
Sea
BRAMBLETOWN
Snake
Riv
er
386
Bolivia
3014 m
Markham
Jewel
Ross
a
b
c
d
e
Charleston
From Brambletown to Ross in a straight line
From Charleston to Markham in a straight line
From Shelly Beach to Baletta in a straight line
From Charleston to Ross in a straight line
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 of
Brambletown.
12 multiple choice
The 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
13 multiple choice
4 cm
2 cm
A 1:27 scale model of a truck is made from clay. What is the length of the tray on the
original truck, if it is 27 cm length on the model?
A 1 cm
B 100 cm
C 270 cm
D 540 cm
E 729 cm
14 multiple choice
A scale factor of 0.2 is:
A a reduction with a scale of 1 cm = 2 cm
B an enlargement with a scale of 1 cm = 0.2 cm
C an enlargement with a scale of 1 cm = 5 cm
D a reduction with a scale of 1 cm = 5 cm
E a reduction with a scale of 1 cm = 20 cm
Ch 08 FM YR 12 Page 387 Friday, November 10, 2000 11:36 AM
387
Chapter 8 Geometry: similarity and mensuration
Similar triangles
Similar triangles can be used to find the height of trees and buildings or widths 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 identified:
1. All three corresponding angles
are equal (AAA).
2. All three corresponding pairs 1
of sides are in the same
ratio (linear scale factor)
(SSS).
2
3
sf =
2
4
1
--2
=
2
--4
=
4
--2
=2
=
3
--6
= 0.5
6
3. Two corresponding pairs of sides
are in the same ratio and the
included angles are equal
(SAS).
3
sf =
2
6
6
--3
4
As in the previous section, we use the known values of a pair of corresponding sides
to determine the scale factor for the similar triangles.
OA′
length of side of image
Scale factor, k = ---------- = ---------------------------------------------------------------------------------------------------length of corresponding side of original
OA
WORKED Example 26
For the similar triangles in the diagram, find
a the scale factor
b the value of the pronumeral, x.
THINK
WRITE
a
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°.
A
A
B
4 100°
30°
6
B'
C
A'
B
Original
4 100°
30°
50°
C
6
A'
6
30°
100°
C'
x
B'
6
30°
100°
x
Image
50°
C'
Continued over page
Ch 08 FM YR 12 Page 388 Friday, November 10, 2000 11:36 AM
388
Further Mathematics
THINK
2
WRITE
Scale factor,
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.
length of side of image
k = ----------------------------------------------------------------------------------------------length of corresponding side of original
A′B′
= -----------AB
=
6
--4
= 1.5
b
1
2
Use the scale factor to find the
unknown length, x.
Transpose and evaluate.
b Scale factor, k = 1.5
Write answer in the correct
units and level of accuracy.
A′C′
1.5 = -----------AC
x
1.5 = --6
x = 1.5 × 6
x=9
The scale factor is 1.5 and the unknown length, x,
is 9 units.
WORKED Example 27
For the given similar triangles, find the value of the
pronumeral, x.
D
3.5
B
4.0
x
C
7
All measurements in m
A
THINK
1
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.
WRITE
B
.0
For clear analysis separate the two
triangles. Note that the lengths of the
sides AE and AD are the sum of the
given values.
C
7m
A
.5
D
Image
m
7
A
2
Original
m
4
E
E
(7 + x) m
AD = 4.0 + 3.5
= 7.5 m
AE = (7 + x) m
Ch 08 FM YR 12 Page 389 Friday, November 10, 2000 11:36 AM
389
Chapter 8 Geometry: similarity and mensuration
THINK
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.
4
5
Use the scale factor to find the
unknown length.
Transpose and evaluate.
Write answer in the correct units and
level of accuracy.
WRITE
Scale factor,
length of side of image
k = ----------------------------------------------------------------------------------------------length of corresponding side of original
AD
= -------AB
7.5
= ------4.0
k = 1.875
Scale factor,
length of side of image
k = ----------------------------------------------------------------------------------------------length of corresponding side of original
AE
1.875 = -------AC
7+x
1.875 = -----------7
1.875 × 7 = 7 + x
13.125 = 7 + x
x = 13.125 − 7
x = 6.125
The value of x is 6 1--8- 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 widths of rivers. This problem is overcome by setting up a triangle
similar to the feature to be examined, as shown in the next example.
WORKED Example
28
s
n's
Find the height of the tree shown in the diagram at right.
Give the answer to 1 decimal place.
Su
Shadow
(140 cm)
THINK
1
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.
WRITE
Girl
(168 cm)
14 metres
Original
168 cm
xm
140 cm
Image
14 m
Continued over page
ray
Ch 08 FM YR 12 Page 390 Friday, November 10, 2000 11:36 AM
390
Further Mathematics
THINK
WRITE
2
For clear analysis separate the two triangles.
3
4
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.
Transpose and evaluate.
5
Write answer in the correct units.
height of tree (image)
Scale factor, k = ------------------------------------------------------height of girl (original)
14
x
k = ------- = ---------1.4
1.68
x
10 = ---------1.68
x = 10 × 1.68
= 16.8 m
Height of the tree is 16.8 metres.
remember
remember
Similar triangles
length of side of image
OA′
1. Scale factor, k = ---------- = ----------------------------------------------------------------------------------------------length of corresponding side of original
OA
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 1
same ratio (linear scale factor) (SSS).
2
3
2
4
6
sf =
1
—
2
=
2
—
4
(c) Two corresponding pairs of sides are in the
same ratio and the included angles are equal
(SAS).
=
3
—
6
= 0.5
3
6
4
8
Ch 08 FM YR 12 Page 391 Friday, November 10, 2000 11:36 AM
391
Chapter 8 Geometry: similarity and mensuration
8F
26a
1 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
b
c
mm
320
240 mm
25°
m
m
0
48
m
0
64
5.6
25°
m
d
5
4.6
Example
10
8.8
4.4
9.2
WORKED
Similar triangles
4.5
.2
11
9
e
f
10.5
0.5
10.5
1
0
.
7
26b
2
2 For the given similar triangles, find the value of the pronumeral, a.
a
b
c
22.5 mm
mm
am
12 m
14.4 m
71°
e
7
f
4
16
67°
6
13
9.6
9
x
x
3 For the given similar triangles, find the value of the pronumeral, a.
a
b
c
6
7.5
a
2
a
4.5
10
.5
27
7.8
a x
x
Example
6
12
14
WORKED
a
3.2
8
a
38°
a
d
12
12
3
12
Similar
triangles
59 cm
38°
20
56°
m
cm
cm
25
56°
15
45 cm
am
75
15 mm
62° 62°
62°
m
62°
Cabri Geom
etry
Example
3.5
4
7
WORKED
3
14
8
a
Ch 08 FM YR 12 Page 392 Friday, November 10, 2000 11:36 AM
392
Further Mathematics
e
f
15.
2
18
17.2
80°
32
4m
m
m
am
10
8m
68 m
a
a
17 m
d
m
80°
m
142 m
43
WORKED
Example
28
4 Find the height of the flagpole shown in the diagram
at right (to the nearest centimetre).
Guy wire
0.9 m
1m
9m
5 Find the length of the bridge, AB , needed to span
the river, using similar triangles as shown (to the
nearest decimetre).
B
(Not to scale
All measurement
are in metres)
2.5 m
A 12.5 m
4.3 m
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?
Lake
7 Find the width of the lake (to the nearest metre) using
these surveyor’s notes at right.
20
b
25 m
In the given diagram, the length of
side b is closest to:
A 24
B 22
C 16
D 15
E 9.6
2m
1.2 m
1.1 m
16
12
8 multiple choice
Not to scale
9 multiple choice
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
Work
8.2
10 multiple choice
The height of the player, y, as shown is closest to:
A 190 cm
B 180 cm
C 170 cm
D 160 cm
y
x
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.
ET
SHE
B
A
0.9 m
5m
10 m
E 1.6 m
E 150 cm
Ch 08 FM YR 12 Page 393 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
393
Area and volume scale factors
An unknown area or volume of a figure can be found without the need to use known
formulas such as in exercises 8B and 8D. We have seen that two figures 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 figures.
Area of similar figures
If the lengths of similar figures 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.
Different length ratios (or scale factors) of a square
1 cm
length of blue square
2 cm
-------------------------------------------------- = ----------- = 2 = k
length of red square
1 cm
area of blue square
4 cm 2
--------------------------------------------- = -------------2- = 4 = 2 2 = k 2
area of red square
1 cm
Area
=
1 cm2
1 cm
2 cm Area = 4 cm2
length of green square
3 cm
----------------------------------------------------- = ----------- = 3 = k
length of red square
1 cm
2 cm
area of green square
9 cm 2
------------------------------------------------ = -------------2- = 9 = 3 2 = k 2
area of red square
1 cm
3 cm
Area = 9 cm2
3 cm
Different length ratios (or scale factors) of a circle
2 cm
radius length of blue circle
------------------------------------------------------------------ = ----------- = 2 = k
radius length of red circle
1 cm
1 cm
Area = πr 2 = 1π cm2
area of blue circle
4 π cm 2
------------------------------------------- = -----------------2- = 4 = 2 2 = k 2
area of red circle
1 π cm
2 cm
Area = π r 2 = 4π cm2
radius length of green circle
3 cm
--------------------------------------------------------------------- = ----------- = 3 = k
radius length of red circle
1 cm
3 cm
Area = π r 2 = 9π cm2
area of green circle
9 π cm 2
---------------------------------------------- = -----------------2- = 9 = 3 2 = k 2
area of red circle
1 π cm
From 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 of image
Area scale ratio or factor (asf) = -----------------------------------area of original
= square of linear scale factor (lsf)
= (lsf)2
= k2
Ch 08 FM YR 12 Page 394 Friday, November 10, 2000 11:36 AM
394
Further Mathematics
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.
WORKED Example 29
For the 2 similar triangles shown, find the
area, x cm2, of the small triangle.
Area = x
THINK
1
2
2.4 cm
Use the area scale factor to find
the unknown area.
4
Transpose the equation to get
unknown by itself.
Write your answer.
=
2
1
---
 2
=
1
--4
area of small triangle (image)
Area scale factor = -------------------------------------------------------------------------area of large triangle (original)
x cm 2
1
--- = -------------------4
100 cm 2
1
x = --4- × 100
x = 25
The area of the small triangle is 25 cm2.
WORKED Example 30
x
For the two similar shapes shown, find the
unknown length, x cm.
2 cm
10 cm 2
THINK
1
4.8 cm
length of small triangle (image)
Determine a scale factor, in this
Linear scale factor = ------------------------------------------------------------------------------length of large triangle (original)
instance the linear scale factor,
2.4 cm
from the two corresponding lengths
k = ---------------4.8 cm
given. It is preferred that the
1
= --2unknown triangle is the image.
Determine the area scale factor.
Area scale factor = k2
3
5
WRITE
Area = 100 cm2
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.
250 cm 2
WRITE
area of image (large trapezium)
Area scale factor = -------------------------------------------------------------------------------area of original (small trapezium)
250 cm 2
k2 = -------------------210 cm
= 25
Ch 08 FM YR 12 Page 395 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
THINK
2
WRITE
Determine the linear scale
factor.
3
Use the linear scale factor to
find the unknown length.
4
Transpose the equation to get
unknown by itself.
Write your answer.
5
395
Linear scale factor =
k2
k = 25
k=5
length of image (large trapezium)
Linear scale factor = ------------------------------------------------------------------------------------length of original (small trapezium)
x cm
5 = -----------2 cm
x=5×2
x = 10
The length, x, is 10 cm.
Volume of similar figures
If the lengths of similar figures 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.
Volume = 1 × 1 × 1
= 1 cm 3
A cube
length of large (blue) cube
2 cm
--------------------------------------------------------------- = ----------- = 2 = k
length of small (red) cube
1 cm
1 cm
1 cm
1 cm
volume of large cube
8 cm 2
--------------------------------------------------- = -------------2- = 8 = 2 3 = k 3
volume of small cube
1 cm
Volume =
2×2×2
= 8 cm3
2 cm
2 cm
2 cm
Volume =
1×1×3
= 3 cm3
A rectangular prism
length of small prism
3 cm
1
--------------------------------------------------- = ----------- = --- = k
length of large prism
6 cm
2
cm 3
1 cm
3 cm
3
volume of small prism
3
1
1
------------------------------------------------------ = ----------------3- = --- =  --- = k 3
 2
volume of large prism
8
24 cm
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 of image
Volume scale factor (vsf) = -------------------------------------------volume of original
= cube of linear scale factor (lsf)
= (lsf)3
= k3
1 cm
Volume =
2×2×6
= 24 cm3
2 cm
6 cm
2 cm
Ch 08 FM YR 12 Page 396 Friday, November 10, 2000 11:36 AM
396
Further Mathematics
WORKED Example 31
For the two similar figures shown, find the volume of the smaller cone.
THINK
1
Volume of
large cone
= 540 cm3
6 cm 9 cm
WRITE
Separate the two figures to
clarify the details of the
similar figures.
Volume =
540 cm3
6 cm
9 cm
Volume =
x cm3
2
3
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.
length of small triangle (image)
Linear scale factor = ------------------------------------------------------------------------------length of large triangle (original)
6 cm
k = ----------9 cm
=
2
--3
Volume scale factor = k3
Determine the volume scale
factor.
4
Use the volume scale factor to
find the unknown length.
5
Transpose the equation to get
the unknown by itself.
6
Write your answer.
k=
3
2
---
 3
k=
8
-----27
volume of small cone (image)
Volume scale factor = --------------------------------------------------------------------------volume of large cone (original)
x cm 3
8
------ = -------------------27
540 cm 3
8
- × 540
x = ----27
x = 160
The volume of the smaller cone is 160 cm3.
We can use the relationship between linear, area and volume scale factors to find any
unknown in any pair of similar figures as long as a scale factor can be established.
1. Given linear scale factor (lsf) = k
For example:
=2
2. Given
volume scale factor = k2
area scale factor = k2
= 22 = 4
= 23 = 8
area scale factor (asf) = k2 linear scale factor =
For example:
=4
k=
3. Given volume scale factor (vsf) = k3 linear scale factor =
For example:
=8
k
k=
2
volume scale factor = k3
4 =2
3
3
k
3
8 =2
= 23 = 8
area scale factor = k2
= 22 = 4
Ch 08 FM YR 12 Page 397 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
397
WORKED Example 32
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
1
2
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.
Determine the area scale
factor. For ease of calculation,
change volume scale factor to
linear and then to area scale
factor.
3
Use the area scale factor to
find the total surface area.
4
Transpose the equation to get
unknown by itself.
Write your answer.
5
volume of larger prism (image)
Volume scale factor = ---------------------------------------------------------------------------------volume of smaller prism (original)
64 m 3
k3 = ------------8 m3
k3 = 8
Linear scale factor =
3
k3 = k
k= 3 8 =2
Area scale factor = k2
= 22
=4
area of larger prism (image)
Area scale factor = --------------------------------------------------------------------------area of smaller prism (original)
x m2
4 = ---------------22.5 m
x = 4 × 2.5
x = 10
The total surface area of the larger triangular prism is 10 m2.
remember
remember
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 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 of image
Area scale ratio or factor (asf) = -----------------------------------area of original
= square of linear scale factor (lsf)
= k2
Volume scale factor
volume of image
Volume scale ratio or factor (vsf) = -------------------------------------------volume of original
= cube of linear scale factor (lsf)
= k3
Ch 08 FM YR 12 Page 398 Friday, November 10, 2000 11:36 AM
398
Further Mathematics
Area and volume scale
factors
8G
1 Complete the following table of values.
Mat
d
hca
Linear scale factors
k
Area and
volume
scale
factors
Area scale factors
k2
Volume scale factors
k3
2
8
16
3
125
100
64
0.027
36
0.1
100
0.16
400
540 mm2
48
x mm2
m
29
2 Find the unknown area of the following pairs of similar figures.
a
b
12 cm2
8c
Example
cm
WORKED
15 mm
c
x cm2
22.5 mm
7m
2m
x m2
d
122.5 m 2
21 mm
Surface area
= x mm2
WORKED
Example
30
14 mm
Surface area
= 100 mm2
3 a Find the unknown length of the following pairs of similar figures.
i
ii
25 cm
xm
Area =
6.25 m2
1.7 m Area =2
1.0 m
Area =
750 cm2
x
Area = 3000 cm2
Ch 08 FM YR 12 Page 399 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
399
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.
WORKED
Example
31
4 Find the unknown volume in the following pairs of similar objects.
a
b
Volume of small pyramid
= 40 cm3
x cm3
7 cm
2400 cm 3
12 cm
14 cm
c
2 cm
d
45 cm
Volume
= 1200 cm3
Volume of large sphere
= 8 litres
WORKED
Example
32
30 cm
5 a For the 2 similar triangular pyramids with volumes of 27 m3 and 3 m3, find the
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,
find 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, find the volume (to
the 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 and
3375 cm2, find 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 Find the volume of the
small cone.
b Find the volume of the larger
triangular pyramid
Area = 45 cm2
Area =
5 cm2
Volume of
large cone
= 270 cm3
TSA of small pyramid
= 200 cm2
Volume of small pyramid
= 1000 cm3
TSA of large pyramid
= 288 cm2
Ch 08 FM YR 12 Page 400 Friday, November 10, 2000 11:36 AM
400
Further Mathematics
c
Find total surface area of
the small prism
Area = 12
d Find the diameter of the
small cylinder.
12 cm
cm2
x cm
TSA =
78 cm2
Area =
6 cm2
TSA =
x cm2
Volume
= 1280 cm3
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)?
Volume
= 20 cm3
10 cm
5 cm
12 cm
8 cm
Ch 08 FM YR 12 Page 401 Friday, November 10, 2000 11:36 AM
401
Chapter 8 Geometry: similarity and mensuration
8 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.
TSA of
large cone
= 840 cm2
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 its
representation 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 smaller
statue stood 15 cm tall. What is the height of the other statue (to the nearest
centimetre)?
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 of
the small sphere to the large sphere:
a for total surface area
b for volume.
17 A cone is half-filled with ice-cream. What is the ratio of ice cream to empty space?
3h
19 multiple choice
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
h
18 multiple choice
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
20 multiple choice
A 1:100 scale model of a building is a cube with sides of 100 cm. The volume of the
real building is:
B 1 000 000 m3
C 100 000 m3
A 10 000 000 m3
3
3
D 10 000 m
E 1000 m
Ch 08 FM YR 12 Page 402 Friday, November 10, 2000 11:36 AM
402
Further Mathematics
summary
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.
Equal sides
Area and perimeter
• Perimeter is the distance around a closed figure.
• Circumference is the perimeter of a circle.
C = 2 × π × radius
= 2π r
• 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 = l 2
2. Area of a rectangle: A = l × w
3. Area of a parallelogram: A = b × h
4. Area of trapezium: A = 1--2- (a + b) × h
5. Area of a circle: A = π r 2
6. Area of a triangle: A = 1--2- × 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 = 2π r(r + h)
4. Cones: TSA = π r(r + s) where s is the slant height
5. Spheres: TSA = 4π r 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 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
Ch 08 FM YR 12 Page 403 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
403
• Volume of a prism, Vprism = area of uniform cross-section × height
V=A×H
• Volume of a pyramid, Vpyramid =
V=
1
--3
1
--3
× area of cross-section at the base × height
×A×H
• The height of a pyramid, H, is sometimes call the altitude.
• Volume of a sphere is Vsphere =
4
--3
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.
SCALE 1:50 000
METRES 1000
0
1
2
3 KILOMETRES
• A simple conversion scale, for example 1 cm = 100 m, means 1 cm on the map
represents 100 metres in real life.
Kilometres 0
1
2
3
4
5
6
7
8 Kilometres
Ch 08 FM YR 12 Page 404 Friday, November 10, 2000 11:36 AM
404
Further Mathematics
Similar figures
B'
• Two objects that have the same shape but different size
are said to be similar.
4
• For 2 figures to be similar, they must have the following
properties:
(a) The ratios of the corresponding sides must be equal. A' 2
A′B′
B′C′
C′D′
A′D′
------------ = ------------ = ------------ = ------------ = common ratio
AB
BC
CD
AD
(b) The corresponding angles are equal.
∠A = ∠A′ ∠B = ∠B′ ∠C = ∠C′ ∠D = ∠D′
C'
2
B' 125°
6
D'
2
3
A 1
D
C'
60°
C
B
125°
A
85°
A'
C
B 1
60°
85°
D
D'
Scale factor, k
length of image
A′B′
A′B′
B′C′
C′A′
• Scale factor, k = ----------------------------------------- = ------------ = ------------ = ------------ = -----------length of original
AB
AB
BC
CA
where for enlargements, k is greater than 1 and
for 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.
B'
B
3
3
9
9
A1C
A' 3 C'
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 of image
• Area scale ratio or factor (asf) = -----------------------------------area of original
= square of linear scale factor (lsf)
= k2
Volume scale factor
volume of image
• Volume scale ratio or factor (vsf) = -------------------------------------------volume of original
= cube of linear scale factor (lsf)
= k3
Ch 08 FM YR 12 Page 405 Friday, November 10, 2000 11:36 AM
Chapter 8 Geometry: similarity and mensuration
405
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
8A
x°
32°
2 A triangle LABC has the following values given. 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
8B
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
8B
220
80
200
All
measurements
in cm
50
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 × π × 20 × (40)
B 2 × π × 40 × (40)
C 2 × π × 40 × (100)
D 2 × π × 40 × (60)
E 2 × π × 20 × (60)
8C
5 The net of an object is shown in the diagram. An appropriate name
for the object is:
A rectangular prism
B rectangular pyramid
C triangular prism
D triangular pyramid
E trapezium prism
8C
6 The volume of a sphere with a diameter of 15 cm is closest to:
A 560π cm3
B 900π cm3
C 4500π cm3
3
3
D 4500π cm
E 36 000π cm
8D
7 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
V
8D
O
Ch 08 FM YR 12 Page 406 Friday, November 10, 2000 11:36 AM
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Further Mathematics
8E
8 A map ratio scale of 1:150 000 expressed as a simple conversion scale is:
A 1 cm = 15 m
B 1 cm = 150 m
C 1 cm = 1500 m
D 1 mm = 1.5 km
E 1 cm = 15 km
8E
9 In the triangle shown, the value of c is:
A 3
B 6
C 9
D 12
E 4
10 The circumference of the larger cone is closest to:
113 mm
151 mm
226 mm
302 mm
459 mm
24 mm
189 mm
63 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:
B 45 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
6
6–x
--- = ----------4
x
4
6–x
--- = ----------6
x
6
x–6
--- = ----------4
x
12
6–x
------ = ----------6
x
6
2+x
--- = -----------4
x
C 34 cm
34 cm
D 30 cm
45 cm
A
B
C
D
E
A 67 cm
8F
7.8
h cm
8F
2.6
c
51 cm
E 26 cm
6
6-x
8E
3
6
x
4
12
Ch 08 FM YR 12 Page 407 Friday, November 10, 2000 11:36 AM
407
Chapter 8 Geometry: similarity and mensuration
8F
13 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):
1000 m
1002 metres
1000 metres
499 metres
501 metres
500 metres
1m
2m
A
B
C
D
E
w
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
B
C
D
E
8G
1:27
1:26
27:1
1:9
1:3
Short answer
1 For each of the figures, find the value of the pronumeral.
a
b
a
c
8A
a
b
b
40°
c
8B
2 Find the outer perimeter and area of the flower.
r = 11 mm
r = 22 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.
4
6m
10 m
8C
4m
3m
5m
6m
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?
Ch 08 FM YR 12 Page 408 Friday, November 10, 2000 11:36 AM
408
Further Mathematics
8D
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?
8E
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.
8F
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)?
8F
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)?
10 m 1 m
8G
6m
9 A plan is drawn to scale of 1:50 000. Find:
a the length in centimetres on the plan that represents 1 km
b the area in hectares of a region represented by 4 cm2 on the plan
c the area on the plan of a region of 25 hectares.
Analysis
A 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?
CHAPTER
test
yourself
8
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.
4.5 cm
6 cm
6 cm
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?
5 cm