1.
syllabussyllabusrrefefererenceence
Strand:
Applied geometry
Core topic:
Elements of applied geometry
In thisIn this chachapterpter
4A Changing units and
calculating perimeters
4B Calculating areas
4C Total surface area
4D Volume and capacity
4
Length, area
and volume
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102 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Introduction
Peta is an expert in the craft of origami (the Japanese art of paper folding). She deftly
folds and creases paper, the end product being a crane, an elegant swan, a gift box . . .
the list is endless. In fact, many two-dimensional and three-dimensional objects can be
folded from a single piece of paper without any cutting. In creating these works of art,
it is necessary to have an understanding of the spatial relationships of geometry. It is
essential to fold the paper at the correct angles and to have the side lengths in the
correct proportions.
We are surrounded by geometric shapes —
in our daily lives, in nature, architecture, etc.
Throughout this chapter you’ll discover prop-
erties of geometry by conducting investi-
gations. Each investigation is designed to
reinforce the particular geometry concept of
that section of study.
MQ Maths A Yr 11 - 04 Page 102 Wednesday, July 4, 2001 4:11 PM
3.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 103
1 What do you understand by the word length?
2 Give some units you would use to measure length.
3 What do you understand by the term perimeter?
4 Calculate the perimeter of the following shapes:
a b c d
5 Deﬁne the term area.
6 Give some units you would use to measure area.
7 Calculate the areas of the ﬁgures in question 4 above.
8 Identify the shapes of the following ﬁgures:
a b
c d
e f
9 Use your calculator to determine answers to the following (correct to 2 decimal
places, if necessary).
a 1.462
b c 4.33
d
10 Distinguish between the terms volume and capacity.
11 Give some units you could use to measure
a volume b capacity.
12 Convert the following quantities to the units indicated.
a 6 mm → cm b 0.25 m → mm c 40 cm2
→ m2
d 0.45 km → m
e 300 mL → L f 5000 cm3
→ m3
g 25 L → kL h 5 cm3
→ mm3
i 100 cm3
→ L j 25 kL → m3
Objects all around us can be classiﬁed as two-dimensional (2-D) or three-dimensional
(3-D). A two-dimensional object can be described by two measurements (for example
a length and a width), whereas a three-dimensional object requires three
measurements to describe its shape (for example a length, a width and a height).
6 cm
10 m
4 m
4.33 m
5 m
3 m
SkillS
HEET 4.1
E
XCEL Spread
sheet
Rounding
3.07 28.763
MQ Maths A Yr 11 - 04 Page 103 Wednesday, July 4, 2001 4:11 PM
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104 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
It’s important to realise that sometimes it is not possible
to construct a 3-D object that has been drawn on paper in
two dimensions. Look carefully and you will see that this is
the case in the situation at right. Try drawing the shape and
you will understand the problem!
Paper folding 1
Constructing a 3-D package from a 2-D shape
Resources: Cardboard, scissors, protractor, ruler.
For our ﬁrst investigation, let’s look at the relative positions, shapes and sizes of
the resulting faces when a 2-D piece of cardboard is folded into a 3-D shape to
form an unusual postage box.
1 Trace the above shape onto a piece of thin cardboard. Basically, it consists of
two rectangles.
2 Cut along all the blue lines.
3 The lines marked – – – – represent ‘valley’ folds. Crease ﬁrmly along these
lines with the fold pointing down like a valley.
4 The line XY marked – – • – – • represents a ‘mountain’ fold. Crease ﬁrmly
along this line, with the fold pointing upwards like a mountain.
5 Rotate the top rectangle through 90º at the point Y, along the crease line XY,
in an anticlockwise direction, until it lies on top of the lower rectangle.
6 Fold the top and bottom ﬂaps in along the crease lines in a 3-D manner. Fold
in the side ﬂaps along the crease lines in a 3-D manner.
7 This container, on a larger scale, would be a suitable package for posting
rectangular objects through the mail.
8 Open the package and
a measure angle ZXY
b measure angle a.
9 Would it be possible to vary these angles and still form a rectangular
container? Investigate the consequences.
inv
estigat
ioninv
estigat
ion
w2 w2
w1w1
a
X
Z Y
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5.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 105
Two-dimensional objects
As mentioned previously, the shape of 2-D objects can be speciﬁed by two measure-
ments. These objects lie in one plane. Among other properties, they possess a perimeter
and an area.
Perimeter
The perimeter represents the distance around the boundary of a ﬁgure. (We are
assuming that all the ﬁgures we are dealing with are closed; that is, they begin and end
at the same point.) Any line inside the boundary is ignored when calculating the perim-
eter. The units used to measure perimeter are those of linear measure: millimetre (mm),
centimetre (cm), metre (m) and kilometre (km).
We’ll ﬁrst discuss converting measurements from one unit to another. You are fami-
liar with the following conversions.
10 millimetres = 1 centimetre
100 centimetres = 1 metre
1000 metres = 1 kilometre
Using these conversion factors, we can now construct a ‘conversion ladder’.
The smallest unit (mm) is at the narrow top of the ladder; working down the rungs
we approach the largest unit (km) which is at the wide base of the ladder. The con-
version factors are placed on the rungs in between the units. In changing measurements
from a smaller unit to a larger unit, we divide by the relevant conversion factor(s)
because we know that our answer must be a lesser amount. When converting from a
larger to a smaller unit, we multiply by the relevant conversion factor(s) since we know
that we require a greater amount as the result.
10 Consider using this container to post a book. On your ﬂat shape, identify:
a on which part the book would lie
b which sections would dictate the maximum length, width and thickness of
the book.
11 Why is the width w1 greater than the width w2?
12 Taking the measurements of your text book, design and construct a package of
this style which could be used to post your book.
SkillS
HEET 4.2
mm
10
cm
100
m
1000
km
Multiply
when
going
to a
smaller
unit
Divide
when
going
to a
larger
unit
Smallest unit
× ÷
Largest unit
MQ Maths A Yr 11 - 04 Page 105 Wednesday, July 4, 2001 4:11 PM
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106 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
We shall now review familiar formulas used to determine the perimeter of common
shapes.
Shape Perimeter
Square P = 4S
Rectangle P = 2(L + W)
Other polygons P = sum of lengths of all sides
Circle C = 2πr or
C = πD
Sector C = length of arc + 2 radii
C = × 2πr + 2r
Complete each of the following.
a 30 mm = cm b 4800 m = km c 6.5 m = cm d 8400 mm = m
THINK WRITE
a Changing millimetres to centimetres:
divide by 10.
a 30 mm = 30 ÷ 10 cm
= 3 cm
b To change metres to kilometres: divide
by 1000.
b 4800 m = 4800 ÷ 1000 km
= 4.8 km
c To change metres to centimetres:
multiply by 100.
c 6.5 m = 6.5 × 100 cm
= 650 cm
d To change millimetres to metres: divide
by 10 (to change to centimetres) then
divide by 100 (to change to metres).
d 8400 mm = 8400 ÷ 10 cm
= 840 ÷ 100 m
= 8.4 m
1WORKEDExample
EXCE
L Spreadshe
et
Length
con-
versions
Mat
hcad
Unit
con-
versions
SkillS
HEET 4.3
SkillS
HEET 4.4
S
W
L
r D
Arc
length
Arc
θ° r
θ°
360°
-----------
MQ Maths A Yr 11 - 04 Page 106 Wednesday, July 4, 2001 4:11 PM
7.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 107
Perimeter of composite ﬁgures
In many instances, ﬁgures are not of one distinct shape; they may be composed of sev-
eral shapes. The perimeter of such shapes is still the distance around the boundary of
the composite ﬁgure. Remember to ignore any lines inside the ﬁgure. It is often helpful
to start at one point in the ﬁgure, work your way around the boundary in a clockwise or
anticlockwise direction, identifying shapes and adding the lengths of all sides, until you
reach your starting point.
Find the perimeter around the following piece of pizza.
THINK WRITE
Identify the shape: sector.
Perimeter means ‘distance around
outside’; that is, arc plus 2 radii.
Write the appropriate formula. P = × 2πr + 2r
Identify the values of the variables. θ° = 40°, r = 10 cm
Substitute values in the formula.
Calculate the answer, not forgetting
units.
P = + (2 × 10)
P = 6.98 + 20
P = 26.98 cm
10 cm
40°
1
2
3
θ°
360°
-----------
4
5
6
40
360
--------- 2× π× 10×
2WORKEDExample
Find the perimeter of this shape.
THINK WRITE
Start at X and travel in a clockwise
direction until reaching X again.
Identify the sides as 3 straight lines and
one semicircle.
P = Side 1 + Side 2 + Side 3
+ circumference of circle
Write the formulas. P = S1 + S2 + S3 + πD
Identify the values of the variables. S1 = 20, S2 = 10, S3 = 20, D = 10
Substitute the values of the variables in
the formula.
Calculate the answer, not forgetting
units.
P = 20 + 10 + 20 + ( × π × 10)
P = 50 + 15.7
P = 65.7 cm
X20 cm
10 cm
1
1
2
---
2
1
2
---
3
4
5
1
2
---
3WORKEDExample
MQ Maths A Yr 11 - 04 Page 107 Wednesday, July 4, 2001 4:11 PM
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108 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Changing units and
calculating perimeters
1 Copy and complete each of the following.
2 Richard is planning to have a garage built. The garage is 5.2 m long, 2.4 m wide and
2.5 m high. All builders, however, work in millimetres. What are the dimensions of the
garage, in millimetres?
3 Find the perimeters of the following ﬁgures (to the nearest whole units).
a b c
d e f
4 Find the perimeters of the following ﬁgures.
a b c
a 70 mm = cm b 600 cm = m c 5000 m = km
d 9 cm = mm e 12 m = cm f 9 km = m
g 86 mm = cm h 9.2 km = m i 2400 m = km
j 6.4 cm = mm k 11.25 m = cm l 2.2 cm = mm
remember
1. Recall unit conversions for length.
2. Multiply when changing to a smaller unit and divide when changing to a larger
unit.
3. The perimeter is the distance around a closed, 2-dimensional ﬁgure.
4. When ﬁnding the perimeter of a composite ﬁgure, start at any place on the
perimeter and continue in a clockwise or anticlockwise direction until reaching
the starting point. Ignore any lines inside the perimeter.
5. Don’t forget to include units in the answer.
remember
4A
WWORKED
Example
1
WORKED
Example
2
7m
12 m
5 m
4 m
23.7 cm
17.8cm
15.4cm
27.5 cm
13.5 mm
7.5 m
11.5 m
5 m
4m
120m
210m
90 m
WORKED
Example
3
12 m
25 m
10 m
14 m
20 m
2 m
3.5m
12 m
17m
MQ Maths A Yr 11 - 04 Page 108 Wednesday, July 4, 2001 4:11 PM
9.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 109
d e f
5
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
6
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
Paper folding 2
Resources: A4 paper, scissors.
Having looked at some 2-D shapes, let’s investigate a ‘Geometry jigsaw’.
1 Cut a square from a sheet of unlined A4 paper. Scribble a pattern on one side
of the paper, so that you can distinguish the ‘top’ of the jigsaw from the
‘underneath’.
2 Fold the square sheet in half along its diagonal. Unfold it, and cut it along the
crease. You should now have two triangles.
3 Take one of these triangles, fold it in half and cut it along the crease line.
4 Take your second triangle from above and lightly crease it to ﬁnd the midpoint
of the longest side. Fold it so that the vertex of the right angle touches that
midpoint and cut it along the crease. This forms a trapezium and a small
triangle.
5 Take the trapezium, fold it in half and cut along the crease line. You should
now have two smaller trapeziums.
125 mm
24 mm
90 mm48 mm
21 cm
16 cm
8cm
10cm
12 cm
10 cm
20 m
22 m
7 m
13m
11m
34m
44m
mmultiple choiceultiple choice
12 cm
7 cm
2 cm
3 cm
mmultiple choiceultiple choice
35.2m
20.5 m
x
inv
estigat
ioninv
estigat
ion
MQ Maths A Yr 11 - 04 Page 109 Wednesday, July 4, 2001 4:11 PM
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110 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
6 Fold the acute base angle of one of the trapeziums to the adjacent right base
angle and cut along the crease. This should result in a square and a small
triangle.
7 Take your other trapezium and fold its right base angle to the opposite obtuse
angle. Cut along the crease. This should result in a parallelogram and a small
triangle.
8 You should now have 7 shapes: 5 triangles,
1 square and 1 parallelogram.
9 See if you can now assemble this Geometry jigsaw
to form the original square from which it was cut.
10 See how many different shapes you can form using
some or all of the 7 shapes.
This investigation should enhance your understanding
of geometric shapes.
Task 1
Take your 7 shapes. Use the smallest triangle as the basic unit of area. Arrange
your pieces in order of increasing area. Give the area of each piece in terms of
‘small triangular units’. Some shapes may have the same area. Copy and complete
the following table:
Task 2
Your pieces can be ﬁtted together in different combinations to form squares of vari-
ous sizes. Experiment with your pieces to see whether you can form a square using
the following number of pieces. Copy and complete the table below.
Shape Sketch of shape Area (e.g. 3 ‘triangular units’)
1
2
3
4
5
6
7
Pieces Sketch Pieces Sketch
1 5
2 6
3 7
4
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11.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 111
Area
Area represents the amount of space within the boundary of a closed ﬁgure. The units
used to measure area are those of square measure: mm2
, cm2
, m2
and km2
. There are
two units of area which are not square units — the hectare (ha) in the metric system
and the acre in the imperial system.
By constructing a conversion ladder, as we did before for linear measure, we ﬁnd:
To obtain the conversion factors for square measures,
it is necessary to square the linear measure conversion
factors. As previously, we multiply when converting to
a smaller unit and divide when converting to a larger unit.
One hectare is equivalent to the area of a square of side
length 100 metres.
So 1 ha = 100 m × 100 m
That is 1 ha = 10 000 m2
Square measure
Resources: Paper, pencil, ruler.
How is it possible to determine unit conversions for square measure knowing the
relevant conversion for linear measure?
1 Draw a square of side 1 cm.
2 You are aware that each centimetre can be divided into 10 millimetres. Mark
these millimetre divisions on each side of the square, then join opposite side
markings.
3 This creates a grid of smaller squares. How many of these smaller squares are
there? Each smaller square is 1 mm2
while the large square is 1 cm2
. So how
many mm2
are there in 1 cm2
?
4 Use the reasoning applied above (it is not really practicable to draw a square of
side length 1 metre) to determine the number of cm2
in 1 m2
.
Converting units in square measure is simply a matter of applying the above
technique.
inv
estigat
ioninv
estigat
ion
mm2
10 × 10 = 100
cm2
100 × 100 = 10 000
m2
1000 × 1000 = 1000 000
km2
× ÷
Math
cad
Unit
conversions
100 m
100 m1 ha
MQ Maths A Yr 11 - 04 Page 111 Wednesday, July 4, 2001 4:11 PM
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112 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
We’ll now review familiar formulas used to ﬁnd the areas of common shapes.
Shape Area
Square A = S2
Rectangle A = L × W
Parallelogram A = b × h
where the height measurement must be at a right
angle to the base measurement.
Trapezium A = (a + b)h
where the height measurement must be at a right
angle to the base measurement.
Triangle A = bh
where the height measurement must be at a right
angle to the base measurement.
Complete each of the following:
a 250 mm2
= _____ cm2
b 5 km2
= _____ ha
THINK WRITE
a Change mm2
to cm2
, so divide by 100. a 250 mm2
= 250 ÷ 100 cm2
250 mm2
= 2.5 cm2
b Change km2
to m2
(multiply by
1 000 000).
Change m2
to ha (divide by 10 000).
b 5 km2
= 5 × 1 000 000 m2
5 km2
= 5 000 000 m2
5 km2
= 5 000 000 ÷ 10 000 ha
5 km2
= 500 ha
4WORKEDExample
EXCE
L Spreadshe
et
Area of a
square
S
EXCE
L Spreadshe
et
Area of a
rectangle
L
W
Area of a
parallelogram
b
h
b
h
a
1
2
---
Area of a
triangle
Vic Gen fig ch 6. 57a
b
h
G
h
b
h
b
1
2
---
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C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 113
Triangle
(Heron’s
formula)
A =
where s = (a + b + c)
(Use when height measurement is unknown.)
Circle A = π r2
Sector
A = × πr2
Shape Area
a
bc
s s a–( ) s b–( ) s c–( )
1
2
---
E
XCEL Spread
sheet
Area of
a circle
r
D
Cabri Geo
metry
Area of
a circler
θ°
θ°
360°
-----------
Find the area of the triangle at right.
THINK WRITE
Write the formula. A = × b × h
Substitute for the base and the height. = × 12.8 × 9.4
Calculate the area. = 60.16 cm2
9.4 cm
12.8 cm1
1
2
---
2
1
2
---
3
5WORKEDExample
Find the area of each of the following shapes.
a b
THINK WRITE
a Write the formula. a A = b × h
Substitute the base and height. A = 14 × 9
Calculate the area. A = 126 m2
b Write the formula. b A = × (a + b) × h
Substitute the sides and height. A = × (5.9 + 11.4) × 7.2
Calculate the area. A = 62.28 cm2
9 m
14 m
7.2 cm
5.9 cm
11.4 cm
1
2
3
1
1
2
---
2
1
2
---
3
6WORKEDExample
E
XCEL Spread
sheet
Area of a
triangle
Math
cad
Substitu-
tion 2
Cabri Geo
metry
Area of a
trapezium
MQ Maths A Yr 11 - 04 Page 113 Wednesday, July 4, 2001 4:11 PM
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114 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Mat
hcad
Area of a
triangle
Find the area of this shape.
THINK WRITE
Identify the shape (in this case it is a
triangle with no height measurement)
and write down the appropriate formula
for the area (Heron’s formula).
A =
where s = (a + b + c)
Identify the values of the pronumerals. a = 7, b = 8, c = 10
To ﬁnd s, substitute a, b and c values
into the formula and simplify.
s = (7 + 8 + 10)
= × 25
= 12.5
Substitute the values of a, b, c, and s
into the formula for the area.
A =
Simplify.
(a) Evaluate the brackets ﬁrst.
(b) Multiply the values together.
(c) Take the square root.
A =
=
= 27.810 744 33
(Round off the answer to 1 decimal
place and include the units.)
A = 27.8 cm2
7 m 8 m
10 m1 s s a–( ) s b–( ) s c–( )
1
2
---
2
3 1
2
---
1
2
---
4 12.5 12.5 7–( ) 12.5 8–( ) 12.5 10–( )
5
12.5 5.5 4.5 2.5×××
773.4375
7WORKEDExample
SkillS
HEET 4.5 A minute hand moving from the number 12 to the number 4 position
sweeps out a sector. If the hand is 10 cm long, what is the area of this
sector?
THINK WRITE
Write down the formula for the area of
the sector.
Area of sector =
Identify the value of the radius. r = 10 cm
Calculate the angle of the sector:
The angle between the numbers on a
clock = 360° ÷ 12
= 30°.
From 12 to 4 there are 4 intervals
between the numbers. So to ﬁnd the
angle of a sector, multiply 30° by 4.
θ° = 30° × 4
= 120°
Substitute the values of r and θ into the
formula and evaluate.
Area of sector = × π × 102
= 104.719 755 1 cm2
Write an answer sentence with the number
rounded off appropriately and units given.
The minute hand as it rotates through an angle
of 120° sweeps an area of 104.7 cm2
.
4
12 1
2
3
10 cm1
θ°
360°
----------- πr2×
2
3
4 120
360
---------
5
8WORKEDExample
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C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 115
Area of composite ﬁgures
The term composite means made up of distinct parts. Composite ﬁgures in geometry
are ﬁgures comprising a number of distinct shapes. Depending upon the composite
ﬁgure, to ﬁnd the overall area you may need to add these individual shapes or subtract
one from another.
For example, the composite ﬁgure in the diagram at right has been
formed using a semicircle and a square.
The area of this shape can be found as follows:
Area of total ﬁgure = Area of a semicircle (A1) + Area of a square (A2)
When ﬁnding the area of a composite ﬁgure, follow the steps given
below.
1. Identify the basic shapes that make up the total ﬁgure and number them.
2. Write the expression for the total area in terms of individual shapes.
3. Calculate the area of each individual shape.
4. Add or subtract areas to ﬁnd the total area of the given shape.
In the diagram above, we added the areas of the square and the
semicircle. If the diagram had been shown as at right, we would
have subtracted the area of the semicircle from that of the square.
A1
A2
A1
A2
A clock has a minute hand that is 6 cm long and an hour
hand that is 3 cm long. In one full revolution of each hand,
the minute hand would sweep out a larger circle than the
hour hand. What is the difference in the area they cover
(to the nearest square centimetre)?
THINK WRITE
The area required is the area
between two circles. Write down
the appropriate formula.
A = outer area − inner area
A = πR2
− πr2
Identify the value of R (radius of
larger circle) and the value of
r (radius of smaller circle).
R = 6, r = 3
Substitute the values of the
pronumerals into the formula and
evaluate.
A = π × 62
− π × 32
= 113.097 − 28.274
= 84.823 cm2
Write an answer sentence with the
value rounded to the nearest
square centimetre.
The difference in area covered by the two hands
is approximately 85 cm2
.
3 cm
6 cm
R
r
1
2
3
4
9WORKEDExample
MQ Maths A Yr 11 - 04 Page 115 Wednesday, July 4, 2001 4:11 PM
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116 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Calculating areas
1 Copy and complete each of the following.
a 70 mm2
= _____ cm2
b 6000 cm2
= _____ m2
c 3 m2
= _____ cm2
d 2.5 km2
= _____ m2
e 4.5 ha = _____ m2
f 3 km2
= _____ ha
2 Find the area of each of the ﬁgures below.
a b c
d e f
g h i
remember
1. The area is the space enclosed by the boundaries of a two-dimensional shape.
2. Area is measured in square units. To change from one unit to another, square
the appropriate linear measure conversion factor. Multiply by the conversion
factor when changing to a smaller unit and divide when converting to a larger
unit.
3. Remember these area formulas:
Square A = S2
Rectangle A = L × W
Parallelogram A = b × h
Trapezium A = (a + b)h
Triangle A = bh (if perpendicular height is known).
If three sides are known, use Heron’s formula
A = where s = (a + b + c)
Circle A = πr2
Sector A = × πr2
4. For composite ﬁgures, identify shapes of parts of the ﬁgure. After calculating
these individual areas, add or subtract them to give the total area.
5. Remember to provide units in the answer.
1
2
---
1
2
---
s s a–( ) s b–( ) s c–( ) 1
2
---
θ°
360°
-----------
remember
4B
Work
SHEET 4.1 WORKED
Example
4
Area of
triangle
8 cm
29 mm
3.6 km
3 m
9 m
27mm
38 mm
47 cm
62 cm
WORKED
Example
5 4.2 m
9.7 m
8.4 km
6.3 km
3.7 m
MQ Maths A Yr 11 - 04 Page 116 Wednesday, July 4, 2001 4:11 PM
17.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 117
3 Look at the ﬁgure at right.
a Find the area of the outer rectangle.
b Find the area of the inner rectangle.
c Find the shaded area by subtracting the area of the
inner rectangle from the area of the outer rectangle.
4 Find the shaded area in each of the following.
a b
c d
5
The area of the triangle at right is:
j k l
m n o
p q r
s t u
A 36 cm2
B 54 cm2
C 108 cm2
D 1620 cm2
WORKED
Example
6a
12.8 km
16.9 km
38 mm
87 mm
8 m
80 cm
WORKED
Example
6b
1 m
12 m
9 m
2.8 m
3.65 m
0.4 m
3.6 cm
9.5 cm
5.4 cm
WORKED
Example
7
38 cm
27 cm15 cm
12 mm
8 mm7 mm
65 m
16 m58 m
WORKED
Example
8
45°
6 cm 120° 2 m
300°
5 mm
WORKED
Example
9 3 m 12 m
20 m
7 m
10 m
8 m14 m
16 m
5 cm
5 cm
9 cm
3 m
9 m
10 m
8 m
5 m
8 m
12 m
12 m
8 m
mmultiple choiceultiple choice
15 cm
12 cm
9 cm
MQ Maths A Yr 11 - 04 Page 117 Wednesday, July 4, 2001 4:11 PM
18.
118 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
6
Which of the two statements is correct
for the two shapes at right?
Statement 1. The rectangle and parallelogram have equal areas.
Statement 2. The rectangle and parallelogram have equal perimeters.
7
The area of the ﬁgure at right is:
8 Len is having his lounge room carpeted. Carpet costs $27.80/m2
. The lounge is
rectangular with a length of 7.2 m and a width of 4.8 m.
a Calculate the area of the lounge room.
b Calculate the cost of carpeting the room.
9 A rectangular garden in a park is 15 m long and 12 m wide. A concrete path 1.5 m
wide is to be laid around the garden.
a Draw a diagram of the garden and the path.
b Find the area of the garden.
c What are the dimensions of the rectangle formed by the path?
d Find the area of concrete needed for the path.
10
Examine the diagram at right.
a The circles cover an area of approximately:
b The shaded area is approximately:
11 A family-size pizza is cut into 8 equal slices. If the diameter of the pizza is 33 cm,
ﬁnd (to the nearest square centimetre) the area of the top part of each slice.
12 The collectable plate shown at right is 22 cm in diameter
and has a golden ring that is 0.5 cm wide.
Find (to 1 decimal place) the area of the golden ring if its
outer edge is 1 cm from the edge of the plate.
A Statement 1 B Statement 2 C Both statements D Neither statement
A 54 m2
B 165 m2
C 225 m2
D 255 m2
A 402 cm2
B 201 cm2
C 804 cm2
D 805 cm2
E 603 cm2
A 219 cm2
B 421 cm2
C 622 cm2
D 823 cm2
E 220 cm2
mmultiple choiceultiple choice
19 cm
38 cm 38 cm
19 cm
mmultiple choiceultiple choice
17 m
7 m
15 m
15 m
mmultiple choiceultiple choice
32 cm
22 cm
1 cm
0.5 cm
WORKED
Example
9
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19.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 119
13 Early-model vehicles
had a single wind-
screen-wiper blade
to remove water
from the windscreen.
(The bus at left has two single blades of
this type.)
Using the dimensions given in the diagram:
a what area (to the nearest whole number)
did the blade cover?
b what percentage (to 1 decimal place) of
the windscreen was cleared?
Calculate the area (or shaded area) of each of the ﬁgures drawn below.
Where necessary, give your answer correct to 1 decimal place.
1 2 3
4 5
6 7 8
9 10
140°
45 cm
120 cm
60 cm
1
12 cm
6.3 m
8 cm
30°
62 mm
91 mm
10 cm
4 cm
24 cm
25 cm
30 cm
25 cm
20 cm
40 m
20 m
40 cm
12 cm
76 mm
32
m
m
15cm
12 cm
6cm
MQ Maths A Yr 11 - 04 Page 119 Wednesday, July 4, 2001 4:11 PM
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120 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Maximising an area of land
Farmer Brown needs to build a paddock for her sheep to graze. She has 1000 m of
fencing with which to build this paddock.
1 If farmer Brown builds the paddock 100 m long and 400 m wide, the area will
be 40 000 m2
. If she builds it 200 m long and 300 m wide, the area will be
60 000 m2
. What dimensions should farmer Brown choose for her paddock so it
has the maximum possible area?
a Set up a spreadsheet with the headings:
LENGTH WIDTH AREA
b Enter an initial value of 50 m for the length of the paddock, then provide a
formula and copy it down to generate length measurements in increments of
50 m; for example:
50
100
150
•
•
•
450
c Since the length of the fencing is 1000 m,
length + width = 500 m
Provide a formula in the width column incorporating the values in the length
column, then copy this formula down the column.
d Enter a formula under the AREA heading to calculate the area of the ﬁgure.
Copy this formula down the column.
e What length and width provide the greatest area? What shape is the paddock?
2 If one side of the paddock is a river, only three sides need to be fenced. If
farmer Brown still uses 1000 m of fencing, what dimensions should she now
choose for her paddock to maximise its area?
EXCE
L Spreadshe
et
Maximising
area of
land
Area of a
rectangle
Maximum
area
investigat
ioninvestigat
ion
MQ Maths A Yr 11 - 04 Page 120 Wednesday, July 4, 2001 4:11 PM
21.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 121
a Set up a spreadsheet similar to the one above. In this case:
width + length + width = 1000 m
Provide formulas in the columns, then copy them down.
b What length and width provide the greatest area in this case? Is the result the
same as for the previous case?
3 Write a paragraph outlining recommendations for farmer Brown. Justify your
conclusions with mathematical evidence.
4 Investigate further to see whether you could write your conclusions in a general
form for any given length of fencing.
Effect of scale factors on
perimeter and area
Resources: Paper, scissors.
A variety of shapes can result when paper is folded.
Applying a scale factor to a ﬁgure affects its perimeter and area differently. Let’s
investigate.
1. Cut two strips of paper: one should be 1 cm
wide, the other should be 2 cm wide.
2. Taking the ends of the strips, fold them as you
would in knotting a piece of string.
3. Gently ﬂatten the knot and sharply crease the edges.
Trim off the ends of the paper strips beyond the knot.
1 What shape does each knot form?
2 Measure the interior angles of each knot. What do you conclude?
3 Measure the side lengths of each knot. What do you ﬁnd?
4 The two knots are similar in shape. What is the scale factor S?
5 Using your measurements, calculate the perimeter of each ﬁgure. What is the
relationship between your two answers?
6 Look at the ﬂat surface of the knot. Because we don’t know a formula for the
area of this shape, we can break it up into familiar shapes whose areas we can
identify. Locate a trapezium and a triangle on each of the ﬁgures. Take the
required measurements and calculate the areas of the smaller knot and the
larger knot. What is the relationship between the two areas?
7 From your two previous answers, what would you expect the perimeter and
area to be if you were to construct a knot from a 3-cm strip of paper?
8 Deduce a general statement indicating what happens to the perimeter and area
of a ﬁgure when we apply a scale factor S.
9 Open each knot out ﬂat.
a What is the overall shape of the paper?
b What shapes do you ﬁnd within the creases?
c Compare the perimeters and areas of your two ﬂat shapes. Do your answers
support your conclusion in question 8?
inv
estigat
ioninv
estigat
ion
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122 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Patchwork designs
Resources: Graph (or unlined) paper, pencil, ruler.
In this investigation we look at two common patchwork designs for quilts, and
consider the shapes within blocks and the requirements for completing the quilts.
Patchwork quilts are frequently constructed
from a set of ‘blocks’ of the same pattern.
The eight-pointed star block is shown at right.
Each block is constructed from a set of
smaller shapes sewn together to create the
pattern. A border of a complementary colour
usually frames the blocks.
1 Identify the shapes in the block above.
How many of each shape are required to form
the pattern?
2 Each block has dimensions 50 cm square. Sketch a design for a single-bed quilt
173 cm by 218 cm, including the border. How many blocks and how many
pieces of each shape would be required?
3 Trace four blocks. Place them together at a point and shade or colour them to
illustrate the overall effect of your quilt.
4 Experiment with other patterns incorporating the eight-pointed star block. The
quilt displayed provides some suggestions.
investigat
ioninvestigat
ion
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23.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 123
Three-dimensional objects
As mentioned previously, the shape of 3-D objects can be described by three measure-
ments. These objects can be classiﬁed as prisms or non-prisms.
Prisms
Prisms are 3-D ﬁgures that have a constant cross-section in one direction. This con-
stant cross-section is parallel to the face which is called the base of the prism. The
name of the prism comes from the shape of its base.
Common examples of prisms are:
Non-prisms
Non-prisms do not have a constant cross-section in any direction. The two types of 3-D
object that we will study in this category are pyramids and spheres.
Pyramids
In pyramids, the cross-section parallel to the base reduces in size as the cross-section
progresses from the base to the apex.
Common examples of pyramids are:
Another interesting block is the cube lattice
shown at right.
5 Trace a set of four of these. Place them together at
a point and experiment with different shading or
colouring of the faces to produce different 3-D
effects. Note that dark colours recede and light
colours come to the fore in terms of vision.
6 Identify the number and types of different shapes
in each block. Draw a design for a single-bed quilt 173 cm by 218 cm
(including borders) using blocks of this type. Determine the number of pieces
of each shape required.
7 Design a patchwork block of your own. Draw a completed quilt incorporating
your block.
Cube
Rectangular prism
Cylinder
Square-based
pyramid
Triangle-based
pyramid Cone
MQ Maths A Yr 11 - 04 Page 123 Wednesday, July 4, 2001 4:11 PM
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124 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Spheres
A sphere has no ﬂat faces. When spheres are sliced, the ﬂat surface exposed is always
circular.
Common examples are:
Two properties which three-dimensional ﬁgures possess are surface area and volume.
Surface area
The surface area of a 3-D object represents the total area of all its exposed faces. To
ﬁnd the surface area, we must calculate the area of each face of the object, as identiﬁed
by its net, then add all of these areas to ﬁnd the total. The units used for total surface
area (TSA) are the same as those used for area.
Object Net TSA
Cube TSA = 6S2
Rectangular
prism
TSA = area of
TSA = 6 rectangles
TSA = 2(WH + LW + LH)
Cylinder TSA = area of 2 circles
+ curved surface
= 2πr2
+ 2πrH
Sphere
Hemisphere
S
1 3
2
6
S
S
4 5
W
L
H W
H
H
W
L
L
H
H
r
H
r
r
2πr
MQ Maths A Yr 11 - 04 Page 124 Wednesday, July 4, 2001 4:11 PM
25.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 125
Square-based
pyramid
TSA= area of square
+ area of 4
triangles
= b2
+ 4 × ( bh)
Cone Curved surface of a cone is
formed by removing a sector
out of a circle.
TSA = area of base (circle)
+ area of curved surface
= πr2
+ πrS
Sphere Not shown TSA = 4πr2
Hemisphere
Open
Not shown TSA = 2πr2
Closed Not shown TSA = 2πr2
+ area ﬂat circle
TSA = 2πr2
+ πr2
TSA = 3πr2
Object Net TSA
b
h b
h
1
2
---
S = Slant
heightH
r
Arc length Slant height
= radius of
sector
Circumference of
base = arc length
of sector
r = S
Minor
sector
Major
sector
Math
cad
TSA
r
r
MQ Maths A Yr 11 - 04 Page 125 Wednesday, July 4, 2001 4:11 PM
26.
126 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Find the total surface area of the object shown.
THINK WRITE
Identify the shape. Shape: rectangular prism
Write down the formula for the TSA of
a rectangular prism.
TSA = area 6 rectangles
TSA = 2(WH + LW + LH)
Allocate a value to the pronumerals. W = 9, H = 17, L = 19
Substitute the values of the
pronumerals into the formula.
TSA= 2(9 × 17 + 19 × 9 + 19 × 17)
Evaluate (brackets ﬁrst, then multiply
by 2).
= 2(153 + 171 + 323)
= 2 × 647
= 1294
Write the answer, including units. TSA= 1294 cm2
17 cm
19 cm
9 cm1
2
3
4
5
6
10WORKEDExample
Find the surface area of an open cylindrical can that is 12 cm high and
8 cm in diameter (to 1 decimal place).
THINK WRITE
Form a net of the open cylinder,
transferring all the dimensions to each
of the surfaces.
(Note that the cylinder has no top
surface.)
Identify the different-sized common
ﬁgures and set up a sum of the surface
areas. The length of the rectangle is the
circumference of the circle.
TSA = A1 + A2
A1 = 2πr × H
= 2 × π × 4 × 12
= 301.59 cm2
A2 = π × r2
= π × 42
= 50.27 cm2
Sum the areas. TSA = A1 + A2
= 301.59 + 50.27
= 351.86 cm2
Write your answer. The total surface area of the open cylindrical
can is 351.9 cm2
.
8 cm
12cm
1 2 rπ
A1
12cm
A2
4 cm
2
3
4
11WORKEDExample
MQ Maths A Yr 11 - 04 Page 126 Wednesday, July 4, 2001 4:11 PM
27.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 127
Find the surface area of the square pyramid at right.
THINK WRITE
Calculate the area of the square base. A = S2
= 62
= 36 cm2
Calculate the area of a triangular side.
Note each side is identical and the
height of each triangular side is 5 cm.
A = × b × h
= × 6 × 5
= 15 cm2
Calculate the total surface area.
(Note: There are 4 identical triangular
sides.)
SA = 36 + 4 × 15
= 96 cm2
6 cm4cm
5cm
1
2
1
2
---
1
2
---
3
12WORKEDExample
Find the total surface area of a size 7 basketball with
a diameter of 25 cm. Give your answer to the nearest 10 cm2
.
THINK WRITE
Use the formula for the total
surface area of a sphere. Use
the diameter to ﬁnd the radius
of the basketball and
substitute into the formula.
Diameter = 25 cm
Radius = 12.5 cm
TSA of sphere = 4πr2
= 4 × π × 12.52
= 1963.495
Write your answer. Total surface area of the ball is approximately 1960 cm2
.
1
2
13WORKEDExample
A die used in a board game has a total surface area of 1350 mm2
.
Find the linear dimensions of the die (to the nearest millimetre).
THINK WRITE
A die is a cube. We can
substitute into the total
surface area of a cube to
determine the dimension of
the cube. Divide both sides
by 6.
TSA = 6 × S2
TSA = 1350 mm2
1350 = 6 × S2
S2
=
= 225
Take the square root of both
sides to ﬁnd S.
S =
= 15 mm
Write your answer. The dimensions of the die are:
15 mm × 15 mm × 15 mm
1
1350
6
------------
2 225
3
14WORKEDExample
MQ Maths A Yr 11 - 04 Page 127 Wednesday, July 4, 2001 4:11 PM
28.
128 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
The diagram shows the proposed shape for a new container for takeaway
Chinese food. The shape will be used if the TSA of the container is less
than 750 cm2
. If the TSA is greater than or equal to 750 cm2
then
production and manufacturing costs are too expensive and the takeaway
shop will have to stay with the old cylindrical container. Next time I order
my Chinese takeaway, could it come in the new design?
THINK WRITE
Identify the distinct shapes that make up
the total object: these are a square-based
pyramid and a cube. The base of the
pyramid and one face of the cube are not
on the surface and therefore their area
should not be included.
TSA= square pyramid (no base)
+ 5 faces of a cube
Calculate the TSA of a square-based
pyramid with no base.
(a) Alter the general square-based
pyramid formula so as not to include
the square base.
(b) Allocate a value to the pronumerals.
(c) Substitute the values into the formula
and evaluate.
TSA of square-based pyramid:
A = 4 × × b × h
b = 10, h = 10
A = 4 × × 10 × 10
= 200
Calculate the TSA of a cube (5 faces
only).
(a) Alter the general cube formula to
include 5 faces instead of 6.
(b) Allocate a value to the length.
(c) Substitute the value of the side length
into the formula and evaluate.
TSA of a cube (5 faces only):
A = 5S2
S = 10
A = 5 × 102
= 500
Add the individual TSA together to ﬁnd
the TSA of the whole object.
TSA= 200 + 500
= 700 cm2
Write an answer sentence. Next time I order Chinese takeaway there
could be a newly designed container with a
surface area of only 700 cm2
.
10
cm
10 cm
1
2
1
2
---
1
2
---
3
4
5
15WORKEDExample
remember
1. The TSA is the sum of the areas of the outside surfaces of a 3-dimensional object.
2. Formulas for all types of objects are not possible. For those objects without a
formula you will need to follow these steps.
(i) Draw the net of the object.
(ii) Work out the different shapes that make up the net.
(iii)Calculate their individual areas.
(iv) Add all the individual parts together.
3. Do not include in the TSA the surfaces of contact of the distinct shapes that
make up a composite ﬁgure.
remember
MQ Maths A Yr 11 - 04 Page 128 Wednesday, July 4, 2001 4:11 PM
29.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 129
Total surface area
1 Name each solid in the top row then match it with a net in the bottom row.
2 Draw the net of each of the following solids.
3 Identify the solids from the nets below. Draw the solid for each.
4 Find the total surface area of each of the ﬁgures below.
5 Oliver is making a box in the shape of a rectangular prism. The box is to be
2.5 m long, 1.2 m wide and 0.8 m high. Calculate the surface area of the box.
a b c
i ii iii
a b c
a b c
a b c
d e f
4C
5 cm
9 cm
32 cm
WORKED
Example
10
3.9 cm
4.1 cm
4 cm
20 cm
13 cm
14 cm
42 mm
7 mm
7 mm
MQ Maths A Yr 11 - 04 Page 129 Wednesday, July 4, 2001 4:14 PM
30.
130 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
6 Calculate the surface area of an open box in the shape of a cube, with a side length
of 75 cm. (Hint: Since the box is open there are only ﬁve faces.)
7 A room is in the shape of a rectangular prism. The ﬂoor is 5 m long and 3.5 m
wide. The room has a ceiling 2.5 m high. The ﬂoor is to be covered with slate
tiles, the walls are to be painted blue and the ceiling is to be painted white.
a Calculate the area to be tiled.
b Each tile is 0.25 m2
. Calculate the number of tiles needed.
c Calculate the area to be painted blue.
d Calculate the area to be
painted white.
e One litre of paint covers
an area of 12 m2
. How
many litres of paint are
needed to paint the
room?
8 Find the total surface area
of the following cylinders.
a
b
c
9 Calculate the surface area of
the square pyramid at right.
10 A triangle-based pyramid has four equal faces as shown at right.
Calculate the surface area to the nearest cm2
.
(Recall Heron’s formula.)
WORKED
Example
11
Radius = 410 mm
Length = 1.5 m
90 cm
28 cm
250 mm
250 mm
(Answer to nearest cm2)
WORKED
Example
12
10 cm
13cm
4 cm
MQ Maths A Yr 11 - 04 Page 130 Wednesday, July 4, 2001 4:14 PM
31.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 131
11
Two cubes are drawn such that the side length on the second cube is double the side
length on the ﬁrst cube. The surface area of the larger cube will be:
A twice the surface area of the smaller cube
B four times the surface area of the small cube
C six times the surface area of the small cube
D eight times the surface area of the small cube.
12 Calculate the surface area of the triangular prism at right.
13 Calculate the surface area of the following cones.
14 Find the total surface area of the following spheres and hemispheres.
15 A cube has a total surface area of 24 cm2
. What is the length of each side?
16 Another cube has a total surface area twice that of the one in question 15. Is the side
length of this cube twice that of the one in the previous question? Explain.
17 Calculate the surface area of this prism.
18 A concrete swimming pool is a
rectangular prism 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
.) Remember there is no top
on the pool.
a b c
a b c
mmultiple choiceultiple choice
4 cm
3 cm
5 cm
2 cm
2.4 cm
2.9 cm
40 cm
32 cm
10 cm
18 cm
WORKED
Example
13
43 mm
(Answer in cm2.)
5 m
Open
(Answer to
nearest mm2.)
Closed
6.3 mm
WORKED
Example
14
WORKED
Example
15 3.2 m
1m
2 m
4 m
6 m
MQ Maths A Yr 11 - 04 Page 131 Friday, July 6, 2001 8:40 AM
32.
132 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
19 What is the total area of canvas needed for the tent
(including the base) shown in the diagram at right?
Give the answer to the nearest m2
.
20
A ball used in a game of pool has a diameter
of 48 mm. The total surface area of the
ball is closest to:
A 1810 mm2
B 2300 mm2
C 7240 mm2
D 28 950 mm2
E 115 800 mm2
21
The total surface of a golf ball of radius 21 mm is closest to:
22
The formula for the total surface area for the object shown is:
A abh B 2 × bh + ab + 2 × ah
C 3( bh + ab) D bh + 3ab
E bh + 3ab
23
The total surface area of a poster tube that is 115 cm long and 8 cm in diameter is
closest to:
24 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 ﬂat, round ends that have 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?
A 550 mm2
B 55 cm2
C 55 000 mm2
D 0.055 m2
E 5.5 cm2
A 3000 cm2
B 2900 cm2
C 1500 mm2
D 6200 m2
E 23 000 cm2
4.5 m
6.5 m
2.5 m
1.5m
1.0m
mmultiple choiceultiple choice
mmultiple choiceultiple choice
mmultiple choiceultiple choice
h
b
a
1
2
---
1
2
---
1
2
---
1
2
---
mmultiple choiceultiple choice
MQ Maths A Yr 11 - 04 Page 132 Wednesday, July 4, 2001 4:14 PM
33.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 133
1 Calculate the area of a rectangle with a length of 0.4 m and a width of 1.1 m.
2 Calculate the area of a triangle with a base of 12.3 m and a height of 4.8 m.
3 Calculate the area of the trapezium at right.
Name the solids below.
4 5 6
7 Find the surface area of the cube shown at left.
8 Find the surface area of a rectangular prism
with a length of 8 cm, a width of 5 cm
and a height of 6 cm.
9 Find the surface area of the
triangular prism below.
10 Find the surface area of the
square pyramid below.
2
32 m
96 m
56 m
10 cm 8 cm
20 cm
6 cm
6cm
8 cm
9 cm
MQ Maths A Yr 11 - 04 Page 133 Wednesday, July 4, 2001 4:14 PM
34.
134 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Volume and capacity
Volume
The volume of a 3-D object represents the amount of space contained in, or occupied
by, the object. The units used to measure volume are those of cubic measure: mm3
,
cm3
, m3
.
Constructing our conversion ladder, as before, will enable us to convert from one
unit to another.
To obtain the conversion factors for cubic measure, the linear measure conversion
factors are cubed. The same procedure applies as before: multiply when converting to a
smaller unit and divide when converting to a larger unit.
The capacity of a 3-D object refers to the quantity of solid, liquid or gas it could
hold. The units used to measure capacity are millilitres (mL), litres (L) and kilolitres
(kL). The conversion ladder for capacity units is as follows:
In capacity units, 1 mL represents the amount
of liquid which a 1-cm cube could hold. Two useful
conversion relationships are:
1 cm3
≡ 1 mL
and, for larger quantities 1 m3
≡ 1 kL
mm3
10 × 10 × 10 = 1000
cm3
100 × 100 × 100 = 1000 000
m3
× ÷
mL
1000
L
1000
kL
× ÷
EXCE
L Spreadshe
et
Capacity
Mat
hcad
Unit con-
versions
Convert 1.12 cm3
to mm3
.
THINK WRITE
The conversion from centimetres to
millimetres is 1 cm = 10 mm.
The conversion factor for cm3
to mm3
is to multiply by 103
or 1000; that is,
1 cm3
= 1000 mm3
.
1.12 cm3
= 1.12 × 1000 mm3
= 1120 mm3
Write the answer in correct units.
1
2
3
16WORKEDExample
1 cm
1 cm
1 cm
Holds 1 mL
MQ Maths A Yr 11 - 04 Page 134 Wednesday, July 4, 2001 4:14 PM
35.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 135
In our calculations of volume, we shall consider only three classes of 3-D ﬁgures:
prisms, pyramids and spheres.
Volume of prisms
The volume of a prism is given by the following formula:
Volume of a prism = cross-sectional area × height of the prism
The height is the dimension perpendicular to the cross-sectional area.
Shape Cross-sectional shape Volume
Cylinder
Area = πr2
V = area of a circle
× height
= πr2
× H
Triangular
prism
Area = bh
V = area of a triangle
× height
= bh × H
Note: Lower-case h
represents the height of
the triangle.
Rectangular
prism
Area = L × W
V = area of a rectangle
× height
= L × W × H
Cube
Area = S2
V = area of a square
× height
= S2
× H
= S2
× S
= S3
(since in a cube, H = S)
Math
cad
Capacity
Convert:
a 400 cm3
into mL b 1200 cm3
into mL and into L c 2 kL into m3
.
THINK WRITE
a Since 1 cm3
is equivalent to 1 mL, then 400 cm3
is
equivalent to 400 mL.
a 400 cm3
= 400 mL
b Each 1 cm3
will hold 1 mL of liquid. Therefore,
1200 cm3
will hold 1200 mL of liquid.
b 1200 cm3
= 1200 mL
To change mL to L, divide by 1000 (since there
are 1000 mL in 1 L).
= 1.2 L
c One kL is equivalent to 1 m3
. Therefore, 2 kL is
equivalent to 2 m3
.
c 2 kL = 2 m3
1
2
17WORKEDExample
H
r
r
b
Hh b
h
1
2
---
1
2
---
E
XCEL Spread
sheet
VolumeWL
H
W
L
Math
cad
Volume
formulasS
H
S
MQ Maths A Yr 11 - 04 Page 135 Wednesday, July 4, 2001 4:14 PM
36.
136 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Volume of pyramids
As we have seen previously, a pyramid has a ﬂat base at one end, and tapers to a point
at the other. Some examples of pyramids are shown below. A cone is really a circle-
based pyramid.
Cone Square pyramid Rectangular pyramid Triangular pyramid
Comparing volumes of
pyramids and prisms
Resources: Set of 3-D volumetric shapes of pyramids and prisms with same base
area and height; water (or rice).
For the following investigation, the volumes of pairs of open 3-D containers are
compared by considering the amount of water (or rice) each can hold. Each 3-D
pair should have the same base area and the same perpendicular height.
Consider the following pairs of containers:
square-based pyramid and cube
rectangle-based pyramid and rectangular prism
triangle-based pyramid and triangular prism
cone and cylinder
1 Fill the ﬁrst container with the water (or rice), then pour the contents into the
second container. Continue reﬁlling the ﬁrst container and pouring the contents
into the second until the second container is full. How many times was it
necessary to do this?
Find the volume of the shape shown correct to 1 decimal place.
THINK WRITE
Identify the shape. Triangular prism
Write down the appropriate formula for the
volume.
V = bh × H
Allocate values to the pronumerals keeping
in mind that b and h are the base and height
of a triangular cross-section or base of the
prism, while H is the height of the prism.
b = 2.6, h = 2.3, H = 3.2
Substitute and evaluate, rounding the answer
to 1 decimal place.
V = × 2.6 × 2.3 × 3.2
= 9.568
≈ 9.6 m3
2.6 m
3.2 m
2.3m
1
2 1
2
---
3
4 1
2
---
18WORKEDExampleinv
estigat
ioninv
estigat
ion
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37.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 137
A pyramid does not have a uniform cross-section. The cross-sectional area becomes
smaller as it nears the apex (point). The internal capacity or volume of a tapered object
is a fraction of the volume of a prism. Mathematicians found this fraction to be a third
( ). They deﬁned the base of a pyramid to be the ﬂat end opposite the apex. To calcu-
late the volume of a pyramid we ﬁnd the area of the ﬂat end, multiply this by the height
of the pyramid (which must be perpendicular to the base) and then multiply by (or
divide by 3).
Volume of a pyramid = × area of base × height of object
The following table shows the formulas for the volume of some common pyramids.
2 From your results, how would you compare the volume of a pyramid with that
of a prism of the same base area and height?
3 Having previously considered the general formula for the volume of a prism,
suggest a general formula for the volume of a pyramid.
4 Draw labelled diagrams and deduce speciﬁc formulas for the volume of each of
the following:
a square-based pyramid b rectangle-based pyramid
c triangle-based pyramid d cone (circle-based pyramid).
Shape
Flat end (base)
shape Volume
Cone V = × area of a circle
× height
V = πr2
× H
Square
pyramid
V = × area of a square
× height
V = S2
× H
Rectangular
pyramid
V = × area of a
rectangle × height
= L × W × H
Triangular
pyramid
V = × area of a triangle
× height
V = ( bh) × H
Note: Lower-case h
represents the height of
the triangle.
1
3
---
1
3
---
1
3
---
Math
cad
Mensura-
tion
r
H r
1
3
---
1
3
---
H
S
S
1
3
---
1
3
---
H
L
W
W
L
1
3
---
1
3
---
Math
cad
Volume
formulasH
h
b
h
b
1
3
---
1
3
---
1
2
---
MQ Maths A Yr 11 - 04 Page 137 Wednesday, July 4, 2001 4:14 PM
38.
138 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Volume of spheres
The volume of a sphere of radius r is given by the following formula:
Volume of a sphere = πr3
A hemisphere is half of a sphere. Its volume, therefore, is half of the volume of a
sphere.
Volume of hemisphere = (volume of sphere)
= × πr3
= πr3
Find the volume of the pyramid at right (to the nearest m3
).
THINK WRITE
Write the equation. V = × area of base × height
The pyramid has a square base. It is a
square pyramid. The area of the base is
given by S2
.
Area of base = S2
Calculate the volume. Volume = × S2
× H
Volume = × 302
× 40
Volume = 12 000 m3
Write your answer. The volume of the square pyramid is 12 000 m3
.
30 m 30 m
Height of pyramid = 40 m
1
1
3
---
2
3
1
3
---
1
3
---
4
19WORKEDExample
Find the volume of the cone at right, correct to 2 decimal places.
THINK WRITE
Write the formula. V = πr2
H
Substitute the radius and height. = × π × 3.22
× 8.5
Calculate the volume. = 91.15 cm3
3.2 cm
8.5 cm
1
1
3
---
2
1
3
---
3
20WORKEDExample
Mat
hcad
Volume
formulas
r 4
3
---
1
2
---
r
1
2
---
4
3
---
2
3
---
MQ Maths A Yr 11 - 04 Page 138 Wednesday, July 4, 2001 4:14 PM
39.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 139
Note that the volume and capacity of a 3-D object do not depend on whether the object
is open or closed. An open rainwater tank could hold the same quantity of water as a
closed one. The surface area, however, varies depending on whether the object is open
or closed.
Cross-sections of solid 3-D shapes
Resources: Plasticine (or play dough), thread.
In this activity we investigate the surfaces exposed when solid 3-D shapes are cut
in different directions.
1 Using plasticine, mould 3-D solids in the shape of a:
a cube b rectangular prism c triangular prism
d cylinder e sphere.
2 Investigate slicing these solids in various directions with a piece of thread to
see whether it is possible to obtain sections with faces in the shape of a:
a square b rectangle c triangle
d ellipse e circle.
3 Copy and complete the following table, showing the direction of the sectional
cut required for each particular face to be exposed. Not all face shapes are possible.
Find the volume of a sphere with a radius of 9.5 cm, correct to the nearest cm3
.
THINK WRITE
Write the formula. V = πr3
Substitute the radius. = × π × 9.53
Calculate the volume. = 3591 cm3
1
4
3
---
2
4
3
---
3
21WORKEDExample
inv
estigat
ioninv
estigat
ion
Section
face
Solid
Cube Rectangular
prism
Triangular
prism
Cylinder Sphere
Square
Rectangle
Triangle
Ellipse
Circle
MQ Maths A Yr 11 - 04 Page 139 Wednesday, July 4, 2001 4:14 PM
40.
140 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Volume of composite objects
Often, an object can be identiﬁed as comprising two or more different common prisms,
pyramids or spheres. Such ﬁgures are called composite objects. The volume of a com-
posite object is found by adding the volumes of the individual common ﬁgures or
deducting volumes. A grain silo can be modelled as the sum of a cylinder and a large
cone, less the tip of the large cone.
The volume of a composite object is equal to the sum (or difference) of the
individual common prisms, pyramids or spheres.
Vcomposite = V1 + V2 + V3 + . . . (or Vcomposite = V1 − V2)
Find the capacity of the object shown at right (to the nearest litre).
THINK WRITE
The object is a composite of a cylinder
and a square prism.
The volume of the composite object is
the sum of volumes of the cylinder plus
the prism.
Vcomposite = volume of cylinder + volume of
square prism
= (πr2
× Hc) + (S2
× Hs)
= (π × 62
× 20) + (182
× 25)
= 2261.946 711 + 8100
= 10 361.946 711 cm3
Convert to litres using the conversion of
1 cm3
= 1 mL
1000 mL = 1 L
10 362 cm3
= 10.362 litres
Write your answer. The capacity of the object is 10 litres.
25cm
12 cm
18 cm
20cm
1 r = 6 cm
18 cm
25cm
18 cm
H=20cm
2
3
22WORKEDExample
MQ Maths A Yr 11 - 04 Page 140 Wednesday, July 4, 2001 4:14 PM
41.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 141
Volume and capacity
1 Convert the volumes to the units speciﬁed.
2 Convert the following units as indicated.
3 Calculate the volume and capacity of each of the prisms below.
a 0.35 cm3
to mm3
b 4800 cm3
to m3
c 56 000 cm3
to litres
d 1.5 litres to cm3
e 1.6 m3
to litres f 0.0023 cm3
to mm3
g 0.000 57 m3
to cm3
h 140 000 mm3
to litres i 250 000 mm3
to cm3
a 750 cm3
= mL b 800 cm3
= L
c 2500 cm3
= mL d 40 000 cm3
= L
e 6 m3
= cm3
= mL = L f 12 m3
= L
g 4.2 m3
= kL h 7.5 m3
= kL = L
i 5.2 mL = cm3
j 6 L = cm3
k 20 L = mL = cm3
l 5.3 kL = m3
a b c
remember
1. Volume is measured in cubic units: mm3
, cm3
, m3
.
2. Capacity is measured in mL, L, kL.
3. To convert volume units into capacity units:
1 cm3
= 1 mL
1 m3
= 1 kL
4. The volume of a prism is found using the formula
V = area of base × height of prism
5. The volume of a pyramid is found using the formula
V = area of base × height of pyramid
6. The volume of a sphere is calculated using
V = πr3
and for a hemisphere,
V = πr3
7. The volume of a container is not dependent on whether it is open or closed.
8. The volume of a composite object can be found by adding or subtracting
volumes of individual prisms, pyramids or spheres.
1
3
---
4
3
---
2
3
---
remember
4D
Work
SHEET 4.2WORKED
Example
16
WORKED
Example
17
5 cm
2.4 m
13 m
MQ Maths A Yr 11 - 04 Page 141 Wednesday, July 4, 2001 4:14 PM
42.
142 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
4 For each of the triangular prisms below ﬁnd:
i the area of the base of the prism ii the volume of the prism.
a b
c d
5
The shape at right could be described as a:
6
The area of the base of a prism is 34.67 cm2
, and the height is 3.6 cm. The volume of
the prism is:
7
The dimensions of a rectangular prism are all doubled. The volume of the prism will
increase by a factor of:
8 A refrigerator is in the shape of a rectangular prism. The internal dimensions of the
prism are 60 cm by 60 cm by 140 cm.
a Find the volume of the refrigerator in cm3
.
b The capacity of a refrigerator is measured in litres. If 1 cm3
= 1 mL, ﬁnd the
capacity of the refrigerator in litres.
d e f
g h i
A cube B square prism
C rectangular prism D both B and C
A 38.27 cm2
B 38.27 cm3
C 124.12 cm2
D 124.812 cm3
A 2 B 4 C 6 D 8
4.2 m
3.2 m
50 mm
9 mm
9 mm
12.5 m
20.5 m
16.5 m
6 cm
12 cm
12 m
3 m
13 cm
27 cm
WORKED
Example
18
8 cm
3 cm 5 cm
6 cm
8 cm
12 cm
3.4 m
2.7 m 1.5 m
3.2 m
12.5 m
7.8 m
mmultiple choiceultiple choice
mmultiple choiceultiple choice
mmultiple choiceultiple choice
MQ Maths A Yr 11 - 04 Page 142 Wednesday, July 4, 2001 4:14 PM
43.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 143
9 A semitrailer is 15 m long, 2.5 m wide and 2.7 m high in the shape of a rectangular
prism. Find the capacity of the semitrailer. (Ignore the thickness of the walls.)
10 A petrol tanker is shown at right.
The tank is cylindrical in shape. The radius of
the tank is 2 m and the length is 12 m. Ignoring
the thickness of the material, calculate:
a the volume of the tank, correct to 3 decimal
places
b the capacity of the tank, to the nearest
100 litres. (1 m3
= 1000 L).
11 At right is a diagram of a concrete slab for a house.
a Calculate the area of the slab.
b The slab is to be 10 cm thick. Calculate the volume of
concrete needed for the slab. (Hint: Write 10 cm as 0.1 m.)
c Concrete costs $145.50/m3
to lay. Calculate the cost of this slab.
12 A ﬂat rectangular roof is 14 m long and 8 m wide. When it rains, the water is col-
lected in a cylindrical tank.
a Calculate the volume of water collected
on the roof when 25 mm of rain falls.
b How many litres of water does the roof
collect?
c The cylindrical tank has a radius of 1.8 m
and is 2.4 m high. What is the capacity of
the tank, in litres?
d By how much does the depth of water in
the tank rise when the rain falls? Answer
in centimetres, correct to 1 decimal place.
13 For each of the following pyramids, calculate the volume by ﬁrst calculating the area
of the base shape.
a b
c d
12 m
2 m
10 m
10 m
2.5 m
15 m
WORKED
Example
19
8 cm
6 cm
15 cm 8 cm
14 cm
10 m
12 m
6 m
12 cm
5 cm
8 cm
6 cm
MQ Maths A Yr 11 - 04 Page 143 Wednesday, July 4, 2001 4:14 PM
44.
144 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
14 Find the volume of each of the following cones, correct to the nearest whole number.
a b
c d
15 A cone has a base with a diameter of 9 cm and a height of 12 cm. Calculate the
volume of that cone, correct to 1 decimal place.
16 Calculate the volume of each of the following spheres, correct to 1 decimal place.
a b
c d
17 Calculate the volume of a sphere with a diameter of 2.3 cm. Answer correct to 2
decimal places.
18
Which of the following solids could not be described as a pyramid?
A B
C D
19
A triangular pyramid and a square pyramid both have a base area of 20 cm2
and a
height of 15 cm. Which has the greater volume?
A the triangular pyramid B the square pyramid
C both have equal volume D this can’t be calculated
WORKED
Example
20
5 cm
10 cm
12 cm
12 cm
8 mm
33 mm
42 cm
42 cm
WORKED
Example
21
6 cm 8 cm
12.5 m
3.2 m
mmultiple choiceultiple choice
mmultiple choiceultiple choice
MQ Maths A Yr 11 - 04 Page 144 Wednesday, July 4, 2001 4:14 PM
45.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 145
20
A spherical balloon has a volume of 500 cm3
. It is then inﬂated so that the diameter of
the balloon is doubled. The volume of the balloon will now be:
21 In each of the following, the prism’s front face is made up of a composite ﬁgure. For
each:
i calculate the area of the front face ii ﬁnd the volume of the prism.
a b
c d
22 Find the volume of the solid at right. Answer correct to
1 decimal place.
23 A hollow rubber ball is to be made with a radius of 8 cm,
and the rubber to be used is 1 cm thick.
a What would be the radius of the hollow inside?
b Calculate the volume of the ball.
c Calculate the volume of space inside the ball.
d Calculate the amount of rubber (in cm3
) needed to make the ball.
24 The ﬁgure at right is a truncated cone, that is, a cone with
the top cut off.
a Calculate the volume of the cone before it was truncated.
b The portion cut off was itself a cone. Calculate its volume.
c Calculate the volume of the truncated cone.
25 Use the same method as in question 24 to ﬁnd the volume of
the truncated pyramid shown at right.
26 The ﬁgure at right is of an ice-cream cone, containing a spherical
scoop of ice-cream (a whole sphere).
a Calculate the volume of the cone.
b Calculate the volume of the scoop of ice-cream.
c Calculate the total volume of the shape. (Hint: Only half the
sphere sits above the cone.)
A 1000 cm3
B 2000 cm3
C 3000 cm3
D 4000 cm3
mmultiple choiceultiple choice
10 cm
10 cm20 cm
4 cm
16 cm
8 m
9 m
4 m
6 cm
20 cm
15 cm
8 cm
12 cm
3 m
18 m
12 m
12 m
6 m
WORKED
Example
22 4 cm
12 cm
6 cm
3 cm
15 cm
6 cm
1 cm
5 cm
3 cm
3 cm
8 cm
2.5 cm
MQ Maths A Yr 11 - 04 Page 145 Wednesday, July 4, 2001 4:14 PM
46.
146 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
The optimum swimming pool
Resources: Pen, paper, calculator.
You are considering having a swimming pool constructed in your back yard. In
order to get the best value for money, you need to research the advantages of
differently shaped pools. Approach this investigation in a formal manner and
summarise your research under the headings:
Aim
Procedure
Results
Conclusion(s).
For the purpose of this investigation, you are to consider four factors:
1. A major cost in the construction of the pool lies in tiling the interior surface.
For this reason, you wish your pool to have a minimum surface area.
2. Your back yard allows you only a 10-metre by 5-metre area of land.
3. The depth of the pool must lie within the range 1.5 metres and 2 metres.
4. The pool must (subject to the above restrictions) have a maximum water
capacity.
It is obviously not possible to satisfy all these requirements with one particular
shape and size of pool. There must be compromises. Your task is to investigate
differently shaped pools and decide on a shape and size which best satisﬁes the
above requirements.
1 Aim
Begin by summarising the above information to deﬁne the aim of your
investigation.
2 Procedure
Explain how you intend to collect data that would enable you to make a
decision in light of the above restrictions.
3 Results
In order to approach this in a methodical manner, draw up a table with the
following headings.
4 Conclusions
Study the two right-hand columns of your table. Decide on a shape that offers
the best compromise between surface area and volume. Write your
recommendations. (You may consider that there are two shapes which would be
just as suitable.)
investigat
ioninvestigat
ion
Shape Surface area Volume
Within the above
restrictions, draw at
least 5 shapes here.
Label ﬁgures with
dimensions.
Calculate the surface
area of the base and
walls of each shape (the
top is open).
Determine the volume
of water each pool
would contain.
MQ Maths A Yr 11 - 04 Page 146 Wednesday, July 4, 2001 4:14 PM
47.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 147
Minimising surface area
Go to the Maths Quest CD and download the spreadsheet ‘Volume’.
Task 1
A cylindrical drink container is to have a capacity of 1 litre
(volume = 1000 cm3
). We are going to calculate the most cost-efﬁcient dimensions
to make the container. To do this, we want to make the container with as little
material as possible; in other words we want to minimise the surface area of the
cylinder. The spreadsheet should look as shown below.
1. In cell B3 enter the volume of the cylinder, 1000.
2. In cell A6 enter a radius of 1. In cell A7 enter a radius of 2 and so on up to a
radius of 20.
3. The formula that has been entered in cell B6 will give the height of the cylinder
corresponding to the radius for the given volume.
4. The surface area of each possible cylinder is in column D. Use the charting
function on the spreadsheet to graph the surface area against the radius.
5. What is the most cost efﬁcient dimensions of the drink container?
Task 2
Use one of the other worksheets to ﬁnd the most efﬁcient dimensions to make a
rectangular prism of volume 1000 cm3
and a cone of volume 200 cm3
.
E
XCEL Spread
sheet
Volume
inv
estigat
ioninv
estigat
ion
MQ Maths A Yr 11 - 04 Page 147 Wednesday, July 4, 2001 4:14 PM
48.
148 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Upkeep of above-ground circular pools
Resources: Computer spreadsheet.
Above-ground pools are a popular alternative to in-ground pools. They can be
dismantled when the children outgrow them and the lawn can be re-established.
The cost of maintaining a pool depends on the volume of water it contains (and
also on its use). The chemicals added to the water help to kill germs and to
maintain a healthy environment.
In this investigation we shall ignore the contribution of frequency of use and
consider only the effects of the diameter and depth of a circular pool on the upkeep
cost. As the diameter (and radius) of a circular pool increases, the volume of water
it contains also increases. If the radius doubles, does the volume also double (for a
given depth)? This is the basis of our investigation. Let us consider two situations:
1. circular swimming pools of varying radii and the same depth
2. circular swimming pools of varying depths and the same radius.
Part 1
1 Set up a spreadsheet with the following headings:
Radius Depth Surface area Volume
2 Enter values for the radius from 5 m to 15 m in steps of 1 m.
3 Under the heading ‘Depth’, enter a ﬁgure of 2 m down the entire column.
4 Enter the formula for surface area (area of a circle) in column 3 and copy it
down the column.
5 In column 4, enter the formula for volume (of a cylinder) then copy it down the
column.
6 Enter the graphing section of the spreadsheet and plot Radius on the x-axis and
Volume on the y-axis. Add suitable headings and print out a copy.
7 What are your conclusions about the variation of volume with radius for a
given depth?
Part 2
1 Amend your spreadsheet, keeping the following headings:
Radius Depth Surface area Volume
2 Enter a value of 10 m for the radius down the entire length of column 1.
3 Under the heading ‘Depth’, enter a ﬁgure of 1 m to 2 m in steps of 0.1 m.
Your spreadsheet should automatically recalculate the surface area and volume
for these new ﬁgures.
4 Enter the graphing section of the spreadsheet and plot Depth on the x-axis and
Volume on the y-axis. Add suitable headings and print a copy.
5 What are your conclusions about the variation of volume with depth for a given
radius?
Part 3
The cost of upkeep for a pool depends largely on the volume of water it contains.
Write a report to outline your ﬁndings on pool maintenance costs for circular
pools.
EXCE
L Spreadshe
et
Upkeep
circular
pools
investigat
ioninv
estigat
ion
MQ Maths A Yr 11 - 04 Page 148 Wednesday, July 4, 2001 4:14 PM
49.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 149
The optimum size for
a rainwater tank
Resources: Spreadsheet.
It is becoming more common these days to have a rainwater tank in the back yard
of a suburban residential property. Because space is limited in these situations,
certain restrictions must be placed on size.
Kirsten and Daniel have recently purchased their ﬁrst home. They wish to install
a cylindrical rainwater tank. Their constraints are as follows:
1. the base diameter plus the height of the tank must not exceed 5 metres
2. within the above restrictions, the volume of the tank must be as large as
possible.
1 Set up a spreadsheet with the following headings:
2 In a systematic manner, investigate the volume obtained from a variety of
combinations of diameter and height. Remember that the sum of these two
measurements must not exceed 5 metres.
3 What sized tank would you recommend for Kirsten and Daniel?
4 How much rainwater would it hold?
5 Prepare a report for Kirsten and Daniel, supporting your recommendations with
mathematical evidence.
Top-dressing lawns
Resources: Pen, paper, calculator.
Len owns a landscaping business. At the moment he has four small jobs on his
books. Each one requires top-dressing a lawn. They are all jobs where the owners
have removed sporting facilities and now want to establish a lawn.
Job 1 A tennis court has been dismantled and the area requiring top-dressing is
14 m by 6 m.
Job 2 A 10-m diameter circular pool has been removed.
Job 3 A children’s sand pit 10 m square is no longer required.
Job 4 A triangular play area 12 m by 16 m by 20 m is to be top-dressed and
turned into lawn.
Without thinking too much about the jobs, Len quoted to supply 4 m3
of top-soil
for each job.
E
XCEL Spread
sheet
Rain-
water
tank
inv
estigat
ioninv
estigat
ion
Diameter Radius Height Volume
E
XCEL Spread
sheet
Top-
dressing
lawns
inv
estigat
ioninv
estigat
ion
MQ Maths A Yr 11 - 04 Page 149 Wednesday, July 4, 2001 4:14 PM
50.
150 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Draw up the table below and complete the blank spaces.
Which job receives the greatest depth of soil? Justify your answer with
mathematical evidence.
Developing islands and canals
Resources: Pen, paper, calculator.
A tourist enterprise is considering developing land by creating artiﬁcial islands
surrounded by canals. It is envisaged that pleasure cruises and water sports would
take place on the canals, while tourist accommodation would be established on the
islands. The land in question has an area of 2 km square.
The constraints for the project are:
1. The canal must go completely around the perimeter of the land.
2. The islands created must be circular and each must be no smaller than 5000 m2
.
3. There must be at least 4 islands in the development.
4. In order that the pleasure cruisers can navigate the islands, each canal must
have a minimum width of 50 metres and a depth of 10 metres across its entire
width.
1 Draw a sketch of the area of land and investigate options within the above
constraints. Organise your investigations in the form of a table.
The developer has to bear in mind that the islands provide accommodation for
guests, while the canals provide entertainment.
2 Consider the results from your table above. Recommend a development which
you consider would provide an estate with the optimum balance between land
and water. Provide a plan (with measurements indicated) and justify your
decision with sound mathematical reasoning.
Job Diagram Area Volume of
topsoil
Depth of
topsoil
1 4 m3
2 4 m3
3 4 m3
4 4 m3
inv
estigat
ioninv
estigat
ion
Sketch Total area of islands
Volume of water
in canals
Draw at least 4 sketches
showing proposals for
the positions and shapes
of the islands.
Calculate the total area
of the exposed land on
the islands.
Find the volume of soil
that would be removed
to form the canals. This
approximates the
volume of water that
would ﬁll the canals.
MQ Maths A Yr 11 - 04 Page 150 Wednesday, July 4, 2001 4:14 PM
51.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 151
Units of measurement
• Measurements of length
10 mm = 1 cm
100 cm = 1 m
1000 m = 1 km
• Measurements of area
100 mm2
= 1 cm2
10 000 cm2
= 1 m2
1 000 000 m2
= 1 km2
10 000 m2
= 1 ha
• Measurements of volume
1000 mm3
= 1 cm3
1 000 000 cm3
= 1 m3
• Measurements of capacity
1000 mL = 1 L
1000 L = 1 kL
• Conversion of volume to capacity
1 cm3
= 1 mL
1 m3
= 1 kL
Perimeter
• Perimeter is the distance around an enclosed ﬁgure.
1. Perimeter formulas for common shapes encountered are
Square P = 4S
Rectangle P = 2(L + W)
Circle C = 2πr or πD
Sector C = × 2πr + 2r
Other ﬁgures P = sum of lengths of all sides.
2. Perimeter is measured in linear measure.
Area
• Area is the amount of space within the boundary of a closed ﬁgure.
• Area formulas for common shapes encountered are:
Square A = S2
Rectangle A = L × W
Parallelogram A = base × perpendicular height
Trapezium A = (a + b) × h
Triangle A = bh
(when 3 sides are known) A = (Heron’s formula)
where s = (a + b + c)
summary
θ°
360°
-----------
1
2
---
1
2
---
s s a–( ) s b–( ) s c–( )
1
2
---
MQ Maths A Yr 11 - 04 Page 151 Wednesday, July 4, 2001 4:14 PM
52.
152 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Circle A = πr2
Sector A = × πr2
Composite ﬁgures A = sum or difference of areas of individual shapes
• Area is measured in square measure.
Total surface area (TSA)
• The surface area of a 3-D object represents the total area of all its exposed surfaces.
• Total surface area formulas for common shapes encountered are:
Cube TSA = 6S2
Rectangular prism TSA = 2(WH + LW + LH)
Cylinder TSA = 2πr2
+ 2πrH
Square pyramid TSA = S2
+ 4 × ( bh)
Cone TSA = πr2
+ πrS
Sphere TSA = 4πr2
Open hemisphere TSA = 2πr2
Closed hemisphere TSA = 3πr2
Composite ﬁgures TSA = sum of areas of all exposed faces
Volume
• Volume represents the amount of space contained in, or occupied by, an object.
• Volume formulas for common shapes encountered are:
Prisms V = area of base × height
Pyramids V = area of base × height
Spheres V = πr3
Hemispheres V = πr3
Composite ﬁgures V = sum or difference of volumes of individual shapes
• While volume is represented in cubic measure, capacity is represented in mL, L or
kL.
• The volume of an object does not depend on whether it is open or closed.
θ°
360°
-----------
1
2
---
1
3
---
4
3
---
2
3
---
MQ Maths A Yr 11 - 04 Page 152 Wednesday, July 4, 2001 4:14 PM
53.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 153
1 Copy and complete each of the following.
a 190 mm = _____ cm b 190 mm2
= _____ cm2
c 190 mm3
= _____ cm3
d 500 mL = _____ L e 500 mL = _____ kL f 50 m3
= _____ L
g 0.2 m3
= _____ cm3
h 0.2 m2
= _____ cm2
i 0.2 m = _____ cm
j 120 cm3
= _____ mL k 120 cm3
= _____ L l 0.3 kL = _____ cm3
2 Find: i the area and ii the perimeter of the following shapes.
3 Calculate i the area and ii the perimeter of the following shapes. Give your answers correct
to 1 decimal place.
4
Examine the diagram at right.
a The circles cover an area of approximately:
b The shaded area is approximately:
5 Draw the net of each of the following solids.
6 Name the solid shape for which the net is given at right.
a b c
d e
a b c
A 402 cm2
B 201 cm2
C 804 cm2
D 805 cm2
E 603 cm2
A 219 cm2
B 421 cm2
C 622 cm2
D 823 cm2
E 220 cm2
a b c
4A
CHAPTER
review
4B
4D
4A
4B
GC pro
gram
Mensuration
20 mm
14 mm
7 cm
20°
12 cm
13 cm
5 cm 9 cm
>>
>>
>
>
16 cm24 cm
21.3cm
4 cm
18 cm 2.5 m
1.5 m
3.2 m
5.5 m
4A
4B
30 cm
15 cm
10 cm
13 cm
25 cm 10 m
6 cm
4Bmmultiple choiceultiple choice
32 cm
4C
4C
MQ Maths A Yr 11 - 04 Page 153 Wednesday, July 4, 2001 4:14 PM
54.
154 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
7 Find the surface area of each of the following solids.
a b c d
8 Calculate the surface area of each of the ﬁgures below, by calculating the area of each face
separately and adding them.
9 Calculate the surface area of each of the following cylinders.
10 Find the total surface area of the following pyramids.
11 Find the total exterior surface area of the following objects (to the nearest whole number).
12 At right is a diagram of an Olympic swimming pool.
a Calculate the area of one side wall.
b Use the formula V = A × h, to calculate the volume
of the pool.
c How many litres of water will it take to ﬁll the pool?
(1 m3
= 1000 L)
d The walls and ﬂoor of the pool need to be painted.
Calculate the area to be painted.
a b c
a b c
a b c
a b c
4C
4.2 cm
3.9 m
2.1 m
0.8 m
1.8 m
0.9 m
4.6 m
4C
4 m
3 m 3.5 m
5 m
4 cm
12 cm
6 cm
10 cm
5 cm
3 m
2 m
15 m
5 m
12 m
4C
6 cm
13 cm
3.8 m
1.6 m
20 cm
32.5 cm
4C
8.4 m
10.5 m
14 mm
42 mm
20 cm
18 cm
4C
12 cm
Closed
9 mm
10 m
Open
2 m
22 m
50 m
50.01 m
1 m
MQ Maths A Yr 11 - 04 Page 154 Wednesday, July 4, 2001 4:14 PM
55.
C h a p t e r 4 L e n g t h , a r e a a n d v o l u m e 155
13 At right are the plans for a garage that Rob is building. All the
walls are bricked. (Note: The garage has an iron roof and is closed
at one end.) Calculate the area that will need to be bricked.
14 Use the formulas to calculate the volume and capacity of each of
the following cubes, rectangular prisms and cylinders.
15 A prism has a base area of 45 cm2
and a height of 13 cm. Calculate the volume.
16 Use the formula V = × A × h to calculate the volume and capacity of each of the pyramids
below.
17 Calculate the volume of each of the pyramids, cones and spheres below.
18 Find the volume and capacity of each of the following shapes correct to 1 decimal place.
a b c
d e f
a b c
a b c
d e f
a b c
4CD
2.5 m 6 m
3 m
4D
6.5 cm
29 mm 11.6 m
4.6 m
3.8 m
3 cm
8 cm
41 cm
3 cm
13 cm
32 mm
18 mm
4D
4D
1
3
---
A = 16 cm2
9 cm
A = 126 mm2
19 mm
A = 6.9 m2
2.3 m
4D
25 m
36 m 7.9 m
3.2 m
2.6 m
19 mm
52 mm
23.5 mm
19.5 mm
23 mm 70 cm
4D
1.4m
2 m
1.8 m
60 cm
113cm
64 cm 22 cm
25 cm
10 cm
15
cm
MQ Maths A Yr 11 - 04 Page 155 Monday, September 24, 2001 7:14 AM
56.
156 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
19 The diagram at right shows 3 tennis balls packed in a cylindrical container. Find:
a the volume of each ball
b the volume of the cylinder
c the volume of space that remains free.
20 Calculate the area of a circle with a diameter of 8.6 cm, correct to 1 decimal place.
21 Calculate the area of the annulus (ring) shown at right,
correct to 2 decimal places.
22 Calculate the area of the sector at right,
correct to 1 decimal place.
23 Calculate the area of the ﬁgure at right.
24 Calculate the shaded area in the ﬁgure drawn at right,
correct to 2 decimal places.
25 Calculate the area in the ﬁgure at right,
correct to 2 decimal places.
26 Calculate the surface area of a closed cylinder with a radius of 10 cm and a height of 23 cm.
Give your answer correct to the nearest whole number.
27 Calculate the surface area of a sphere with a radius of 1.3 m. Give your correct to 3 decimal
places.
28 Calculate the volume of the prism
drawn at right.
29 Calculate the volume of the solid at right, correct to the
nearest whole number.
H
C
7 cm
9 cm
3 cm
13.2 cm
85°
10 cm
28 cm
9cm
29cm
9.7 cm
4.6cm
5 cm
13.7 cm
9.1 cm
13.4 cm
20.3cm
testtest
CHAPTER
yyourselfourself
testyyourselfourself
4
4 cm
8 cm
MQ Maths A Yr 11 - 04 Page 156 Wednesday, July 4, 2001 4:14 PM
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