New Century Mathematics (Second Edition)
S3 Question Bank
3B Chapter 9 Area and Volume (III)
Level 2
<code=10169592>
<bk=3B><ch=9><ex=9A><type=L2><mark=6><title=10169592><content>
A
8.5 cm
4 cm E
D
12 cm
C
B
The figure shows a right pyramid ABCDE of height 4 cm. AD = 8.5 cm and CD = 12 cm.
(a) Find the diagonal BD of the base of the pyramid.
(b) Find the volume of the pyramid ABCDE.
(6 marks)
Solution:
(a) Let AN be the height of the pyramid.
i.e. AN = 4 cm
A
8.5 cm
4 cm E
D
N
B
12 cm
C
In △AND,
ND = AD 2  AN 2
= 8.52  42 cm
= 7.5 cm
BD = 2  ND = 2  7.5 cm = 15 cm
(b) In △BCD,
BC = BD 2  CD 2
= 152  12 2 cm
= 9 cm
∴
1M
1A
1A
1A
1
3
Volume of the pyramid = (12 9) 4 cm 3
= 144 cm3
1M
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.31
<code=10169596>
<bk=3B><ch=9><ex=9A><type=L2><mark=6><title=10169596><content>
A
8 cm
9.6 cm
E
D
Q 5.6 cm
B
C
P
The figure shows a right rectangular pyramid ABCDE.
(a) Find AC.
(b) Find the total surface area of ABCDE.
(6 marks)
Solution:
(a) CQ =
1
1
CD = 5.6 cm = 2.8 cm
2
2
1M
In △ACQ,
AC = AQ 2  CQ 2
= 9.6 2  2.8 2 cm
= 10 cm
1M
1A
(b) In △ACP,
PC = AC 2  AP 2
= 10 2  8 2 cm
= 6 cm
BC = 2  PC = 2  6 cm = 12 cm
∴ Total surface area of ABCDE
= 2  area of △ACD + 2  area of △ABC + area of rectangle BCDE


1
2
1
2
1A
1M


=  2  5.6 9.6  2  12 8  12 5.6  cm2
= 216.96 cm2
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.32
<code=10169642>
<bk=3B><ch=9><ex=9A><type=L2><mark=8><title=10169642><content>
8.5 cm
15 cm
The figure shows an octahedron ABCDEF which is formed by two identical regular pyramids. The
length of AE is 8.5 cm and the height of the octahedron is 15 cm.
(a) Find the length of CE.
(b) Find the volume of the octahedron.
(8 marks)
Solution:
(a) Refer to the notations in the figure.
8.5 cm
15 cm
AN =
1
1
AF = 15 cm = 7.5 cm
2
2
In △ANE,
NE = AE 2  AN 2
= 8.52  7.52 cm
= 4 cm
∴ CE = 2  NE = 2  4 cm = 8 cm
(b) Let x cm be the length of each side of square BCDE.
In △CDE,
CD2 + DE2 = CE2
x 2 + x 2 = 82
2x2 = 64
x2 = 32
∴ Volume of the octahedron = 2  volume of pyramid ABCDE
© OXFORD UNIVERSITY PRESS 2017
1A
1M
1A
1M
1A
1M
3B Chapter 9 Level 2 P.33
1
3
= 2  32 7.5 cm 3
1M
= 160 cm3
1A
<end>
<code=10169702>
<bk=3B><ch=9><ex=9A><type=L2><mark=7><title=10169702><content>
B
10 cm
C
D
A
V
The figure shows a paper cup in the shape of an inverted right pyramid VABCD with a square base
of side 10 cm. The capacity of the cup is 400 cm3. Find the total area of the paper used to make the
cup.
(7 marks)
Solution:
Refer to the notations in the figure.
B
P
A
10 cm
Q
D
C
V
Let h cm be the height VQ of the pyramid.
1
10 2 h
3
= 400
1M
h = 12
1A
In △VPQ,
PQ =
1
1
BC = 10 cm = 5 cm
2
2
VP = VQ 2  PQ 2
= 122  52 cm
= 13 cm
∴ The required area = 4  area of △VAB
1
2
1M
1M
1A
= 4  10 13 cm2
1M
= 260 cm2
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.34
<code=10169848>
<bk=3B><ch=9><ex=9A><type=L2><mark=6><title=10169848><content>
V
39 cm
C
V
V
25 cm
D
P
V
C
Q
B
D
B
A
P
Q
A
V
Fig. A
Fig. B
The cardboard in Fig. A is a net of the right pyramid in Fig. B. ABCD is a rectangle. The total
surface area of the pyramid is 2 739 cm2. It is given that VP is 4 cm longer than VQ. Find the
volume of the pyramid, correct to 3 significant figures.
(6 marks)
Solution:
Let x cm be the length of VQ.
Then VP = (x + 4) cm.
2  area of △VAD + 2  area of △VAB + area of ABCD = 2 739 cm2
1
1
2  25 ( x  4)  2  39 x  39 25 = 2 739
2
2
1M
25x + 100 + 39x + 975 = 2 739
64x = 1 664
x = 26
∴ VQ = 26 cm
Let VN be the height of the pyramid.
1A
V
C
B
D
N
P
Q
A
In △VQN,
QN =
1
1
AD = 25 cm = 12.5 cm
2
2
VN = VQ 2  QN 2
= 26 2  12.52 cm
= 519.75 cm
© OXFORD UNIVERSITY PRESS 2017
1M
1M
3B Chapter 9 Level 2 P.35
∴
Volume of the pyramid
1
3
= (39 25)  519.75 cm 3
1M
= 7 410 cm3, cor. to 3 sig. fig.
<end>
1A
<code=10169866>
<bk=3B><ch=9><ex=9A><type=L2><mark=6><title=10169866><content>
The figure shows a frustum ABCDEFGH of a right pyramid. Its upper base and lower base are
squares of sides 2 cm and 6 cm respectively. The height of pyramid VEFGH is 4 cm.
(a) Find the height of the pyramid VABCD.
(b) Find the volume of the frustum ABCDEFGH.
(6 marks)
Solution:
Refer to the notations in the figure.
(a) Let h cm be the height of the pyramid VABCD.
1
1
FG = 2 cm = 1 cm
2
2
1
1
SR = AB = 6 cm = 3 cm
2
2
PQ =
∵
∴
△VPQ ~ △VSR
VP
PQ
=
VS
SR
4 cm
h cm
1M
(AAA)
1M
=
1 cm
3 cm
1M
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.36
h = 12
The height of the pyramid VABCD is 12 cm.
∴
1A
(b) Volume of the frustum ABCDEFGH
= volume of pyramid VABCD – volume of pyramid VEFGH
1
3
2
=  6 12 
= 138
1

2 2 4  cm3
3

1M
2
cm 3
3
1A
<end>
<code=10169870>
<bk=3B><ch=9><ex=9B><type=L2><mark=6><title=10169870><content>
8 cm
The volume of the right circular cone in the figure is 96 cm3.
(a) Find the slant height of the cone.
(b) Find the total surface area of the cone in terms of .
(6 marks)
Solution:
(a) Let r cm be the base radius of the cone.
1
πr 2 8
3
= 96
1M
∴
r2 = 36
r=6
Slant height = 62  82 cm
= 10 cm
(b) Total surface area of the cone
= (π 6 10  π 62 ) cm2
= 96 cm2
<end>
© OXFORD UNIVERSITY PRESS 2017
1A
1M
1A
1M
1A
3B Chapter 9 Level 2 P.37
<code=10169879>
<bk=3B><ch=9><ex=9B><type=L2><mark=5><title=10169879><content>
A
40 cm
B
10 cm
P
D
15 cm
Q
C
E
In the figure, BCED is a frustum of a right circular cone. Find the curved surface area of the
frustum in terms of .
(5 marks)
Solution:
(AAA)
1M
∵ △ABP ~ △ADQ
∴
BP
AB
=
DQ
AD
BP
15 cm
=
40 cm
(40  10) cm
1M
BP = 12 cm
∴ Curved surface area of the frustum
= curved surface area of cone ADE – curved surface area of cone ABC
= [  15  (40 + 10) –   12  40] cm2
= 270 cm2
<end>
© OXFORD UNIVERSITY PRESS 2017
1M
1M
1A
3B Chapter 9 Level 2 P.38
<code=10169880>
<bk=3B><ch=9><ex=9B><type=L2><mark=8><title=10169880><content>
18 cm
The figure shows a paper cup in the shape of an inverted right circular cone. The capacity of the
cup is 1 080 cm3.
(a) Find the slant height of the cup.
(b) Emily claims that if the cup is cut along a slant height to form a sector, the angle of the sector
is an acute angle. Do you agree? Explain your answer.
(8 marks)
Solution:
(a) Let h cm be the height of the cup.
2
1
 18 
π 
 h
3
 2 
= 1 080
1M
27h = 1 080
h = 40
2
∴
18
Slant height =    402 cm
 2
= 41 cm
1A
(b) Let  be the angle of the sector.
Arc length of the sector = circumference of the base of the cup

2 π 41 =   18
360
=
∴
<end>
1M
1M+1M
9π
360
41π
= 79.024, cor. to 5 sig. fig.
< 90
The claim is agreed.
© OXFORD UNIVERSITY PRESS 2017
1A
1M
1A
3B Chapter 9 Level 2 P.39
<code=10169926>
<bk=3B><ch=9><ex=9B><type=L2><mark=9><title=10169926><content>
Q
O
Q
120
P
P
Fig. A
Fig. B
Fig. A shows a sector. It is folded into a paper cup (as shown in Fig. B) in the shape of an inverted
right circular cone, whose slant height is 9 cm longer than the base radius.
(a) Find the curved surface area of the cup.
(b) Find the capacity of the cup.
(Give the answers correct to 3 significant figures.)
(9 marks)
Solution:
(a) Let r cm be the base radius of the cup.
Then PQ = (r + 9) cm.
1M
∵ Arc length of the sector = circumference of the base of the cup
∴
∴
120
2 π ( r  9) = 2r
360
r 9
=r
3
r + 9 = 3r
2r = 9
r = 4.5
Curved surface area of the cup =   4.5  (4.5 + 9) cm2
= 191 cm2, cor. to 3 sig. fig.
(b) Consider △POQ in Fig. B.
OP = PQ 2  OQ 2
= ( 4.5  9) 2  4.52 cm
= 162 cm
∴
1
3
Capacity of the cup = π 4.52  162 cm3
© OXFORD UNIVERSITY PRESS 2017
1M+1M
1A
1M
1A
1M
1M
3B Chapter 9 Level 2 P.40
= 270 cm3, cor. to 3 sig. fig.
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.41
<code=10169998>
<bk=3B><ch=9><ex=9B><type=L2><mark=8><title=10169998><content>
The figure shows a solid formed by a right circular cone and a cylinder with a common base. The
ratio of the curved surface area of the cone to that of the cylinder is 2 : 5.
(a) Find the height of the cylinder.
(b) If the height of the solid is 41 cm, find the volume of the solid in terms of .
(8 marks)
Solution:
(a) Let h cm and r cm be the height and the base radius of the cylinder respectively.
Curved surface area of the cone =   r  20 cm2
1M
= 20r cm2
Curved surface area of the cylinder = 2rh cm2
1M
∵
Curved surface area of the cone
Curved surface area of the cylinder
20πr
2
=
2πrh
5
10
2
=
h
5
∴
∴
=
1M
h = 25
The height of the cylinder is 25 cm.
1A
(b) Height of the cone = (41 – 25) cm = 16 cm
Base radius of the cone = 202  162 cm = 12 cm
∴
1
3
1M


2
2
Volume of the solid =  π 12 16  π 12 25  cm3
= 4 368 cm3
2
5
1M+1M
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.42
<code=10170010>
<bk=3B><ch=9><ex=9B><type=L2><mark=8><title=10170010><content>
5 cm
18 cm
10 cm
7.5 cm
A cylindrical vessel of base radius 5 cm contains water to a depth of 7.5 cm. All the water is
poured from the vessel into an empty cup in the shape of an inverted right circular cone as shown
in the figure. The base diameter of the cup is 18 cm, and the depth of water in the cup is
10 cm.
(a) Find the radius of the circular water surface in the cup.
(b) Find the additional volume of water required to fill up the cup in terms of .
(8 marks)
Solution:
(a) Let r cm be the radius of the circular water surface in the cup.
Volume of water in the cup = volume of water in the vessel
1
π r 2 10 = π 52 7.5
3
1M+1M
r2 = 56.25
r = 7.5
The radius of the circular water surface is 7.5 cm.
∴
1A
(b) Refer to the notations in the figure.
18 cm
Q
D
B
P
C
A
10 cm
V
∵
△VAP ~ △VCQ
∴
VP
AP
=
VQ
CQ
10 cm
VQ
(AAA)
1M
7.5 cm
= 1
2
18 cm
1M
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.43
∴
VQ = 12 cm
The required volume of water = capacity of the cup – volume of water in the cup
1
18
2
1A
1M

1


2
=  3 π  2  12  3 π 7.5 10 cm3




= 136.5 cm3
1A
<end>
<code=10170032>
<bk=3B><ch=9><ex=9C><type=L2><mark=8><title=10170032><content>
The figure shows a solid glass souvenir consisting of two hemispheres. The difference in their radii
is 1 cm and the curved surface area of the large hemisphere is 128 cm2.
(a) Find the radius of the large hemisphere.
(b) Find the volume of the souvenir in terms of .
(c) If the production cost of the souvenir is $0.03 per cm3, is $260 enough to produce 5 such
souvenirs? Explain your answer.
(8 marks)
Solution:
(a) Let r cm be the radius of the large hemisphere.
1
4 πr 2
2
= 128
1M
∴
r2 = 64
r=8
The radius of the large hemisphere is 8 cm.
1A
(b) The required volume = volume of the large hemisphere + volume of the small hemisphere
1
4
1
4

3
3
=   π 8   π (8  1)  cm3
2 3
2 3

1M+1M
= 570 cm3
1A
(c) Total cost of producing 5 such souvenirs = $0.03  5  570
= $268.61, cor. to the nearest $0.01
> $260
∴ $260 is not enough to produce 5 such souvenirs.
<end>
© OXFORD UNIVERSITY PRESS 2017
1M
1M
1A
3B Chapter 9 Level 2 P.44
1.5 cm
<code=10170129>
<bk=3B><ch=9><ex=9C><type=L2><mark=6><title=10170129><content>
The figure shows a sausage, which is in the shape of a cylinder with two hemispheres at both ends.
The diameter of each hemisphere is 1.5 cm. The total surface area of the sausage is 20.25 cm2.
(a) Find the length of the whole sausage.
(b) Find the volume of the whole sausage in terms of .
(6 marks)
Solution:
(a) Let x cm be the height of the cylinder.
∵ Total surface area of the sausage
= 2  curved surface area of each hemisphere + curved surface area of the cylinder
= surface area of the sphere + curved surface area of the cylinder
2
 1 .5 
4 π 
  π 1.5 x
 2 
∴
= 20.25
1M+1M
2.25  1.5 x = 20.25
1.5x = 18
x = 12
1.5 
 1.5
 12 
 cm
2 
 2
∴
Length of the whole sausage = 
= 13.5 cm
1A
(b) Volume of the whole sausage
= 2  volume of each hemisphere + volume of the cylinder
= volume of the sphere + volume of the cylinder
4
3
2

1.5 
 1.5 
  π 
 12
 2 
 2 


=  π 
 3
= 7.312 5 cm3
<end>
cm3
1M+1M
1A
<code=10170151>
<bk=3B><ch=9><ex=9C><type=L2><mark=7><title=10170151><content>
A solid metal sphere of diameter 12 cm is melted and recast into 54 identical solid hemispheres.
(a) Find the radius of each hemisphere.
(b) Alex wants to paint the 54 hemispheres. Given that a bottle of paint is just enough to paint the
original sphere, are three bottles of paint enough to paint all the 54 hemispheres? Explain your
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.45
answer.
(7 marks)
Solution:
(a) Let r cm be the radius of each hemisphere.
4
 12 
π 

3
 2 
3
1
2
4
3


3
= 54   π r 
1M+1M
∴
288 = 36r3
r3 = 8
r=2
The radius of each hemisphere is 2 cm.
1A
1
2


2
2
(b) Total surface area of the 54 hemispheres = 54  4 π 2  π 2  cm2
1M
= 648 cm2
2
12 
 cm2 = 144 cm2
 2 
Surface area of the original sphere = 4 π 
∵
∴
<end>
1M
Total area that three bottles of paint can cover = 3  144 cm2
= 432 cm2
< 648 cm2
Three bottles of paint are not enough to paint all the 54 hemispheres.
1M
1A
<code=10170159>
<bk=3B><ch=9><ex=9C><type=L2><mark=7><title=10170159><content>
6 cm
15 cm
14 cm
In the figure, a cylindrical cup of base radius 6 cm and height 15 cm contains some water. When
three spherical stones of radius 3 cm each are immersed in the water inside the cup, the water level
rises to 14 cm.
(a) Find the original water level.
(b) If the three stones are replaced by a new spherical stone of radius 4.5 cm and it is immersed in
water, will the water overflow? Explain your answer.
(7 marks)
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.46
Solution:
(a) Let h cm be the original water level.
Total volume of the three stones = volume of water that has risen
4

3  π 33  =   62  (14 – h)
3

1M+1M
3 = 14 – h
h = 11
The original water level is 11 cm.
∴
1A
(b) Total volume of water and the new stone


4
3


2
3
=  π 6 11  π 4.5  cm3
1M
= 517.5 cm3
Capacity of the cup =   62  15 cm3
= 540 cm3
∵ 517.5 < 540
1M
1M
1A
∴ The water will not overflow.
<end>
<code=10170231>
<bk=3B><ch=9><ex=9C><type=L2><mark=8><title=10170231><content>
The figure shows a right circular cone and a sphere. The base radius of the cone is equal to the
diameter of the sphere. The ratio of the total surface area of the cone to the surface area of the
sphere is 8 : 3. If the volume of the sphere is 36 cm3, find the volume of the cone in terms of .
(8 marks)
Solution:
Let r cm be the radius of the sphere.
4
πr 3
3
= 36
1M
r3 = 27
r=3
∴ Base radius of the cone = 2  3 cm = 6 cm
© OXFORD UNIVERSITY PRESS 2017
1A
1M
3B Chapter 9 Level 2 P.47
Let  cm be the slant height of the cone.
∵
Total surface area of the cone
Surface area of the sphere
=
8
π 6   π 6 2
=
2
3
4 π 3
∴
1M+1M
6  + 36 = 96
6  = 60
 = 10
Height of the cone = 2  6 2 cm = 10 2  6 2 cm = 8 cm
∴
8
3
1A
1M
1
3
Volume of the cone = π 6 2 8 cm3 = 96 cm3
1A
<end>
<code=10170248>
<bk=3B><ch=9><ex=9D><type=L2><mark=3><title=10170248><content>
Let a, b and c be the linear measurements of a solid. Jessie claims that the total surface area of the
solid is 2a2 + 2ab + bc. Kelvin claims that the total surface area of the solid is a2b + abc. It is
known that the statement made by one of them must be wrong. Whose statement must be wrong?
Explain your answer.
(3 marks)
Solution:
The dimension of the quantity represented by 2a2 + 2ab + bc is 2, while the dimension of the
quantity represented by a2b + abc is 3.
1A
2
1M
∴ a b + abc cannot represent the total surface area of the solid.
1A
∴ Kelvin’s statement must be wrong.
<end>
<code=10170254>
<bk=3B><ch=9><ex=9E><type=L2><mark=6><title=10170254><content>
C
B
pathway
G
A
H
D
I
garden
F
E
In the figure, ABCDE is a plot of land in the shape of a regular pentagon. A part of the land is used
to build a garden EFGHI in the shape of a regular pentagon. The remaining part of the land is used
to build a pathway. It is given that AF : FE = 1 : 2 and the area of the pathway is 12 m2 greater than
that of the garden. Find the area of the garden.
(6 marks)
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.48
Solution:
AE
AF  FE
=
FE
FE
AF
1
=
FE
1
= 1
2
3
=
2
1M
1A
Let x m2 be the area of the garden.
Area of the plot of land = (x + x + 12) m2
= (2x + 12) m2
∵ EFGHI is similar to ABCDE.
∴
2 x  12
x
3
=  
1M
2
 2
1M+1M
2 x  12
x
=
9
4
8x + 48 = 9x
x = 48
∴ The area of the garden is 48 m2.
<end>
1A
<code=10170296>
<bk=3B><ch=9><ex=9E><type=L2><mark=7><title=10170296><content>
The figure shows a vertical right conical vessel containing some water. The ratio of the curved
surface area of the vessel to area of the surface of the vessel in contact with water is 36 : 25.
(a) Find the ratio of the base radius of the vessel to the radius of the water surface.
(b) It is given that the vessel contains 600 cm3 of water. Alex claims that if 300 cm3 of water is
added into the vessel, the water will overflow. Do you agree? Explain your answer.
(7 marks)
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.49
Solution:
Refer to the notations in the figure.
(a) Let x and y be the base radius of the vessel and the radius of the water surface respectively.
∵ Cones VAB and VCD are similar.
2
36
25
x
6
=
y
5
∴
 x
 
 y
∴
The required ratio is 6 : 5.
=
1M
1A
(b) Let z cm3 be the capacity of the vessel.
z
600
6
=  
 5
3
1M
z
600
=
216
125
z = 1 036.8
∴ Volume of the remaining space in the vessel = (1 036.8 − 600) cm3
= 436.8 cm3
> 300 cm3
∴ Water will not overflow.
∴ The claim is disagreed.
<end>
1A
1A
1M
1A
<code=10170327>
<bk=3B><ch=9><ex=9E><type=L2><mark=9><title=10170327><content>
In the figure, B and E are points on AC and AD respectively such that EB // DC. It is given that
AB : AC = 3 : 5. N is the point on AD such that BN  AD.
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.50
(a) Find
(i) area of △ABE : area of △ABD,
(ii) area of △ABE : area of quadrilateral BCDE.
(b) If the area of BCDE is 640 cm2, find the area of △BDE.
(9 marks)
Solution:
(a) (i) ∵
∴
△ABE ~ △ACD
(AAA)
AE
AB
3
=
=
AD
AC
5
1M
1A
1
AE BN
2
=
1
AD BN
2
Area of △ ABE
Area of △ ABD
1M
AE
AD
3
=
5
=
∴
The required ratio = 3 : 5
1A
(ii) Let x be the area of △ABE and y be the area of BCDE.
Area of △ ABE
Area of △ ACD
3
=  
2
 5
1M
x
9
=
x y
25
25x = 9x + 9y
16x = 9y
x
9
=
y
16
∴
The required ratio is 9 : 16.
1A
(b) From the result of (a)(ii),
area of △ABE =
9
640 cm2 = 360 cm2
16
1M
From the result of (a)(i),
5
3
area of △ABD = 360 cm2 = 600 cm2
∴
Area of △BDE = (600 − 360) cm2 1M
= 240 cm2
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.51
<code=10170358>
<bk=3B><ch=9><ex=9E><type=L2><mark=7><title=10170358><content>
pyramid P
frustum Q
pyramid S
In the figure, a right pyramid S is divided into a right pyramid P and a frustum Q. The heights of P
and Q are 12 cm and 4 cm respectively.
(a) John claims that volume of P : volume of Q = 3 : 4. Do you agree? Explain your answer.
(b) If the total area of all lateral faces of Q is 259 cm2, find the total area of all lateral faces of S.
(7 marks)
Solution:
(a) Let x cm3 and y cm3 be the volumes of P and Q respectively.
Volume of S = (x + y) cm3
1M
∵ Pyramids S and P are similar.
∴
x
xy
= 
x
xy
=
12 

 12  4 
3
1M
27
64
64x = 27x + 27y
37x = 27y
x
27
=
y
37
3
4
1M
The claim is disagreed.
1A

∴
1A
(b) Let A cm2 be the total area of all lateral faces of S.
Total area of all lateral faces of P = (A – 259) cm2
A  259  12  2
=

A
 12  4 
A  259
9
=
A
16
1M
16A − 4 144 = 9A
7A = 4 144
A = 592
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.52
∴ The total area of all lateral faces of S is 592 cm2.
<end>
1A
<code=10170412>
<bk=3B><ch=9><ex=9E><type=L2><mark=10><title=10170412><content>
6 cm
3 cm
In the figure, a large right circular cone is divided into a small right circular cone of height 6 cm
and a frustum of height 3 cm. The volume of the frustum is 3.04 cm3.
(a) Find the volume of the large cone in terms of .
(b) Katy claims that the total surface area of the large cone is smaller than 40 cm2. Do you agree?
Explain your answer.
(10 marks)
Solution:
(a) Let x cm3 be the volume of the large cone.
Volume of the small cone = (x – 3.04) cm3
1M
∵ The large cone and the small cone are similar.
∴
3
x  3.04 π
6 
= 

x
 6 3
x  3.04 π
8
=
x
27
1M
27x – 82.08 = 8x
19x = 82.08
x = 4.32
∴
The volume of the large cone is 4.32 cm3.
1A
(b) Let r cm be the base radius of the large cone.
1
π r 2 (6  3)
3
= 4.32
1M
r2 = 1.44
r = 1.2
Slant height of the large cone = 1.2 2  (3  6) 2 cm = 82.44 cm
∴ Total surface area of the large cone = (π 1.2  82.44  π 1.2 2 ) cm2
= 38.8 cm2, cor. to the nearest 0.1 cm2
∵ 38.8 cm2 < 40 cm2
© OXFORD UNIVERSITY PRESS 2017
1A
1M
1M
1A
1M
3B Chapter 9 Level 2 P.53
∴ The claim is agreed.
<end>
1A
<code=10170463>
<bk=3B><ch=9><ex=9E><type=L2><mark=11><title=10170463><content>
20 cm
24 cm
X
Y
The figure shows two similar solids X and Y. Each of them consists of a right circular cone and a
cylinder with a common base. The ratio of the volume of solid X to that of solid Y is 125 : 64. The
total curved surface area of solid X is 1 275 cm2. Find the volume of solid Y in terms of .
(11 marks)
Solution:
Let R cm and r cm be the base radii of the cylindrical parts of solids X and Y respectively.
3
125
 R
  =
64
 r 
1M
5
R
=
r
4
1A
Let S cm2 be the total curved surface area of Y.
1 275π
S
5
=  
 4
2
1M
1 275π
S
=
S = 816
∴ The total curved surface area of Y is 816 cm2.
2r  24 + r  20
1M+1M
68r = 816
r = 12
Height of the conical part of Y = 20 2  12 2 cm
= 16 cm
∴
1
3
1A
= 816
1A
1M


2
2
Volume of solid Y =  π 12 16  π 12 24  cm3
= 4 224 cm3
25
16

1M+1M
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.54
<code=10170549>
<bk=3B><ch=9><ex=9E><type=L2><mark=8><title=10170549><content>
In the figure, a solid chocolate in the shape of a regular pyramid is melted to form three identical
solid chocolates in the shapes of regular pyramids. The base area of the large chocolate is 360 cm2,
while the base area of each small chocolate is 160 cm2. It is given that the height of the large
chocolate is 18 cm.
(a) Find the height of each small chocolate.
(b) Is the large chocolate similar to each small chocolate? Explain your answer.
(8 marks)
Solution:
(a) Let h cm be the height of each small chocolate.
1

3  160 h 
3

1
3
= 360 18
1M+1M
∴
h = 13.5
The height of each small chocolate is 13.5 cm.
1A
(b) From the result of (a),
 height of the large chocolate 
 height of each small chocolate 


2
2
16
18 cm 
 =
9
 13.5 cm 
= 
Base area of the large chocolate
9
360 cm 2
=
=
2
Base area of each small chocolate 160 cm
4
16 9

∵
9
4
∴ The large chocolate is not similar to each small chocolate.
<end>
© OXFORD UNIVERSITY PRESS 2017
1M+1A
1A
1M
1A
3B Chapter 9 Level 2 P.55
<code=10170560>
<bk=3B><ch=9><ex=9A><type=L2><mark=8><title=10170560><content>
18 cm
In the figure, the slant edge VP of a right pyramid VPQRS is 18 cm long and its base is a square of
side 16 cm. Find
(a) the height of the pyramid,
(b) the volume of the pyramid,
(c) the total surface area of the pyramid.
(Give the answers correct to 3 significant figures if necessary.)
(8 marks)
Solution:
(a) Refer to the notations in the figure.
18 cm
In △VPA,
PA =
1
1
PQ = 16 cm = 8 cm
2
2
1M
1M
VA  VP 2  PA 2
 18
2
2
 8 cm
 260 cm
AN =
1
1
PS = 16 cm = 8 cm
2
2
1M
VN = VA 2  AN 2
 ( 260 ) 2  8 2 cm
14 cm
∴
The height of the pyramid is 14 cm.
1
3
(b) Volume of the pyramid  16 2 14 cm 3
1 190 cm 3
© OXFORD UNIVERSITY PRESS 2017
, cor. to 3 sig. fig.
1A
1M
1A
3B Chapter 9 Level 2 P.56
(c) Total surface area of the pyramid = 4  area of △VPQ + area of square PQRS


1
2
1M


2
2
=  4  16  260  16  cm
= 772 cm 2 , cor. to 3 sig. fig.
1A
<end>
<code=10170572>
<bk=3B><ch=9><ex=9A><type=L2><mark=8><title=10170572><content>
8 cm
The figure shows a regular pyramid VABCDEF with VO as its height. The diagonals AD, BE and
CF of the base divide the base into 6 identical equilateral triangles. It is given that VA = 17 cm and
AB = 8 cm.
(a) Find the area of △OAB.
(b) Find the volume of the pyramid.
(Give the answers correct to 3 significant figures.)
(8 marks)
Solution:
(a) Let OP be the height of △OAB.
8 cm
PB =
1
1
AB = 8 cm = 4 cm
2
2
1M
∵ △OAB is an equilateral triangle.
∴ OA = OB = AB = 8 cm
In △OPB,
1M
OP  OB 2  PB 2
1M
 8
2
 4
2
cm
 48 cm
∴
1A
1
Area of △OAB  AB OP
2
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.57
1
 8  48 cm 2
2
 27.7 cm 2
, cor. to 3 sig. fig.
27.713
1A
(b) In △VAO,
VO  VA 2  OA 2  17 2  8 2 cm =
∴
15 cm
1A
1
Volume of the pyramid  (6 27.713) 15 cm 3
3
831 cm 3
1M
, cor. to 3 sig. fig.
1A
<end>
<code=10170585>
<bk=3B><ch=9><ex=9B><type=L2><mark=7><title=10170585><content>
108
The figure shows a right circular cone formed by folding a piece of paper in the shape of a sector.
If the radius of the sector is 10 cm and the angle of the sector is 108, find
(a) the base radius of the circular cone,
(b) the volume of the circular cone.
(Give the answer correct to the nearest cm3.)
(7 marks)
Solution:
(a) Let r cm be the base radius of the circular cone.
Slant height of the circular cone = 10 cm
1M
∵ Area of the sector = curved surface area of the circular cone
∴
∴
108
π 10 2 = r(10)
360
1M+1M
r=3
The base radius of the circular cone is 3 cm.
1A
(b)
3 cm
Height of the circular cone
 10 2  3 2
 91
∴
cm
1M
cm
1
3
Volume of the circular cone  π 3 2  91 cm 3
90 cm 3
, cor. to the nearest cm3
1M
1A
<end>
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.58
<code=10170637>
<bk=3B><ch=9><ex=9B><type=L2><mark=5><title=10170637><content>
A paper cup is in the shape of an inverted right circular cone of base diameter 6 cm and height
12 cm. Suppose the cup is held vertically and it is originally filled up with water. If some water
flows out of the cup until the depth of water drops 6 cm as shown in the figure, find the volume of
the water flowing out in terms of π.
(5 marks)
Solution:
Refer to the notations of the figure.
BC =
1
6 cm = 3 cm
2
AV = BV – BA = (12 – 6) cm = 6 cm
(AAA)
∵ △VAD ~ △VBC
∴
∴
AD
BC
r
6
=
3
12
r = 1.5
Volume of the water flowing out
= volume of cone VEC – volume of cone VFD
1
3
=  π 3 2 12 
1M
AV
=
BV
1

π 1.5 2 6  cm 3
3

© OXFORD UNIVERSITY PRESS 2017
1A
1M
1M
3B Chapter 9 Level 2 P.59
= 31.5π cm 3
<end>
1A
<code=10170684>
<bk=3B><ch=9><ex=9C><type=L2><mark=4><title=10170684><content>
15 cm
The figure shows a container formed by removing a hemisphere of radius 14 cm from a
hemisphere of radius 15 cm with the same centre. Find the total surface area of the container in
terms of .
(4 marks)
Solution:
Total surface area of the container
= curved surface area of the large hemisphere +
curved surface area of the small hemisphere +
area of the ring
1M
1
2
1
2


=  4 π 15 2  4 π 14 2  ( π 15 2  π 14 2 ) cm 2
1M+1M
= [450 + 392 + 29] cm2
= 871π cm 2
<end>
1A
<code=10170700>
<bk=3B><ch=9><ex=9C><type=L2><mark=3><title=10170700><content>
The figure shows a test tube consisting of a cylinder of height 10 cm and a hemisphere with a
common base. The radius of the hemisphere is 1 cm. Find the capacity of the test tube in terms
of .
(3 marks)
Solution:
Capacity of the test tube = volume of the cylinder + volume of the hemisphere


1
2
4
3


2
3
3
=  π 1 10   π 1  cm
© OXFORD UNIVERSITY PRESS 2017
1M+1M
3B Chapter 9 Level 2 P.60


=  10π 
2 
π  cm 3
3 
2
= 10 3 π cm
3
1A
<end>
<code=10170747>
<bk=3B><ch=9><ex=9D><type=L2><mark=3><title=10170747><content>
Refer to the solid as shown. Write down the dimensions of the quantities represented by the
expressions below.
(a) abk 
abh
3
(b) 4(a + b + e + k)
(c) a(b + c + 2k) + b(d + 2k)
(3 marks)
Solution:
(a) Its dimension is 3.
1A
(b) Its dimension is 1.
1A
(c) Its dimension is 2.
<end>
1A
© OXFORD UNIVERSITY PRESS 2017
3B Chapter 9 Level 2 P.61
<code=10170757>
<bk=3B><ch=9><ex=9E><type=L2><mark=5><title=10170757><content>
In the figure, a right circular cone is cut by two planes parallel to its base into three parts P, Q and
R. Find the ratio of volume of P to that of Q to that of R.
(5 marks)
Solution:
Let VP, VQ and VR be the volumes of P, Q and R respectively.
∵ Cone P is similar to the cone formed by P and Q.
∴
VP
V P  VQ
2
=  
1M
3
 4
1M
VP
V P  VQ
∵
∴
8VP = VP + VQ
VQ = 7VP
Cone P is similar to the original cone.
3
VP
2
=  
VP  VQ  VR
 6
VP
VP  7VP  VR
27VP = 8VP + VR
VR = 19VP
∴ The required ratio = VP : VQ : VR
= VP : 7VP : 19VP
= 1 : 7 : 19
<end>
© OXFORD UNIVERSITY PRESS 2017
=
1
8
1A
=
1
27
1A
1A
3B Chapter 9 Level 2 P.62