1. No category

Surface Area
and Volume
6
pa
ge
s
How much paint
will I need?
pl
e
Chapter Contents
Sa
m
6:01 Review of surface area
MS4·2, MS5·2·2
6:02 Surface area of a pyramid
MS5·3·1
6:03 Surface area of a cone
MS5·3·1
Investigation: The surface area of a cone
MS5·3·1
6:04 Surface area of a sphere
Investigation: The surface area of a sphere
Fun Spot: How did the raisins win the
war against the nuts?
MS5·2·2
6:05 Volume of a pyramid
Investigation: The volume of a pyramid
MS5·2·2
6:06 Volume of a cone
6:07 Volume of a sphere
MS5·2·2
Investigation: Estimating your surface
area and volume
6:08 Practical applications of surface
area and volume
MS5·2·2, MS5·3·1
Maths Terms, Diagnostic Test, Revision
Assignment, Working Mathematically
Learning Outcomes
MS4·2
Calculates surface area of rectangular and triangular prisms and volume of right prisms
and cylinders.
MS5·2·2 Applies formulae to find the surface area of right cylinders and volume of right pyramids, cones
and spheres and calculates the surface area of and volume of composite solids.
MS5·3·1 Applies formulae to find the surface area of pyramids, right cones and spheres.
Working Mathematically Stages 5·3·1–5
1 Questioning, 2 Applying Strategies, 3 Communicating, 4 Reasoning, 5 Reflecting
173
6:01 Review of
Surface Area
Outcomes MS4·2, MS5·2·2
In Year 9, the surface areas of prisms, cylinders and composite solids were calculated by adding the
areas of the faces (or surfaces).
The following formula may be needed.
L
8 Surface area of a rectangular prism:
A = 2LB + 2LH + 2BH
r
s
2␲r
3 Find the surface area
of this solid.
14 cm
8·7 m
3·7 m
6·8 m
14
6 cm
cm
8 cm
The hole is 4 cm deep
Sa
m
2·1 m
Solutions
pa
2 Find the surface area
of this cylinder.
pl
e
2·9 m
B
9 Surface area of a cylinder:
A = 2πrh + 2πr2
Worked examples
1 Find the surface area
of this prism.
H
ge
Area formulae
1 square: A = s2
2 rectangle: A = LB
3 triangle: A = 1--- bh
2
4 trapezium: A = 1--- h(a + b)
2
5 parallelogram: A = bh
6 rhombus: A = 1--- xy
2
7 circle: A = πr2
1 Surface area = 2LB + 2LH + 2BH
= 2 × 2·9 × 3·7 + 2 × 2·9 × 2·1 + 2 × 3·7 × 2·1
= 49·18 m2
2 For a cylinder:
Surface area = curved surface area + area of circles
Surface area = 2πrh + 2πr2
= (2π × 4·35 × 6·8) + (2 × π × 4·352)
= 304·75 m2 (correct to 2 dec. pl.)
8·7 m
4·35 m
3 Surface area = 2 × (14 × 14) − π × 42 + 4 × (14 × 6) + 2π × 4 × 4 + π × 42
= 728 + 32π
= 829 cm2 (to nearest cm2)
174
New Signpost Mathematics Enhanced 10 5.1–5.3
6·8 m
h
Foundation Worksheet 6:01
Exercise 6:01
1
Surface area review MS4·2, MS5·2·2
1 Find the surface area of the
following rectangular prisms.
a
12
Find the surface area of the following prisms.
a
b
15 cm
3
10 cm
8
7 cm
2 Find the surface area of the
following triangular prisms.
a
8 cm
14 cm
8 cm
c
12 cm
6
d
16 cm
10 cm
8 cm
18 cm
8 cm
33·6 cm
6 cm
14 cm
3
8
3 Find the surface area of the
following cylinders.
a
6
15 cm
5
5 cm
Find the surface area of the following solids.
a
b
6 cm
s
8 cm
c
7 cm
9 cm
pa
x cm
6 cm
Right-angled
triangles
14 cm
Find the surface area of the following solids. All measurements are in metres.
a
b
c
3
7
12
10
9
3
3
10 6
4
14
5
b
7 cm
Sa
4
7 cm
c
a
x cm
m
9 cm
■ Caution!
c2 = a2 + b2
11 cm
12 cm
8 cm
6 cm
In each of the following questions use Pythagoras’ theorem to
calculate the unknown length x, correct to two decimal places,
and then calculate the surface area.
a
b
pl
e
3
ge
2
8
9
8
12
9
Find the surface area of the following solids. All measurements are in centimetres.
a
b
c
3
10
8
5
12
8
5
4
4
3
6
12
3
8
8
20
Chapter 6 Surface Area and Volume
175
6:02 Surface Area of a Pyramid
Outcome MS5·3·1
Prep Quiz 6:02
ABCDE is a rectangular pyramid.
E
16
D
xc
C
m
2·4 cm
1 Are the triangular faces of a square
pyramid congruent?
2 Are the triangular faces of a rectangular
pyramid congruent?
3 Find x
X
O
A
1·8 cm
B
Y
10
Find:
5 OX
6 EX
7 OY
8 EY
Find the area of:
9 ΔEBC
10 ΔABE
ge
s
4 The net of a square
pyramid is shown.
Find the area of
the net.
6
pa
8
12
m
Worked examples
pl
e
To calculate the surface area of a pyramid with a polygonal base, we add the area of the
base and the area of the triangular faces.
Sa
Calculate the surface area of the following square and rectangular pyramids.
1
2
3
E
D
T
13 cm
O
B
LM = 8 cm
MN = 6 cm
TO = 7 cm
C
10 cm
P
N
10 cm
A
AO = 8 cm
BC = 6 cm
O
L
M
Solutions
1 Surface area = (area of square) + 4 × (area of a triangular face)
10 × 13
= (10 × 10) + 4 × ⎛ ------------------⎞
⎝ 2 ⎠
= 100 + 260
■ In right square pyramids,
= 360 cm2
all the triangular faces are
congruent. This simplifies the
calculation of the surface area.
176
New Signpost Mathematics Enhanced 10 5.1–5.3
2 As the perpendicular height of the triangular face
E
is not given, this must be calculated.
In the diagram, AM is the perpendicular height of
M
B
the face.
In ΔAOM
(Pythagoras’ theorem)
AM2 = AO2 + OM2
AO = 8 cm
BC = 6 cm
1
2
2
=8 +3
(Note: OM = --- CD)
2
= 64 + 9
= 73
AM = 73
Surface area = (area of square) + 4 × (area of a triangular face)
D
O
C
A
In ΔTOA,
TA2 = AO2 + OT2
= 32 + 72
= 58
∴ TA = 58 cm
pa
P
pl
e
OA = 12MN
N
O
3 cm
L
A
4 cm
B
M
m
In ΔTOB,
TB2 = BO2 + OT2
= 42 + 72
= 65
∴ TB = 65 cm
OB = 12LM
T
7 cm
3 The perpendicular heights TA and TB must
be calculated, as these are not given.
ge
s
6 × 73
= ( 6 × 6 ) + 4 × ⎛ -------------------⎞
⎝ 2 ⎠
= 36 + 12 73
= 138·5 cm2 (correct to 1 dec. pl.)
Sa
Now,
Surface area = (area of rect. LMNP) + 2 × (area of ΔTMN) + 2 × (area of ΔTLM)
MN × TB
LM × TA
= ( LM × MN ) + 2 × ⎛ -----------------------⎞ + 2 × ⎛ ----------------------⎞
⎝
⎠
⎝
⎠
2
2
2 × 6 × 65 2 × 8 × 58
= 8 × 6 + ----------------------------- + ----------------------------2
2
= 48 + 6 65 + 8 58
= 157·3 cm2 (correct to 1 dec. pl.)
■ In right rectangular
pyramids, the opposite
triangular faces are
congruent.
Chapter 6 Surface Area and Volume
177
Foundation Worksheet 6:02
Exercise 6:02
The following diagrams represent the nets of pyramids.
Calculate the area of each net.
a
b
12 cm
6 cm
1
6
5
6 cm
17·86 cm
6 cm
2 Find the surface area of each
square or rectangular pyramid.
a
10
18 cm
32 cm
6 cm
6 cm
6
6
12 cm
6 cm
2
Surface area of a pyramid MS5·3·1
1 Calculate the area of each net.
a
Calculate the surface area of the following square and rectangular pyramids.
a
b
c
7·5 cm
10 cm
17 cm
ge
pa
2 cm
4 cm
Use Pythagoras’ theorem to calculate the perpendicular
height of each face and then calculate the surface area
of each pyramid. Give the answers in surd form.
a
b
T
E
D
O
A
4
Sa
C
m
pl
e
3
cm
cm
6 cm
20
9·5
6 cm
s
7·5 cm
X
B
AB = 10 cm
EO = 12 cm
P
N
X
O
L
LM = 10 cm
TO = 10 cm
M
Calculate the surface area of the following pyramids. Give all answers correct to one decimal
place where necessary.
a
b
c
A
A
M
W
F
E
O
H
G
Y
WXYZ is a square.
MO = 3 cm, XY = 11 cm.
New Signpost Mathematics Enhanced 10 5.1–5.3
C
B
O
X
EFGH is a rectangle.
AO = 8 cm, HG = 12 cm,
GF = 9 cm.
178
Z
O
E
D
BCDE is a rectangle.
AO = 4 cm, ED = 10 cm,
DC = 6 cm.
5
Find the surface area of:
a a square pyramid, base edge 6 cm, height 5 cm
b a rectangular pyramid, base 7 cm by 5 cm, height 10 cm
6
Find the surface area of the following solids. Give all answers correct to three significant figures.
a
b
c
2 cm
A
M
H
O
F
A
F
C
3·5 cm
D
E
G
E
O
C
B
D
ABCD is a square.
AB = 5 m, MO = 3 m,
CG = 2 m
4 cm
B
s
AO = OB = 9 cm
CD = DE = FE = FC = 6 cm
Find the surface area of a pyramid that has
a regular hexagonal base of edge 6 cm and a
height of 8 cm.
8
A square pyramid has to have a surface area of 2000 cm2.
If the base edge is 20 cm, calculate:
a the perpendicular height, x cm, of one of the
triangular faces
b the perpendicular height, h cm, of the pyramid
pl
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pa
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7
x cm
m
h cm
20 cm
Sa
20 cm
6:03 Surface Area of a Cone
Outcome MS5·3·1
Prep Quiz 6:03
s
What fraction of a circle is each of the following sectors?
5
6
7
120°
1 Area = ?
2 Circumference = ?
2πr
3 Simplify: --------2πs
r
4 Simplify: - × πs 2
s
O
O
O
3 cm
circumference = 9 cm
8
6 cm
O
9
2πr
O
10 Evaluate πrs if
r = 3·5 and s = 6·5.
Answer correct to
1 dec. pl.
circumference = 2πs
Chapter 6 Surface Area and Volume
179
Investigation 6:03
The surface area of a cone
2
3
10 cm
B
10 cm
4
A= 12 π(10)2
A
d
r
A= πr 2
pl
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sticky tape
B
10 cm
10 cm
pa
BA
A
ge
1
s
The surface area of a cone comprises two parts: a circle and a curved surface.
The curved surface is formed from a sector of a circle.
• This investigation involves the making of two cones
The centre of AB on the
and the calculation of their surface area.
semicircle is the point of the cone.
Step 1
Draw a semicircle of radius 10 cm.
Step 2
Make a cone by joining opposite sides of the
semicircle, as shown below.
Step 3
Put the cone face down and trace the circular base.
Measure the diameter of this base.
Step 4
Calculate the area of the original semicircle plus the area of the circular base.
This would be the total surface area of the cone if it were closed.
m
• Repeat the steps above, making the original sector a quarter circle of radius 10 cm.
What is the surface area of a closed cone of these dimensions?
Sa
A cone may be thought of as a pyramid with a circular base.
Consider a cone of slant height s and base radius r.
• Imagine what would happen if we cut along a straight line
joining the vertex to a point on the base.
• By cutting along this line, which is called the slant height,
we produce the net of the curved surface. The net of the
curved surface is a sector of a circle, radius s.
A
h
s
r
B
2πr
r
s
A
s
B
2πr
180
A
B
B
curved surface
New Signpost Mathematics Enhanced 10 5.1–5.3
base
If you bring the two ‘B’s
together, the sector
will bend to form
a cone.
To calculate the area of a sector, we must find what fraction it is of the complete circle. Normally
this is done by looking at the sector angle and comparing it to 360°, but it can also be done by
comparing the length of the sector’s arc to the circumference of the circle. Hence, if the sector’s arc
length is half the circumference of the circle, then the sector’s area is half the area of the circle.
(See Prep Quiz 6:03.)
length of sector’s arc
∴ Area of sector = --------------------------------------------------------- × area of circle
circumference of circle
2πr
= --------- × πs 2
2πs
= πrs
Now, since the area of the sector = area of the curved surface,
curved surface area = πrs
h
s
s
Surface area of a cone:
surface area = πrs + πr2
where r = radius of the cone and
s = slant height of the cone
Note: s = h 2 + r 2
ge
r
pa
Worked examples
1 Find the surface area of a cone with a radius of 5 cm and a slant height of 8 cm.
2 Find the surface area of a cone with a radius of 5 cm and a height of 12 cm.
Surface area = πrs + πr2
= π × 5 × 8 + π × 52
= 40π + 25π
= 65π cm2
= 204·2 cm2
(correct to 1 dec. pl.)
2
Sa
m
1
pl
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Solutions
■ The height of a cone is
the perpendicular height.
Exercise 6:03
1
Find the curved surface area of the following
cones, giving answers in terms of π.
a
b
8 cm
10 cm
10 cm
First the slant height must
be
calculated.
s
12
Now, s2 = 52 + 122
(Pythag. theorem)
5
= 169
∴ s = 13
surface area = πrs + πr2
= 65π + 25π
= 90π cm2
= 282·7 cm2
(correct to 1 dec. pl.)
Foundation Worksheet 6:03
Surface area of a cone MS5·3·1
1 For each cone shown, find the:
i radius
ii height
iii slant height
a
b
8
O
10 cm
10
6
7·5
12·5
O
20
2 For each of the cones in question 1 find:
i the curved surface area
ii the surface area
3 Use Pythagoras’ theorem to find the slant height if:
a radius = 3 cm; height = 4 cm
b diameter = 16 cm; height = 15 cm
Chapter 6 Surface Area and Volume
181
c
d
e
1·2 m
12 cm
40 cm
16 cm
2
r = 40 cm
Find the surface area of each of the cones in question 1 giving all answers in terms of π.
Calculate the surface area of the following
cones, giving all answers in terms of π.
a radius 8 cm and height 6 cm
b radius 1·6 m and height 1·2 m
c diameter 16 cm and height 15 cm
d diameter 1 m and height 1·2 m
Don’t forget,
use the slant
height, not the
vertical height,
of the cone.
ge
s
3
20 cm
In each of the following, find the surface area of the cone correct to four significant figures.
a radius 16 cm and height 20 cm
b radius 5 cm and slant height 12 cm
c radius 12·5 cm and height 4·5 cm
d diameter 1·2 m and height 60 cm
e diameter 3·0 m and slant height 3·5 m
5
Find the surface area of the following solids. Give all answers correct to one decimal place.
a
b
c
AB = 12 cm
A
O
6
7
182
C
10 cm
7 cm
7 cm
12 cm
Note: This is half a cone
Sa
B
m
BO = 12 cm
BC = 5 cm
OD = 10 cm
9 cm
pl
e
pa
4
D
A cone is to be formed by joining the radii of the
sector shown. In the cone that is formed, find:
a the slant height
b the radius
c the perpendicular height
15 cm
120°
15 cm
a A cone with a radius of 5 cm has a surface area of 200π cm2. What is the perpendicular
height of the cone?
b A cone cannot have a surface area greater than 1000π cm2. What is the largest radius,
correct to one decimal place, that will achieve this if the slant height is 20 cm?
New Signpost Mathematics Enhanced 10 5.1–5.3
6:04 Surface Area of
a Sphere
Investigation 6:04
Outcome MS5·3·1
The surface area of a sphere
Carry out the experiment outlined below to
demonstrate that the reasoning is correct.
Step 1
Cut a solid rubber ball or an orange into
two halves.
The faces of the two hemispheres are
identical circles.
It is impossible to draw
the net of a sphere in
two dimensions...
ge
pa
Step 3
Using thick cord, cover the circular face
of one of the hemispheres as shown,
carefully working from the inside out.
Mark the length of cord needed. Call this length A.
Length A covers the area of the circle, ie πr2.
s
Step 2
Push a long nail through the centre of a
hemisphere, as shown.
pl
e
Step 4
Put a second mark on the cord at a point that is double the length A.
The length of the cord to the second mark is 2A.
2A covers the area of two identical circles, ie 2πr2.
1
Sa
m
Step 5
Turn the other hemisphere over and use the cord of
length 2A to cover the outside of the hemisphere.
It should fit nicely.
It seems that 2A covers half of the sphere.
It would take 4A to cover the whole sphere.
ie the surface area of a sphere = 4A
Surface area = 4πr2
2
3
4
5
Chapter 6 Surface Area and Volume
183
6:08 Practical Applications Outcomes MS5·2·2, MS5·3·1
of Surface Area and Volume
Worked example
A buoy consists of a cylinder with two hemispherical ends,
as shown in the diagram. Calculate the volume and surface
area of this buoy, correct to one decimal place.
1·1 m
2·4 m
1·1 m
Solution
s
Since the two hemispheres have the same radius, they will form a sphere if joined.
=
4
--- πr3 + πr2h
3
4
--- π(1·1)3 + π
3
× 1·12 × 2·4
= 14·7 m3 (correct to 1 dec. pl.)
■ The height of
the hemisphere
is equal to its
radius.
pa
=
ge
Volume = ⎛ volume of ⎞ + ⎛ volume of ⎞
⎝ sphere ⎠ ⎝ cylinder ⎠
of buoy
This means the
hemispheres and the
cylinder have the
same radius.
Sa
m
pl
e
Surface = surface area + curved surface
area
of sphere
area of cylinder
2
= 4πr + 2πrh
= 4π × 1·12 + 2 × π × 1·1 × 2·4
= 31·8 m2 (correct to 1 dec. pl.)
Exercise 6:08
1
A swimming pool is rectangular in shape and has uniform depth.
It is 12 m long, 3·6 m wide and 1·6 m deep. Calculate:
a the cost of tiling it at $75/m2
b the amount of water in litres that needs to be added to raise the
level of water from 1·2 m to 1·4 m
2
The tipper of a trick is a rectangular prism in shape. It is 7 m long, 3·1 m wide and 1·6 m high.
a Calculate the volume of the tipper.
b If the truck carries sand and 1 m3 of sand
7m
weighs 1·6 tonnes, find the weight of sand
carried when the truck is three-quarters full.
c Calculate the area of sheet metal in the tipper.
3·1
m
■ 1 m3 = kL
1·6
m
Chapter 6 Surface Area and Volume
195
A large cylindrical reservoir is used to store water. The reservoir is 32 m in diameter and has a
height of 9 m.
a Calculate the volume of the reservoir to the nearest cubic metre.
b Calculate its capacity to the nearest kilolitre below its maximum capacity.
c In one day, the water level drops 1·5 m. How many kilolitres of water does this represent?
d Calculate the outside surface area of the reservoir correct to the nearest square metre.
Assume it has no top.
4
Assuming that the earth is a sphere of radius
6400 km, find (correct to 2 sig. figs):
a the volume of the earth in cubic metres
b the mass of the earth if the average density
is 5·4 tonnes/m3
c the area of the earth’s surface covered by
water if 70% of the earth is covered by water.
5
A bridge is to be supported by concrete supports. Calculate the volume of concrete needed for
each support.
The supports are trapezoidal prisms, as pictured in
6·4 m
the diagram.
3·6 m
Give the answer to the nearest cubic metre.
pl
e
pa
ge
s
3
Sa
m
6·2 m
1·8 m
6
A house is to be built on a concrete slab
which is 20 cm thick. The cross-sectional
shape of the slab is shown in the diagram.
Calculate the volume of concrete needed
for the slab.
6·8 m
2·1 m
10·8 m
4·7 m
7
2·1 m
A steel tank is as shown in the diagram.
10·5 m
3·5 m
2·4 m
196
8·5 m
Given that the dimensions are external
dimensions and that the steel plate is 2 cm
thick, calculate the mass of the tank if the
density of the steel is 7·8 g/cm3. Give the
answer correct to one decimal place.
New Signpost Mathematics Enhanced 10 5.1–5.3
8
Calculate the volume of a concrete beam that has
the cross-section shown in the diagram. The beam is
10 m long.
100 mm
300 mm
Calculate the mass of this beam if 1 m3 of concrete
weighs 2·5 tonnes.
1300 mm
700 mm
300 mm
100 mm
s
The large tank in the photo consists
of two cones and a cylinder. If the
diameter of the cylinder is 5·2 m and
the heights of the bottom cone, cylinder
and top cone are 2·8 m, 8·5 m and
1·8 m respectively, calculate the volume
of the tank correct to one decimal place.
10
pl
e
pa
ge
9
1800 mm
A storage bin has been made from a square prism and a
square pyramid. The top 1·8 m of the pyramid has been
removed. Calculate the volume of the bin.
Sa
m
3·6 m
A
B
AB = 4·5 m
BC = 3·6 m
11
C
1·2 m
A glasshouse is in the shape of a square
pyramid. Calculate the area of the four
triangular faces to the nearest square
metre if the side of the square is 20 m
and the height of the pyramid is 17 m.
Chapter 6 Surface Area and Volume
197
12
a The solid shown is known as a frustrum.
It is formed by removing the top part of the cone.
1m
2·7 m
2·5 m
i By comparing the values of
tan θ in two different triangles,
find the value of x.
ii Find the volume of the frustrum.
xm
␪
1m
2·7 m
␪
2·5 m
1 23
Literacy in Maths
s
2·4 m
ge
slant height (of a cone)
slant
height
• The distance from a point on the
circumference of the circular base
to the apex of the cone.
surface area
• The sum of the areas of the faces or surfaces
of a three-dimensional figure
(or solid).
volume
• The amount of space (cubic units) inside a
three-dimensional shape.
pl
e
m
Sa
2·0 m
Maths terms 6
composite solid
• A solid that is formed
by joining simple
solids.
prism
• A solid that has two identical ends
joined by rectangular
faces.
pyramid
• A solid that has a base from
which triangular faces rise to
meet at a point.
Maths terms 6
198
2·5 m
pa
b A storage bin for mixing cement is formed from
two truncated cones (frustrums). Calculate the
volume of this bin.
2·7 m
1m
New Signpost Mathematics Enhanced 10 5.1–5.3
Diagnostic Test 6
Surface Area and Volume
• Each part of this test has similar items that test a certain question type.
• Errors made will indicate areas of weakness.
• Each weakness should be treated by going back to the section listed.
These questions can be used to assess all of outcome MS5·3·1 and parts of outcome MS5·2·2.
Section
6:02
12 cm
cm
12·36 cm
5
·6
12 cm
12
10 cm
1 Calculate the surface area of the following pyramids.
a
c
b
6 cm
12 cm
10 cm
8 cm
10 cm
2·6 m
pa
2·4 cm
6·5 cm
2·6 m
1·8 cm
5 cm
3 Calculate the surface area of:
a a sphere of radius 5 cm, correct to 2 dec. pl.
b a sphere of diameter 16·6 cm, correct to 2 dec. pl.
c a hemisphere of radius 3 cm, correct to 2 dec. pl.
4 Calculate the volume of the following solids.
a
c
b
pl
e
6:04
m
Sa
5·3 m
4·6 m
5·4 m
5·4 m
6:03
ge
s
2 Calculate the surface area of the following cones. Give answers correct
to one decimal place.
a
b
c
5·4 m
6:05
7·6 cm
16·4 c
m
8·7 cm
8·4 m
5 Calculate the volume of the following solids correct to one decimal place.
a
b
c
4·3 cm
6·5 cm
6·1 cm
6:06
12·3 cm
2·6 cm
2·6 cm
6 Calculate the volume of the following solids correct to one decimal place:
a a sphere of radius 5 cm
b a sphere of diameter 8·6 cm
c a hemisphere of diameter 15 cm
6:07
Chapter 6 Surface Area and Volume
199
Revision
Chapter 6 Revision Assignment
6 cm
1 Calculate the volume
and surface area of
the pyramid shown.
9 cm
9 cm
2 A cone has a diameter of 16 cm and a
height of 15 cm. Calculate the surface area
of the cone in terms of π.
3 Calculate the surface area of a hemisphere
that has a diameter of 16 cm. Give your
answer correct to two significant figures.
6 Calculate the surface
area of the solid shown.
Give the answer correct
to three significant
figures.
17·0 cm
7 A tank for holding liquid chemicals
consists of a cylinder with two
hemispherical ends as shown in the
diagram. Calculate its volume (correct
to 3 sig. figs.)
3·62 m
ge
8 An octahedron is a double
pyramid with all its
edges equal in length.
Calculate the surface
area and volume
6 cm
of an octahedron
with all its edges
6 cm in length.
pl
e
12·5 cm
Sa
m
10·2 cm
1·80 m
pa
7·6 cm
4·1 cm
s
4 A spherical shaped tank is to hold 100 m3.
What radius to the nearest millimetre will
give a volume closest to 100 m3?
5 Calculate the volume of the solid pictured.
Give the answer correct to three significant
figures.
8·6 cm
• Engineers solve many surface area and volume problems in the design and construction
of buildings.
200
New Signpost Mathematics Enhanced 10 5.1–5.3
Chapter 6 Working Mathematically
1 The value of a library is depreciated at a
rate of 15% pa. If the library is presently
valued at $800 000, what will its value be
after 4 years?
A
S
X
C
T
4
2
s
3 ABCD is a right triangular A
pyramid. Its base is an
equilateral triangle
of side 72 units B
and the edges AB,
X
AC and AD are all
6 units long. Find
C
the height, AX and
the volume of the pyramid.
6 One large sheet of
paper was ruled
up and folded.
It was then cut
along the fold
shown in the
diagram on the
top to form an
8-page booklet.
On the diagrams
to the right, put
the page number
on each quarter
(as has been done
for 2 and 4).
pl
e
pa
D
ge
2 ΔABC is equilateral with
a side of length a units.
AT, BS and RC are
perpendicular heights R
of the triangle that
meet at X. Find the
lengths:
B
a BX
b XT
5 The dot indicates the position of one chess
queen on a chess board. How many more
queens can you place on the board so that
none of the queens threatens another?
2
Front
Back
Sa
m
4 Three positive whole numbers are
multiplied in pairs. The answers obtained
are 756, 1890 and 4410. What are the
numbers?
4
4
Revision
• The glass pyramids outside
the Louvre, Paris.
Chapter 6 Surface Area and Volume
201