7 Areas and Volumes (III)

7 Areas and Volumes (III) 4. Volume of the prism    8  6   7 cm3

1
2 
Quick Review  168 cm3
Total surface area of the prism
Let’s Try (p. 7.3)  1  
120  (8  6  10)  7    8  6   2 cm 2
1. CD   2  9 cm  2  
360
 6 cm  216 cm 2

120
Area of the sector     92 cm 2 5. Volume of the cylinder    6 2  14 cm3
360
 27 cm 2  504 cm3
Total surface area of the cylinder
1  (2    6  14  2    6 2 ) cm 2
2. (a) Base area of the prism  12  5 cm 2
2  240 cm 2
 30 cm 2
Volume of the prism  30  30 cm 3 6. ∵ ABCD ~ EFGH
 900 cm 3 EH HQ
∴ 
AD DP
(b) Total area of all lateral faces of the prism x 8

 (12  5  13)  30 cm 2 9 6
x  12
 900 cm 2
Total surface area of the prism  (900  30  2) cm 2
7. ∵ △ABC ~ △ADE
 960 cm 2
BC AB
∴  (corr. sides, ~△s)
DE AD
3. ∵ △ABC ~ △EFG y 10  5
EH FG 
∴  (corr. sides, ~△s) 12 10
AD BC y 15
EH 12 cm 
 12 10
6 cm 10 cm y  18
EH  7.2 cm

Review Exercise 7 (p. 7.5)

Activity
135
1. AB   2  8 cm Activity 7.1 (p. 7.9)
360 1. 6
 6 cm
135 2. (a) (i) 6V
Area of the sector     82 cm 2 (ii) 2Ah
360
 24 cm 2 (b) From (a)(i) and (ii),
6V  2 Ah
2. Reflex AOB  360  40 1
∴ V  Ah
 320 3
320
AB   2  9 cm
360 Activity 7.2 (p. 7.26)
 16 cm 1. (a) The curved surface area of the cone is equal to the
area of the sector VAB.
320
Area of the sector     92 cm 2
360 
(b)   2
 72 cm 2 360

3. Volume of the cuboid  5  3  8 cm 3

2. (a) (i)
   2 
AB 

 120 cm 3
Total surface area of the cuboid (ii)
AB  2360
r
 (5  8  5  3  3  8)  2 cm 2
 158 cm 2

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Junior Secondary Mathematics in Action 3B Full Solutions

(b) From (a)(i) and (ii),

 To Learn More
 2   2 r
360
 2 r To Learn More (p. 7.58)

360 2  Area of △ ABM BM
1. 
 r Area of △ AMC MC
∴  Area of △ ABM 3
360  
18 cm 2 1
3. Curved surface area of the right circular cone Area of △ ABM  3  18 cm 2
  54 cm 2
   2
360
r Area of △ ABM AM
   2 2. 
 Area of △ BCM BC
  r
Area of △ ABM AM

60 cm 2 AD
Activity 7.3 (p. 7.46) Area of △ ABM 2
1. The area of the curved surface of the hemisphere is twice 
60 cm 2
23
that of its flat surface.
2
Area of △ ABM   60 cm 2
2. (a) r 2 5
 24 cm 2
(b) 2r 2
Classwork
3. 4r 2
Classwork (p. 7.10)
Activity 7.4 (p. 7.57) 1
1. Ratio of 1. Volume of the pyramid   36  5 cm3
lengths of 3
Similar plane  60 cm3
corresponding Ratio of areas
figures
sides (or line
segments)
1
1  2. Volume of the pyramid   54  7 cm3
 (kb)( kh)  3
2   k2
Triangles k, k  126 cm3
1 
 bh 
2 
1
(kr ) ( (kr )2 ) 3. Volume of the pyramid   65  6 cm3
Circles k  k2 3
(r ) ( r 2 )
 130 cm3
2
 2 
2.   Quick Practice
 
 1 
Quick Practice 7.1 (p. 7.11)
Activity 7.5 (p. 7.60) 1
Volume of the pyramid   (15  7)  12 cm
3
1. Ratio of 3
lengths of Ratio of
Similar Ratio of  420 cm3
corresponding base
solids volumes
sides (or line areas
segments)
Quick Practice 7.2 (p. 7.11)
((ka)3 )
Cubes k k2  k3 Let h cm be the height of the pyramid.
(a 3 )
1
( (kr ) 2 (kh))  (4  4)  h  48
( (kr ) 2 ) 3
(r 2 h)
Cylinders k, k (r 2 ) 16
 k3 h  48
 k2 3
h9
∴ The height of the pyramid is 9 cm.
2. (a) 2
Quick Practice 7.3 (p. 7.12)
(b) 3
(a) In △ABD,
BD2  BA2  AD 2 (Pyth. theorem)
BD  12  16 cm
2 2

 20 cm
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 7 Areas and Volumes (III)

1 (b) In △VON,
OB  BD
2 VN 2  VO 2  ON 2 (Pyth. theorem)
1 2
  20 cm  20 
2 VN  ( 189) 2    cm
 10 cm  2 
In △VOB,  189  102 cm
VO 2  OB 2  VB 2 (Pyth. theorem)  17 cm
1
VO  ( 181) 2  10 2 cm Area of △VBC   BC  VN
2
 9 cm 1
∴ The height of the pyramid is 9 cm.  12 17 cm 2
2
1  102 cm 2
(b) Volume of the pyramid   (16  12)  9 cm 3
3
 576 cm 3 (c) Total surface area of pyramid VABCD
 2  area of △VAB  2  area of △VBC  base area
Quick Practice 7.4 (p. 7.15)  (2  150  2  102  20  12) cm 2
(a) ∵ △VAE  △VBE (RHS)  744 cm 2
∴ AE = BE (corr. sides,  △s)
AB Quick Practice 7.6 (p. 7.24)
∴ BE 
2 1
12 Volume of the cone     6 2  13 cm 3
 cm 3
2  156 cm 3
 6 cm
AB = DA (definition of square)
= 12 cm Quick Practice 7.7 (p. 7.25)
In △VBE, (a) In △OVA,
OV 2  OA2  AV 2 (Pyth. theorem)
VE 2  BE 2  VB 2 (Pyth. theorem)
OA  82  7 2 cm
VE  10 2  6 2 cm
 8 cm  15 cm
1  3.87 cm (cor. to 3 sig. fig.)
Area of △VAB   AB  VE
2 ∴ The base radius of the cone is 3.87 cm.
1
  12  8 cm 2 1
2 (b) Volume of the cone     ( 15) 2  7 cm3
 48 cm 2 3
 110 cm3 (cor. to 3 sig. fig.)
(b) Total surface area of pyramid VABCD
 4  area of △VAB  base area Quick Practice 7.8 (p. 7.25)
 (4  48  12 12) cm 2 Let h cm be the height of the cone.
Volume of the cone  volume of the cylinder
 336 cm 2
1
   82  h    82  7
3
Quick Practice 7.5 (p. 7.16) h  21
(a) In △VOM, ∴ The height of the cone is 21 cm.
VM 2  VO 2  OM 2 (Pyth. theorem)
2 Quick Practice 7.9 (p. 7.27)
 12 
VM  ( 189) 2    cm Curved surface area of the cone    4  6 cm 2
 2
 24 cm 2
 189  62 cm
Base area of the cone    4 2 cm 2
 15 cm
1  16 cm 2
Area of △VAB   AB  VM
2 ∴ Total surface area of the cone  (24  16 ) cm 2
1  40 cm 2
  20  15 cm 2
2
 150 cm 2

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Junior Secondary Mathematics in Action 3B Full Solutions

Quick Practice 7.10 (p. 7.28) (c) Volume of frustum ABCDHEFG

(a) In △VOA,  volume of pyramid VABCD
VO 2  OA2  VA2 (Pyth. theorem)  volume of pyramid VEFGH
OA  17  15 cm 2 2  (168.75  50) cm3
 8 cm  118.75 cm3
∴ The base radius of the cone is 8 cm.
Quick Practice 7.13 (p. 7.38)
(b) Total surface area of the cone  (  8  17    8 2 ) cm 2 (a) (i) Let NA = x cm.
 200 cm 2 ∵ △VNA ~ △VMB (AAA)
NA VN
∴  (corr. sides, ~△s)
Quick Practice 7.11 (p. 7.29) MB VM
(a) Let r cm be the base radius of the cone. x 12

1 10 12  12
100     r 2  12
3 24 x  120
r 2  25 x5
r 5 ∴ NA  5 cm
∴ The base radius of the cone is 5 cm. 1
(ii) Volume of cone VAD     5  12 cm
2 3
3
(b) Denote the centre of the base of the cone by O.
 100 cm 3
1
Volume of cone VBC     10 2  (12  12) cm 3
3
 800 cm 3
Volume of the frustum  (800  100 ) cm3

In △VOA,  700 cm3

AV 2  VO 2  OA2 (Pyth. theorem)
(b) (i) In △VNA,
AV  12  5 cm
2 2
VA2  VN 2  NA2 (Pyth. theorem)
 13 cm
VA  122  52 cm
 13 cm
(c) ∵ Area of sector VAB = curved surface area of the cone
 In △VMB,
∴    132    5  13
360 VB 2  VM 2  MB 2 (Pyth. theorem)
  138.46...... VB  (12  12)2  102 cm
∵   140
 26 cm
∴ Daisy’s claim is agreed.
(ii) Curved surface area of cone VAD    5  13 cm
2

Alternative Solution  65 cm2

∵ Arc length of sector VAB
Curved surface area of cone VBC    10  26 cm
2
= circumference of the base of the cone
  260 cm 2
∴  2  13  2  5
360 Curved surface area of the frustum
  138.46......  (260  65 ) cm2
∵   140  195 cm2
∴ Daisy’s claim is agreed. Total surface area of the frustum
 (195    52    102 ) cm2
Quick Practice 7.12 (p. 7.36)
1  320 cm2
(a) Volume of pyramid VEFGH   (5  5)  6 cm 3
3
 50 cm3 Quick Practice 7.14 (p. 7.43)
4
Volume of the sphere     43 cm 3
1 3
(b) Volume of pyramid VABCD   (7.5  7.5)  (6  3) cm3
3  268 cm 3 (cor. to 3 sig. fig.)
 168.75 cm3

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 7 Areas and Volumes (III)

Quick Practice 7.15 (p. 7.44) 1

Let r cm be the radius of the sphere.
(b) Inner curved surface area   4  82 cm 2
2
∵ Volume of the sphere  112 cm3  128 cm 2
4 3
Area of the ring  ( 10    8 ) cm
2 2 2
∴ r  112
3
84  36 cm 2
r3  Total surface area of the mould

r  2.99 (cor. to 3 sig. fig.)  (200  128  36 ) cm 2
∴ The radius of the sphere is 2.99 cm.  364 cm 2

Quick Practice 7.16 (p. 7.45) Quick Practice 7.21 (p. 7.56)
Volume of the hemisphere  1  4    33 mm 3 Let x cm2 be the area of the smaller octagon.
2 3 x 4
2

 18 mm 3  
98  7 
Volume of the cylinder    32  22 mm 3
x 16

 198 mm 3 98 49
Volume of the bullet  (18  198 ) mm 3 16
x  98
 216 mm 3 49
 32
∴ The area of the smaller octagon is 32 cm2.
Quick Practice 7.17 (p. 7.45)
Let r cm be the base radius of the vessel.
Quick Practice 7.22 (p. 7.56)
∵ Volume of the sphere
Let x cm be the length of a side of the larger hexagon.
= volume of water above the original water level 2
4  x  25
∴    53    r 2  2   
3 21
  9
250 x 25
r2  
3 21 9
r  9.13 (cor. to 3 sig. fig.) 5
x  21
∴ The base radius of the vessel is 9.13 cm. 3
 35
Quick Practice 7.18 (p. 7.47) ∴ The length of a side of the larger hexagon is 35 cm.
2
 20 
Surface area of the sphere  4      cm 2
 2  Quick Practice 7.23 (p. 7.57)
(a) Let x cm2 be the area of quadrilateral ABED.
 400 cm 2
∵ △DEF ~ △ABF (AAA)
2
Area of △ DEF  DF 
Quick Practice 7.19 (p. 7.47) ∴  
Total surface area of the hemisphere Area of △ ABF  AF 
2
= curved surface area + area of the flat surface 15  1 
 
1  x  15  3  1 
   4 r 2   r 2  cm 2
2  15 1

 (2 r   r ) cm 2
2 2
x  15 16
 3 r 2 cm 2 240  x  15
∵ Total surface area of the hemisphere  75 cm2 x  225
3 r 2  75 ∴ The area of quadrilateral ABED is 225 cm2.
∴
r 2  25
(b) Let y cm2 be the area of △CEB.
r 5
∵ △CEB ~ △DEF (AAA)
2
Area of △CEB  CB 
Quick Practice 7.20 (p. 7.48) ∴  
(a) Let r cm be the outer radius of the mould. Area of △ DEF  DF 
∵ Outer curved surface area  200 cm2 y  AD 
2

 
1 15  DF 
∴  4πr 2  200π
2 y  3
2

r 2  100  
15  1 
r  10 y
∴ Inner radius of the mould  (10  2) cm 9
15
 8 cm y  135

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Junior Secondary Mathematics in Action 3B Full Solutions

Area of parallelogram ABCD Alternative Solution

 area of △CEB  area of quadrilateral ABED 1 2 1 1
 r h    R2H
 (135  225) cm 2 3 18 3
 360 cm 2 h 1 R2
  2
H 18 r
2
Quick Practice 7.24 (p. 7.61) h 1 R
Let x cm2 be the surface area of the larger sphere.   
H 18  r 
2
x  16  1
   9
40.5  9  18
x 256 1
 
40.5 81 2
256 r 1 h 1 r h
x  40.5 ∵  ,  and 
81 R 3 H 2 R H
 128 ∴ The two cones are not similar.
∴ The surface area of the larger sphere is 128 cm2. ∴ Percy’s claim is disagreed.

Quick Practice 7.25 (p. 7.63)

Consolidation Corner
Let A1 and A2 be the total areas of all lateral faces of pyramid A
and the whole pyramid respectively.
2 Consolidation Corner (p. 7.13)
A1  6  1
  1. (a) V  Ah
A2  6  3  3
2
2 1
    8  6 cm3
3 3
4  16 cm3

9
9 1
∴ A2  A1 (b) V  Ah
4 3
Total area of all lateral faces of pyramid A A1 1
∴    15  7 m3
Total area of all lateral faces of frustum B A2  A1 3
A1  35 m3

9
A1  A1
4 1
(c) V Ah
A 3
 1
5 3V
A1 h
4 A
4 3(40)
  cm
5 12
∴ The ratio of the total area of all lateral faces of pyramid A  10 cm
to that of frustum B is 4 : 5.

Quick Practice 7.26 (p. 7.64) 1

(d) V  Ah
r
2
1 3
(a)    3V
R
  9 A
h
r 1
 3(65)
R 3  cm 2
13
∴ r : R  1: 3
 15 cm 2

3 3
 r  1 1 1 1 
(b) ∵      2. (a) Volume of the pyramid     5  9   10 cm3
 R   3 27 3 2 
volume of the smaller cone 1  75 cm3
and 
volume of the larger cone 18
3
Volume of the smaller cone  r  (b) Volume of the pyramid 
1
∴    (11  7)  15 cm3
Volume of the larger cone  R  3
∴ The two cones are not similar.  385 cm3
∴ Percy’s claim is disagreed.

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 7 Areas and Volumes (III)

3. (a) 1
Area of △VAB   AB  VH
2
1
  18  13 cm 2
2
 117 cm 2
∵ △VBK  △VCK (RHS)
∴ BK = CK (corr. sides,  △s)
BC
With the notations in the figure, ∴ BK 
in △VOH, 2
1 10
OH  BC  cm
2 2
1  5 cm
  16 cm In △VBK,
2
 8 cm VK 2  BK 2  VB 2 (Pyth. theorem)
VO 2  OH 2  VH 2 (Pyth. theorem) VK  ( 250) 2  52 cm
VO  17 2  82 cm  15 cm
 15 cm 1
Area of △VBC   BC  VK
∴ The height of the pyramid is 15 cm. 2
1
  10  15 cm 2
1 2
(b) Volume of the pyramid   (16 16)  15 cm3
3  75 cm 2
 1280 cm3 Total surface area of the pyramid
 2  area of △VAB  2  area of △VBC  base area
Consolidation Corner (p. 7.17)  (2  117  2  75  18  10) cm 2
1. (a) Total surface area of the pyramid  564 cm 2
 4  area of △VBC  base area
 (4  80  13 13) cm 2 2.
 489 cm 2

(b) Total surface area of the pyramid

 2  area of △VAD  2  area of △VCD  base area
 (2  100  2  208  32  10) cm2
 936 cm2

1 With the notation in the figure,

(c) Area of △VAB   AB  VH in △VOE,
2
1
1 OE  AB
  12  10 cm 2 2
2
1
 60 cm 2  3m
2
Total surface area of the pyramid
 1.5 m
 4  area of △VAB  base area
VE 2  VO 2  OE 2 (Pyth. theorem)
 (4  60  12  12) cm 2
 384 cm 2 VE  52  1.52 m
 27.25 m
(d) ∵ △VAH  △VBH (RHS) 1
Area of △VBC   BC  VE
∴ AH = BH (corr. sides,  △s) 2
AB 1
∴ AH    3  27.25 m 2
2 2
18 3
 cm  27.25 m 2
2 2
 9 cm ∴ Total surface area of the pyramid
In △VAH,  4  area of △VBC  base area
VH 2  AH 2  VA2 (Pyth. theorem)  3 
 4 27.25  3  3  m 2
VH  ( 250)  9 cm 2 2  2 
 13 cm  40.3 m 2 (cor. to 3 sig. fig.)

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Junior Secondary Mathematics in Action 3B Full Solutions

Consolidation Corner (p. 7.25) Consolidation Corner (p. 7.30)

1 1. (a) Curved surface area  r
1. (a) V  r 2 h
3    3  8 cm 2
1
    32  4 cm 3  24 cm 2
3
Total surface area  r  r
2
 12 cm 3
 (24    32 ) cm2
1  33 cm2
(b) V  r 2 h
3
1 (b) Curved surface area  r
    52  6 m3
3 curved surface area
 50 m 3 
r
48
 m
1  6
(c) V  r 2 h
3 8m
3V
Total surface area  r  r
2
h 2
r
3(16 )  (48    6 2 ) m 2
 cm
  22  84 m 2
 12 cm
(c) Curved surface area  r
1 curved surface area
(d) V  r 2 h r
3 
3V 105
r  cm
h  15
 7 cm
3(245 )
 cm
Total surface area  r  r
2
 15
 7 cm  (105    7 2 ) cm 2
 154 cm 2
2. (a) Let h cm be the height of the cone.
h 2  7 2  252 (Pyth. theorem)
(d) Total surface area   r   r
2

h  252  7 2
total surface area   r 2
 24 
r
∴ The height of the cone is 24 cm.
198    92
 cm
1  9
(b) Volume of the cone     7 2  24 cm3  13 cm
3
 392 cm3 Curved surface area  r
   9  13 cm2
1  117 cm2
3. Capacity of the cup     6 2  12 cm3
3
 144 cm3 2. Let  cm be the slant height of the cone.
2
 62  42 (Pyth. theorem)
Capacity of the cylindrical container    12  20 cm
2 3

 2880 cm3  6 4
2 2

2880  52
Number of cups of water needed 
144 ∴ The slant height of the cone is 52 cm .
 20
Curved surface area of the cone    4  52 cm 2
∴ 20 cups of water are needed to completely fill up the
cylindrical container.  4 52 cm 2
Base area of the cone    4 2 cm 2
 16 cm 2
∴ Total surface area of the cone
 (4 52  16 ) cm 2
 141 cm 2 (cor. to the nearest cm 2 )

8 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

240 2. (a) Let M and N be the mid-points of FG and BC

3. (a) Circumference of the base   2    12 cm respectively.
360
 16 cm
Let r cm be the base radius of the cone.
2    r  16
r 8
∴ The base radius of the cone is 8 cm.

(b) In △VOA,
VO 2  OA2  VA2 (Pyth. theorem)
VO  122  82 cm
 80 cm
Capacity of the cone ∵ △VQN ~ △VPM (AAA)
1 VQ QN
    82  80 cm3 ∴  (corr. sides, ~△s)
3 VP PM
 599.45 cm3 VQ 9 cm

8 cm 3 cm
 600 cm3
VQ  24 cm
∴ Kevin’s claim is disagreed.

Consolidation Corner (p. 7.38) 1

(b) Volume of pyramid VABCD   (18  18)  24 cm3
1 3
1. (a) Volume of cone VSR     32  4 cm3
3  2592 cm3
 12 cm3 1
Volume of pyramid VEFGH   (6  6)  8 cm
3

1 3
Volume of cone VPQ     62  (4  4) cm3
3  96 cm3
 96 cm3 Volume of the frustum  (2592  96) cm
3

Volume of the frustum  2496 cm3

 (96  12 ) cm 3
 84 cm 3
Consolidation Corner (p. 7.45)
3
4 8
(b) (i) In △VHR, 1. Volume of the sphere       cm3
3 2
VR 2  VH 2  HR 2 (Pyth. theorem)
256
 cm3
VR  42  32 cm 3
 5 cm
In △VKQ, 2. Let r cm be the radius of the hemisphere.
∵ Volume of the hemisphere = 150 cm3
VQ 2  VK 2  KQ 2 (Pyth. theorem)
1 4
VQ  (4  4) 2  62 cm
∴     r 3  150
2 3
 82  62 cm 225
r3 

 10 cm
r  4.15 (cor. to 3 sig. fig.)
(ii) Curved surface area of cone VSR ∴ The radius of the hemisphere is 4.15 cm.
   3  5 cm2
 15 cm2 3. Let r cm be the radius of the hemisphere.
∵ Volume of the hemisphere = volume of the sphere
Curved surface area of cone VPQ
1 4 4
   6  10 cm2 ∴     r 3     103
2 3 3
 60 cm2 r 3  2000
Curved surface area of the frustum
r  12.6 (cor. to 3 sig. fig.)
 (60  15 ) cm2
∴ The radius of the hemisphere is 12.6 cm.
 45 cm2
Total surface area of the frustum
 (45    32    6 2 ) cm 2
 90 cm 2

9 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

Consolidation Corner (p. 7.49)  y

2
32
1. Let r cm be the radius of the volleyball. 2. (a)   
8
  18
∵ Surface area of the volleyball  1385 cm2
y 16
∴ 4r 2  1385 
8 9
1385
r2  4
4 y  8
r  10.5 (cor. to 3
32
3 sig. fig.) 
3
∴ The radius of the volleyball is 10.5 cm.

2. Curved surface area of the hemisphere  y 125

2

1 4
2 (b)   
  4      cm2 18
  405
2 2
y 25
 8 cm2 
18 81
2
Area of the flat surface of the hemisphere      cm2
4 5
y  18
2 9
 4 cm2  10
Total surface area of the hemisphere  (8  4 ) cm
2
∴
 12 cm2 3. (a) ∵ △CBA ~ △CDE (AAA)
Perimeter of △CDE ED
∴ 
Perimeter of △ ABC AB
3. (a) Let r cm be the inner radius of the sphere.
Perimeter of △CDE 3
∵ Inner surface area of the sphere  729 cm2 
32 cm 2
∴ 4r 2  729 3
Perimeter of △CDE   32 cm
r 2  182.25 2
r  13.5  48 cm
∴ The inner radius of the sphere is 13.5 cm.
(b) ∵ △CBA ~ △CDE (AAA)
(b) Outer radius of the hollow sphere  (13.5  1.5) cm 2
Area of △ ABC  AB 
 15 cm ∴  
Area of △CDE  ED 
Volume of the hollow sphere 2
Area of △ ABC 2
4 4   
     153     13.53  cm3 180 cm 2  3
3 3 
Area of △ ABC 4
 3831 cm3 (cor. to the nearest cm3 ) 
180 cm 2 9
4
Area of △ ABC   180 cm 2
Consolidation Corner (p. 7.59) 9
2
x 8  80 cm 2
1. (a)  
252  12 
x 2
2 Consolidation Corner (p. 7.65)
  3
252  3   x  125
1. (a)   
4 8 64
x   252
9 x 3 125

 112 8 64
5
x  8
x  28 
2 4
(b)    10
40  35 
2
x 4
   x
2
36
40  5  (b)   
16 9 81
x  40
25 x 36

 25.6 9 81
x 6

9 9
x6

10 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

2. (a) Let h cm be the height of the smaller cone. 1 1 

3 5. Volume of the pyramid     9  8   15 cm3
 h 8 3 2 
  
15
  27  180 cm3
h 3 8

15 27 6. Let h cm be the height of the pyramid.
2 1
h   15  (28  16)  h  3584
3 3
 10 448h
   3584 
∴ The height of the smaller cone is 10 cm. 3
h  24
(b) Let A1 cm2 and A2 cm2 be the curved surface areas of ∴ The height of the pyramid is 24 cm.
the smaller cone and the larger cone respectively.
A1  10 
2 7. Let x cm be the side length of the square base.
  1
A2  15   ( x  x)  6  50
2
3
2   x 2  25 
 
3 x5
4 ∴ The side length of the square base is 5 cm.

9
9 8. Total surface area of the pyramid
∴ A2  A1
4  4  area of △VBC  base area
Curved surface area of the smaller cone  (4  28  7  7) cm2
∴
Curved surface area of the frustum  161 cm2
A1

A2  A1
9. Total surface area of the pyramid
A1  2  area of △VAD  2  area of △VCD  base area

9
A1  A1  (2  168  2  300  30  14) cm 2
4
A  1356 cm2
 1
5
A1
4 1
4 10. Area of △VAB   AB  VH
 2
5 1
∴ The ratio of the curved surface area of the   13  20 cm 2
2
smaller cone to that of the frustum is 4 : 5.  130 cm 2
∴ Total surface area of the pyramid
Exercise  4  area of △VAB  base area
Exercise 7A (p. 7.18)  (4  130  13  13) cm2
Level 1  689 cm2
1
1. Volume of the pyramid   36  5 cm3
3
1
 60 cm3 11. Area of △VAB   AB  VM
2
1
1   40  17 cm 2
2. Volume of the pyramid   (8  8)  6 cm3 2
3
 340 cm 2
 128 cm3
1
Area of △VBC   BC  VN
2
1 1
3. Volume of the pyramid   (12  7)  16 cm3   16  25 cm 2
3 2
 448 cm3  200 cm 2
∴ Total surface area of the pyramid
1  2  area of △VAB  2  area of △VBC  base area
4. Volume of the pyramid   (4  3)  7 cm3
3  (2  340  2  200  40  16) cm2
 28 cm3  1720 cm2

11 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

12. (a) In △VOE, Level 2

1 15. (a) In △VAB,
OE  AB
2 VA2  AB 2  VB 2 (Pyth. theorem)
1 VA  10  62 cm
2
  18 cm
2
 8 cm
 9 cm
VE 2  VO 2  OE 2 (Pyth. theorem)
(b) In △BAC,
VE  122  92 cm AB 2  AC 2  BC 2 (Pyth. theorem)
 15 cm
AC  10  62 cm 2

 8 cm
1 1 1
(b) Area of △VBC   BC  VE 
Volume of the pyramid     6  8   8 cm
3
2 3 2 
1
  18  15 cm 2  64 cm3
2
 135 cm 2
∴ Total surface area of the pyramid 16. (a) In △BDA,
 4  area of △VBC  base area
 ( 4  135  18  18) cm 2
 864 cm 2

13. (a) ∵ △VAE  △VBE (RHS)

∴ AE = BE (corr. sides, △s) BD 2  BA2  AD 2 (Pyth. theorem)
AB
∴ AE  BD  ( 32)  ( 32) 2 cm
2

2
20  8 cm
 cm 1
2 OD  BD
 10 cm 2
In △VAE, 1
  8 cm
VE 2  AE 2  VA2 (Pyth. theorem) 2
 4 cm
VE  262  102 cm In △VOD,
 24 cm

1
(b) Area of △VAB   AB  VE
2
1
  20  24 cm 2
2 VO 2  OD 2  VD 2 (Pyth. theorem)
 240 cm 2 VO  5  42 cm
2

∴ Total surface area of pyramid VABCD  3 cm

 4  area of △VAB  base area
 (4  240  20  20) cm2 1
(b) Volume of the pyramid   ( 32  32 )  3 cm 3
 1360 cm2 3
 32 cm 3
14. Total surface area of the gift box
 4  area of each lateral face  base area 17. (a) In △ABC,
 (4  12  4  4) cm 2 AC 2  AB 2  BC 2 (Pyth. theorem)
 64 cm 2
AC  4  4 cm
2 2

Total cost to make the gift box

 $0.2  64  32 cm
 $12.8  5.66 cm (cor. to 3 sig. fig.)
 $10
∴ The claim is disagreed.

12 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

(b) In △VEC, 20. (a) ∵ △VAH  △VBH (RHS)

VE  EC  VC
2 2 2
(Pyth. theorem) ∴ AH = BH (corr. sides,  △s)
AB
∴ HB 
2
 32 
VE  102    cm 2
 2  10
 cm
32 2
 100  cm  5 cm
4
In △VHB,
 92 cm
VB 2  VH 2  HB 2 (Pyth. theorem)
 9.59 cm (cor. to 3 sig. fig.)
VB  12  5 cm
2 2

1  13 cm
(c) Volume of the pyramid   (4  4)  92 cm3 ∵ △VBG  △VCG (RHS)
3
∴ BG = CG (corr. sides,  △s)
 51.2 cm3 (cor. to 3 sig. fig.)
BC
∴ BG 
2
18. (a) In △ABC, 8
AC 2  AB 2  BC 2 (Pyth. theorem)  cm
2
AC  162  122 cm  4 cm
 20 cm In △VBG,
1 VG 2  BG 2  VB 2 (Pyth. theorem)
OC  AC
2 VG  13  42 cm
2

1
  20 cm  153 cm
2
 10 cm  12.4 cm (cor. to 3 sig. fig.)
In △VOC,
VO 2  OC 2  VC 2 (Pyth. theorem) 1
(b) Area of △VAB   AB  VH
VO  12.5  10 cm 2 2 2
1
 7.5 cm   10  12 cm 2
2
 60 cm 2
1
(b) Volume of the pyramid   (16  12)  7.5 cm3 1
3 Area of △VBC   BC  VG
2
 480 cm3
1
  8  153 cm 2
2
19. (a) In △ABC,
 4 153 cm 2
AB 2  BC 2  AC 2 (Pyth. theorem) ∴ Total surface area of the pyramid
AB  32  242 cm
2
 2  area of △VAB  2  area of △VBC 
 448 cm base area
 21.2 cm (cor. to 3 sig. fig.)  (2  60  2  4 153  10  8) cm 2
 299 cm 2 (cor. to 3 sig. fig.)
(b) In △VEC,
1 21. (a) With the notations in the figure,
EC  AC
2
1
  32 cm
2
 16 cm
VE 2  EC 2  VC 2 (Pyth. theorem)
VE  26  162 cm
2


 420 cm
1
 20.5 cm (cor. to 3 sig. fig.) (i) MH  AB
2
1
1   12 cm
(c) Volume of the pyramid   ( 448  24)  420 cm3 2
3  6 cm
 3470 cm3 (cor. to 3 sig. fig.)

13 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

In △VMH, 1
(c) Volume of the pyramid   144  8 cm3
VH  VM  MH
2 2 2
(Pyth. theorem) 3
VH  8  6 cm
2 2  384 cm3
 10 cm
1 23. (a) Let h cm be the height of the pyramid.
∴ Area of △VBC   BC  VH ∵ Volume of the pyramid = 400 cm3
2
1 1
  8  10 cm 2 ∴  (10 10)  h  400
2 3
h  12
 40 cm 2
∴ The height of the pyramid is 12 cm.
1
(ii) MG  BC
2 (b)
1
  8 cm
2
 4 cm
In △VGM,
VG 2  VM 2  MG 2 (Pyth. theorem)
VG  8  4 cm
2 2

 80 cm
1 With the notations in the figure,
∴ Area of △VAB   AB  VG in △VOE,
2
1 VE 2  OE 2  VO 2 (Pyth. theorem)
  12  80 cm 2 2
2  10 
VE     122 cm
 6 80 cm 2  2
 53.7 cm 2 (cor. to 3 sig. fig.)  13 cm
1
Area of △VBC   BC  VE
(b) Total surface area of pyramid VABCD 2
 2  area of △VAB  2  area of △VBC  base area 1
  10  13 cm 2
2
 (2  6 80  2  40  12  8) cm 2
 65 cm 2
 283 cm 2 (cor. to 3 sig. fig.) ∴ Total surface area of the pyramid
 4  area of △VBC  base area
22. (a) Base area of the pyramid  12  12 cm
2
 (4  65  10 10) cm 2
 144 cm2  360 cm 2
∵ Total surface area of the pyramid
 4  area of the lateral face VBC  base area 24. (a) In △VBM,
∴ 4  area of the lateral face VBC  144 cm2 VM 2  BM 2  VB 2 (Pyth. theorem)
 384 cm 2
2
8
∴ 4  area of the lateral face VBC  240 cm2 VM  82    cm
2
1
∴ Area of lateral face VBC   240 cm2  48 cm
4
 60 cm2  6.93 cm (cor. to 3 sig. fig.)

1
1 (b) Area of each lateral face   8  48 cm 2
(b) ∵ Area of △VBC   BC  VE 2
2
1  4 48 cm 2
60 cm2   12 cm  VE Total surface area of the tetrahedron
2
VE  10 cm  4  area of each lateral face
∴
 4  4 48 cm2
In △VHE,
VH 2  HE 2  VE 2 (Pyth. theorem)  111 cm2 (cor. to 3 sig. fig.)
2
 12 
VH  102    cm
 2
 8 cm

14 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

25. (a) In △ABC, 27. Let A and h be the base area and the height of the
AC  AB  BC
2 2 2
(Pyth. theorem) rectangular tank respectively, then
volume of the water
AC  5  5 cm
2 2
 capacity of the tank  volume of the pyramid
 50 cm 1
 Ah  Ah
In △VBA, 3
VA2  VB 2  AB 2 (Pyth. theorem) 2
 Ah
3
VA  12  5 cm2 2
∴ After the pyramid is taken out,
 13 cm 2
Similarly, VC = 13 cm Ah
Construct VE  AC . the water level  3
A
2
 h
3
∴ Percentage decrease in the water level
2
h h
 3  100%
h
 2
∵ △VAE  △VCE (RHS)  1    100%
 3
∴ AE = CE (corr. sides,  △s)
1
AC  33 %
∴ AE  3
2
50
 cm Exercise 7B (p. 7.31)
2 Level 1
In △VEA, 1
1. Volume of the cone     6 2  10 cm3
VE 2  AE 2  VA2 (Pyth. theorem) 3
 50 
2  120 cm3
VE  132    cm
 2 
1
313 2. Volume of the cone     32  7 cm 3
 cm 3
2  21 cm3
1
Area of △VAC   AC  VE
2 2
1  10 
1 313 3. Volume of the cone        12 m 3
  50  cm2 3  2
2 2
 100 m 3
 44.2295 cm2
 44.2 cm2 (cor. to 3 sig. fig.) 4. (a) Let r cm be the base radius of the cone.
92  r 2  152 (Pyth. theorem)
(b) Total surface area of the pyramid
 area of △VAC  2  area of △VBC  base area r  152  92
   12
1  1
 44.2295  2    5 12    5  5 cm 2 ∴ The base radius of the cone is 12 cm.
 2  2 
 117 cm (cor. to 3 sig. fig.)
2
1
(b) Volume of the cone     12 2  9 cm 3
3
26. Let x be the length of the side of the base of pyramid P,  432 cm 3
then the length of the side of the base of pyramid Q = 2x.
∵ Volume of pyramid P  volume of pyramid Q
5. (a) Curved surface area of the cone    9  12 cm 2
1 2 1
∴  x  hP   (2 x) 2  hQ  108 cm 2
3 3
hP
4 (b) Total surface area of the cone  (108    9 2 ) cm 2
hQ
∴ hP : hQ  4 : 1  189 cm 2

6. (a) Curved surface area of the cone    15  18 cm 2

 270 cm 2

15 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

(b) Total surface area of the cone  (270    152 ) cm 2 10. Let h cm be the height of the cone.
 495 cm 2 1
   52  h  100
3
12
 16  h
7. (a) Curved surface area of the cone       21 m 2 
 2
 3.82 (cor. to 3 sig. fig.)
 168 m 2
∴ The height of the cone is 3.82 cm.

(b) Total surface area of the cone 11. Let r cm be the base radius of the cone.
   r  8  50
 16  
2

 168       m 2 25
  2   r
4
 232 m 2  1.99 (cor. to 3 sig. fig.)
∴ The base radius of the cone is 1.99 cm.
8. (a) Let  cm be the slant height of the cone.
2
 162  122 (Pyth. theorem) 12. (a) Let  cm be the slant height of the cone.
 16  12 2 2   3      32  150
150  9
 20 
∴ The slant height of the cone is 20 cm. 3
 12.9155
(b) Total surface area of the cone  12.9 (cor. to 3 sig. fig.)
 (  12  20    122 ) cm2 ∴ The slant height of the cone is 12.9 cm.
 384 cm2
(b) Height of the cone

9. (a) When r = 5 cm and h = 12 cm,  12.91552  32 cm (Pyth. theorem)

 12.5623 cm
 52  122 cm (Pyth. theorem)
 12.6 cm (cor. to 3 sig. fig.)
 13 cm
1
volume of the cone     52  12 cm3 (c) Volume of the cone
3
1
 100 cm3     32  12.5623 cm 3
3
curved surface area of the cone    5  13 cm
2
 118 cm 3 (cor. to 3 sig. fig.)
 65 cm2
13. (a) Let r cm be the base radius of the cone.
  r 2  32
(b) When r = 6 cm and  = 10 cm,
r  32 (or 4 2 )
h  102  62 cm (Pyth. theorem)
 8 cm ∴ The base radius of the cone is 32 cm

1 (or 4 2 cm) .
volume of the cone     62  8 cm3
3
 96 cm3 (b) Height of the cone
 92  ( 32) 2 cm (Pyth. theorem)
curved surface area of the cone    6  10 cm
2
 7 cm
 60 cm2
(c) Volume of the cone
(c) When h = 15 cm and  = 17 cm, 1
  32  7 cm 3
r  172  152 cm (Pyth. theorem) 3
224
 8 cm  cm 3
3
1
volume of the cone     82  15 cm3
3 14. (a) Let r cm be the base radius of the cone.
 320 cm3 2    r  12
r 6
curved surface area of the cone    8  17 cm
2
∴ The base radius of the cone is 6 cm.
 136 cm2

16 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

(b) Slant height of the cone

 6  8 cm (Pyth. theorem)
2 2
Level 2
19. (a)

AB 
108
 2    15 cm
360
 10 cm
 9 cm
Total surface area of the cone
 (  6  10    6 2 ) cm2
Let r cm be the base radius of the vessel.
∵ Circumference of the base of the vessel  AB

 96 cm 2
∴ 2    r  9
r  4.5
15. Let r m be the base radius of the tent. ∴ The base radius of the vessel is 4.5 cm.
  r 2  4
r2 (b) Height of the vessel

Slant height of the tent  22  2.52 m (Pyth. theorem)  152  4.52 cm (Pyth. theorem)
 10.25 m  204.75 cm
1
Total surface area of the tent  (  2  10.25  4 ) m
2
Capacity of the vessel     4.5 2  204.75 cm3
3
 32.7 m 2 (cor. to 3 sig. fig.)  303 cm3 (cor. to 3 sig. fig.)

16. Height of the upper cone

20. (a) Let 3k cm and 4k cm be the base radius and the
 782  302 mm (Pyth. theorem) height of the cone respectively, where k ≠ 0.
 72 mm 1
   (3k ) 2  4k  768
Volume of the sand in the upper cone at the beginning 3
1 k 3  64
    302  72 mm3
3 k 4
 21 600 mm3 ∴ The base radius and the height of the cone are
Time taken  1 h  60  60 s  3600 s 12 cm and 16 cm respectively.
∵ Volume of the sand in the lower cone after 1 hour Slant height of the cone
 volume of the sand in the upper cone at the beginning  122  162 cm (Pyth. theorem)
∴ x  3600  21 600
 20 cm
21 600
x
3600
 18.8 (cor. to 3 sig. fig.) (b) Curved surface area of the cone    12  20 cm2
 240 cm2
1 Total surface area of the cone  (240    12 2 ) cm 2
17. Volume of wine in the conical glass     42  7 cm3
3  384 cm 2
112
 cm3
3 21. (a) Let r cm be the radius of the semi-circle.
Let h cm be the height of the wine in the cylindrical glass. Circumference of the base of the paper cup
∵ Volume of wine in the cylindrical glass  2    5 cm
 volume of wine in the conical glass
∴   32  h 
112
 10 cm
Length of AB 
 180
 2    r cm
3 360
h  4.15 (cor. to 3 sig. fig.)  r cm
∴ The height of the wine in the cylindrical glass is ∴ r  10
4.15 cm. r  10
∴ The radius of the semi-circle is 10 cm.
18. Let r cm be the base radius of the cylinder.
∵ Volume of the cylinder  volume of the cone
(b) Height of the paper cup
1
∴   r 15    18 2  20
2
 10 2  52 cm (Pyth. theorem)
3
r 2  144  75 cm
Capacity of the paper cup
r  12 1
∴ The base radius of the cylinder is 12 cm.     5 2  75 cm3
3
 227 cm3 (cor. to 3 sig. fig.)

17 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

22. Let r m be the base radius of the heap of rice. (b) Let be the angle of the sector.
∵ Circumference of the base = 2.8 m ∵ Curved surface area of the cover
∴ 2    r  2.8 = area of the sector
r  1.4 
∴    452  1620
360
Height of the heap of rice  1.82  1.4 2 m (Pyth. theorem)
  288
 1.28 m ∴ The angle of the sector is 288.
Volume of the heap of rice  1    1.42  1.28 m3
3 25. (a) Volume of the solid
 2.3221 m3 1 
     1.52  2    1.52  20  m3
Weight of the heap of rice  2.3221  250 kg 3 
 580.525 kg  146 m3 (cor. to 3 sig. fig.)
 600 kg
∴ Peter’s claim is disagreed. (b) Slant height of the cone
 1.52  22 m (Pyth. theorem)
23. (a) Curved surface area of the vessel
 area of the sector  2.5 m
Total surface area of the solid
90
    82 cm2  (2   1.5  20   1.5  2.5    1.5 2 ) m 2
360
 207 m 2 (cor. to 3 sig. fig.)
 16 cm2
 50.3 cm2 (cor. to 3 sig. fig.)
26. Let h cm be the height of the cone.
∵ Volume of the cone  volume of the cylinder
(b) 1
∴    52  h    52  4
3
h  12
∴ The height of the cone is 12 cm.
Slant height of the cone
 122  52 cm (Pyth. theorem)
 13 cm
Total surface area of the cone
 (  5 13    52 ) cm 2
Let r cm be the base radius of the vessel.  90 cm 2
∵ Curved surface area of the vessel  16 cm 2 Total surface area of the cylinder
∴   r  8  16  (2    5  4  2    52 ) cm 2
r2  90 cm 2
∴ The base radius of the vessel is 2 cm. ∵ Total surface area of the cone
Height of the vessel  8  2 cm (Pyth. theorem)
2 2 = total surface area of the cylinder
∴ Alan’s claim is disagreed.
 60 cm
1 27. (a) Height of the paper cup
Capacity of the vessel     2  60 cm
2 3

3  82  4.82 cm (Pyth. theorem)

 32.4462... cm3  6.4 cm
 32 cm3 Capacity of the paper cup
∴ David’s claim is agreed. 1
    4.82  6.4 cm3
3
24. (a) Slant height of the cover  49.152 cm3
 362  27 2 cm (Pyth. theorem)  154 cm3 (cor. to 3 sig. fig.)
 45 cm
Curved surface area of the cover    36  45 cm 2 (b) Capacity of the container
 1620 cm 2   102 15 cm3
 1500 cm3
Capacity of the container
Capacity of the paper cup
1500 cm3

49.152 cm3
 30.5 (cor. to 3 sig. fig.)

18 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

∴ The maximum number of cups of water pouring ∴ Volume of the frustum

into the container without any overflow is 30. = volume of pyramid VABC
 volume of pyramid VDEF
28. (a) Base radius of the cone  3r cm  (500  108) cm3
h
Height of the cone  cm  392 cm3
2
1 h
∴ Volume of the cone     (3r ) 2  1
3 2 4. Volume of pyramid VABCD   (12  8)  24 cm 3
3 2 3
 r h cm 3

2  768 cm 3
Height of pyramid VEFGH  (24  12) cm
(b) ∵ Volume of the cylinder = 600 cm3  12 cm
∴ r 2 h  600 Volume of pyramid VEFGH  1  (6  4)  12 cm3
3 2 3
∴ Volume of the cone  r h  96 cm3
2
3 ∴ Volume of frustum ABCDHEFG
 (600) = volume of pyramid VABCD
2
 volume of pyramid VEFGH
 900 cm3
 (768  96) cm3
 672 cm3
Exercise 7C (p. 7.39)
Level 1
1 1
1. (a) Volume of pyramid VEFGH   (10  10)  6 cm3 5. (a) Volume of cone VAD     7 2  6 cm3
3 3
 200 cm3  98 cm3

1 1
(b) Volume of pyramid VABCD   (15  15)  9 cm3 (b) Volume of cone VBC     142  12 cm3
3 3
 675 cm3  784 cm3

(c) Volume of frustum ABCDHEFG (c) Volume of frustum ABCD

= volume of pyramid VABCD = volume of cone VBC – volume of pyramid VAD
– volume of pyramid VEFGH
 (784  98 ) cm3
 (675  200) cm3
 686 cm3
 475 cm3

1 1
2. (a) Volume of pyramid VEFGH   (5  5)  9 cm3 6. Capacity of the paper cup     5 2  8 cm3
3 3
200
 75 cm3  cm3
3
Volume of water in the cup
(b) Volume of pyramid VABCD 1
1     32  4.8 cm3
  (12.5 12.5)  (13.5  9) cm3 3
3  14.4 cm3
 1171.875 cm3 Volume of water needed to fill up the cup
= capacity of the paper cup – volume of water in the cup
(c) Volume of frustum ABCDHEFG  200 
= volume of pyramid VABCD   14.4  cm3
 3 
– volume of pyramid VEFGH
 (1171.875  75) cm3  164 cm3 (cor. to 3 sig. fig.)
 1096.875 cm3
(a) Curved surface area of cone VCD    6  10 cm
2
7.
3. Height of pyramid VABC  (12  8) cm  60 cm2
 20 cm Curved surface area of cone VAB    3  5 cm
2

1  15 cm2
Volume of pyramid VABC   75  20 cm3
3
 500 cm3
1
Volume of pyramid VDEF   27  12 cm3
3
 108 cm3
19 © Pearson Education Asia Limited 2022
Junior Secondary Mathematics in Action 3B Full Solutions

Curved surface area of the frustum 9. (a)

= curved surface area of cone VCD
 curved surface area of cone VAB
 (60  15 ) cm2
 45 cm2

(b) Total surface area of the frustum

 (45    6 2    32 ) cm 2 Let MB = x cm.
 90 cm 2 ∵ △VNC ~ △VMB (AAA)
MB VB
∴  (corr. sides, ~△s)
NC VC
Level 2
8. (a) x 13  26

5 13
x 39

5 13
x  15
∴ The radius of the lower base is 15 cm.

(b) Curved surface area of cone VDC

12    5  13 cm2
QN  cm  6 cm
2  65 cm2
6 Curved surface area of cone VAB
PM  cm  3 cm
2   15  (13  26) cm2
Let VP = x cm, then VQ = (x + 4) cm.
∵ △VQN ~ △VPM (AAA)  585 cm2
Curved surface area of the frustum
VQ QN
  (585  65 ) cm 2
∴ VP PM (corr. sides, ~△s)
x4 6  520 cm 2
 Total surface area of the frustum
x 3
x4  (520    52    152 ) cm 2
2
x  2420 cm 2 (cor. to 3 sig. fig.)
x  4  2x
x4 10.
VQ  (4  4) cm
 8 cm
∴ The height of pyramid VABCD is 8 cm.

(b) Volume of the frustum

 volume of pyramid VABCD
 volume of pyramid VEFGH
1 1 
   (12  12)  8   (6  6)  4  cm3
 3 3  Let MV = x cm, then NV = (30 + x) cm.
 336 cm3 ∵ △CNV ~ △AMV (AAA)
CN NV
∴  (corr. sides, ~△s)
AM MV
15 30  x

9 x
5 30  x

3 x
5 x  90  3 x
2 x  90
x  45
Volume of cone VAB  1    92  45 cm3
3
 1215 cm3
Volume of cone VCD  1    152  (30  45) cm3
3
 5625 cm3
20 © Pearson Education Asia Limited 2022
 7 Areas and Volumes (III)

Capacity of the bucket 12. (a) (i)

= volume of cone VCD  volume of cone VAB
 (5625  1215 ) cm3
 4410 cm 3

11. EF  36 cm
 6 cm
AB  64 cm
 8 cm
Let PD = x cm.
Let K, L, M and N be the mid-points of EG, AC, FG and
∵ △VPD ~ △VQC (AAA)
BC respectively.
PD VP
∴ QC  VQ (corr. sides, ~△s)

x 6

3 63
x 6

3 9
x2
∴ The radius of the upper base is 2 cm.
1 1
KM  EF (ii) Volume of cone VAD     2 2  6 cm 3
2 3
1  8 cm 3
  6 cm
2 1
Volume of cone VBC     32  (6  3) cm3
 3 cm 3
1  27 cm 3
LN  AB
2 Volume of the frustum
1
  8 cm  (27  8 ) cm3
2
 59.7 cm3 (cor. to 3 sig. fig.)
 4 cm
Let VK = x cm, then VL = (x + 6) cm.
∵ △VLN ~ △VKM (AAA) (b) (i) In △VPD,
∴ VL LN (corr. sides, ~△s) VD 2  VP 2  PD 2 (Pyth. theorem)

VK KM
VD  62  22 cm
x6 4
  40 cm
x 3
3 x  18  4 x  6.32 cm (cor. to 3 sig. fig.)
x  18 In △VQC,
VL  (18  6) cm VC 2  VQ 2  QC 2 (Pyth. theorem)
 24 cm
Volume of the frustum VC  (6  3) 2  32 cm
= volume of pyramid VABCD  90 cm
 volume of pyramid VEFGH
 9.49 cm (cor. to 3 sig. fig.)
1 1 
   64  24   36 18  cm3 (ii) Curved surface area of cone VAD
3 3 
 296 cm 3    2  40 cm 2
 2 40 cm 2
Curved surface area of cone VBC
   3  90 cm 2
 3 90 cm 2
Curved surface area of the frustum
 (3 90  2 40 ) cm2
Total surface area of the frustum
 [(3 90  2 40 )    22    32 ] cm 2
 90.5 cm2 (cor. to 3 sig. fig.)

21 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

13. (a) Volume of water can be added without overflow

 capacity of the vessel  volume of water contained
 (1500  768 ) cm3
 732 cm3
 2299.64... cm3
 2300 cm3
Let VB = x cm. ∴ 2300 cm3 of water added to the vessel will lead to
∵ △VBM ~ △VDN (AAA) outflow.
VB BM ∴ Brian’s claim is disagreed.
∴  (corr. sides, ~△s)
VD DN
15.
x 12

x  10 18
x 2

x  10 3
3x  2 x  20
x  20
∴ VB  20 cm Let MA = x cm.
∵ △VMA ~ △VNC (AAA)
(b) Curved surface area of cone VCD ∴ MA  VM (corr. sides, ~△s)
   18  (20  10) cm2 NC VN
x 8
 540 cm2 
6 10
Curved surface area of cone VAB
10 x  48
   12  20 cm2
x  4.8
 240 cm2
VA  82  4.82 cm (Pyth. theorem)
Curved surface area of the frustum
 (540  240 ) cm2  87.04 cm
 300 cm 2
VC  10 2  6 2 cm (Pyth. theorem)
Total surface area of the frustum  136 cm
 (300    122    182 ) cm2 Curved surface area of the paper cup
 768 cm2    6  136 cm 2
Cost to paint the whole frustum
 6 136 cm 2
 $0.02  768
Curved surface area of the paper cup that is wet
 $48.3 (cor. to 3 sig. fig.)
   4.8  87.04 cm 2
 4.8 87.04 cm 2
14.
Inner surface area of the paper cup that is dry
= curved surface area of the paper cup
– curved surface area of the paper cup that is wet
 (6 136  4.8 87.04 ) cm 2
 79.1 cm 2 (cor. to 3 sig. fig.)

Let MA = x cm. 16.

∵ △VMA ~ △VNC (AAA)
MA VM
∴  (corr. sides, ~△s)
NC VN
x 16

15 20
x 4

15 5
x  12
1
Capacity of the vessel     152  20 cm3
3
With the notations in the figure, let r cm be the radius of
 1500 cm3
the opening of the vessel,
Volume of water contained  1    122  16 cm3 then the height of the vessel is 2r cm.
3 Let k cm be the radius of the water surface and d cm be the
 768 cm3 depth of water.

22 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

∵ △ABE ~ △CDE (AAA) 3

1 4  14 
(b) Volume of the hemisphere        cm3
∴ AE  AB (corr. sides, ~△s) 2 3  2
CE CD
686
2r r  cm3
 3
d k
rd Total surface area of the hemisphere
k = curved surface area + area of flat surface
2r
d 1  14 
2
 14  
2
    4            cm 2
2  2  2  2  
(a) When d = 20 + 1.2 = 21.2,
 147 cm 2
21.2
k
2
 10.6 3. (a) Let r cm be the radius of the sphere.
∴ Diameter of the water surface now  2  10.6 cm ∵ Surface area of the sphere = 144 cm2
 21.2 cm ∴ 4r 2  144
r 2  36
(b) When d = 20, r6
20 ∴ The radius of the sphere is 6 cm.
k
2
 10 (b) Let r cm be the radius of the sphere.
∴ Original radius of the water surface = 10 cm ∵ Volume of the sphere = 4500 cm3
Volume of the stone 4 3
∴ r  4500
 volume of water above the original water level 3
1 1  r 3  3375
     10.62  21.2     102  20  cm3
 3 3  r  15
 400 cm3 (cor. to the nearest cm3 ) ∴ The radius of the sphere is 15 cm.

4. (a) Let r cm be the radius of the hemisphere.

Exercise 7D (p. 7.49)
∵ Volume of the hemisphere = 144 cm3
Level 1
4 1 4 3
1. (a) Volume of the sphere    123 cm3 ∴  r  144
3 2 3
 2304 cm3 r 3  216
r 6
Surface area of the sphere  4   122 cm2
Diameter  2r cm
 576 cm2
 2(6) cm
 12 cm
3
4  10 
(b) Volume of the sphere       cm3
3  2 (b) Let r cm be the radius of the hemisphere.
500 ∵ Curved surface area of the hemisphere
 cm3 = 128 cm2
3
1
2
∴  4r 2  128
 10  2
Surface area of the sphere  4      cm
2

 2 r 2  64
 100 cm 2 r 8
Diameter  2r cm
1 4  2(8) cm
2. (a) Volume of the hemisphere      183 cm3
2 3  16 cm
 3888 cm3
Total surface area of the hemisphere 5. (a) Let r cm be the radius of the sphere.
= curved surface area + area of flat surface ∵ Area of the largest cross-section = 81 cm2
1 
   4    182    182  cm 2 ∴ r 2  81
2  r 2  81
 972 cm 2 r 9
∴ The radius of the sphere is 9 cm.

23 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

4 Let d cm be the diameter of the larger sphere.

(b) Volume of the sphere     93 cm3 3
3 4 d 32
    
 972 cm3 3 2 3
32 d 3
6. (a) Let r cm be the radius of the hemisphere. 
3 6
∵ Circumference of the flat surface = 44 cm
d 3  64
∴ 2r  44
d 4
r  22
∴ The diameter of the larger sphere is 4 cm.
∴ The radius of the hemisphere is 22 cm.
11. Volume of the solid part of the sphere
(b) Total surface area of the hemisphere
= curved surface area + area of flat surface 4 4 
     (8  1)3     83  cm3
1 
   4    222    222  cm 2 3 3 
2  868
 cm3
 1452 cm 2 3

7. (a) Let r cm be the radius of the balloon. 1

∵ Surface area of the balloon = 196 cm2 12. (a) Volume of the cone     7 2  28 cm3
3
∴ 4r 2  196 1372
 cm3
r 2  49 3
r7 Let r cm be the radius of the sphere.
∴ The radius of the balloon is 7 cm. 4 3 1372
r 
3 3
(b) Volume of air inside the balloon r 3  343
4
    73 cm3 r 7
3
∴ The radius of the sphere is 7 cm.
 1437 cm3 (cor. to the nearest cm3 )
(b) Surface area of the sphere  4    7 cm
2 2

8. Let r cm be the radius of the watermelon.

 196 cm 2
∵ Total surface area of the watermelon = 243 cm2
1
∴  4r 2  r 2  243 13. (a) Volume of the solid
2
 volume of the cylinder
3r 2  243 + volume of the hemisphere
r 2  81  1 4 
    42  9      43  cm3
r 9  2 3 
∴ The radius of the watermelon is 9 cm. 560
 cm3
3
Total surface area of the cube  6  10  10 cm
2
9.
 600 cm2 (b) Total surface area of the solid
Let r cm be the radius of the sphere. = base area of the cylinder
4r 2  600 + curved surface area of the cylinder
150 + curved surface area of the hemisphere
r2   1 
     42  2    4  9   4    42  cm 2
r  6.91 (cor. to 3 sig. fig.)  2 
∴ The radius of the sphere is 6.91 cm.  120 cm 2

10. Total volume of the 8 smaller lead spheres 1 4 3

14. (a) Volume of the hemisphere       5  cm
3
3
4 2  2 3 
 8       cm3
3 2 250
 cm 3
32 3
 cm3
3 1  3
Volume of the cone      5  12  cm
2

3 
 100 cm3
 250 
Volume of the solid    100  cm3
 3 
550
 cm 3
3

24 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

(b) Total surface area of the solid Volume of water in the container
= curved surface area of the hemisphere  500  3
+ curved surface area of the cone  1000   cm
 3 
1 
   4    52    5  13  cm 2 Let h cm be the depth of water if the ball is taken out from
 2 
the container.
 115 cm 2 500
10  10  h  1000 
3
Level 2 500
15. Volume of water above the original water level 1000 
2 h 3
 64  2 100
      cm3
 2  3  4.76 (cor. to 3 sig. fig.)
2048 ∴ The depth of water is 4.76 cm.
 cm3
3
Let r cm be the radius of the sphere. 19. Capacity of the can    52  (5  2) cm3
∵ Volume of the sphere  250 cm3
volume of water above the original water level 4
4 3 2048 Volume of the ball     53 cm3
∴ r  3
3 3 500
 cm3
r 3  512 3
r 8  500 
Volume of water in the can   250   cm
3
∴ The radius of the sphere is 8 cm.  3 
250
3  cm3
4  2.2  3
16. Volume of the whole chocolate ball      cm3
3  2  Let d cm be the depth of water if the ball is taken out from
1331 the can.
 cm3 250
750   52  d 
3
 1331 
Volume of chocolate    2.5  cm3 10
 750  d
3
 1331   10
Cost of a chocolate ball  $  2.5   3  0.8 ∴ The depth of the water is cm .
 750   3
 $10.0 (cor. to 3 sig. fig.)
20. The remaining capacity of the container
2
4
17. Let r cm be the radius of the metal ball.       (10  7.5) cm 3
∵ Volume of the metal ball = 500 cm3 2
4 3  10 cm 3
∴ r  500 ∴ Total volume of 60 marbles
3
375 4
r3   60     0.53 cm3
 3
 10 cm3
375
r3  the remaining capacity of the container

2 ∴ The water will not overflow.
 375  ∴ Harry’s claim is disagreed.
Surface area of the metal ball  4     3
2
 cm
  
1 4
 304.65 cm 2 21. Capacity of the hemispherical part      1.53 cm3
2 3
Cost of painting the metal ball  $0.02  304.65
9
 $6.093  cm3
4
 $6  9  3
∴ Percy’s claim is disagreed. Volume of water in the cylindrical part   60   cm
 4 
Let d cm be the depth of water in the cylindrical part.
18. Capacity of the cubic container  10  10  10 cm
3
9
 1000 cm3   1.52  d  60 
4
3
4  10  9
Volume of the ball       cm 3 60 
3  2 d 4
500   1.52
 cm 3  7.488
3

25 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

∴ Depth of water in the test tube Let h cm be the height of the cone.
 (1.5  7.488) cm 1
   6 2  h  96 
 8.99 cm (cor. to 3 sig. fig.) 3
h8
∴ The height of the cone is 8 cm.
22. (a) Volume of the hemisphere  1  4    63 cm3
2 3 (b) Slant height of the cone
 144 cm3
 62  82 cm (Pyth. theorem)
Volume of the cylinder    3  6 cm
2 3
 10 cm
 54 cm3 Total surface area of the solid
Volume of the solid  (144  54 ) cm
3
= curved surface area of the cone
+ curved surface area of the hemisphere
 198 cm3
 1 
    6  10   4    62  cm 2
 2 
(b) Curved surface area of the hemisphere
 132 cm 2
1
  4    62 cm 2
2
 72 cm 2 25. (a) (i) Radius of the larger sphere
Curved surface area of the cylinder  3  2r cm
 2    3  6 cm2  6r cm
Volume of the larger sphere
 36 cm2 4
Total area of the flat surfaces of the solid     (6r )3 cm3
3
   6 2 cm2  288 r 3 cm3
 36 cm2 4
Total surface area of the solid (ii) Volume of the smaller sphere   r 3 cm3
3
= curved surface area of the hemisphere
∵ Sum of the volumes of two spheres
+ curved surface area of the cylinder
+ total area of the flat surfaces of the solid = 7812 cm3
4 3
 (72  36  36 ) cm 2 ∴  r  288 r 3  7812
3
 144 cm 2
868 3
 r  7812
3
23. Let x mm be the length of the capsule. r 3  27
Sum of the curved surface areas of the two hemispheres
r 3
1  4 
2

 2    4       mm 2
2   2   (b) Sum of the surface areas of the two spheres
 16 mm 2  [4    32  4    (6  3) 2 ] cm 2
Curved surface area of the cylinder  1332 cm 2
4
 2     ( x  2  2) mm2
2 26. (a) Let r cm be the external radius of the ceramic.
 4 ( x  4) mm2 ∵ Area of the ring = 75 cm2
∵ Total surface area of the capsule = 60 mm2 ∴ r 2    112  75
∴ Sum of the curved surface areas of the two r 2  121  75
hemispheres + curved surface area of the cylinder
= 60 mm2 r 2  196
16  4 ( x  4)  60 r  14
4 ( x  4)  44 ∴ The external radius of the ceramic is 14 cm.
x  4  11
1
x  15 (b) External curved surface area   4    142 cm 2
2
∴ The length of the capsule is 15 mm.  392 cm 2
1
Internal curved surface area   4    112 cm 2
24. (a) Volume of the hemisphere  1  4    63 cm3 2
2 3
 242 cm 2
 144 cm3
Volume of the cone  2  144 cm3
3
 96 cm3

26 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

Total surface area of the ceramic (b) Surface area of the sphere  4r 2 cm2
 external curved surface area Slant height of the cone
+ internal curved surface area
+ area of the ring  r 2  (4r ) 2 cm (Pyth. theorem)
 (392  242  75 ) cm2
 17 r cm
 709 cm2
Curved surface area of the cone    r  17r cm 2

27. (a) Let r cm be the inner radius of the cake mould.  17 r 2 cm 2
∵ Capacity of the mould = 144 cm3 Base area of the cone  r 2 cm2
1 4 Total surface area of the cone  ( 17r 2  r 2 ) cm 2
∴   r 3  144
2 3  (1  17 )r 2 cm 2
r 3  216
∵ (1  17 )r 2  4r 2
r6 ∴ The total surface area of the cone is larger than
∴ The inner radius of the cake mould is 6 cm.
the surface area of the sphere.
∴ James’s claim is agreed.
(b) Volume of the outer hemisphere
1 4 Exercise 7E (p. 7.65)
     (6  0.1)3 cm3
2 3 Level 1
 151.3207 cm3 Area of figure A  5 
2

Volume of the steel 1.  

Area of figure B  3 
 (151.3207  144 ) cm3
25
 7.3207 cm3 
9
Cost to produce one mould ∴ The ratio of the area of figure A to that of figure B is
 $0.5  7.3207 25 : 9.
 $11.5 (cor. to 3 sig. fig.)
2
Area of figure A  4 
2.  
28. (a) ∵ Volume of the sausage = 153 cm3 Area of figure B  6 
1 4 2
∴   r 3  r 2  h  153 2
2 3  
3
2 3
r  r 2  (5r )  153 4
3 
9
2  3
  5 r  153 ∴ The ratio of the area of figure A to that of figure B is
3  4 : 9.
r 3  27
2
r 3  x 81
3.   
∴ h  5r  y  121
 5 3 x 81

 15 y 121
9

1 11
(b) Volume of each part of the sausage   153 cm3
2 ∴ x : y  9 : 11
 76.5 cm3
Length of the part consisting of cylindrical part only 2
76.5 x 100
 cm 4.   
  32 y
  144
 8.5 cm x 25

Length of the part consisting of hemispherical and y 36
cylindrical parts 5
 [(3  15)  8.5] cm 
6
 9.5 cm ∴ x: y  5:6

29. (a) Let h cm be the height of the cone. 2

∵ Volume of the cone = volume of the sphere x 3
5.  
1 2 4 7 2
∴ r h  r 3 x 9
3 3 
h  4r 7 4
∴ The height of the cone is 4r cm. x  15.75

27 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions
2 3
6.  y 140 y  15 
   11.  
 13  240 81  9 
y 7 y 5
3
  
13 12
81  3 
7 125
y  13
12 y  81
27
 9.93 (cor. to 3 sig. fig.)
 375
2
Total surface area of A  8  2
7. (a)   w  16 
Total surface area of B  7  12.  
81  18 
64
 w 8
2
49  
∴ The ratio of the total surface area of A to that of 81  9 
B is 64 : 49. 64
w   81
3
81
Volume of A  8   64
(b)  
Volume of B  7 
512 3
  y 800
343 13.   
∴ The ratio of the volume of A to that of B is  18  1500
512 : 343. y 3 8

2
18 15
Total surface area of A  20  8
8. (a)   y 3  18
Total surface area of B  16 
15
2
5  14.6 (cor. to 3 sig. fig.)
 
4
25 2
 z 48
16 14.   
6 20
∴ The ratio of the total surface area of A to that of
B is 25 : 16. z 12

6 5
3
Volume of A  20  12
(b)   z 6
Volume of B  16  5
5
3  9.30 (cor. to 3 sig. fig.)
 
4
125 15. Let x cm be the perimeter of the smaller star.
  x  4
2
64   
∴ The ratio of the volume of A to that of B is 45
  9
125 : 64. x 4

3 45 9
x 54
9.    2
y
  128  45 x
3
x 3 27  30

y 64 ∴ The perimeter of the smaller star is 30 cm.
3
 2
4 Area of painting A  42 
16.  
∴ x : y  3: 4 Area of painting B  48 
 0.766 (cor. to 3 sig. fig.)
x
3
512  76.6%
10.    ∴ The area of painting A is 76.6% of the area of
y
  216
painting B.
x 3 64

y 27
4

3
∴ x: y  4:3

28 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

17. Let x kg be the weight of the larger horse. 20. Let 1 : n be the scale of the map.
∵ Weight of the horse is proportional to its volume. Area of the playground on the map  1 
2

3  
x 5 Actual area of the playground n
∴  
0.8  3  49 cm 2  1 
2

x 125  
 784 m 2  n 
0.8 27 7 cm 1
125 
x  0.8 28 m n
27
28  100
 3.70 (cor. to 3 sig. fig.) n
7
∴ The weight of the larger horse is 3.70 kg.
 400
∴ The scale of the map is 1 : 400.
Level 2
18. (a) Let h cm be the height of the smaller cone. 2
3 Area of AEFG  AE 
 h 54 21.  
   Area of ABCD  AB 
16
  128
2
3
h 3 27  
 5
16 64
9
3 
h   16 25
4
Let 9k and 25k be the areas of AEFG and ABCD
 12
respectively, where k ≠ 0.
∴ The height of the smaller cone is 12 cm. Area of the yellow region = 9k
Area of the blue region = 25k – 9k
Base radius of the smaller cone 12 3 = 16k
(b)  
Base radius of the larger cone 16 4 Area of the yellow region 9k

∴ The ratio of the base radius of the smaller cone Area of the blue region 16k
to that of the larger cone is 3 : 4.
9

2 16
Curved surface area of the smaller cone  3  9
(c)    ∴ The ratio of the area of the yellow region to that of
Curved surface area of the larger cone  4  16 the blue region is 9 : 16.
∴ The ratio of the curved surface area of the
smaller cone to that of the larger cone is 9 : 16. 22. (a) In △ADE and △ABC,
ADE  ABC corr. s, BC // DE
19. (a) Curved surface area of A  18 AED  ACB corr. s, BC // DE
Curved surface area of B 50 DAE  BAC common 
9 ∴ △ADE ~ △ABC AAA

25
∴ The ratio of the curved surface area of A to that (b) Let DB = k and AD = 4k, where k ≠ 0.
of B is 9 : 25. AB = AD + DB = 4k + k = 5k
∴ AD : AB  4k : 5k
2  4:5
 Height of B  25 ∵ △ADE ~ △ABC
(b)   
 Height of A  9 Area of △ ADE  AD 
2

∴  
Height of B 5 Area of △ ABC  AB 

Height of A 3 4
2

Let V cm3 be the volume of B.  

5
3
V 5 16
  
54  3  25
V 125 Let 16t and 25t be the areas of △ADE and △ABC

54 27 respectively, where t ≠ 0.
125 Area of the orange region 16t
V  54 
27 Area of the yellow region 25t  16t
 250 16t

∴ The volume of B is 250 cm3. 9t
16

9
∴ The ratio of the area of the orange region to that
of the yellow region is 16 : 9.

29 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

23. (a) Let x cm2 be the area of △CFD. ∵ △BDF ~ △BCA (AAA)
∵ △CFD ~ △CAB (AAA) Area of △ BDF  BD 
2

2 ∴  
∴ Area of △CFD  CD  Area of △ BCA  BC 
 
Area of △CAB  CB  2
2
x 3
2  
  3
8 2 4
x 9 
 9
8 4
∴ X : (X + Y + Z)  4 : 9
x  18
∴ The area of △CFD is 18 cm2. ∴ CD : CB  k : 3k
 1: 3
(b) Let y cm2 be the area of quadrilateral CEGD. ∵ △CED ~ △CAB (AAA)
∵ △CFD ~ △EFG (AAA) 2
Area of △CED  CD 
Area of △CFD  DF 
2
∴  
 Area of △CAB  CB 
∴ 
Area of △ EFG  GF  2
1
18  2 1
2  
   3
18  y  1  1
18 
9 9
18  y
∴ Z : (X + Y + Z)  1 : 9
18  162  9 y
(ii) ∵ X : Z : ( X  Y  Z )  4 :1: 9
9 y  144
∴ Let Z = k, X = 4k and X + Y + Z = 9k, where
y  16 k ≠ 0.
∴ The area of quadrilateral CEGD is 16 cm2. ∴ Y  9 k  k  4k
 4k
Area of △ DEF EF ∴ Y : (X + Y + Z) = 4k : 9k  4 : 9
24. (a) 
Area of △CDF FC
Area of △ DEF 2 (b) Area of △CED = 10 cm2

27 cm 2 1 Z = 10 cm2
2 k = 10 cm2
Area of △ DEF   27 cm 2
1 Area of parallelogram AFDE  4k
 54 cm 2  4  10 cm 2
 40 cm 2
(b) Let x cm2
be the area of quadrilateral ABED.
∵ △ABC ~ △DEC (AAA)
26. Let h be the height of the smaller pyramid.
2
Area of △ ABC  BC  Then, the height of the frustum is h.
∴   Let V1 and V2 be the volumes of the smaller pyramid and
Area of △ DEC  CE 
2
the whole pyramid respectively.
x  54  27  1  2  1  3
  V1  h 
54  27  2 1   
V2  h  h 
2
x  81  4  1
  
81 3 8
x  81 16 ∴ V2  8V1

81 9 Volume of the smaller pyramid V1
∴ 
x  81  144 Volume of the frustum V2  V1
x  63 V1
∴ The area of quadrilateral ABED is 63 cm2. 
8V1  V1
V1
25. (a) (i) Let DC = k and BD = 2k, where k ≠ 0. 
BC = BD + DC = 2k + k = 3k 7V1
∴ BD : BC  2k : 3k 1

 2:3 7
∴ The ratio of the volume of the smaller pyramid to that
of the frustum is 1 : 7.

30 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

Area of wet surface before pouring water 29. (a) Let A0, A1 and A2 be the total areas of lateral faces of
27. pyramid VABC, frustum ABCFDE and pyramid
Area of wet surface after pouring water VDEF respectively.
2 2
 5  A0  VC 
   
 5 1  A2  VF 
25 A0  2 
2
 
36 
A0  A1  2  1 
Let 25k and 36k be the areas of wet surface before and
A0 4
after pouring water respectively, where k ≠ 0. 
Percentage increase in the area of the wet surface A0  A1 9
36k  25k 9 A0  4 A0  4 A1
  100%
25k 5 A0  4 A1
 44% A0 4

A1 5
28. (a) Let r0 and r1 be the radii of the sphere before and ∴ The ratio of the total area of lateral faces of
after heating respectively; and y be the surface area of pyramid VABC to that of frustum ABCFDE is
the sphere before heating. 4 : 5.
Surface area of the sphere after heating
= (1 + 21%)y (b) Let x cm3 be the volume of pyramid VDEF.
= 1.21y 3
x  VF 
2
 r1  1.21y  
   16  VC 
 r0  y
x  2 1
3

r1  
 1.1 16  2 
r0
x 27
∴ r1  1.1r0 
16 8
∴ Percentage increase in the radius of the sphere 27
r r x  16
 1 0  100% 8
r0  54
1.1r0  r0 ∴ The volume of pyramid VDEF is 54 cm3.
  100%
r0
3
 10%  Radius of the lower base of frustum BCED 
30. (a)  
 Radius of the upper base of frustum BCED 
3
(b) Let V0 and V1 be the volumes of the sphere before and  base radius of cone ABC 
after heating respectively.  
 base radius of cone ADE 
V1
 1.13 volume of cone ABC
V0 
volume of cone ADE
V1  1.331V0 27

Percentage increase in the volume of the sphere 8
V V Radius of the lower base of frustum BCED 3 27
 1 0  100% 
V0 Radius of the upper base of frustum BCED 8
1.331V0  V0 3
  100% 
V0 2
 33.1% ∴ The ratio of the radius of the lower base to that
of the upper base of frustum BCED is 3 : 2.

(b) Let x cm2 be the curved surface area of cone ADE.

2
x 2
 
x  20  3 
x 4

x  20 9
9 x  4 x  80
5 x  80
x  16
∴ The curved surface area of cone ADE is 16 cm2.

31 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions
2
 x  16 33. Let V cm3 be the final volume of water in the vessel.
31. (a)    V 8
3

 y 1  
V  300  5 
x 4
 V 512
y 1 
V  300 125
∴ x : y  4 :1 125V  512V  153 600
387V  153 600
3 3
 x  4 64 V  397 cm3 (cor. to 3 sig. fig.)
(b) ∵     
 y 1 1 ∴ The final volume of water in the vessel is 397 cm3.
volume of cuboid A 32
and 
volume of cuboid B 1 34. (a) Total volume of the two stone fish
3  20  12  1.2 cm3
Volume of cuboid A  x 
∴  
Volume of cuboid B  y   288 cm3
∴ The two cuboids are not similar.
2
∴ Eric’s claim is disagreed.  Length of stone fish A  36
(b)  Length of stone fish B   144
Alternative Solution  
2
Let m cm and n cm be the heights of cuboid A and  Length of stone fish A  1
cuboid B respectively.   
 Length of stone fish B  4
x 2 m  32  y 2 n
Length of stone fish A 1
m y2 
 32  2 Length of stone fish B 2
n x 3
2 Volume of stone fish A  1  1
 y   
 32    Volume of stone fish B  2  8
x
1
1 Volume of stone fish A  288  cm 3
 32  1 8
16
2  32 cm3

1 Volume of stone fish B  (288  32) cm
3

x 4 m 2 x m  256 cm3
∵  ,  and 
y 1 n 1 y n
∴ The two cuboids are not similar. 35. (a) ∵ The two frustums are similar.
∴ Eric’s claim is disagreed. AB CD
∴ 
CD EF
32. (a) Capacity of the vessel
r 18
1 
    8 2  10 cm3 18 30
3
640
r  10.8
 cm3
3
(b) Volume of the cone with base radius 30 cm
1
(b) Let h cm be the depth of water in the vessel.     302  45 cm3
3 3
 h 500
    13 500 cm3
 10  640
3 Let x cm3 be the volume of the cone with base radius
3 18 cm.
 h 75 3
   x  CD 
10
  32  
13 500  EF 
h 3 75 3
 x  18 
10 32  
13 500  30 
75
h3 10 27
32 x  13 500
125
 9.07 (cor. to 3 sig. fig.)
 2916
∴ The depth of water in the vessel is 9.07 cm. ∴ The volume of the cone with base radius 18 cm
is 2916 cm3.
Volume of the larger frustum
 (13 500  2916 ) cm3
 10 584 cm3

32 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

Let y cm3 be the volume of the smaller frustum. ∴ The ratio of the areas of the smaller plane figure to
3
y  CD  the larger plane figure is 9 : 16.
 
10 584  EF 
9. Volume of the frustum
3
y  18   volume of cone VBC  volume of cone VAD
  
10 584  30  1 1 
27      6 2  10     32  5  cm3
y  10 584  3 3 
125
 330 cm3 (cor. to 3 sig. fig.)
 2286.144
∴ The volume of the smaller frustum is
Revision Exercise 7 (p. 7.74)
2286.144 cm3
Level 1
1
Check Yourself (p. 7.73) Volume of the pyramid   (10 10) 12 cm
3
1.
1. (a)  3
 400 cm 3
(b) 

(c) 
Total surface area of the hemisphere
= curved surface area + area of the flat surface
1 
   4    r 2    r 2  cm 2
 2 
 (2 r 2   r 2 ) cm 2
 3 r 2 cm 2

(d)  Construct VM ⊥ BC.

In △VNM,
1 VM 2  VN 2  NM 2 (Pyth. theorem)
2. Volume of the pyramid   (4  4)  9 cm3
3 2
 48 cm3  10 
VM  12 2    cm
 2
1  13 cm
3. Volume of the cone     6 2  8 cm3 1
3 Area of △VBC   BC  VM
2
 302 cm3 (cor. to 3 sig. fig.)
1
  10  13 cm2
2
4  65 cm2
4. Volume of the sphere     53 cm 3
3 ∴ Total surface area of the pyramid
 524 cm 3 (cor. to 3 sig. fig.) = 4  area of △VBC + base area
= (4  65 + 10  10) cm2
2
5. Total surface area of the pyramid = 360 cm
= total area of all lateral faces + base area
 (2  106.25  2  130  12.5  16) cm 2
1 1 
 672.5 cm 2 2. Volume of the pyramid    12  9  18 cm3
3 2 
 324 cm3
6. Total surface area of the cone
= curved surface area + base area
 (  7  10    7 2 ) cm 2 3. Height of the cone  102  82 cm (Pyth. theorem)
 374 cm 2
(cor. to 3 sig. fig.)  6 cm
1
Volume of the cone     82  6 cm3
7. Total surface area of the hemisphere 3
= curved surface area + area of the flat surface  128 cm3
1 8
2
8 
2
Total surface area of the cone
   4          cm 2
 2 2  2   = curved surface area + base area
= (  8  10 +   82) cm2
 151 cm (cor. to 3 sig. fig.)
2

= 144 cm2
3
Area of the smaller plane figure  3 
2
1 4 6
8.   4. Volume of the hemisphere        cm3
Area of the larger plane figure  4  2 3 2
9  18 cm3

16
33 © Pearson Education Asia Limited 2022
Junior Secondary Mathematics in Action 3B Full Solutions

Total surface area of the hemisphere Volume of the solid

= curved surface area + area of the flat surface = volume of the cylinder + volume of the hemisphere
1 6
2
6 
2
 1 4 
   4            cm 2     12  8      13  cm3
2  2    2 3 
 2
26
 27 cm 2  cm 3

5. (a) Let r cm be the radius of the sphere. Total surface area of the solid
= base area of the cylinder
∵ Surface area of the sphere = 210 cm2
+ curved surface area of the cylinder
∴ 4 r 2  201 + curved surface area of the hemisphere
201  1 
r2 
=   1  2   1 8   4    12  cm 2
2
4 2
 
r  3.9994
= 19 cm
2
 4.00
(cor. to 3 sig. fig.)
10. Height of the cone
∴ The radius of the sphere is 4.00 cm.
2
 10 
(b) Volume of the sphere  132    cm (Pyth. theorem)
4 2
    3.99943 cm3  12 cm
3
Volume of the solid
 268 cm3 (cor. to 3 sig. fig.)
= volume of the cone + volume of the cylinder
1  10 
2
 10 
2

6. (a) Let cm be the slant height of the cone. =       12       10 cm3
∵ Curved surface area of the cone = 175 cm2  3 2 2 
∴   7   175 = 350 cm3
 25
Total surface area of the solid
∴ The slant height of the cone is 25 cm.
= curved surface area of the cone
+ curved surface area of the cylinder
(b) Height of the cone + base area of the cylinder
 252  7 2 cm (Pyth. theorem)  10  10   2
2
10
 24 cm =     13  2     10       cm
 2 2  2  
Volume of the cone
1 = 190 cm2
    7 2  24 cm3
3
 392 cm3 11. Base radius of the cone = radius of the hemisphere
6
 cm
7. Volume of the frustum 2
 volume of pyramid VEFGH  volume of pyramid VABCD  3 cm
1 1  Height of the cone = (7 – 3) cm = 4 cm
   162  18   82  9  cm3
3 3  Slant height of the cone  32  42 cm (Pyth. theorem)
 1344 cm3  5 cm
Volume of the solid
= volume of the cone + volume of the hemisphere
Curved surface area of cone VAB    2  6 cm
2
8.
1 1 4 
=    3  4      33  cm3
2
 12 cm 2
3 2 3 
Curved surface area of cone VCD    6  (6  12) cm
2
= 30 cm3
 108 cm 2
Total surface area of the solid
Curved surface area of the frustum
= curved surface area of the cone
 (108  12 ) cm2 + curved surface area of the hemisphere
 96 cm2  1 
=   3 5   4    32  cm 2
Total surface area of the frustum  2 
 (96    2 2    6 2 ) cm2 = 33 cm2
 427 cm (cor. to 3 sig. fig.)
2

9. Base radius of the cylinder = radius of the hemisphere

= (9 – 8) cm
= 1 cm

34 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

12. Volume of the solid 17. Let A0 and A1 be the original area and the new area of the
= volume of the original cylinder heptagon respectively; and x be the original length of each
– volume of the cone removed side of the heptagon.
 1  New length of each side = (1  20%)x
=    5  12     5  9  cm
2 2 3
= 0.8x
 3  2
A1  0.8 x 
= 707 cm (cor. to 3 sig. fig.)
3  
A0  x 
Slant height of the conical part A1  0.64 A0
 52  92 cm (Pyth. theorem) Percentage decrease in the area of the heptagon
 106 cm A  A1
 0  100%
Total surface area of the solid A0
= curved surface area of the conical part A0  0.64 A0
+ curved surface area of the cylindrical part   100%
A0
+ base area of the cylinder
 36%
= (  5  106  2    5 12    5 ) cm
2 2

= 617 cm2 (cor. to 3 sig. fig.) 18. (a) Let V0 be the volume of the size 4 football.
Volume of the size 5 football
13. Height of the can  3  4  2 cm = (1 + 72.8%)V0
= 1.728V0
 24 cm 3
Volume of the empty space in the can  Radius of size 5 football  1.728V0
  
= capacity of the can  Radius of size 4 football  V0
– total volume of the tennis balls
Radius of size 5 football 3
 4   1.728
    42  24  3     43  cm3 Radius of size 4 football
 3  6

 128  cm3 5
∴ The ratio of the radius of the size 5 football to
14. Let h cm be the rise in the water level. that of the size 4 football is 6 : 5.
∵ Volume of water above the original water level
2
= volume of the sphere
(b) Surface area of size 5 football  6 
 
4 Surface area of size 4 football  5 
∴  r 2h   r 3
3 36
4r 
h 25
3 ∴ The ratio of the surface area of the size 5 football
4r to that of the size 4 football is 36 : 25.
∴ The rise in the water level is cm .
3
19. Let hA cm and hB cm be the heights of cone A and cone B;
15. Let x cm2 be the area of the smaller figure. and V cm3 be the volume of cone B respectively.
2 2
x 4  hB  27
    
405  9  h  75
 A
x 16
 hB 9
405 81 
16 hA 25
x  405
81 3

 80 5
∴ The area of the smaller figure is 80 cm2. 3
V 3
 
16. Let x cm be the internal diameter of the smaller life buoy. 250  5 
3 V 27
 x  540 
   250 125
50
  2500
27
x 3 27 V  250
 125
50 125  54
x 3 ∴ The volume of cone B is 54 cm3.

50 5
3 Capacity of the larger bowl  6 cm 
3
x   50 20. 
5 
Capacity of the smaller container  2 cm 
 30
 27
∴ The internal diameter of the smaller life buoy is
∴ A participant must carry water at least 27 times so
30 cm.
that the larger bowl is fully filled.

35 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

Level 2 ∴ Total surface area of the cone

21. With the notations in the figure, = curved surface area + base area
 (  5  10    5 2 ) cm 2
 75 cm 2

(b) In △VOB,
VO 2  OB 2  VB 2 (Pyth. theorem)
VO  102  52 cm
 75 cm
(a) In △VMB, 1
Volume of the cone     52  75 cm3
3
VM 2  MB 2  VB 2 (Pyth. theorem)
 227 cm3 (cor. to 3 sig. fig.)
2
 10 
VM  132    cm
2
23. (a)

Length of AB 
120
 2    6 cm
 12 cm 360
∴ The height of △VAB with the base AB is 12 cm.  4 cm
Let r cm be the base radius of the cone.
(b) In △VMN, ∵ Circumference of the base of the cone = 4 cm
VN 2  MN 2  VM 2 (Pyth. theorem) ∴ 2r  4
2 r2
 10 
VN  122    cm ∴ The base radius of the cone is 2 cm.
2
 119 cm (b) Height of the cone  62  22 (Pyth. theorem)
Volume of the pyramid
 32
1
  (10  10)  119 cm 3 1
3 Volume of the cone     22  32 cm3
3
 363.62... cm3
 23.7 cm3 (cor. to 3 sig. fig.)
 360 cm3
∴ Barry’s claim is agreed.
24. Volume of water in the hemispherical part
1 4
22. With the notation in the figure,      1.53 cm3
2 3
9
 cm3
4
Volume of water above the hemispherical part
 9  3
  21   cm
 4 
(a) ∵ △VAO  △VBO (RHS) 
75
cm3
∴ AO = BO (corr. sides,  △s) 4
AB Let h cm be the depth of water in the cylindrical part.
∴ OB  75
2   1.52  h 
10 4
 cm 25
2 h
 5 cm 3
25
∵ OVA  OVB (corr. s,  △s) ∴ The depth of water in the cylindrical part is cm .
3
1
∴ OVB  AVB Depth of water = depth of water in the cylindrical part
2 + radius of the hemisphere
1
  60  25 
2    1.5  cm
 30  3 
In △VOB, 59
 cm
OB 6
sin 30 
VB
5 cm
VB 
sin 30
 10 cm

36 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

25. Let h cm be the height of the cylinder. 29. (a) ∵ Volume of the hemisphere
∵ Curved surface area of the cylinder = volume of the cylinder
= curved surface area of the sphere
∴ Volume of the hemisphere  1  36 cm3
∴ 2rh  4r 2 2
h  2r  18 cm3
Let r cm be the radius of the hemisphere.
Volume of the cylinder  r 2 h
1 4
 r 2 (2r ) 18    r 3
2 3
 2r 3 r 3  27
4
Volume of the sphere  r 3 r 3
3 ∴ The radius of the hemisphere is 3 cm.
∵ Volume of the cylinder  volume of the sphere
∴ Ben’s claim is disagreed. (b) Volume of the cylinder
= volume of the hemisphere
4
26. Total volume of the 23 solid spheres  23     23 cm3 = 18 cm3
3
Base radius of the cylinder
736
 cm3 = radius of the hemisphere
3
= 3 cm
Total volume of the solid spheres and water
Let h cm be the height of the cylinder.
 736 
  1500  cm 3 18    32  h
 3 
h2
 2270.737... cm 3 ∴ The height of the cylinder is 2 cm.
Capacity of the conical vessel
1 (c) Total surface area of the solid
    122  15 cm3
3 = curved surface area of the hemisphere
 720 cm3 + curved surface area of the cylinder
+ base area of the cylinder
 2261.947... cm3
1 
 total volume of the 23 solid spheres and water    4    32  2    3  2    32  cm 2
2 
∴ Water will overflow.
 39 cm 2
27. Volume of the steel part
= volume of the cylinder + volume of the hemisphere 30. (a) With the notations in the figure,
 1 4 
=    5  20      103  cm3
2

 2 3 
3500
= cm3
3
Let r cm be the radius of the sphere.
∵ Volume of the sphere  volume of the steel part
4 3500
∴    r3 
3 3
r 3  875 ∵ △VGD ~ △VHC (AAA)
VG GD
r  9.56 (cor. to 3 sig. fig.) ∴  (corr. sides, ~△s)
∴ The radius of the sphere is 9.56 cm. VH HC
12  h 2

28. Let r cm be the internal radius of the bowl. 12 3
1 4 3 36  3h  24
144   r
2 3 3h  12
r 3  216 h4
r 6
External radius of the bowl  (6  0.5) cm (b) Volume of the frustum
= volume of cone VBC – volume of cone VAD
 6.5 cm
∴ Volume of the material required to make the bowl 1 1 
=     3 12     2  (12  4) cm
2 2 3

1 4  3 3 
      6.53  144  cm3
 2 3  76
= cm3
 123 cm3 (cor. to 3 sig. fig.) 3

37 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions
2
31. (a) Curved surface area of cone VAB  3 
33. (a)  
Curved surface area of cone VCD  3  4 
9

49
Let 9k m2 and 49k m2 be the curved surface areas of
cone VAB and cone VCD respectively, where k ≠ 0.
Curved surface area of frustum ACDB
= (49k – 9k) m2
= 40k m2
∴ The ratio of the curved surface area of the small
circular cone to that of the frustum
 9k m 2 : 40k m 2
For similar solids, the ratio of their heights is equal to
the ratio of the lengths of the corresponding edges.  9k : 40k
With the notations in the figure, let PV = x cm.  9 : 40
PV DC

QV HG (b) Let x be the required number of cans of green paint.
x 16 ∵ Number of cans of paint is proportional to the
 area of the surface to be painted.
x9 8
x 40
x 
2 2 9
x9
40
x  2 x  18 x 2
9
x  18  8.89
∴ The height of the original pyramid VABCD is
9
18 cm.
∴ Eric’s claim is agreed.
(b) Height of pyramid VEFGH  (18  9) cm  9 cm
34. (a) Let A0 cm2, A1 cm2 and A2 cm2 be the curved surface
Capacity of the container areas of solids A, B and C respectively.
 volume of pyramid VABCD Let h cm be the height of solid A.
– volume of pyramid VEFGH 2
1 1 
A0  h 
   (16  16)  18   (8  8)  9  cm3  
 3 3  A0  A1  h  h 
2
 1344 cm3 A0 1
 
A0  A1  2 
32. (a) Volume of water in the conical vessel A0 1
2 
1  14  A0  A1 4
       12 cm3
3 2 4 A0  A0  A1
 196 cm 3
A1  3 A0
∴ A0 : A1  1 : 3
(b) (i) Let h cm be the water level of the cylindrical 2
A0  h 
vessel.  
2 A0  A1  A2  h  h  h 
 14 
196       h A0 1
2
 2  
h4 A0  A1  A2  3 
∴ The water level of the cylindrical vessel is A0 1

4 cm. A0  3 A0  A2 9
(ii) Area of surface in contact with water
A0 1
= curved surface area in contact with water 
+ base area 4 A0  A2 9
 9 A0  4 A0  A2
 14  
2
 14 
 2       4       cm2 A2  5 A0
  2  2  
∴ A0 : A2  1 : 5
 105 cm2 ∴ A0 : A1 : A2  1 : 3 : 5
∴ The ratio of the curved surface areas of solids
A to B to C is 1 : 3 : 5.

38 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

(b) Let V0 cm3, V1 cm3 and V2 cm3 be the volumes of ∵ △DFH ~ △DAG (AAA)
solids A, B and C respectively. 2
Area of △ DAG  DA 
V0  h 
3
∴  
  Area of △ DFH  DF 
V0  V1  h  h  2
Area of △ DAG  2 
3  
V0 1 23 cm 2 1
 
V0  V1  2  Area of △ DAG  4  23 cm 2
V0 1  92 cm 2

V0  V1 8 ∴ Area of quadrilateral AGHF
8V0  V0  V1 = area of △ DAG  area of △ DFH
V1  7V0  (92  23) cm 2
∴ V0 : V1  1 : 7  69 cm 2
2
V0  h 
  36. (a) △APB ~ △CPD (AAA)
V0  V1  V2  h  h  h 
3
V0 1 (b) ∵ △APB ~ △CPD
 
V0  V1  V2  3  Area of △CPD  CD 
2

∴  
V0 1 Area of △ APB  AB 

V0  7V0  V2 27 Area of △CPD  5 
2

 
V0 1 45 cm 2 3

8V0  V2 27 25
Area of △CPD   45 cm 2
27V0  8V0  V2 9
V2  19V0  125 cm 2
∴ V0 : V2  1 : 19
∴ V0 : V1 : V2  1 : 7 : 19 (c) ∵ △APB ~ △CPD
∴ The ratio of the volumes of solids A to B to C is CP PD CD 5
∴    (corr. sides, ~△s)
1 : 7 : 19. AP PB AB 3
If we take AP and CP as the bases of △APB and
35. (a) ∵ AD  BC △CPB respectively, the two triangles have the same
∴ AF  DF  BE  EC height.
AF  DF  2 EC Area of △CPB CP
∴ Area of △ APB  AP
AF  DF  2 AF
DF  AF Area of △CPB 5

∵ △DFH ~ △BCH (AAA) 45 cm 2 3
FH DF 5
∴  (corr. sides, ~△s) Area of △CPB   45 cm 2
CH BC 3
AF  75 cm 2

BC If we take PB and PD as the bases of △APB and
EC △APD respectively, the two triangles have the same

BC height.
1 Area of △ APD PD
 ∴ Area of △ APB  PB
2
∴ FH : HC  1 : 2 Area of △ APD 5

45 cm 2 3
(b) If we take FH and HC as the bases of △DFH and 5
Area of △ APD   45 cm 2
△CDH respectively, the two triangles have the same 3
height.  75 cm 2
Area of △ DFH FH ∴ Area of the trapezium
∴ Area of △CDH  HC = area of △APB + area of △CPB
Area of △ DFH 1 + area of △CPD + area of △APD

46 cm 2 2  (45  75  125  75) cm2
1  320 cm2
Area of △ DFH   46 cm 2
2
 23 cm 2

39 © Pearson Education Asia Limited 2022

Junior Secondary Mathematics in Action 3B Full Solutions

37. With the notation in the figure, 38. (a) Let V cm2 be the new volume of the liquid in the
funnel.
3
V  12  3 
 
128  12 
3
V 3
 
128  4 
27
V   128
64
(a) ∵ △VAM  △VBM (RHS)  54
∴ AM = BM (corr. sides,  △s) Volume of liquid in the cylindrical container
AB  (128  54) cm3
∴ MB 
2
 74 cm3
4
 cm Base area of the cylindrical container
2 74
 2 cm  cm 2
5
In △VMB,
 14.8 cm 2
VM  VB 2  MB 2 (Pyth. theorem)
 42  22 cm (b) Capacity of the cylindrical container
 12 cm  14.8  8 cm3
1  118.4 cm3
∴ Area of △VAB   4  12 cm 2
2  128 cm3
 2 12 cm 2 ∴ The water will overflow.
∴ Total surface area of the tetrahedron
39. (a) The frustum can be formed by cutting away a smaller
 4  2 12 cm2 cone from a larger cone.
Let V0 cm3 and V1 cm3 be the volumes of the larger
 27.7 cm2 (cor. to 3 sig. fig.) cone and the smaller cone respectively.
3
V0  6 
(b) ∵ △ABC is an equilateral triangle.  
V1  3 
∴ ABC  60
∵ O is the incentre of △ABC.
V1  105
8
1 V1
∴ OBM  ABC V1  105  8V1
2
1 V1  15
  60
2 V0  15  105  120
 30 Let h0 cm and h1 cm be the heights of the larger cone
In △OMB, and the smaller cone respectively.
OM 1
tan OBM  120     62  h0
MB 3
OM h0  10
tan 30 
2 cm h0 6
OM  2 tan 30 cm 
h1 3
In △VMO,
10 6
VO 2  OM 2  VM 2 (Pyth. theorem) 
h1 3
VO  ( 12) 2  (2 tan 30) 2 cm h1  5
VO  12  4 tan 30 cm 2
∴ h  h0  h1
∴ Volume of the tetrahedron  10  5
1 5
  base area  height
3
1 Alternative Solution
  2 12  12  4 tan 2 30 cm3
3 The frustum can be formed by cutting away a smaller
 7.54 cm3 (cor. to 3 sig. fig.) cone from a larger cone.
Let hL cm and hS cm be the heights of the larger cone
and the smaller cone respectively.
hL 6

hS 3
∴ hL  2hS

40 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

∵ Volume of the frustum  105 cm3 Challenging Questions (p. 7.80)

∴ Volume of the larger cone
1. Let V0 cm3 and V1 cm3 be the volume of water in the
– volume of the smaller cone  105 cm3 hourglass and the volume of the empty space in the upper
1 1 cone respectively.
∴    6 2  (2hS )     32  hS  105
3 3
24hS  3hS  105
hS  5
∴ h  h L  hS 3
 2 hS  hS
V0  5
 
V0  V1  10 
 hS
3
5 V0 1
 
V0  V1  2 
(b) Slant height of the larger cone V0 1

V0  V1 8
 10 2  6 2 cm (Pyth. theorem)
8V0  V0  V1
 136 cm V1  7V0
Slant height of the smaller cone When all the water has flowed down to the lower cone, the
 52  32 cm (Pyth. theorem) volume of the empty space in the lower cone is also equal
to V1 cm3.
 34 cm
Curved surface area of the frustum
 curved surface area of the larger cone
 curved surface area of the smaller cone
Let d cm be the depth of the water in the lower cone.
 (  6  136    3  34) cm 2 V1  10  d 
3

 
 165 cm (cor. to 3 sig. fig.)
2
V0  V1  10 
3
7V0  10  d 
1  
40. (a) Volume of the whole pyramid   (6  6)  18 cm 3 V0  7V0  10 
3 3
 216 cm 3
7  10  d 
 
3
Let V cm be the volume of the pyramid that is above 8  10 
the water surface. 10  d
3
0.956 47 
V  18  3  10
  9.5647  10  d
216  18 
3 d  0.4353
V 5
   0.435 (cor. to 3 sig. fig.)
216  6 
∴ The depth of the water in the lower cone is 0.435 cm.
125
V  216
216 2. (a) Volume of water above the original water level
 125    52  7 cm3
∴ Volume of the pyramid that is below the water
 175 cm3
surface
= (216 – 125) cm3 ∵ Volume of the larger cylinder
 volume of the smaller cylinder  175 cm3
3
= 91 cm
∴ Volume of the smaller cylinder
(b) Volume of water  (6  6  3  91) cm  17 cm
3 3
 175 cm3  volume of the larger cylinder
17  (175π  135π ) cm3
Depth of water  cm
66  40 cm3
 0.472 cm (cor. to 3 sig. fig.)

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Junior Secondary Mathematics in Action 3B Full Solutions

 Base radius of the smaller cylinder 

3
Base radius of the container
(b) ∵   ∵
Base radius of the smaller cylinder
 Base radius of the larger cylinder 
volume of the smaller cylinder height of the container
 
volume of the larger cylinder height of the smaller cylinder
∴ The smaller cylinder is similar to the container.
40 cm3
 ∵ The smaller cylinder is similar to the larger
135 cm3
cylinder.
8 ∴ The container and the two cylinders are of

27 similar shapes.
Base radius of the smaller cylinder 2
∴  Multiple Choice Questions (p. 7.81)
Base radius of the larger cylinder 3
1. Answer: A
Curved surface area of the smaller cylinder 1
Volume of the pyramid   (3  3)  10 cm 3
Curved surface area of the larger cylinder 3
 base radius of the smaller cylinder 
2
 30 cm 3
  
 base radius of the larger cylinder 
2 2. Answer: A
2 Let  cm be the slant height of the cone.
 
3   9    135
4   15

9 Height of the cone
∴ The ratio of the curved surface area of the  152  92 cm (Pyth. theorem)
smaller cylinder to that of the larger cylinder is
4 : 9.  12 cm

(c) Let r1 cm and r2 cm be the base radii of the smaller 3. Answer: C

cylinder and the larger cylinder respectively. Let r cm be the radius of the hemisphere.
r1 2 1 4
From (b),    r 3  18
r2 3 2 3
Consider the base radii of the container and the r 3  27
smaller cylinder, r 3
base radius of the container r r Curved surface area of the hemisphere
 1 2
base radius of the smaller cylinder r1 1
  4    32 cm 2
r2 2
1
r1  56.5 cm 2 (cor. to 1 d.p.)
3
1
2 4. Answer: C
5 4
 a    r3
2 3
Let h1 cm and h2 cm be the heights of the smaller 4
 r 3
cylinder and the larger cylinder respectively. 3
h1 r1 1
 b     r 2  2r
Then, 3
h2 r2
2
2  r 3
 3
3 c    r 2  2r
Consider the heights of the container and the smaller
cylinder,  2r 3
height of the container h h ∴ cab
 1 2
height of the smaller cylinder h1
h2 5. Answer: D
1
h1 Total volume of the spheres  4  4    33 cm3
3
3  144 cm3
1
2 Let r cm be the base radius of the cone.
5 1
    r 2  3  144
2 3
 r 2  144
∴ The base area of the cone is 144 cm2.

42 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

6. Answer: B 10. Answer: C

Height of the cone = (14 – 8) cm = 6 cm
Slant height of the cone
 82  6 2 cm (Pyth. theorem)
 10 cm
Curved surface area of the cone
   8  10 cm2
 80 cm2 With the notations in the figure, let VE = x cm.
Curved surface area of the hemisphere ∵ △VEC ~ △VFB (AAA)
1
  4    82 cm 2 VE EC
2 ∴  (corr. sides, ~△s)
VF FB
 128 cm 2
x 2
Total surface area of the solid 
9 4
 (80  128 ) cm2 x  4 .5
 208 cm2 ∴ VE  4.5 cm
∴ Volume of the frustum
7. Answer: B = volume of cone VAB  volume of cone VDC
Let x cm2 be the area of the shaded region. 1 1 
∵ ABCE ~ EFGH      42  9     22  4.5  cm3
2
3 3 
Area of ABCD  AB 
∴    42 cm3
Area of EFGH  EF 
2
x  72  2  6  2  11. Answer: D
 
72  6  Let d cm be the depth of the water.
2 3
x  72  5  d  9
    
72 3 6 1
 (6  6)  6
25 3
x  72   72 3
9 d  1
  
x  72  200 6 8
x  128 d 1

∴ The area of the shaded region is 128 cm2. 6 2
d 3
8. Answer: C ∴ The depth of the water is 3 cm.
Let x cm2 be the total area of all lateral faces of the larger
pyramid.
2
x  30 
 
A  10 
x
9
A
x  9A
∴ The total area of all lateral faces of the larger is
9A cm2.

9. Answer: A
Let x and y be the heights of the smaller pyramid and the
larger pyramid respectively.
3
x 1
  
y
  27
x 3 1

y 27
x 1

y 3
∴ The required ratio is 1 : 3.

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Junior Secondary Mathematics in Action 3B Full Solutions

Exam Corner

Exam-type Questions (p. 7.84)

Suggested Solutions Marks

1.

With the notations in the figure,

r h

15 60
h
r 
4
∵ Volume of the water = 2304 cm3
2
1 h
∴       h  2304  + 
3 4
h3  110 592
h  48
∴ The depth of the water is 48 cm. 

Alternative Solution
Let h cm be the depth of the water.
3
 h  2304
  1  + 
 60     152  60
3
3
 h  64
  
60
  125
h 4
 
60 5
h  48
∴ The depth of the water is 48 cm. 
2. (a) Let h cm be the height of the cylinder.
∵ Volume of the cylinder = volume of the hemisphere
1 4
∴   6 2  h      63 
2 3
36h  144
h4
∴ The height of the cylinder is 4 cm. 

(b) Total surface area of the solid

 1 
  2    6  4    62   4    62  cm 2  + 
 2 
 156 cm2 

44 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

Suggested Solutions Marks

3. (a) Let r cm be the radius of the smaller sphere.
1
Then, radius of the larger sphere   8r cm  4r cm .
2
Surface area of the smaller sphere
 4    r 2 cm 2
 4 r 2 cm 2

Surface area of the larger sphere
 4    (4r ) 2 cm 2
 64 r 2 cm 2
∵ Sum of the surface areas of two spheres = 612 cm2
∴ 4 r 2  64 r 2  612
68 r 2  612
r2  9
r 3

Surface area of the smaller sphere
 4  32 cm 2
 36 cm 2


Alternative Solution
Let r1 cm and r2 cm be the radii of the smaller sphere and the larger sphere;
S1 cm2 and S2 cm2 be the surface areas of the smaller sphere and the larger
sphere respectively.
r1 1

2r2 8 
r1 1

r2 4
2
S1  1 
 
S2  4  
1

16
Surface area of the smaller sphere
1
 612  cm 2
16  1
 36 cm 2 

(b) Volume of the smaller sphere

4
    33 cm3
3
 36 cm3

Volume of the larger sphere
4
    (4  3)3 cm3
3
 2304 cm3
∴ Sum of the volumes of the two spheres
 (36  2304 ) cm3
 2340 cm3 

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Junior Secondary Mathematics in Action 3B Full Solutions

Suggested Solutions Marks

Alternative Solution
Let V1 cm3 and V2 cm3 be the volumes of the smaller sphere and the larger
sphere respectively.
3
V1  1 
  
V2  4 
1

64
∵ Surface area of the smaller sphere = 36 cm2
∴ 4 r12  36
r1  3
Sum of the volumes of the two spheres
4 64  1
    33  cm3
3 1
 2340 cm3 
4. With the notations in the figure,
r 48  30

36 48
r  13.5 
h 24

48  30 13.5

h  32
1  242  322 cm (Pyth. theorem)
 40 cm 
2  13.52  (48  30) 2 cm (Pyth. theorem)
 22.5 cm
Area of the wet curved surface of the container
 (  24  40    13.5  22.5) cm 2 
 (960  303.75 ) cm 2

 656.25 cm 2 
5. (a) Let V1 cm3 and V2 cm3 be the volumes of the larger pyramid and the smaller
pyramid respectively.
3
V1  3 
  
V2  2 
V1 27

V2 8
27
V1  V2
8
∵ Total volume of the two pyramids = volume of the prism
∴ V1  V2  250  7 
27
V  V2  1750
8
35
V2  1750
8
V2  400
∴ The volume of the smaller pyramid is 400 cm3. 

46 © Pearson Education Asia Limited 2022

 7 Areas and Volumes (III)

Suggested Solutions Marks

(b) Side length of the base of the smaller pyramid
 100 cm
 10 cm
Let h cm be the height of the smaller pyramid.
1
100  h  400 
3
h  12
Total surface area of the smaller pyramid
 1 
 10  
2
 
 4    10  122      100  cm 2 
  2  2   
   
 360 cm 2

Let A cm2 be the total surface area of the larger pyramid.

2
A 3
  
360  2 
9
A   360
4
 810
∴ The total surface area of the larger pyramid is 810 cm2. 
6. Answer: C
Let x m2 be the actual area of the theme park.
2
 1  50 cm 2
  
 40 000  x m2
2
 1  50  100  100 m 2
  
 40 000  x m2
x  0.005  40 000 2
 8  106
∴ The actual area of the theme park is 8 106 m2 .
7. Answer: A
Let h be the heights of the two cones.
Volume of the cone with radius R
1
    R2  h
3
1
  R2h
3
Volume of the cone with radius r
1
   r2  h
3
1 2
 r h
3
1 1
 R2h  2   r 2h
3 3
R  2r
2 2

2
R 2
  
r 1
R 2

r 1
∴ R:r  2 :1

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Junior Secondary Mathematics in Action 3B Full Solutions

Suggested Solutions Marks

8. Answer: A
If we take AD and AE as the bases of △ADC and △ACE respectively, the two
triangles have the same height.
Area of △ ACE AE
∴ 
Area of △ ADC AD
Area of △ ACE 3

200 cm 2 3 2
3
Area of △ ACE   200 cm 2
5
 120 cm 2
∵ △ADC ~ △AEB (AAA)
2
Area of △ AEB  AE 
∴  
Area of △ ADC  AD 
2
Area of △ AEB  3 
 
200 cm 2  3 2 
9
Area of △ AEB   200 cm 2
25
 72 cm 2
Area of △BEC  (120  72) cm
2

 48 cm 2

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