Mathematics Stage 5
MS5.3.1 Surface area and volume
Part 1
Surface areas
Number: 43696
Title: MS5.3.1 Surface area and Volume
This publication is copyright New South Wales Department of Education and Training (DET), however it may contain
material from other sources which is not owned by DET. We would like to acknowledge the following people and
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Extracts from Mathematics Syllabus Years 7-10 © Board of Studies, NSW 2002
Unit Overview
pp iii- iv. Part 1 p 4.
Part 2 p 4
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Contents – Part 1
Introduction – Part 1 ..........................................................3
Indicators ...................................................................................4
Preliminary quiz.................................................................5
Surface areas of prisms ....................................................9
Lengths along pyramids ..................................................15
Surface area of a pyramid ...............................................21
Surface area of spheres ..................................................25
Hemispheres ...........................................................................30
Surface area of a cone ....................................................33
Suggested answers – Part 1 ...........................................39
Exercises – Part 1 ...........................................................45
Part 1
Surface areas
1
2
MS 5.3.1 Surface area and volume
Introduction – Part 1
Area is the amount of surface a particular shape has. You have known
about areas since early primary school. In your Stage 4 course you
extended this to finding areas of the surface of solid shapes, such as
prisms. This is known as surface area and consists of adding the area of
the faces and other surfaces of a three-dimensional shape.
Don’t confuse area with volume, which is the space occupied by a solid
shape. Areas are two-dimensional and involve surfaces that can be
flattened out. On the other hand, volumes are three-dimensional
measures. While areas are measured in square units such as
mm 2 , cm 2 , m 2 or hectares, volumes are measured in cubic units such as
mm 3 , cm 3 or m 3 .
This simple example shows the difference. The large cube has a volume
27units 3 and surface area 54 units 2 . Yet when this large cube is broken
3 units
up into 27 unit cubes, the volume hasn’t changed; it is just arranged
differently. But the total surface area has changed. Each single cube has
surface area 6 units 2 , so the total surface now becomes 162 units 2 !
Same volume but a massive increase in surface area.
its
3 units
Part 1
Surface areas
3
un
3
There are many instances where the relationship between surface area
and volume are important. For example, a lump of coal can take a while
to set alight yet this same lump, finely powdered, can form an explosive
mixture that will ignite and burn instantly.
In this part you will extend your knowledge of surface areas to cover the
surface areas of some relatively common shapes such as pyramids, right
cones and spheres.
Indicators
By the end of Part 1, you will have been given the opportunity to work
towards aspects of knowledge and skills including:
•
identifying the perpendicular and slant height of pyramids and right
cones
•
using Pythagoras’ theorem to find slant height, base length or
perpendicular height of pyramids and right cones
•
devising and using methods to calculate the surface area of pyramids
•
developing and using a formula to calculate the surface area of cones
•
using the formula to calculate the surface area of spheres
•
finding the dimensions of solids given their surface area by
substitution into a formula to generate an equation
By the end of Part 1, you will have been given the opportunity to work
mathematically by:
•
applying Pythagoras’ theorem to problems involving surface area
•
solving problems involving the surface area of solids.
Source:
4
Adapted from outcomes of the Mathematics Years 7–10 syllabus
<www.boardofstudies.nsw.edu.au/writing_briefs/mathematics/mathematics_
710_syllabus.pdf > (accessed 04 November 2003).
© Board of Studies NSW, 2002.
MS 5.3.1 Surface area and volume
Preliminary quiz
Before you start this part, use this preliminary quiz to revise some skills
you will need.
Activity – Preliminary quiz
Try these.
1
Use Pythagoras’ theorem to calculate the lengths of the sides marked
with a letter.
a
48
m
m
m
m
55
X
___________________________________________________
___________________________________________________
___________________________________________________
b
m
9.7 c
k
6.
5
cm
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
5
2
Calculate the areas of these shapes.
4.5 cm
a
12.2 cm
___________________________________________________
___________________________________________________
b
3.
7.
7
m
8.5 m
6
m
___________________________________________________
___________________________________________________
3
a
Calculate the missing length on this triangle.
g
39 cm
89 cm
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
b
Calculate the area of the triangle.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
6
MS 5.3.1 Surface area and volume
4
A dodecahedron is a solid shape with each face a regular pentagon.
a
How many faces does a
dodecahedron have? ____________
b
How many vertices (corners) in a
dodecahedron? ________________
c
If each face has an area of 15 cm 2 , calculate the total surface
area of this shape.
___________________________________________________
___________________________________________________
The following diagram shows the net of this dodecahedron.
3
4
B
C
A
6
2
5
F
1
E
D
d
When this dodecahedron is built, which face is opposite?
i
6?
ii
D?
iii 2?
Check your response by going to the suggested answers section.
Part 1
Surface areas
7
8
MS 5.3.1 Surface area and volume
Surface areas of prisms
You can use various area formulas to find the amount of surface of
different flat shapes. Most things you deal with normally are solid, or
3-dimensional shapes. The surface area of any solid is the total area of
all its flat faces and curved surfaces.
Among the simplest 3-D shapes you have dealt with is the cube.
right
side
back
bottom
front
5 cm
5 cm
left
side
5
cm
5 cm
top
5 cm
By cutting along some edges of a hollow cube you can flatten it to see
the full surface area. The diagram on the right (called the net of a cube)
shows how the cube can be ‘flattened’ out to show its faces consist of six
squares.
If the edge of the cube has length 5 cm, the side of each square in the net
is 5 cm long with area 5 × 5 = 25 cm2. As there are six identical squares,
the total surface area of this cube is 6 × 25 = 150 cm2.
Part 1
Surface areas
9
The surface area of a cube
equals the area of the six squares
that cover it. If the area of one
square is a × a = a2, the total
surface area of the cube is 6a2.
a
a
a
While you can develop formulas for the surface areas of some solids, this
is not so for all 3-D shapes. It can sometimes be better to calculate each
surface individually then add their areas together.
Activity – Surface areas of prisms
Try these.
1
Calculate the total surface area of this rectangular prism.
15 cm
__________________________
__________________________
__________________________
12
cm
__________________________
35 cm
2
__________________________
A rectangular prism has sides a units, b units, and c units. Write a
formula for determining the surface area of this prism.
a units
b
un
its
c units
___________________________
___________________________
___________________________
___________________________
___________________________
Check your responses by going to the suggested answers section.
10
MS 5.3.1 Surface area and volume
The surface area of a solid is the sum of the areas of all its surfaces.
Sometimes you may not be given all lengths to calculate it immediately.
Follow through the steps in this example. Do your own working in the
margin if you wish.
Calculate the surface area of the triangular prism.
14 cm
x
12
cm
13 c
m
Solution
You may wish to draw a net and see that the ends, in this case,
are right-angled triangles, and also to determine what
information you have.
x
x
13
12
13
13
13
12
14
x
x
14
One of the lengths (x) in the triangles is not given, but you are
able to calculate it using Pythagoras’ theorem.
Part 1
Surface areas
11
x
Let the missing side be x.
13
x 2 = 132 −12 2
= 25
cm
∴x = 5cm
•
12 cm
x 2 +12 2 = 132
Area of each triangular end:
1
A = ×b×h
2
1
= ×12 × 5
2
= 30
Therefore, area of both triangular ends is 2 × 30 = 60cm 2 .
•
Area of the three rectangular sides:
A = 12 ×14 +13 ×14 + 5 ×14
= 420
Alternatively, A = 30 ×14
= 420 (since12 +13 + 5 = 30)
Therefore, area of rectangular sides is 420cm 2 .
Hence total surface area = 60 + 420
= 480cm 2
Remember that sometimes you will need to calculate a missing side or
length first before being able to calculate the surface area.
12
MS 5.3.1 Surface area and volume
Activity – Surface area of prisms
Try these.
3
The cross section of the prism shown below is a regular hexagon.
15
cm
8c
m
The regular hexagonal face of this prism (shown on the right) can be
divided into six equilateral triangles.
h
4 cm
8 cm
a
Calculate the height of each triangle.
___________________________________________________
___________________________________________________
___________________________________________________
b
Calculate the area of each triangle, correct to one decimal place.
___________________________________________________
___________________________________________________
___________________________________________________
4
a
Calculate the area of the hexagon.
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
13
b
Calculate the area of the rectangular faces of this hexagonal
prism.
___________________________________________________
___________________________________________________
___________________________________________________
c
Calculate the total surface area of the prism.
___________________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
You have been practising calculating the surface area of prisms. Now
check that you can solve these kinds of problems by yourself.
Go to the exercises section and complete Exercise 1.1 – Surface areas of
prisms.
14
MS 5.3.1 Surface area and volume
Lengths along pyramids
A pyramid is a solid shape with a polygonal base and triangular sides
meeting at a common point (vertex). The base must be a polygon. That
is, it needs to be a closed plane figure formed by three or more line
segments that do not cross over each other.
In a rectangular pyramid the base is a rectangle. In a triangular pyramid
the base is a triangle. Can you see how pyramids are named?
The base can be any polygon and
therefore have any number of
straight edges, such as this
hexagonal pyramid.
Nevertheless, regardless of the
polygonal base of the pyramid,
the slanting sides are all
triangles.
Egyptian pyramids have square bases and triangular sides meeting at the
top, the vertex (sometimes also known as the apex).
In this course you will need to consider only right pyramids. A right
pyramid has its apex directly above the centre of the base.
There are some terms relating to pyramids you should know. These are
illustrated on the figure below.
Part 1
Surface areas
15
apex
n
sla
perpendicular height
te
dg
e
n
sla
t he
t
igh
base
base edge
You can use Pythagoras’ theorem to calculate the perpendicular height,
slant height, or slant edge given some edge lengths.
Follow through the steps in this example. Do your own working in
the margin if you wish.
A square pyramid has a base edge 10 cm long and perpendicular
height of 14 cm. Calculate its slant height and slant edge length.
Solution
A diagram showing this information is a useful first step.
u
14 cm
l
10 cm
Label the slant height, l, and slant edge, u.
16
MS 5.3.1 Surface area and volume
Draw the appropriate rightangled triangle to find the slant
height, l.
14 cm
The base length of the triangle is
1
×10cm = 5cm
2
l
Using Pythagoras’ theorem,
l 2 = 14 2 + 5 2
= 221
5 cm
∴l = 221
= 14.87
The diagonal length of the base
can also be found using
Pythagoras’ theorem,
d
d 2 = 10 2 +10 2
= 200
∴d = 200
= 14.14
10 cm
The slant height of this pyramid is 14.87 cm.
10 cm
You will need this value when you calculate the slant edge length.
Part 1
Surface areas
17
Again, draw the appropriate
triangle to find the slant edge
length, u.
Once again you use Pythagoras’
theorem,
14 cm
Can you see that the base length
of this triangle is half the
diagonal length?
u
7.07 cm
u 2 = 14 2 + 7.07 2
= 245.98
∴u = 245.98
= 15.68
The slant edge length of the triangle is 15.68 cm.
In this example, the diagram of the pyramid
gives you an overall view of the whole situation.
By taking out and drawing the relevant triangles
flat on a piece of paper, you can get a better,
and uncluttered, view of each part you are trying
to calculate.
I find this helps me to answer the question.
From this example, and by looking at the diagram of the pyramid, you
should see that:
perpendicular height < slant height < slant edge length.
This should help you to check whether your answers look correct.
18
MS 5.3.1 Surface area and volume
Activity – Lengths along pyramids
Try these.
1
A rectangular pyramid has a slant height of 10 cm along the
triangular face shown, and base lengths of 12 cm and 9 cm.
u
10
h
9c
m
cm
12 cm
a
Calculate the perpendicular height, h, of this pyramid.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
b
Calculate the diagonal length of the base of this pyramid.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
c
Calculate the slant edge length, u, of this pyramid.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
19
You need to know which
shape you are referring to.
height
Don’t confuse the height and
base of a triangle with the
height and base of a pyramid.
height
Check your response by going to the suggested answers section.
base
base
In this exercise you will distinguish between perpendicular height, slant
height and slant edge length.
Go to the exercises section and complete Exercise 1.2 – Lengths along
pyramids.
One final point. In geometry, the word altitude is sometimes used for
the perpendicular height of a triangle or a pyramid. The intersection
between the base and the altitude is called the foot of the altitude. The
length of the altitude is the distance between the base and the vertex.
20
MS 5.3.1 Surface area and volume
Surface area of a pyramid
To calculate the surface area of a pyramid you need to add the area of the
base and the area of the triangular faces.
Sometimes the slant edge of a pyramid may be given to you. This slant
edge is the height of the triangular face. At other times the perpendicular
height of the pyramid may be given and you need to calculate the slant
height before you can calculate the surface area.
Follow through the steps in this example. Do your own working in the
margin if you wish.
A square pyramid has a base length of 56 cm and perpendicular
height of 45 cm.
a
Calculate the slant height, PL.
b
Are the heights of all triangles the same? Explain.
c
Calculate the total surface area of the pyramid.
H
I
O
K
J
P
L
Part 1
Surface areas
21
Solution
The base of the pyramid is a square. Therefore each length is
the same, and so KH = 56 cm. Hence OP = 28 cm (half this
value).
a
To calculate PL, a diagram of
the relevant triangle would be a
useful aid.
P
28 cm
O
Using Pythagoras’ theorem,
45 cm
PL2 = OP 2 + OL2
= 28 2 + 45 2
= 2809
∴PL = 2809
= 53 cm
L
b
The heights of all 4 triangles are the same. The triangles
are congruent having the same base length (56 cm) and the
same perpendicular height (53 cm).
c
The area of each triangle is
1
× base length × perpendicular height
2
1
= × 56 × 53
2
= 1484 cm 2
A=
The pyramid consists of 4 such triangles and a square base.
Its total surface area is
SA = 4 ×1484 + 56 × 56
= 9072 cm 2
Notice the slant height of the pyramid forms the perpendicular height of
the triangle.
22
MS 5.3.1 Surface area and volume
A
If the base is a regular
polygon the slant heights of
all triangular faces will be
the same.
If the base is not a regular
polygon the slant heights will
not all be the same.
For example, the base of this
pyramid is not regular. So,
AB ≠ AC.
B
C
Activity – Surface area of a pyramid
Try these.
1
For the rectangular pyramid shown, calculate:
A
AO = 24 mm
OB = 45 mm
OC = 70 mm
S
R
O
P
B
a
C
Q
the slant height of the pyramid, AB.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
b
the slant height of the pyramid, AC.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
23
c
the area of the rectangular base.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
d
the areas of ∆APQ and ∆AQR .
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
e
the total surface area of the pyramid.
___________________________________________________
___________________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
You have been practising calculating the surface areas of pyramids. Now
check that you can solve these kinds of problems by yourself.
Go to the exercises section and complete Exercise 1.3 – Surface area of a
pyramid.
24
MS 5.3.1 Surface area and volume
Surface area of spheres
Earlier in this course you learned how to calculate the surface area of
cylinders.
r
r
2π r
h
r
In that unit you determined that its surface area consists of:
•
two congruent circles, each with area A = π r 2 , and
•
a rectangle with area A = 2 π rh .
So the total surface area of a closed cylinder is given by
A = 2 π r 2 + 2 π rh .
The ancient Greek mathematician, Archimedes, found out that the
surface area of a sphere is the same as the curved surface area of a
cylinder having the same diameter as the sphere and a height the same
length as the diameter.
r
r
Part 1
Surface areas
2r
25
That is, a sphere has the same surface area as an open cylinder with the
same width and height that wraps around it.
Hence the surface area of a sphere is: Area = 4 π r 2 .
Coincidentally, this area is four times the area of a circle with the same
radius.
The proof of this formula is beyond the scope of this course, but a
connection between the cylinder and the sphere can be demonstrated.
Imagine a sphere having radius r placed inside a cylinder, with the
cylinder just touching the equator of the sphere, and cut off at the height
of the top and bottom of the sphere. The diagram shows a cutaway view.
sphere
hs
hc
cylinder
hc
hs
r
The area of the curved part of the cylinder is 2 π r × 2r = 4 π r 2 .
Now imagine a small horizontal slice through the diagram, such as the
dark coloured slice on the sphere and the cylinder. This cuts a rectangle
out of the rolled out cylinder and a slightly distorted rectangle out of the
sphere. Of course, you need to imagine the slice to be very thin so that
the distortion is very slight.
In the cross-sectional view hc is the height of the slice on the cylinder
and hs is the length of the arc on the sphere cut out by the slice.
The area of the dark coloured rectangle on the cylinder is 2 π rhc . The
area of the slightly distorted dark coloured rectangle stripe on the sphere
is approximately 2 π rhs . These two areas are approximately equal as
hc ≈ hs .
26
MS 5.3.1 Surface area and volume
Of course, hs > hc , and as the strip gets closer to the poles of the sphere,
this difference increases no matter how thin the strips that you take are.
But at the same time as hs increases, the length of the strip decreases
since r decreases. So one compensates for the other.
Since the cylinder and the sphere can be approximated by these
rectangular strips, the area of the sphere and the area of the cylinder
should be approximately equal. In fact, it can be shown mathematically
that they are exactly equal.
This argument is similar to how the ancient Greeks compared slices of
spheres and their wrapping open cylinders to show that their areas and
volumes are the same.
An interesting property of the surface area of the slice taken from the
sphere is that it depends only on the ‘thickness’ of the slice, and not from
where on the sphere it is taken!
This means that the two coloured slices shown have the same surface
areas, as their ‘thicknesses’ are the same.
You can use the formula for the surface area of a sphere: Area = 4 π r 2 to
solve problems.
Follow through the steps in this example. Do your own working in the
margin if you wish.
Part 1
Surface areas
a
A sphere has radius 6.4 cm. Calculate its surface area.
b
A sphere has a surface area of 100cm 2 . Calculate its
diameter.
27
Solution
a
Using the formula for the surface area of a sphere ,
Surfacearea = 4 π r 2
Surfacearea = 4 × π × 6.4 2
= 514.7 cm 2 (correct to 1 decimal place)
b
This time you are give the surface area and need to use the
formula: Surfacearea = 4 π r 2 to calculate the radius, r,
from which you can find the diameter, d.
Surfacearea = 4 π r 2
100 = 4 × π × r 2
100
r2 =
4×π
= 7.957747
∴r = 7.957747
= 2.820948
Now diameter = 2 × r
= 2 × 2.820948
= 5.64cm (2 decimal places)
100
(4 × π )
to keep both the 4 and together. One way to
do this on the calculator is to type:
Remember, when calculating the fraction
100 ÷ ( 4 × π ) =
Also at intermediate steps, keep more decimal
places in your values than what you will
eventually round to.
28
MS 5.3.1 Surface area and volume
Activity – Surface area of spheres
Try these.
1
A ball has diameter of 28 cm. Calculate the surface area of the ball.
_______________________________________________________
_______________________________________________________
_______________________________________________________
2
The Moon is almost a sphere with surface area 38 500 000 km 2 .
Calculate its radius, correct to the nearest kilometre.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
Check your response by going to the suggested answers section.
Drawing a labelled diagram can sometimes be useful in answering
questions.
Part 1
Surface areas
29
Hemispheres
A hemisphere (from Greek hemi = half, sphaira = ball) is literally half of
a sphere or globe. Slicing a sphere into two equal halves (passing
through the centre) creates two hemispheres.
A hemisphere has two surfaces:
•
a circular face with area π r 2 ,
•
a curved surface with area
1
× 4 π r 2 = 2π r 2 .
2
area = πr2
area = 2πr2
∴surface area of hemisphere = 3π r 2
Sometimes only the curved surface area is
required such as when calculating the outer
surface of a curved hemispherical bowl. It is
therefore important to remember
πr 2 + 2πr 2 = 3πr 2
rather than just 3πr 2
This way you will know whether you need to
calculate the whole surface area, or only the
curved part.
Follow through the steps in this example. Do your own working in the
margin if you wish.
A hemisphere has a circumference of 28.6 cm.
30
a
Calculate its radius.
b
Use the radius result to calculate its total surface area.
MS 5.3.1 Surface area and volume
Solution
a
The circumference of the hemisphere is the circumference
of its circular part.
C = 2π r
28.6 = 2 × π × r
28.6
r=
2×π
= 4.55cm
b
The total surface area of a hemisphere is 3π r 2 .
Surfacearea = 3 × π × 4.55 2
= 195.3cm 2
In this next activity think whether you need to calculate the total surface
area of a hemisphere, or just its circular part.
Activity – Surface area of spheres
Try these.
3
The hemispherical dome on top of a telescope observatory has a
circumference of 75.4 metres.
circu
mference = 75.4 m
a
Calculate the radius of the dome, correct to the nearest metre.
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
31
b
Calculate the area of steel in the curved surface of the dome.
___________________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
You can often leave your answers in exact
form,that is in terms of π. This looks a lot neater,
and is easier to do than writing a lot of digits.
For example, consider the surface
area of a sphere with radius 10 cm.
SA = 4πr 2
SA = 4πr 2
= 4 × π × 102
= 4 × π × 102
= 1256.64 cm2
= 400π cm2
10 cm
(correct to 2 decimal places)
You have been practising calculating the surface areas of spheres and
hemispheres. Now check that you can solve these kinds of problems by
yourself.
Go to the exercises section and complete Exercise 1.4 – Surface area of
spheres.
32
MS 5.3.1 Surface area and volume
Surface area of a cone
In earlier parts of your course you defined a cone and calculated its
volume. Here you will develop a formula for the surface area of a cone.
apex or vertex
axis
A right cone can be considered as a
solid object obtained by rotating a
right-angled triangle around one of its
two shorter sides, the cone’s axis. The
disc swept out by the other short side
is called the base, and the endpoint of
the axis that is not on the base is the
cone’s apex or vertex. The hypotenuse
of that triangle sweeps out the curved
surface of the cone.
base
Here are some other features of cones that you should know.
ht
altitude
(perpendicular height)
nt
he
ig
curved surface
radius
sla
vertex
base
right cone
While there are different kinds of cones, in this course you will only
consider right cones. That is, cones where the vertex is vertically above
the centre of the circular base.
A cone, therefore, consists of two surfaces:
Part 1
•
a circular base with area A = π r 2 , and
•
a curved surface.
Surface areas
33
The total surface area of a cone is the sum of these two areas.
But what is the area of the curved surface? You will now develop a
formula for the curved surface of a cone.
Activity – Surface area of a cone
Try these.
You know that the area of a circle is given by A = π r 2 and the
circumference of the circle is given by C = 2 π r .
1
a
What is the area of a semicircle? _________________________
r
b
What is the length of the curved edge of a semicircle?
______________________________
______________________________
r
2
a
______________________________
What fraction of the circle is this sector? __________________
r
b
What is the area of the sector?
___________________________________________________
34
MS 5.3.1 Surface area and volume
c
What is the length of the curved edge of the sector?
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
You have used here the obvious relationship between the part of the
circle and the whole circle.
For parts of a circle (that is, sectors) the fraction of the circumference of
the circle is the same fraction as the fraction of the area of the circle.
This will help you to find the area of the curved surface of a cone, which
is made by joining the edges of a sector.
If you were to take a cone, such as a conical party hat, cut along the slant
edge AC and lay it out flat, the shape formed is a sector.
C
ig
he
nt
l
l
sla
#
C
ht
A
r
l
B
A, B
The area of that sector is the same as the area of the curved surface of the
cone.
The radius of the sector AC (or BC) used for the curved surface is the
slant height of the cone.
It is usual to use r to represent the radius of the base of a cone and use h
for its perpendicular height. It is also standard to use the pronumeral l
(the letter ‘el’) for the slant height of a cone.
Part 1
Surface areas
35
Can you see that the circumference of the base of the cone (2 π r) equals
the length of the arc AB of the sector.
Now,
area of sector ABC
arc length AB of sector ABC
=
area of circle (centre C) circumference of circle (centre C)
area of sector ABC 2 π r
=
πl 2
2π l
2π r
∴ area of sector ABC =
× πl 2
2π l
= π rl
Since the curved surface area equals the area of this sector, then the area
of the curved surface of the cone is also π rl .
So as the area of the curved surface is π rl
and the area of the circular base is π r 2 ,
the total surface area of the closed cone is
πrl
l
Surfaceareaofcone = π r 2 + π rl
r
πr2
Follow through the steps in this example. Do your own working in the
margin if you wish.
A solid cone has a base diameter 40 cm and perpendicular
height 21 cm.
36
a
the radius, r, of the base
b
the area of the base
c
the slant height, l
d
the area of the curved surface
e
the total surface area.
l
h = 21 cm
Calculate, to the nearest whole number,
r
40 cm
MS 5.3.1 Surface area and volume
Solution
a
The radius is half the diameter,
∴r = 40 ÷ 2 = 20cm
b
A = πr2
= π × 20 2
= 1257cm 2 (correct to the nearest square centimetre)
c
You need to use Pythagoras’ theorem as l 2 = r 2 + h 2 .
l 2 = 20 2 + 212
= 841
d
∴l = 841
= 29 cm
Curved surface area = π rl
= π × 20 × 29
e
= 1822 cm 2
Total surface area = 1257 +1822
= 3079 cm 2
A quick diagram in answering these questions is a useful visual aid.
Activity – Surface area of a cone
Try these.
3
A conical party hat has a base diameter 10 cm and perpendicular
height 12 cm.
a
Calculate the slant height of
the party hat.
__________________________
12 cm
__________________________
10 cm
Part 1
Surface areas
37
b
Calculate the area of cardboard in the hat. (Remember, a party
hat does not have a base.)
___________________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
You have been practising calculating the surface area of cones. Now
check that you can solve these kinds of problems by yourself.
Go to the exercises section and complete Exercise 1.5 – Surface area of a
cone.
38
MS 5.3.1 Surface area and volume
Suggested answers – Part 1
Check your responses to the preliminary quiz and activities against these
suggested answers. Your answers should be similar. If your answers are
very different or if you do not understand an answer, contact your teacher.
Activity – Preliminary quiz
1
a
h 2 = 55 2 + 48 2
= 5329
∴h = 5329
= 73mm
b
9.7 2 = k 2 + 6.5 2
∴k 2 = 9.7 2 − 6.5 2
= 5184
∴k = 5184
= 7.2cm
2
a
A = 12.2 × 4.5
= 54.9cm 2
b
3
a
1
×base × height
2
1
A = × 3.6 × 7.7
2
= 13.86m 2
A=
As this is a right-angled triangle, use Pythagoras’ theorem
g 2 = 89 2 − 39 2
= 6400
∴g = 6400
=80cm 2
Part 1
Surface areas
39
b
4
1
×base × height
2
1
A = × 39 × 80
2
= 1560cm 2
A=
a
12
c
12 ×15 = 180cm 2
d
i
F
b
20 (count them from the diagram)
ii
4
ii
A
Activity – Surface areas of prisms
1
There are six surfaces, with opposite faces having the same areas.
A = 2(35 ×12 +12 ×15 + 35 ×15)
= 2250cm 2
2
A = 2(ab + ac + bc)orA = 2ab + 2ac + 2bc
82 = h2 + 4 2
3
a
h2 = 82 − 4 2
= 48
∴h = 48
≈ 6.928cm
1
× base × height
2
1
= × 8 × 6.928
2
= 27.7cm 2
bArea =
4
a
Each hexagon is made up of 6 equilateral triangles.
A = 6 × 27.7
= 166cm 2
b
There are 6 congruent rectangular faces.
A = 6 × 8 ×15
= 720cm 2
cTotalarea = 166 +166 + 720
= 1052cm 2
40
MS 5.3.1 Surface area and volume
Activity – Lengths along pyramids
1
(Diagrams of the relevant triangles will help you answer these
questions.)
h 2 = 10 2 − 6 2
a
b
= 64
∴h = 8cm
d 2 = 12 2 + 9 2
= 225
∴d = 15cm
⎛d ⎞
u = h +⎜ ⎟
⎝2⎠
2
2
c
2
= 8 2 + 7.5 2
= 120.25
∴u = 10.97cm
Activity – Surface area of a pyramid
1
a
AB 2 = AO 2 + OB 2
= 24 2 + 45 2
= 2601
AB = 2601
= 51mm
b
AC 2 = AO 2 + OC 2
= 24 2 + 70 2
= 5476
AC = 5476
= 74mm
c
PQ = 2 × 70 = 140 mm
QR = 2 × 45 = 90 mm
Area = 140 × 90
= 12 600 mm 2
Part 1
Surface areas
41
d
For ∆APQ,
1
area = ×140 × 51
2
= 3570 mm 2
For ∆AQR,
1
area = × 90 × 74
2
= 3330 mm 2
e
The pyramid has a rectangular base, 2 triangular faces like
∆APQ , and 2 triangular faces like ∆AQR .
Total surface area = 12 600 + 2 × 3570 + 2 × 3330
= 26 400 mm 2
Activity – Surface area of spheres
1
The radius is 14 cm.
Surfacearea = 4 π r 2
= 4 × π ×14 2
= 2463 cm 2
2
Surfacearea = 4 π r 2
38500000 = 4 × π × r 2
38500000
r2 =
4×π
= 3063733
∴r = 3063733
= 1750km
3
a
C = 2π r
75.4 = 2 × π × r
75.4
r=
2×π
= 12 m
b
Only the curved surface area is required.
Curvedarea = 2 π r 2
= 2 × π ×12 2
= 904.8 m 2
42
MS 5.3.1 Surface area and volume
Activity – Surface area of a cone
1
2
3
1 2
πr .
2
a
The area of a semicircle is half that of a circle: A =
b
The length of the curved edge of a semicircle is half that of a
circle:
1
L = × 2π r
2
= πr
a
Three quarters
c
L=
a
Using Pythagoras’ theorem: l 2 = 5 2 +12 2 from which you
b
A=
3 2
πr
4
3
× 2π r
4
3
= πr
2
obtain l = 13 cm. Alternatively, you might remember that
{5, 12, 13} form a Pythagorean triad.
b
Part 1
Surface areas
Using the formula for the curved surface of a cone: A = π rl ,
A = π × 5 ×13
.
= 204.2 cm 2
43
44
MS 5.3.1 Surface area and volume
Exercises – Part 1
Exercises 1.1 to 1.5
Name
___________________________
Teacher
___________________________
Exercise 1.1 – Surface areas of prisms
This is a triangular prism standing on its triangular cross section.
5c
cm
1
10 cm
4
m
6 cm
a
Next to this diagram draw, and label, the same triangular prism
standing on the rectangular face with the greatest area.
b
Part 1
Surface areas
Draw and label the net of this triangular prism.
45
c
Identify what further information you need to find before you
can calculate the total surface area of this prism.
___________________________________________________
___________________________________________________
2
Calculate the surface area of this triangular prism.
6c
m
12 cm
10
cm
8c
m
cm
A canvas tent with sewn-in floor is in the shape of a triangular prism
180 cm wide, 120 cm high and 240 cm long.
240
120 cm
3
180 cm
Calculate:
a
the area of the triangular ends.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
46
MS 5.3.1 Surface area and volume
b
the area of canvas in the floor.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
c
the slant height l (use Pythagoras’ Theorem).
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
d
the area of canvas in the sloping sides.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
e
the total area of canvas in the tent.
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
47
Calculate the surface area of this gingerbread house. The house has
a rectangular base.
12
.5
cm
30
cm
10 cm
7.5 cm
4
20 cm
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
5
This diagram shows the framework for the base of a lounge which
Lana wants to re-cover.
20 cm
42.5 c
m
21
0
cm
72 cm
30 cm
40 cm
66.5 cm
48
MS 5.3.1 Surface area and volume
a
The cross section can be divided horizontally into two shapes.
Draw these as two separate shapes and calculate any unknown
edges.
b
The area of a trapezium is: A =
sum of sides
× ⊥ height .
2
( means parallel, and ⊥ means perpendicular.)
Calculate the area of a side of the lounge.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
c
Calculate the total surface area that needs to be covered, if the
base is not covered.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
49
Exercise1.2 – Lengths along pyramids
1
a
Name the base of this pyramid.
___________________________________________________
b
Draw, if necessary, and label the
following on this pyramid:
i
a base edge
ii
a slant height
iii a slant edge.
2
a
The diagram shows the net of a pyramid.
Name the pyramid. ___________________________________
b
50
Use a ruler, to measure to the nearest millimetre,
i
the length of the base edge. _________________________
ii
the slant height. __________________________________
MS 5.3.1 Surface area and volume
3
A regular pentagonal pyramid is a pyramid having a regular
pentagon as its base. The edge length, u, and the slant height, l, can
be calculated using the formulas:
u = h2 +
(
)
(
)
1
1
5 + 5 b 2 and l = h 2 +
5 + 2 5 b2
10
20
where h is the perpendicular height, and b is the side length of the
base.
Use these formulas to calculate the edge length and the slant height
for a pentagonal pyramid having h = 14.0 cm and b = 6.0 cm.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
4
A rectangular pyramid has side lengths AB = 10 cm and BC = 6 cm.
The slant height PN = 9 cm.
P
D
C
O
A
N
M
B
a
Calculate the perpendicular height, OP, leaving your answer as a surd.
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
51
b
Calculate the slant height, PM, leaving your answer in exact
form.
___________________________________________________
___________________________________________________
___________________________________________________
c
Explain why length PN is longer than length PM.
___________________________________________________
___________________________________________________
___________________________________________________
52
MS 5.3.1 Surface area and volume
Exercise1.3 – Surface area of a pyramid
The diagrams show the net of two pyramids. Calculate the area of
each net.
8 cm
1
8 cm
8 cm
a
___________________________________________________
___________________________________________________
___________________________________________________
210 mm
66 mm
65 mm
b
119 mm
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
53
2
Calculate the total surface area of this square prism.
10
.0
cm
12.0 cm
.0
12
cm
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
54
MS 5.3.1 Surface area and volume
3
A tetrahedron is a triangular pyramid where each face is an
equilateral triangle. The diagram shows a tetrahedron and its net.
7.0 cm
a
Calculate the surface area of the tetrahedron.
(Hint: you can do this by using Pythagoras’ theorem first, or by
using the area formula for non-right-angled triangles).
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
b
How can the net of the tetrahedron aid in calculating its surface
area?
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
55
4
a
Use triangle AOC to calculate the perpendicular height of this
pyramid.
A
AC = 170 mm
OB = 65 mm
OC = 154 mm
S
R
O
P
B
C
Q
___________________________________________________
___________________________________________________
___________________________________________________
b
Calculate the slant height of face APQ of this pyramid.
___________________________________________________
___________________________________________________
___________________________________________________
c
Calculate the surface area of this rectangular prism.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
56
MS 5.3.1 Surface area and volume
5
Many minerals form crystals. One of these crystal shapes is the
octahedron. Placing two square-based pyramids back-to-back makes
this solid. Each of the 8 faces in an octahedron is an equilateral
triangle.
Diamond is often found as octahedral crystals. A crystal of diamond
has the side of each triangle 8.0 mm in length.
a
Calculate the height of each triangular face in this crystal.
(Answer correct to one decimal place.)
________________________
________________________
________________________
________________________
________________________
________________________
________________________
b
Calculate the total surface area of this diamond crystal.
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Surface areas
57
6
The Great Pyramid at Khufu was built around 2589 BC. The base of
the pyramid is a square with the length of each side originally
230 metres. It is now 227 metres. Its original height was
146.6 metres tall, but is now only 136.9 metres tall.
a
Calculate the slant height of the pyramid both when it was built
and now.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
b
Calculate the total surface area (including the base) of the Great
Pyramid when it was built and now.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
c
How many square metres more was its surface area when it was
built compared to now?
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
58
MS 5.3.1 Surface area and volume
Exercise1.4 – Surface area of spheres
1
Calculate the surface area, correct to three significant figures, of a
sphere with:
a
radius 4.0 cm.
___________________________________________________
___________________________________________________
___________________________________________________
b
diameter 8.6 m.
___________________________________________________
___________________________________________________
___________________________________________________
2
Calculate the surface area, in terms of π , of a sphere with:
a
radius 62.5 mm.
___________________________________________________
___________________________________________________
___________________________________________________
b
diameter 18.4 m.
___________________________________________________
___________________________________________________
___________________________________________________
3
The Earth can be considered a sphere with diameter 12 760 km.
a
Calculate the surface area of Earth.
____________________________________
____________________________________
____________________________________
Part 1
Surface areas
59
b
Water covers 70.9% of the Earth’s surface. Calculate this area.
___________________________________________________
___________________________________________________
___________________________________________________
c
Asia is the largest continent covering 44.6 million square
kilometres of the Earth surface. What percentage of the Earth’s
surface area does Asia cover?
___________________________________________________
___________________________________________________
4
A hemispherical fish bowl has an outer radius of 9.5 cm. Calculate
the outer surface area of glass in the bowl.
_______________________________________________________
_______________________________________________________
_______________________________________________________
5
A hemispherical canvas tent, complete with base, has a
circumference of 25.1 m.
a
Calculate its radius.
___________________________________________________
___________________________________________________
___________________________________________________
b
Calculate its total surface area.
___________________________________________________
___________________________________________________
___________________________________________________
60
MS 5.3.1 Surface area and volume
6
a
Calculate the surface area of each sphere, leaving the answer in
each case in terms of π .
6 cm
3 cm
___________________________________________________
___________________________________________________
___________________________________________________
b
While the radius of the larger sphere is double that of the
smaller sphere, by how many times has its surface area
increased? __________________________________________
7
(Harder) Archimedes stated: If a sphere is inscribed in a cylinder,
2
then the sphere is of the cylinder in both surface area and volume.
3
Formulas:
Surface area of sphere = 4π r 2
Surface area of cylinder = 2π r 2 + 2 π rh
4
Volume of sphere = π r 3
3
Volume of cylinder = π r 2 h
Use these formulas to show that
Archimedes’ statement is true.
_______________________________________________________
_______________________________________________________
_______________________________________________________
Part 1
Surface areas
61
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
62
MS 5.3.1 Surface area and volume
Exercise 1.5 – Surface area of a cone
Decide for yourself how accurately you should give answers for each
question to be sensible and sufficiently meaningful.
1
3 cm
An ice cream cone has an edge
length of 16 cm and a radius of
3 cm. Calculate the surface area
of the cone.
m
16 c
___________________________
___________________________
___________________________
___________________________
___________________________
2
cm
___________________________
m
18 c
10
A conical party hat has a base
diameter of 10 cm and a slant
height of 18 cm. What is the
surface area of the material?
(Ignore the overlap.)
___________________________
___________________________
___________________________
___________________________
Part 1
Surface areas
63
3
A spinning top is close to being
conical and has a top diameter of 7 cm
and is 9 cm tall. Calculate:
a
the slant height.
________________________________
________________________________
________________________________
________________________________
________________________________
b
the total surface area of the spinning top.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
4
A round garden house has a conical
copper covered roof. The radius of the
house is 3 m and the distance from the
pinnacle to the edge of the roof is 4 m.
a
Calculate the area of copper used to
cover the roof.
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
64
MS 5.3.1 Surface area and volume
b
The copper sheeting used has a mass of 17.8kg/m 2 . Calculate
the mass of copper in the roof.
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
5
(Harder) A right circular cone is formed by bending a semicircular
piece of paper of radius 11 cm. Determine the slant height, l, and the
curved surface area of the cone, correct to one decimal place.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
6
(Harder) Two canvas tents are made differently. One is in the form
of a prism with a rectangular base 2 m by 1.4 m and height 2 m. The
other tent is conical with radius 2 m and height 2 m. Which tent
requires more canvas to make, and by how much?
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
Part 1
Surface areas
65