New Progress in Junior Mathematics 3B
(b)
9
Area and Volume (III)
Total surface area of the hemisphere
1

   4    2.52    2.52  cm 2
2


 58.9 cm 2 (cor. to 1 d.p.)
p.127
p.99
1
Volume of VABCD   (6  6)  5 cm 3
3
 60 cm 3
Type of Measurement
Dimension
1.
Linear
1
2.
Quadratic
2
3.
Cubic
3
4.
Linear
1
p.130
p.108
(a)
(b)
(c)
Area of quadrilateral ABCD  16 
 
Area of quadrilateral PQRS  24 
1
Volume of the right circular cone     62 11 cm 3
3
 132 cm 3

2
4
9
1
Volume of the right circular cone     8.52 12 m3
3
 289 m3
9.1
1
Volume of the right circular cone     7 2  9.6 mm 3
3
 156.8 mm3
p.112
Curved surface area   1013 cm 2
Let s cm2 be the base area of VABCDE .
p.99
1
Volume of VABCDE   Base area  Height
3
1
310   s 15
3
s  62

Base area of VABCDE is 62 cm2.
(a)
In ABC,
AC 2  AB 2  BC 2 (Pyth. theorem)
 130 cm 2
Base area   10 cm 2
9.2
2
 100 cm
Total surface area  (130 100π) cm2
2

AC  6 2  82 cm
 230 cm 2
 10 cm
1
EC  AC  5 cm
2
p.119
In VEC,
VC 2  EC 2  VE 2 (Pyth. theorem)
3
(a)
4
 11 
Volume of the sphere       cm 3
3
2
VE  132  5 2 cm
 696.9 cm3 (cor. to 1 d.p.)
(b)
 12 cm
Volume of the hemisphere
1 4
     23 cm 3  16.8 cm3 (cor. to 1 d.p.)
2 3
(b)
p.121
(a)
Surface area of the sphere
2
 14 
 4      cm 2  615.8 cm 2 (cor. to 1 d.p.)
 2
© Hong Kong Educational Publishing Co.
58
1
Volume of VABCD   (6  8) 12 cm 3
3
 192 cm 3
p.100
9
Area and Volume (III)
9.3
1
Volume of pyramid VEFGH   4.52  5 cm 3
3
9.6
p.101
 33.75 cm 3
1
Volume of pyramid VABCD  10.82  (5  7) cm 3
3
r 8
 466.56 cm 3

Let r cm be the base radius of the circular cone. p.108
1
   r 2 15  320
3
r 2  64
Volume of ABCDEFGH
 Volume of VABCD  Volume of VEFGH
9.7

The base radius of the circular cone is 8 cm.
(a)
In VOA,
p.109
1
1
OA  AB  (12) cm  6 cm
2
2
2
2
(Pyth. theorem)
VO  OA  VA 2
 (466.56  33.75) cm 3
 432.81 cm 3
9.4
VO  10 2  6 2 cm
1
Area of VAB  11 8 cm 2
2
p.102
 8 cm
 44 cm 2
Base area  1111 cm 2

(b)
 121 cm 2
Total surface are of VABCD  (121 44  4) cm2
1
Volume of the cone     62  8 cm 3
3
 96 cm 3
 301.44 cm 3
 297 cm 2
9.5
In VAB, let M be the mid-point of AB.
1
MB  18 cm  9 cm
2
9.8
p.103
(a)


Since VAB is an isosceles triangle,
VM  AB (property of isos. )
MB 2  VM 2  VB 2 (Pyth. theorem)
VM  202  92 cm

 319 cm
1
Area of VAB  18  319 cm 2
2
(b)
 9 319 cm 2
In VBC, let N be the mid-point of BC.
1
NB  12 cm  6 cm
2
Since VBC is an isosceles triangle,
VN  BC (property of isos. )
Volume of the smaller right circular cone

1
   4.5 2  3 cm 3  20.25 cm 3
3
Volume of the larger right circular cone

1
   12 2  (3  5) cm 3  384 cm 3
3

NB 2  VN 2  VB 2 (Pyth. theorem)
VN  202  62 cm
9.9
 364 cm

Area of VBC 
(a)
1
12  364 cm 2
2
Volume of the frustum  (384  20.25) cm
 363.75 cm 3
In VOA,
VA2  VO 2  OA2 (Pyth. theorem)
p.113
VA  4.8 2  3.6 2 m
 6m
 6 364 cm 2

p.110
VAB ~ VCD (AAA)
VA AB

(corr. sides, ~ s)
VC CD
VA
4.5 cm

VA  5 cm 12 cm
12VA  4.5(VA  5 cm)
7.5VA  22.5 cm
VA  3 cm
Total surface area of VABCD
(b)
 (18 12  2  9 319  2  6 364 ) cm 2
Total surface area
   3.6  6 m 2    3.6 2 m 2
 766 cm (cor. to 3 sig. fig.)
2
 34.56 m 2
 109 m2 (cor. to the nearest m2)
9.10 (a)
Let r cm be the base radius.
  r 14  56
r4

59
p.113
The base radius is 4 cm.
© Hong Kong Educational Publishing Co.
3
New Progress in Junior Mathematics 3B
(b)
Mark the points V, O and A as shown in the
9.14
figure.
In VOA,
VO 2  OA2  VA2 (Pyth. theorem)

Base area of the cylindrical container  5 2  cm 2
 180 cm
 25 cm 2
Rise in the water level  121.5  25cm
Height of the right
 4.86 cm
180 cm .
Volume of the right circular cone
1
    42  180 cm 3
3
 225 cm3 (cor. to 3 sig. fig.)
(b)
p.120
VO  142  42 cm
circular cone is
9.11 (a)
Volume of the sphere
4
  4.53   cm 3
3
 121.5 cm 3
1
   (7 cm ) 2  VO  147  cm 3
3
VO  9 cm
9.15 (a)
Let r cm be the radius of the hemisphere. p.122
1
 4    r 2  r 2  147 
2
3r 2  147 
r 2  49
r 7
p.114

(b)
In VOA,
VA2  VO 2  OA2 (Pyth. theorem)
The radius of the hemisphere 7 cm.
Volume of the hemisphere
1 4
     7 3 cm 3  718 cm3 (cor. to 3 sig. fig.)
2 3
VA  92  7 2 cm
 130 cm
9.16 Let r cm be the base radius of the hemisphere.
1 4
1024
   r3 

2 3
3
r 3  512
Curved surface area of the circular cone
   OAVA
   7  130 cm 2
r 8
 7 130  cm 2

Area of the sector  Curved surface area of

  ( 130 ) 
2

The base radius of the hemisphere is 8 cm.
Side length of the cube  8 cm  2
 16 cm
the circular cone

360
Total surface area of the whole solid
 Total surface area of the cube
Base area of the hemisphere
 7 130 
  221
 Curved surface area of the hemisphere
(cor. to 3 sig. fig.)
9.12 Let r cm be the radius of the sphere.
4
   r 3  972
3
r 3  729
1


  6 16 2    8 2   4    8 2  cm 2
2


 1740 cm 2 (cor. to 3 sig. fig.)
p.119
r 9

9.17
The radius of the sphere is 9 cm.
1 4
    83 cm 3
p.120
2 3
1024

 cm 3
3
1
Volume of the right circular cone     8 2  7 cm 3
3
448

 cm 3
3
9.13 Volume of the hemisphere 

p.123
Area of the smaller circle  80 
2
p.131
16
cm 2
25
 51.2 cm 2
 1024 448 

Volume of the whole solid  
 cm 3
3 
 3
1472

 cm3
3
© Hong Kong Educational Publishing Co.
Area of the smaller circle  4 
 
Area of the larger circle  5 
Area of the smaller circle 16

80 cm 2
25
60
Area and Volume (III)
2
3
 Perimeter of the larger rectangle 
25
 
9.18 
16
 Perimeter of the smaller rectangle 
 Slant height of A  125
 
9.21 
512
 Slant height of B 
24 cm
5

Slant height of B 8
p.131
2
25
 Perimeter of the larger rectangle 

 
40 cm
16


8
Slant height of B  24  cm
5
 38.4 cm
Perimeter of the larger rectangle 5

40 cm
4
Perimeter of the larger rectangle  50 cm
9.19 AEF ~ ABC (AAA)
Area of AEF  AE 


Area of ABC  AB 
1
 
2
9.22 Let d1 and A1 be the diameter and surface area
p.132
surface area of the new sphere.
A2  (1  19%) A1
 0.81A1
1
4
Area of BCFE 4  1


Area of ABC
4
3

4
CGH ~ CAD (AAA)


Area of CGH  CG 


Area of CAD  AC 
 d2

d
 1
2

A
  2

A1

0.81A1

A1
d2
 0 .9
d1

2
d 2  0.9d1
Percentage decrease in its diameter 
2




d1  d 2
100%
d1
d1  0.9d1
100%
d1
 10%
1
16
Area of AGHD 16  1

Area of CAD
16
15

16
ABC  CDA (ASA)
3
Area of BCFE
 4
Area of AGHD 15
16
4

5
Area of BCFE : Area of AGHD = 4 : 5


p.135
of the original sphere. Let d 2 and A2 be the diameter and
2
2
1
 
4
p.135
9.23 (a)
Volume of the smaller pyramid
1

Volume of the larger pyramid 26  1

p.136
1
27
3

 Height of the smaller pyramid 
1

 
Height
of
the
larger
pyramid
27


Height of the smaller pyramid 1

Height of the larger pyramid 3
Height of the frustum
3 1

Height of the larger pyramid
3
2
3
Height of the frustum 2

12 cm
3
2
Height of the frustum  12  cm  8 cm
3

2
9.20
Base area of the smaller pyramid  10 cm 
p.134


Base area of the larger pyramid  15 cm 
Base area of the smaller pyramid 4

108 cm 2
9
4
Base area of the smaller pyramid  108  cm 2
9
 48 cm
(b)
2
Area of VAD  1 
 
Area of VBC  3 
2
1
9
Area of ABCD 9  1

Area of VBC
9
8

9


61
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
Example 9.4T
p.102
1
 5  9 cm 2
2
 22.5 cm 2
Base area  5  5 cm 2
Area of VAB 
Example 9.1T
p.99
Let a cm be the length of each side of the base of VABCD.
1
Volume of VABCD   Base area  Height
3
1
72   a 2  6
3

 115 cm 2
Example 9.5T
a 2  36
a6

MB 
1
12 cm  6 cm
2
Since VAB is an equilateral triangle,
p.100
VM  AB .
In ABC ,
MB 2  VM 2  VB 2 (Pyth. theorem)
AC 2  AB 2  BC 2 (Pyth. theorem)
VM  12 2  6 2 cm
AC  8 2  8 2 cm
 108 cm
 8 2 cm
EC 

1
AC  4 2 cm
2
Area of VAB 

VE 2  EC 2  VC 2 (Pyth. theorem)
Total surface of VABC
 4  6 108 cm 2
VE  9  (4 2 ) cm
2
2
 7 cm
 249 cm 2
1
 (8  8)  7 cm 3
3
448

cm 3
3
Volume of VABCD 
Example 9.3T
p.108
Let r cm be the base radius of the right circular cone.
1
   r 2  6  72
3
r 2  36
p.101
r 6

 245.76 cm 3
1
Volume of pyramidVABCD  16.8 2  (8  6) cm 3
3
 1317.12 cm
Volume of ABCDEFGH
 Volume of pyramidVABCD
(cor. to 3 sig. fig.)
Example 9.6T
1
Volume of pyramidVEFGH   9.6 2  8 cm 3
3

1
12  108 cm 2
2
 6 108 cm 2
In VEC ,
(b)
p.103
In VAB , let M be the mid-point of AB.
The length of each side of the base of VABCD is 6 cm.
Example 9.2T
(a)
 25 cm 2
Total surface area of VABCD  (25  22.5  4) cm 2
The base radius of the right circular cone is 6 cm.
Example 9.7T
(a)
3
p.109
1
AB  8 cm
2
In VOB ,
OB 
OB 2  VO 2  VB 2 (Pyth. theorem)
VO  17 2  82 cm
 15 cm
 Volume of pyramidVEFGH
 (1317.12  245.76) cm 3
 1071.36 cm 3
(b)
Volume of the circular cone 
1
   82  15 cm 3
3
 320 cm 3
 1004.8 cm 3
© Hong Kong Educational Publishing Co.
62
Area and Volume (III)
Example 9.8T
(a)
(b)
p.110
VAB ~ VCD (AAA)
VA AB

(corr. sides, ~ s)
VC CD
VA
6 cm

VA  6 cm 18 cm
VA
1

VA  6 cm 3
3VA  VA  6 cm
2VA  6 cm


VO  152  52 cm
 200 cm

Volume of the right circular cone
1
    52  200 cm 3
3
 370 cm 3 (cor. to 3 sig. fig.)
Volume of the smaller circular cone

Example 9.11T
1
   6 2  3 cm 3  36 cm 3
3
Volume of the larger circular cone
1
   18 2  (3  6) cm 3  972 cm 3
3
Volume of the frustum  (972  36) cm 3

Curved surface area of the circular cone    511 m 2
 936 cm 3
Example 9.9T
 55 m 2
Let  be the angle of the sector of the curved surface.

 112 
 55
360
 164 (cor. to 3 sig. fig.)
p.112
1
AB  5 cm
2
In VOB ,
OB 
Example 9.12T
VB 2  VO 2  OB 2 (Pyth. theorem)
VB  12 2  52 cm
 13 cm
(b)
r 6
 65 cm 2

Base area    52 cm 2
 25 cm 2
Total surface area  (65  25) cm2
4
   4 3 cm 3
3
256

 cm 3
3
Volume of the cube  93 cm 3
 283 cm (cor. to 3 sig. fig.)
p.113
Let l cm be the slant height.
  5  l    5 2  100
5l  75
l  15

p.119
Volume of the sphere 
2
(a)
The radius of the sphere is 6 m.
Example 9.13T
 90 π cm 2
Example 9.10T
p.119
Let r m be the radius of the sphere.
4
   r 3  288
3
r 3  216
Area of the curved surface    5 13 cm 2

p.114
Let l m be the slant height.
  5  l    5 2  80
5l  55
l  11

(a)
The height of the right
circular cone is 200 cm .
VA  3 cm
(b)
Mark the points V, O and A as shown in the figure.
VO 2  OA2  VA 2 (Pyth. theorem)

 729 cm 3
The volume of the empty space in the container
256  3

  729 
  cm
3 

(cor. to 3 sig. fig.)
 461 cm3
The slant height is 15 cm.
63
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
Example 9.14T
p.120
Volume of the sphere 
Example 9.17T
p.130
2
Area of ABC  3 
 
Area of PQR  7 
Area of ABC
9

84 cm 2
49
9
Area of ABC 
 84 cm 2
49
108

cm 2
7
4
   7.53 cm 3
3
 562.5 cm 3
Let h cm be the decrease in the water level.
Decrease in the volume of water   10 2  h cm 3


 100h cm 3
Volume of the sphere  Decrease in the volume of water
562.5  100h
h  5.625

The decrease in the water level 5.625 cm.
Example 9.18T
p.131
2
Example 9.15T
(a)
 Perimeter of the larger pentagon 
49

 
Perimeter
of
the
smaller
pentagon
36


35 cm
7

Perimeter of the smaller pentagon 6
6
Perimeter of the smaller pentagon   35 cm
7
 30 cm
p.122
Let r m be the radius of the hemisphere.
1
 4    r 2  4.5
2
r 2  2.25
r  1.5

(b)
The radius of the hemisphere is 1.5 m.
Example 9.19T
Volume of the hemisphere
1 4
     1.53 m 3
2 3
AB : AD : AF  1 : (1  2) : (1  2  3)
 1: 3 : 6
 2.25 m 3
 7.07 m 3 (cor. to 3 sig. fig.)


Example 9.16T
p.122

Let r cm be the radius of the hemisphere.
1 4
    r 3  486
2 3
r 3  729
r 9
The radius of the hemisphere is 9 cm.
1
Curved surface area of the hemisphere   4    9 2 cm 2
2
 162 cm 2
 1 : 9 : 36
Area of ABC：Area of BDEC：Area of DFGE
 1 : (9  1) : (36  9)
 1 : 8 : 27
p.134
2
Base area of the smaller cuboid  12 cm 


Base area of the larger cuboid  30 cm 
28 cm 2
4

Base area of the larger cuboid 25
25
Base area of the larger cuboid 
 28 cm 2
4
 175 cm 2
Let l cm be the slant height of the right circular cone.
l 2  10 2  24 2 (Pyth. theorem)
l  10 2  24 2
 26
The slant height of the right circular cone is 26 cm.
Example 9.21T
Curved surface area of the right circular cone
   10  26 cm 2  260 cm 2
Area of the ring  ( 10 2    9 2 ) cm 2

ABC, ADE and AFG are similar triangles.
Area of ABC：Area of ADE：Area of AFG
 12 : 32 : 6 2
Example 9.20T


p.132
3
 The corresponding slant height in A 
216

 
The
slant
height
in
B
343


 19 cm 2
2
Total surface area  (162  260  19) cm
 441 cm
The corresponding slant height in A 6

28 cm
7
6
The corresponding slant height in A   28 cm
7
 24 cm
2
© Hong Kong Educational Publishing Co.
p.134
64
Area and Volume (III)
Example 9.22T
p.135
pp.104 – 106
9.1
Let l1 and V1 be the length of one edge and the area of one
Level 1
face of the original cuboid respectively. Let l2 and V2 be the
length of one edge and the area of one face of the new cuboid
1.
respectively.
l 2  (1  20%)l1
1
3
Volume of the pyramid   28 15 cm
3
 140 cm 3
 0.8l1
V2  l 2

V1  l1




3
2.
 0.8l1 

 

 l1 
 0.83
3
3.
Percentage decrease in its volume
V1  V2

 100%
V1
 V 
 1  2   100%
 V1 
 [1  (0.8) 3 ]  100%
 48.8%
4.
Example 9.23T
(a)
1
3
Volume of the pyramid   33 13 cm
3
 143 cm 3
1
Volume of VABCD   8 2 18 cm 3
3
 384 cm 3
1 10 8
Area of ABCD  4    cm 2
2 2 2
 40 cm 2
1
Volume of VABCD   40  9 cm 3
3
 120 cm 3
p.136
Volume of the smaller pyramid 64  37

Volume of the larger pyramid
64
5.
1 1

3
Volume of VABCD     (3  6)  5  7 cm
3 2

 52.5 cm 3
27

64
6.
3

 Height of the smaller pyramid 
27

 
Height
of
the
larger
pyramid
64


15 cm
3

Height of the larger pyramid 4

4
Height of the larger pyramid   15 cm
3
 20 cm
Height of the frustum  (20  15) cm
 5 cm
7.
Let h m be the height of VABC.
1
128  h  256
3
h6

The height of VABC is 6 m。
(a)
Let s cm2 be the area of ABCD.
1
 s  7  105
3
s  45

The area of ABCD is 45 cm2.
2
(b)
Area of VAD  15 
 
Area of VBC  20 
9

16
Area of VAD
9


Area of ABCD 16  9
9

7
(b)
9 cm  BC  45 cm 2
BC  5 cm
8.
1
Volume of VABCD   (6  8) 14 cm 3
3
 224 cm 3
3
Volume of ABCDEFGH  (224  28) cm
 196 cm 3
65
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
1
 20  18 cm 2
2
 180 cm 2
13.
Area of VBC 
9.
Total surface area of VABCD
 (20  20  180  4) cm 2
Let a cm be the side length of the square.
1
1000
 a 10 
3
3
a 2  100

 1120 cm 2
a  10
AB  BC  10 cm
In ABC ,
10.
AC 2  AB 2  BC 2 (Pyth. theorem)
1
 5.9  7.5 cm 2
2
 22.125 cm 2
Area of VPQ 
AC  10 2  10 2 cm
 200 cm
Total surface area of VPQRS
 (5.9  5.9  22.125  4) cm 2

 123.31 cm 2
 10 2 cm
1
1
OC  AC  (10 2 ) cm  5 2 cm
2
2
In VOC ,
11.
VC 2  VO 2  OC 2 (Pyth. theorem)
1
Area of VAB   10  15 cm 2
2
 75 cm 2
1
Area of VBC   18  13 cm 2
2
 117 cm 2
VC  10 2  (5 2 ) 2 cm
 150 cm
 12.2 cm (cor. to 3 sig. fig.)
Total surface area of VABCD
 (10 18  75  2  117  2) cm 2
14.
 564 cm 2
Volume of the larger pyramid
1
  18 2  18 cm 3  1944 cm 3
3
Level 2
12.
Volume of the smaller pyramid
1
  6 2  (18  12) cm 3  72 cm 3
3

Volume of ABCDEFGH  (1944  72) cm
 1872 cm 3
Mark the centre of ABCD as O and the mid-point of BC
as E as shown in the figure.
Since VBC is an isosceles triangle, VE  BC .
1
CE  BC  7.5 cm
2
2
15.
Area of VBC 
1
 12 
12  18 2    cm 2
2
 2
 6 288 cm 2
2
Area of VCD 
 8 260 cm 2
In VCE ,

VE 2  CE 2  VC 2 (Pyth. theorem)
Total surface area of VABCD
 (12 16  2  6 288  2  8 260 ) cm 2
VE  152  7.52 cm
 654 cm 2 (cor. to 3 sig. fig.)
 168.75 cm
1
1
OE  AB  (20 cm )  10 cm
2
2
16.
In AFG ,
AG 2  AF 2  FG 2 (Pyth. theorem)
In VOE ,
AG  6 2  6 2 cm
VO 2  OE 2  VE 2 (Pyth. theorem)
 72 cm
1
Area of AFG   6  6 cm 2  18 cm 2
2
1
Area of AGH   72  6 cm 2  3 72 cm 2
2
VO  168.75  10 2 cm
 68.75 cm

1
 16 
16  18 2    cm 2
2
 2
Volume of VABCD
1
  (20  15)  68.75 cm 3
3
 829 cm 3
(cor. to 3 sig. fig.)

Total surface area of pyramid AFGHE
 (6  6  18  2  3 72  2) cm 2
 123 cm 2
© Hong Kong Educational Publishing Co.
66
(cor. to 3 sig. fig.)
3
Area and Volume (III)
17.
(a)
Let h cm be the height of VABCD.
1
 (16 16)  h  1280
3
h  15

(b)
2.
Let h cm be the height of the circular cone.
1
   9 2  h  216
3
h8

The height of VABCD is 15 cm.
Mark the mid-point of AB as N and the centre of
3.
ABCD as O as shown in the figure.
Since VAB is an isosceles triangle, VN  AB .
The height of the circular cone is 8 cm.
Let r cm be the base radius of the circular cone.
1
   r 2  6  50
3
r 2  25
r 5
4.

The base radius of the circular cone is 5 cm.
(a)
Mark points V, O and A as shown in the figure.
In VON ,
VO  15 cm
1
1
ON  BC  (16) cm  8 cm
2
2
2
VN  VO 2  ON 2 (Pyth. theorem)

VN  152  82 cm

(c)
18.
VO 2  OA2  VA 2 (Pyth. theorem)
 17 cm
The height of VAB with base AB is 17 cm.
VO  20 2  12 2 cm
 16 cm

Total surface are of VABCD
1


 16 16  4  16 17  cm 2
2


 800 cm 2
(b)
Height of the circular cone is 16 cm.
1
Volume of the circular cone    12 2 16 cm 3
3
 768 cm 3
 2411.52 cm 3
Volume of the pyramid above the water
1
 4.5  4.5  6 m 3  40.5 m 3
3
1
Volume of the large pyramid   12  12  16 m 3
3
 768 m 3

Volume of the frustum inside the water

5.
h  132  5 2 cm
 12 cm
Volume of the right circular cone
1
    5 2  12 cm 3
3
 1 cm 3
 (768  40.5) m 3
 727.5 m 3
Capacity of the tank  20 18  (16  6) m 3

 3600 m 3
Volume of water inside the tank
Curved surface area of the right circular cone
   5  13 cm 2
 (3600  727.5) m 3  2872.5 m 3
9.2
Let h cm be the height of the right circular cone.
5 2  h 2  13 2 (Pyth. theorem)
 65 cm 2
pp.115  117
Level 1
1.
1
2
3
Volume of the circular cone    12 10 cm
3
 480 cm 3
67
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
6.
Let r m be the base radius of the right circular cone.
r 2  9 2  15 2 (Pyth. theorem)
11.
r  15 2  9 2 m

 12 m
Volume of the right circular cone
1
    12 2  15 m 3
3
 432 m 3
12.
Curved surface area of the right circular cone
   12  15 m 2
 180 m 2
7.
Let r cm be the base radius of the circular cone.
  r 12  84
r 7
13.
The base radius of the circular cone is 7 cm.
Let l cm be the slant height of the circular cone.
  8  l    8 2  176
8l  112
l  14

The slant height of the circular cone is 14 cm.
(a)
Let l cm be the slant height of the circular cone.
1
radius   30 cm  15 cm
2
2
l  20 2  15 2 (Pyth. theorem)
Let l cm be the slant height of the right circular cone.
7 2  24 2  l 2 (Pyth. theorem)
l  20 2  15 2
l  7 2  24 2 cm
 25
 25 cm

Volume of the right circular cone
1
    7 2  24 cm 3
3
 3 cm 3
(b)
The slant height of the circular cone is 25 cm.
Total surface area  ( 15 2   15  25) cm 2
 600 cm 2
 1880 cm 2 (cor. to 3 sig. fig.)
Curved area of the right circular cone
   7  25 cm 2
14.
 175 cm 2
Volume of the smaller circular cone
2
8.
1
 21 
       7 cm 3
3
 2
Let r m be the base radius of the right circular cone.
1
600 m 3     r 2  8 m
3
r   m
 257.25 cm 3

Volume of the frustum  (750  257.25) cm 3
 492.75 cm 3
 1548 cm 3
Let l m be the slant height of the right circular cone.
15 2  8 2  l 2 (Pyth. theorem)
(cor. to the nearest cm3)
l  152  8 2 m
 17 m
Level 2
15.
Curved surface area of the right circular cone
   15  17 m 2
 255 m
9.
10.
(a)
(i)
2    OA  28 cm
14
cm

 4.46 cm
OA 
2
(cor. to 3 sig. fig.)
Curved surface area of the circular cone
14
    9 cm 2
2
 63 cm 2
(ii)
VO 2  OA2  VA 2
(Pyth. theorem)
2
 14 
VO  12 2    cm

 11.1 cm (cor. to 3 sig. fig.)
Curved surface area    2.7  4.5 cm 2
 12.15 cm 2
Base area    2.7 cm 2
(b)
2
 7.29 cm
2
Total surface area  (12.15  7.29) cm
2
 19.44 cm 2
© Hong Kong Educational Publishing Co.
In VOA ,
68
Curved surface area    OA  VB
14
    12 cm 2

 168 cm 2
Area and Volume (III)
(c)
Volume of the circular come
1
    OA2  VO
3
2

18.
Let r be the base radius of the right circular cone, then
2r be the slant height.
Curved surface area    r  (2r )
 2r 2
2
1
 14 
 14 
      12 2    cm 3
3


Area of the sector    ( 2r ) 2 
 232 cm (cor. to 3 sig. fig.)
3

16.
(a)
VAB ~ VCD
AB VB

(corr. sides, ~ s)
CD VD
6 cm
VB

9 cm VB  5 cm
6VB  30 cm  9VB
3VB  30 cm
VB  10 cm


19.

(a)
Area of the sector    8 2 
h cm be the height.
Curved surface area    r  8 cm 2
VA  10 2  6 2 cm
 8 cm


Volume of the smaller circular cone

1
   6 2  8 cm 3
3
 96 cm 3
VA AB

(corr. sides, ~ s)
VC CD
8 cm 6 cm

VC
9 cm
98
VC 
cm
6
 12 cm

Volume of the larger circular cone


(a)
 6.4

(b)
20.
1
   9 2  12 cm 3  324 cm 3
3
Volume of the frustum
(b)
Height of the paper cup 6.4 cm.
1
Volume of the cup     4.8 2  6.4 cm 3
3
 49.152 cm 3
 154 cm 3 (cor. to 3 sig. fig.)
Let r cm be the radius of the water surface.
r cm
18 cm

7.5 cm 20 cm
7.5 18
r
cm
20
 6.75 cm
Let l1 cm be the slant height of the circular cone formed
Let h cm be the height of the circular cone.
1
  112  h  605
3
h  15

Curved surface area = Area of the sector
  r  8  38.4
r  4.8
2
2
4.8  h  8 2 (Pyth. theorem)
h  8 2  4.8 2
 (324  96) cm 3  228 cm 3
17.
216
cm 2
360
 38.4 cm 2
Let r cm be the base radius of the paper cup,
VA 2  AB 2  VB 2 (Pyth. theorem)

4r 2
360
Curved surface area = Area of the sector
4r 2
2r 2 
360
  180
In VAB ,
(b)

360
by the water and l2 cm be the slant height of the
container.
l12  6.75 2  18 2 (Pyth. theorem)
The height of the circular cone is 15 cm.
l1  6.75 2  18 2
 369.5625
Let l cm be the slant height.
l 2  15 2  112 (Pyth. theorem)
l22  7.52  20 2 (Pyth. theorem)
l2  7.5 2  20 2
l  152  112
 456.25
 346
Total surface area  ( 11   11 346 ) cm
2
2
Inner surface area of the container that is not wet
 1022.9 cm 2 (cor. to 1 d.p.)
= Curved surface area of the container – Curved
surface area of the circular cone formed by the water
 (  7.5  456.25    6.75  369.5625 ) cm 2
 95.6 cm 2 (cor. to 3 sig. fig.)
69
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
7.
pp.123 – 125
9.3
(a)
Level 1
1.
4
   4 3 cm 3
3
256

 cm 3
3
Volume 
Let r cm be the radius of the sphere.
4
500
 r3 

3
3
r 3  125
r 5

(b)
Surface area  4    4 cm
2
The radius of the sphere is 5 cm.
Surface area  4    5 2 cm 2
 100 cm 2
2
 64 cm 2
8.
3
2.
Volume 
4
 15 
     cm 3
3
 2
 562.5 cm 3
Let r m be the radius of the hemisphere.
1
 4    r 2  50
2
r 2  25
r 5
2
 15 
Surface area  4      cm 2
 2
 225 cm 2
3.
9.

The radius of the hemisphere is 5 m.
(a)
Let r cm be the radius.
4    r 2  81
1 4
   12 3 m 3
2 3
 1152 m 3
81
4
r  4.5
Volume 
r2 

1

Surface area    4    12 2    12 2  m 2
2

 432 m 2
4
   4.53 cm 3
3
 121.5 cm 3
(b)
Volume 
(a)
Let r cm be the base radius.
2    r  17 
r  8.5
3
4.
Volume 
1 4
 21 
      cm 3
2 3
 2
10.
 771.75 cm 3
2
2
1
 21 
 21  
Surface area    4            cm 2
 2
 2  
 2
2
 330.75 cm

(b)
5.
The base radius is 8.5 cm.
1
 4    8.5 2 cm 2
2
 144.5 cm 2
Curved surface area 
Volume of the white chocolate
4
4

     1.83     1.53  cm 3
3
3


 3.276 cm 3
 10.3 cm 3 (cor. to 3 sig. fig.)
6.
The radius is 4.5 cm.
11.
Let r cm be the radius of the hemisphere.
1 4
128
   r3 

2 3
3
r 3  64
 600 cm 3

r4

The radius of the hemisphere is 4 cm.
© Hong Kong Educational Publishing Co.
4
   2 3 cm 3
3
32

 cm 3
3
Capacity of the container  10  4 15 cm 3
Volume of the sphere 
Volume of the empty space in the container
32 

  600    cm 3
3 

 566 cm 3 (cor. to 3 sig. fig.)
70
Area and Volume (III)
12.
Curved surface area of the hemisphere
1
  4    6 2 cm 2
2
 72 cm 2
16.
Area of the lateral face of the cylinder
 2  6  5 cm 2
 60 cm

The radius of the hemisphere is 10 cm.
Base radius of the cylinder  10 cm  3 cm
 13 cm
2
Base area
   6 2 cm 2
Curved surface area of the hemisphere
1
  4    10 2 cm 2  200 cm 2
2
Area of the lateral face of the cylinder
 36 cm

Total surface area  (72  60  36) cm 2
2
 168 cm 2
13.
 2    13  6 cm 2  156 cm 2
Base area of the cylinder   132 cm 2
Let r cm be the radius of the sphere.
Surface area of the sphere  4    r 2 cm 2
 169 cm 2
Area of the ring  ( 13   102 ) cm2  69 cm2

Total surface area
2
Curved surface area of the circular cone
   7  12 cm 2  84 cm 2

Surface area of the sphere

= Curved surface area of the circular cone
4    r 2  84
Let r cm be the radius of the hemisphere.
1 4
2000
   r3 

2 3
3
r 3  1000
r  10
 (200  156  169  69) cm 2
 594 cm 2
r 2  21
17.
(a)
4
(33  4 3  53 ) cm 3
3
 288 cm 3
Volume of the larger ball 
r  21
 4.58 (cor. to 3 sig. fig.)

14.
Let r cm be the radius of the larger ball.
4
   r 3  288
3
r 3  216
Radius of the sphere 4.58 cm.
4
   64003 km 3
3
 1.10 1012 km 3 (cor. to 3 sig. fig.)
Volume 
r 6

The radius of the larger ball is 6 cm.
2
2
Surface area  4    6400 km
 5.15 108 km 2
(cor. to 3 sig. fig.)
(b)
Total surface area before the recasting process
 4    (32  4 2  5 2 ) cm 2
 200 cm 2
Level 2
15.
Total surface area after the recasting process
 4    6 2 cm 2
 144 cm 2
4
3
3
Volume of the 4 spheres  4     2 cm
3
128

 cm 3
3
Percentage change of the total surface area
144  200

 100%
200
  28%
Let h cm be the decrease in the water level.
Decrease in the volume of water    4 2  h cm 3


Volume of the 4 spheres
18.
= Decrease in the volume of water
128
    42  h
3
8
h
3
(a)
Let r cm be the outer radius of the bowl.
1
 4    (r 2  112 )    (r 2  112 )  517.75
2
3r 2  121  517.75
3r 2  396.75
r 2  132.25
 2.67 (cor. to 3 sig. fig.)

r  11.5
Decrease in the water level is 2.67 cm.

71
The outer radius of the bowl 11.5 cm.
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
(b)
Volume of the bowl
1 4
1 4

     11.53     113  cm 3
2
3
2
3


2.
(a)
2
(b)
1
(c)
3
(a)
2
(b)
3
(a)
Area
The radius of the sphere is 7 m.
(b)
Length
Total surface area  (4    7 2  6  14 2 ) m 2
(c)
Area
 (196  1176) m
Cost of plating  $60  (196  1176)
 $108000 (cor. to 3 sig. fig.)
(d)
Volume
(e)
Length
(f)
Area
 398 cm 3
19.
(a)
Let r m be the radius of the sphere.
4 3
r

3

14 3
6
8r 3  14 3 
3.
14 3
8
r 7
r3 

(b)
4.
2
20.
(a)
Radius of the ice-cream ball 
1 5
5
 cm  cm
2 2
4
Volume of the ice-cream ball
3

4
125
5
     cm 3 
cm 3
3
4
48
 
Level 2
5.
(b)
melts and r cm be the base radius of the
ice-cream circular cone.
r
h
(corr. sides, ~ s)

6 12
2
1
r h
4
(b)
2
2

1.
Area of A  4 cm 


Area of B  8 cm 

2.
1
3.
(c)
2
© Hong Kong Educational Publishing Co.
2
16
9
Area of A：Area of B  16 : 9


3
1
4
Area of A：Area of B  1 : 4
Area of A  12 cm 


Area of B  9 cm 
p.128
Level 1
(b)
2

Height of the ice-cream in the cone is 5 cm.
9.4
pp.137  139
Level 1
1
h 3
1 
     h   h cm 3 
cm 3
3
48
4 
h 3 125


48
48
3
h  125
h5
(a)
3
9.5
Volume of the ice-cream circular cone
1.
(a)
Let h cm be the height of the ice-cream after it
Volume of A  6 cm 


Volume of B  10 cm 
27
125
Volume of A：Volume of B  27 : 125


72
3
Area and Volume (III)
3
4.
2
Volume of A  7 cm  343

 
Volume of B  2 cm 
8

Volume of A：Volume of B  343 : 8
11.
(a)

2
5.
18 cm 2
 3 cm 

 
x cm 2
 4 cm 
9 18

16
x
x  32
20 cm 2
 y cm 

 
45 cm 2
 9 cm 
2
4
 y
  
9
9
y 2

9 3
2
y  9
3
6
12.
Volume of the football
 5  125
  
Volume of the basketball  6 
216

Volume of the football：Volume of the
basketball  125 : 216
Time taken to walk around the garden is proportional to
the perimeter of the garden.
2

 Time taken to walk around garden A 
25

 
Time
taken
to
walk
around
garden
B
49


2
3
7.
Radius of the football：Radius of the
basketball  5 : 6
3
(b)
2
6.
25
 Radius of the football 

 
Radius
of
the
basketball
36


Radius of the football
5

Radius of the basketball 6
x cm 3
 2 cm 

 
64 cm 3
 4 cm 
1
x

8 64
x8



3 minutes
25

 
49
 Time taken to walk around garden B 
3 minutes
5

Time taken to walk around garden B 7
Time taken to walk around garden B

7
 3 minutes  4.2 minutes
5
Level 2
3
8.
 10 cm 
16 cm 3

 
54 cm 3
 y cm 
2
13.
(a)
3
 10 
8
  
y
27
 
10 2

y
3
3
y   10
2
 15
 Perimeter of ABCD  16

 
8
 Perimeter of QBRP 
Perimeter of ABCD
2

Perimeter of QBRP
1

Perimeter of ABCD：Perimeter of QBRP
 2 :1
(b)
Perimeter of the larger rectangle
5 2 cm
9.
Total surface area of the smaller pyramid  2 
 
Total surface area of the larger pyramid  5 
3
14.
Total surface area of the larger pyramid
10.
(a)
25
 40 cm 2  250 cm 2
4
Volume of the bus model  1 
 
Volume of the real bus
8
2
1
Perimeter of the larger rectangle  2  5 2 cm
 10 cm
2
40 cm 2
4

Total surface area of the larger pyramid 25


 Height of the smaller bottle 
216

 
729
 Height of the larger bottle 
Height of the smaller bottle 6

Height of the larger bottle 9

3
Height of the smaller bottle：Height of the
larger bottle  2 : 3
3
0.15 m3
1
1
  
Volume of the real bus  8  512

Volume of the real bus  512  0.15 m3  76.8 m3
2
(b)
Surface area of the smaller bottle  2 
4
  
Surface area of the larger bottle  3 
9
Surface area of the smaller bottle：Surface area of
the larger bottle  4 : 9
73
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
15.
Let r1 cm and r2 cm be the base radius of the metallic
17.
bar before and after heating.
r1 50

r2 52

16.
(a)
1
25  1
1

24
Area of A：Area of B  1 : 24
25
26
(b)
Percentage change of the volume
1
1
  r22  52    r12  50
3
3

100%
1 2
r1  50
3
 r  2 52 
  2  
 1  100%
 r1  50 
 26  2 26 
   
 1  100%
 25  25 
 12.5% (cor. to 3 sig. fig.)
24
49  25
24

24
1

1
Area of B：Area of C  1 : 1
18.

ABC ~ ADE （AAA）

Area of ABC  AB 


Area of ADE  AD 
 1 


1 4 
r22
 49%
r12

r2
7

r1 10
Percentage change of the radius
r r
 2 1  100%
r1
1
25

BDF ~ ADE （AAA）

Area of BDF  BD 


Area of ADE  AD 
2
 4 


 1 4 

2
2
 BD 


 AB  BD 
r

  2  1  100%
r
 1

 7

   1  100%
10


  30%
2
2
16
25
Area of BDF
Area of BFEC
Area of BDF

Area
of

ADE  Area of ABC 



  Area of BDF



Percentage change of the volume
4 3 4 3
r2  r1
3
 3
100%
4 3
r1
3
 r  3 
  2   1 100%
 r1 

16
25  1  16
16

8
2

1

Area of BDF：Area of BFEC  2 : 1

 7  3 
    1 100%
 10 

  65.7%
© Hong Kong Educational Publishing Co.
2
 AB 


 AB  BD 
be the new radius.
4r22
 1  51%
4r12
(b)
Area of B
(2  1  2) 2  12

Area of C (1  2  1  2  1) 2  (2  1  2) 2

Let r1 be the radius of the original sphere and r2

Area of A
12

Area of B ( 2  1  2) 2  12

1
Volume of the bar after heating    r22  52 cm 3
3
1
Volume of the bar before heating    r12  50 cm 3
3

(a)
74
Area and Volume (III)
2
19.
(a)
 Length of the base of the smaller pyramid 
9

 
64
 Length of the base of the larger pyramid 
(b)
Area in contact with the water after addition of water
Area in contact with the water before addition of water
Length of the base of the smaller pyramid 3

Length of the base of the larger pyramid 8
Volume of the smaller pyramid  3 
 
Volume of the larger pyramid  8 

2
9
 12 
  
8
4
 
3

27
512
21.
(a)
= Volume of the larger pyramid 
Increase in the volume of water
Volume of the larger cuboid
 Increase in the volume of water
Volume of the smaller pyramid
= Volume of the larger pyramid 
27
 Volume of the larger pyramid
512
485
 Volume of the larger pyramid
=
512
485
 Volume of the larger pyramid  776 cm 3
512
Volume of the larger pyramid 
 Volume of the smaller cuboid
 1680 cm 3  60 cm 3
 1620 cm 3
3
 Height of the smaller cuboid 
60
1

 

 Height of the larger cuboid  1620 27
Height of the smaller cuboid 1

Height of the larger cuboid
3
512
 776 cm 3
485
2
Surface area of the smaller cuboid  1 
1
  
Surface area of the larger cuboid
9
 3

Ratio of the surface area  1 : 9
 819.2 cm 3
Volume of the smaller pyramid 
27
 819.2 cm 3
512
2
(b)
 43.2 cm 3

smaller pyramid respectively.
1
 90  h2  43.2
3
h2  1.44

20.
(a)
The square of the ratio of the heights of the
their surface areas because the square of
the ratio of their decrease in heights is
h2 3

h1 8
equal to the ratio of their surface areas.
Hence they are still similar.
8
h1   1.44
3
 3.84
Height of the frustum  (3.84  1.44) cm
 2.4 cm
p.143
1.
B
Mark points V, A, B, O and M as shown in the figure,
in which M is the mid-point of AB.
Since VAB is an isosceles triangle,
VM  AB .
Let x cm3 be the extra water needed.
In VOM ,
VO  3 cm
1
OM   8 cm  4 cm
2
VM 2  VO 2  OM 2 (Pyth. theorem)
3
360
 8 
 
360  x  12 
360
8

360  x 27
360  27  360  8  8 x
8 x  360  19
x  855

Surface area of the smaller cuboid
Surface area of the larger cuboid
remaining parts is equal to the ratio of
1.44 3

h1
8

2
1
 0.5 cm   1 

   
9
 1.5 cm   3 
Let h1 and h2 be the height of the larger and the

94
100%  125%
4
 15  10  11.2 cm 3  1680 cm 3
Volume of the frustum
(b)
Percentage increase 
VM  32  42 cm•
 5 cm
1
Area of VAB   8 cm  5 cm  20 cm
2

Total surface area  (8  8  20  4) cm 2  144 cm 2
The extra water needed is 855 cm3.
75
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
2.
B
Curved surface area    16 17 cm 2
2
 136 cm 2
3.
Area of APQ
Area of PBCQ
(b)

Area of APQ
Area of ABC  Area of APQ
4
94
4

5

Area of APQ：Area of PBCQ  4 : 5

B
2
Surface area  4     12  cm 2
 2
 144 cm 2
 452 cm 2 (cor. to 3 sig. fig.)
4.
C
5.
D
pp.144 – 148
Level 1
Volume of the smaller circular cone  3 
 
Volume of the larger circular cone
9

3
1.
1
27
 1200 cm 3
V cm 3
1

Volume of the larger circular cone 27
Volume of the larger circular cone  27V cm
6.
2.
3
1
3
Volume   (9  9) 10 cm
3
 270 cm 3

Let r cm be the base radius of the circular cone and
Mark points V, A, B, C, M and N as shown in the figure,
VM  AB and VN  BC
1
1
MB  AB  (10) cm  5 cm
2
2
1
1
NB  BC  (8) cm  4 cm
2
2
In VBM ,

2
(Pyth. theorem)
 25 
h  52   
 18 
2
VM 2  MB 2  VB 2 (Pyth. theorem)
VM  152  52 cm
 4.80 (cor. to 3 sig. fig.)
8.

The height of the circular cone is 4.80 cm.
(a)
APQ ~ ABC （AAA）
Area of APQ  AP 


Area of ABC  AB 

 2 


 2 1
 200 cm
1
Area of VAB  10  200 cm 2
2
 5 200 cm 2
2
In VBN ,
VN 2  NB 2  VB 2 (Pyth. theorem)
 AP 


 AP  PB 
2
VN  152  4 2 cm
 209 cm
2

Area of VBC 
1
 8  209 cm 2
2
 4 209 cm 2
4
9
Area of APQ：Area of ABC  4 : 9


The side length of the base is 2.5 cm.
in which M and N are the mid-points of AB and BC.
VAB and VBC are isosceles triangles.

h cm be the height.
100
2r  2  5 
360
25
r
18
 25 
h 2     52
 18 
Let a cm be the side length of the base of the pyramid.
1 2
 a  6  12.5
3
a 2  6.25
a  2.5
3.
7.
1 1

3
Volume of the pyramid    15  20   24 cm
3 2


Total surface area of all lateral faces
 (5 200  2  4 209  2) cm 2
 257 cm 2 (cor. to 3 sig. fig.)
© Hong Kong Educational Publishing Co.
76
Area and Volume (III)
4.
Let h cm be the height of the circular cone.
h 2  5 2  132 (Pyth. theorem)
4
  4 3 cm 3
3
256

 cm 3
3
 268 cm 3 (cor. to 3 sig. fig.)
(b)
Volume 
(a)
Let r cm be the radius of sphere C.
Surface area of sphere A  4  9 2 cm 2
h  132  5 2
 12
1
2
3
Volume of the cone     5 12 cm
3
 100 cm 3
5.
9.
 324 cm 2
Surface area of sphere B  4 12 2 cm 2
Let r cm be the base radius of the cone.
1
 r 2  16  400
3
1200
r2 
16
75


r

 900
r 2  225
r  15

75

Base circumference of the cone
(b)
75
 2 
cm  30.7 cm (cor. to 3 sig. fig.)

6.
Let r cm be the base radius of the cone and l cm be the
slant height.
r 2  576
r 2  576
r  24
l 2  24 2  7 2 (Pyth. theorem)
4
  9 3 cm 3
3
 972 cm 3
4
Volume of sphere B   12 3 cm 3
3
 2304 cm 3
4
Volume of sphere C   153 cm 3
3
 4500 cm 3

10.
 600 cm 2
2
Total surface area  (576  600) cm
 1176 cm 2
Base area    4.5 2 cm 2
 20.25 cm 2
Curved surface area  1  4    4.5 2 cm 2
2
 40.5 cm 2
2

Total surface area  (20.25  40.5) cm
11.
(a)
2
(b)
1
(c)
3
(a)
 60.75 cm 2
8.
(a)
Let r cm be the radius of the sphere.
4r 2  64
18 cm 2  6 cm 


x cm 2
 8 cm 
18 9

x 16
16
x
 18
9
 32
2
3
(b)
r 2  16
r4

Total volume of the 3 spheres
 (972  2304  4500) cm 3
 7776 cm 3
 25
Curved surface area    24  25 cm 2
7.
Radius of sphere C is 15 cm.
Volume of sphere A 
l  24 2  7 2

 576 cm 2
4r  324  576
2
16 cm 3
 y cm 

 
54 cm 3
 12 cm 
3
8
 y
  
27
 12 
y
2

12 3
2
y   12
3
8
The radius of the sphere 4 cm.
77
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
12.
Let r1 and r2 be the radii of the 2 spheres respectively.
 r1

r
 2
(b)
2
 16
 

9

r1 4

r2 3
15.
(a)
Ratio of their volumes  r : r
3
1
 4 :3
The following figure shows the base of the right
pyramid, mark the mid-point of AB as D and join
3
2
3
1
Volume of the pyramid   32  24  21 cm 3
3
 5376 cm 3
CD.
3
 64 : 27
13.
(a)
Let h cm be the height of the smaller pyramid.
3
16
h
  
9
250
 
3
8
h
  
 9  125
h 2

9 5
2
h  9
5
 3 .6

(b)
Since ABC is an equilateral triangle, CD  AB .
1
1
AD  AB  (6) cm  3 cm
2
2
AD 2  CD 2  AC 2 (Pyth. theorem)
CD  6 2  32 cm

Surface area of the smaller pyramid
 3.6 2 : 9 2
Surface area of the larger pyramid
 15.6 cm 2 (cor. to 3 sig. fig.)
Level 2
(a)
1
 6  3 3 cm 2
2
 9 3 cm 2
 4 : 25
14.
 3 3 cm
Base area of the pyramid

The height of the smaller pyramid is 3.6 cm.
(b)
Let h cm be the slant height of the pyramid.
Height of the lateral surface  h 2  32
1
9 3  3   6  h 2  32  2  (4  3  2  3  4  2)
2
h  5.04 cm
Mark points V, A, B, C, D, M, N and O as shown
in the figure.
(cor. to 3 sig. fig.)

16.
M is the mid-point of AB and O is the centre of
ABCD.
MB 
1
1
AB  (16) cm  16 cm
2
2
VM  MB  VB
2
2
The height of the smaller pyramid is 19.2 cm.
1
Volume of the smaller pyramid   8  8 19.2 cm 3
3
 409.6 cm 3
1
Volume of the larger pyramid  10 10  24 cm 3
3
 800 cm 3
3

Volume of the frustum  (800  409.6) cm
(Pyth. theorem)
VM  29  16 2 cm
2
 585 cm
1
1
OM  BC  (24) cm  12 cm
2
2
VO 2  OM 2  VM 2 (Pyth. theorem)
 390.4 cm 3
VO  585  12 2 cm
 21 cm

The height of the pyramid is 21 cm.
© Hong Kong Educational Publishing Co.
Let h cm be the height of the smaller pyramid.
h
8

24 10
8
h   24
10
 19.2

Since VAB is an isosceles triangle, VM  AB .
2
(a)
The slant height of the pyramid is 5.04 cm.
78
Area and Volume (III)
(b)

Mark points V, A, B, C, D, E, F, G and H. Join
Lateral surface area of the frustum
 (500  320) cm 2  180 cm 2
HF as shown in the figure.

Total surface area of the frustum
 (10 10  8  8  180) cm 2
 344 cm 2
17.
(a)
Volume of the larger cone  1    16 2  24 cm 3
3
 2048 cm 3
Volume of the cone above the water  24  6 


Volume of the larger cone
 24 
HF 2  HE 2  EF 2 (Pyth. theorem)

HF  10 2  10 2 cm
3
Volume of the cone above the water
3
 18 
    2048 cm 3
 24 
 864 cm 3
 200 cm
VF  VH 2  HF 2 (Pyth. theorem)
2

VF  24 2  200 cm
 776 cm
Volume of the frustum
 (2048  864) cm 3
 1184 cm 3
VE 2  VH 2  HE 2 (Pyth. theorem)
VE  24 2  10 2 cm
(b)
 676 cm
 26 cm
In VEF ,
VE 2  EF 2  676  10 2
Volume of the cylinder below the water level
  16 2  6 cm 3
 1536 cm 3

Volume of the water  (1536  1184) cm 3
 352 cm 3
 676  100
Height of the water level after the cone is
352 
cm
taken out 
  16 2
 1.375 cm
 776
VF  776
VE 2  EF 2  VF 2
2


VEF  90 (converse of Pyth. theorem)
1
Area ofVEF   VE  EF
2
1
  26  10 cm 2
2
 130 cm 2
1
Area of VHE   HE  VH
2
1
  10  24 cm 2
2
 120 cm 2


18.
Let l cm be the slant height of the cone.
Curved surface area of the cone  360 cm 2

rl  360

360
l
r
Curved surface area of the cone = Area of the sector

rl  l 2 
360
2
VHE  VHG and VEF  VGF
Area of VHG  Area of VHE
Area of VGF  Area of VEF


Lateral surface area of the larger pyramid
 (130  2  120  2) cm 2
 500 cm 2
Lateral surface area of the smaller pyramid  VD 


Lateral surface area of the larger pyramid  VH 

 360 

 360
 
 r  360
360 2 
 360π
360r 2
  r2
19.
If the sphere is completely covered by water, then the
2
Lateral surface area of the smaller pyramid  19.2 cm 


500 cm 2
 24 cm 
height of the water level is 6 cm.
Volume of extra water required
3


4
6
 2110  6        400 cm 3
3
2
 


2
Lateral surface area of the smaller pyramid  320 cm 2
 747 cm 3 (cor. to 3 sig. fig.)
79
© Hong Kong Educational Publishing Co.
9
New Progress in Junior Mathematics 3B
20.
(a)
Let h cm be the height of the cone.
1
2 1 4
  9 2  h      93
3
3 2 3
h  12

(b)
(b)
The height of the cone is 12 cm.
Total surface area of the solid
= Curved surface area of the cone + Curved
Percentage change in the volume
4 3 4 3
r2  r1
3
 3
100%
4 3
r1
3
 r  3 
  2   1 100%
 r1 

3
 (1.35  1) 100%
 146%
surface area of the hemisphere
1


    9  9 2  12 2   4    9 2  cm 2
2


22.
 297  cm 2
(a)
(cor. to 3 sig. fig.)
Volume of VABC
27
27


Volume of VDEF 27  37 64
3
(c)

Volume of the solid
 Volume of the cone  Volume of the hemisphere
1 4
1

     9 2  12      9 3  cm 3
3
2
3


 810 cm

3
Let r cm be the radius of the sphere formed.
4 3
r  810
3
r 3  607.5
(b)
Total surface area of VABC  VA 


Total surface area of VDEF  VE 
180 cm 2
3
 
Total surface area of VDEF  4 
r  3 607.5
Total surface area of the sphere  4  (3 607.5 ) 2

27
 VA 

 
VE
64


VA 3

VE 4
VA : AE  3 : (4  3)
 3 :1
Total surface area of VDEF 
Percentage change in the total surface
4  (3 607.5 ) 2  297
100%
297
 3.39% (cor. to 3 sig. fig.)
(a)


Let r1 be the original radius of the balloon and r2
 (320  180  45) cm 2
 185 cm 2
r1
1

r2 1.5  0.9
1
1.35
r2  1.35r1
23  26

Percentage change in the radius
r r
 2 1  100%
r1
1.35r1  r1
 100%
r1
 35%
© Hong Kong Educational Publishing Co.
VABC is a regular tetrahedron.
Area of ABC  1  180 cm 2
4
 45 cm 2
 Total surface area of VABC  Area of ABC
r12
1

r22 225%  81%

16
 180 cm 2
9
Total surface area of ABCDEF
 Total surface area of VDEF
be the radius after the change.
4r12  (1  125%)  (1  19%)  4r22

2
 320 cm 2
area 
21.
2
80
(HKCEE Questions)
Area and Volume (III)
Extended Question
27.
Open-ended Question
Volume of the cylinder    6 2  3 cm 3
28.
 108 cm 3
Volume of the circular cone  1   6 2  12 cm 3
3
 144 cm 3

Capacity of the container  (108  144) cm 3
Let r1 and r2 be the base radii of the 2 right circular
cones formed by A and B respectively.
For A,
πr118  π182 
330
360
r1  16.5

 252 cm 3
Volume of water in the container  1  252 cm 3
2
 126 cm 3
126 cm 3  108 cm 3

The water level is higher than the height of the
Using Pythagoras’ theorem,
cylinder when it is placed vertically.
Height of container A = 182 16.52 cm  51.75 cm
For B,
πr2 20  π 20 2 
r2 
Let h cm be the height of the water level in the cone.
1

Volume of the water   Capacity of the container
2

Volume of the empty space in the cone
 1
 1    Capacity of the container
2


3
Consider   50 .
Let A and B be the name of the containers after
decreasing the angle of the sector by   , and r3 and r4
Volume of the empty space in the cone
Volume of the cone
be their radii respectively.
 12  h  126π cm 3

 
3
 12  144π cm
For A,
πr318  π18 2 
h  0.522

2
 100 
= 20 2  
 cm  16.630 cm
 9 
1
Volume of container A  π16.5 2  51.75 cm 3
3
2
3

Height of container B
1  100 
3
Volume of container B  π
 16.630 cm
3  9 
 2150 cm 3
 126 π cm 3

100
9
 2051 cm 3
1
 252 π cm 3
2
 Height of the empty space in the cone 


Height of the cone


200
360
330  50
360
r3  14
Height of the water level now
 3 cm  0.522 cm
 3.52 cm (cor. to 3 sig. fig.)
For B,
πr4 20  π 20 2 
r2 
200  50
360
25
3
Using Pythagoras’ theorem,
Height of container A = 182 142 cm  128 cm
2
 25 
Height of container B = 20 2    cm  18.181 cm
 3 
1
Volume of container A  π14 2  128 cm 3
3
 2322 cm 3
2
1  25 
Volume of container B  π  18.181 cm 3
3  3 
 1322 cm 3
Value of   50 (or other reasonable answers)
81
© Hong Kong Educational Publishing Co.
9