CHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON
SOLIDS
EXERCISE 124 Page 293
1. Change a volume of 1 200 000 cm3 to cubic metres.
1m3 = 106 cm3 or 1cm3 = 10− 6 m3
Hence,
1 200 000 cm3 = 1 200 000 ×10−6 m3
= 1.2 m3
2. Change a volume of 5000 mm3 to cubic centimetres.
1cm3 = 103 mm3 or 1mm3 = 10− 3 cm3
Hence,
5000 mm3 = 5000 ×10−3 mm3 = 5 cm3
3. A metal cube has a surface area of 24 cm 2 . Determine its volume.
A cube had 6 sides. Area of each side = 24/6 = 4 cm 2
Each side is a square hence the length of a side =
4 = 2 cm
Volume of cube = 2 × 2 × 2 = 8 cm3
4. A rectangular block of wood has dimensions of 40 mm by 12 mm by 8 mm. Determine (a) its
volume in cubic millimetres and (b) its total surface area in square millimetres.
(a) Volume of cuboid = l × b × h = 40 × 12 × 8 = 3840 mm3
(b) Surface area = 2(bh + hl + lb)
= 2(12 × 8 + 8 ×40 + 40 × 12)
= 2(96 + 320 + 480)
= 2 × 896 = 1792 mm 2
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© 2014, John Bird
5. Determine the capacity, in litres, of a fish tank measuring 90 cm by 60 cm by 1.8 m, given
1 litre = 1000 cm3 .
Volume = (90 × 60 × 180) cm3
Tank capacity =
90 × 60 ×180 cm3
= 972 litre
1000 cm3 /litre
6. A rectangular block of metal has dimensions of 40 mm by 25 mm by 15 mm. Determine its
volume in cm3 . Find also its mass if the metal has a density of 9 g/cm3.
Volume = length × breadth × width = 40 × 25 × 15 = 15 000 mm3
= 15 000 × 10−3 cm3 = 15 cm3
Mass = density × volume = 9 g/cm3 × 15 cm3 = 135 g
7. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by 2.5 m
(1 litre = 1000 cm3).
Volume = 50 × 40 × 250 cm3
Tank capacity =
50 × 40 × 250 cm3
= 500 litre
1000 cm3 /litre
8. Determine how many cubic metres of concrete are required for a 120 m long path, 150 mm wide
and 80 mm deep.
Width = 150 mm = 0.15 m and depth = 80 mm = 0.080 m
Hence, volume of path = length × breadth × width = 120 × 0.15 × 0.080 = 1.44 m3
i.e.
concrete required = 1.44 m3
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© 2014, John Bird
9. A cylinder has a diameter 30 mm and height 50 mm. Calculate (a) its volume in cubic
centimetres, correct to 1 decimal place, and (b) the total surface area in square centimetres,
correct to 1 decimal place.
Diameter = 30 mm = 3 cm hence radius, r = 3/2 = 1.5 cm and height, h = 50 mm = 5 cm
(a) Volume = π r 2 h = π × (1.5 ) × 5 = 11.25π = 35.3 cm3 , correct to 1 decimal place
2
(b) Total surface area = 2πrh + 2π r 2
= (2 × π × 1.5 × 5) + (2 × π × 1.52 )
= 15π + 4.5π = 19.5π = 61.3 cm 2
10. Find (a) the volume and (b) the total surface area of a right-angled triangular prism of length
80 cm whose triangular end has a base of 12 cm and perpendicular height 5 cm.
(a) Volume of right-angled triangular prism =
i.e.
1
1
bhl = × 12 × 5 × 80
2
2
Volume = 2400 cm3
(a) Total surface area = area of each end + area of three sides
2
In triangle ABC, AC
=
AB 2 + BC 2
from which,
AC =
AB 2 + BC 2 = 52 + 122 = 13 cm
1 
Hence, total surface area = 2  bh  + (AC × 80) + (BC × 80) + (AB × 80)
2 
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© 2014, John Bird
= (12 × 5) + (13 × 80) + (12 × 80) + (5 × 80)
= 60 + 1040 + 960 + 400
i.e.
total surface area = 2460 cm 2
11. A steel ingot whose volume is 2 m 2 is rolled out into a plate which is 30 mm thick and 1.80 m
wide. Calculate the length of the plate in metres.
Volume of ingot = length × breadth × width
i.e.
2 = l × 0.030 × 1.80
from which,
2
length = 0.030 ×1.80 = 37.04 m
12. The volume of a cylinder is 75 cm3 . If its height is 9.0 cm, find its radius.
Volume of cylinder, V = π r 2 h
i.e.
75 = π r 2 (9.0)
from which,
r2 =
75
π × 9.0
75
= 1.63 cm
π × 9.0
and radius, r =
13. Calculate the volume of a metal tube whose outside diameter is 8 cm and whose inside diameter
is 6 cm, if the length of the tube is 4 m.
Outer diameter, D = 8 cm and inner diameter, d = 6 cm
Area of cross-section of copper =
π D2
4
−
πd2
4
=
π (8)
4
2
−
π (6)
2
4
= 50.2655 – 28.2743 = 21.9912 cm 2
Hence, volume of metal tube = (cross-sectional area) × length of pipe
= 21.9912 × 400 = 8796 cm3
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© 2014, John Bird
14. The volume of a cylinder is 400 cm3. If its radius is 5.20 cm, find its height. Determine also its
curved surface area.
Volume of cylinder, V = π r 2 h
400 = π (5.20) 2 h
i.e.
from which, height, h =
400
= 4.709 cm
π × (5.20) 2
Curved surface area = 2πrh = (2 × π × 5.20 × 4.709)
= 153.9 cm 2
15. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by 12 cm. If the length of the
cylinder is to be 60 cm, find its diameter.
Volume of rectangular piece of alloy = 5 × 7 × 12 = 420 cm3
Volume of cylinder = π r 2 h
Hence,
420 = π r 2 (60)
r2
from which,=
420
7
=
π (60) π
and
radius, r =
7
π
= 1.4927 cm
and diameter of cylinder, d = 2r = 2 × 1.4927 = 2.99 cm
16. Find the volume and the total surface area of a regular hexagonal bar of metal of length 3 m if
each side of the hexagon is 6 cm.
A hexagon is shown below.
In triangle 0BC, tan 30° =
3
x
from which,
x=
3
= 5.196 cm
tan 30°
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© 2014, John Bird
1

Hence, area of hexagon = 6  × 6 × 5.196  = 93.53 cm 2
2

and volume of hexagonal bar = 93.53 × 300 = 28 060 cm3
Surface area of bar = 6 [ 0.06 × 3] + 2 [93.53 ×10−4 ]
in metre units
= 1.099 m 2
17. A block of lead 1.5 m by 90 cm by 750 mm is hammered out to make a square sheet 15 mm
thick. Determine the dimensions of the square sheet, correct to the nearest centimetre.
Volume of block of lead = length × breadth × height
= 150 × 90 × 75 cm3
If length = breadth = x cm and height = 15/10 = 1.5 cm, then
x 2 (1.5 ) = 150 × 90 × 75
from which,
x2 =
150 × 90 × 75
1.5
and
x=
150 × 90 × 75
= 821.6 cm = 8.22 m
1.5
Hence, dimensions of square sheet are 8.22 m by 8.22 m
18. How long will it take a tap dripping at a rate of 800 mm3 /s to fill a three-litre can?
3 litre = 3000 cm3 = 3000 ×103 mm3 =
3 ×106 mm3
Time to fill can =
3750
3 ×106 mm3
= 3750 s =
= 62.5 minutes
3
60
800 mm /s
19. A cylinder is cast from a rectangular piece of alloy 5.20 cm by 6.50 cm by 19.33 cm. If the
height of the cylinder is to be 52.0 cm, determine its diameter, correct to the nearest centimetre.
Volume of cylinder, V = π r 2 h and volume of rectangular prism = 5.20 × 6.50 ×19.33 cm3
i.e.
5.20 × 6.50 ×19.33 = π r 2 (52.0)
from which,
r2 =
5.20 × 6.50 ×19.33
π × 52.0
and radius, r =
473
5.20 × 6.50 ×19.33
= 2.0 cm
π × 52.0
© 2014, John Bird
i.e. diameter = 2 × r = 2 × 2.0 = 4 cm
20. How much concrete is required for the construction of the path shown, if the path is 12 cm
thick?
2
1
Area of path = (8.5 × 2) +  π ( 2 )  + (3.1 × 2) + (2.4 × [2 + 1.2])
4

= 17 + 3.1416 + 6.2 + 7.68 = 34.022 m 2
If thickness of path = 12 cm = 0.12 m then
Concrete required = volume of path = 34.022 × 0.12 = 4.08 m3
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© 2014, John Bird
EXERCISE 125 Page 296
1. If a cone has a diameter of 80 mm and a perpendicular height of 120 mm, calculate its volume in
cm3 and its curved surface area.
Volume of cone =
1 2
1
2
π r h = π ( 40 ) (120 ) = 201 061.9 mm3 = 201.1 cm3
3
3
From diagram below, slant height, l =
(122 + 42 )
= 12.649 cm
Curved surface area = πrl = π(4)(12.649) = 159.0 cm 2
2. A square pyramid has a perpendicular height of 4 cm. If a side of the base is 2.4 cm long find the
volume and total surface area of the pyramid.
A sketch of the pyramid is shown below.
Volume of pyramid =
1
( 2.4 × 2.4 )( 4 ) = 7.68 cm3
3
In the sketch, AB = 4 cm and BC = 2.4/2 = 1.2 cm
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© 2014, John Bird
Length AC =
( 42 + 1.22 )
Hence, area of a side =
= 4.176 cm
1
( 2.4 )( 4.176 ) = 5.01 cm 2
2
Total surface area of pyramid = 4 [5.01] + ( 2.4 ) = 25.81 cm 2
2
3. A sphere has a diameter of 6 cm. Determine its volume and surface area.
3
Volume of sphere =
4 3 4 6
π r = π   = 113.1 cm3
3
3 2
2
6
Surface are of sphere = 4π r 2 = 4π   = 113.1 cm 2
2
4. If the volume of a sphere is 566 cm3, find its radius.
Volume of sphere =
4 3
πr
3
hence, 566 =
=
r3
from which,
4 3
πr
3
3 × 566
= 135.12255
4π
radius, r = 3 135.12255 = 5.131 cm
and
5. A pyramid having a square base has a perpendicular height of 25 cm and a volume of 75 cm3 .
Determine, in centimetres, the length of each side of the base.
If each side of base = x cm then volume of pyramid =
1 2
x (25)
3
and x 2 =
1
1
× A × h = x2h
3
3
3 × 75
=9
25
i.e.
75 =
from which,
length of each side of base =
9 = 3 cm
6. A cone has a base diameter of 16 mm and a perpendicular height of 40 mm. Find its volume
correct to the nearest cubic millimetre.
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© 2014, John Bird
2
1
1  16 
Volume of cone = π r 2 h = π   ( 40 ) = 2681 mm3
3
3  2
7. Determine (a) the volume and (b) the surface area of a sphere of radius 40 mm.
(a) Volume of sphere =
4 3 4
3
π r = π ( 40 ) = 268 083 mm3 or 268.083 cm3
3
3
(b) Surface are of sphere = 4π r 2 = 4π ( 40 ) = 20 106 mm 2 or 201.06 cm 2
2
8. The volume of a sphere is 325 cm3. Determine its diameter.
4
4 d 
Volume of sphere = π r 3 = π  
3
3 2
Hence,
and
4 d 
325 = π  
3 2
3
3
 d  3 × 325
from which,   =
4π
2
3
d 3 3 × 325
= 4.265 cm and diameter, d = 2 × 4.265 = 8.53 cm
=
2
4π
9. Given the radius of the Earth is 6380 km, calculate, in engineering notation (a) its surface area
in km 2 and (b) its volume in km3 .
(a) Surface area of Earth = 4π r 2 = 4π ( 6380 ) = 512 ×106 km 2
2
(b) Volume of earth =
4 3 4
3
π r = π ( 6380 ) = 1.09 ×1012 km3
3
3
10. An ingot whose volume is 1.5 m3 is to be made into ball bearings whose radii are 8.0 cm. How
many bearings will be produced from the ingot, assuming 5% wastage?
Volume of one ball bearing =
4 3 4
3
π r = π (8)
3
3
Let number of ball bearings = x
Volume of x bearings = 0.95 × 1.5 × 106 cm3
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© 2014, John Bird
Hence,
0.95 × 1.5 × 106 =
from which, number of bearings, x =
4
3
π ( 8 ) [ x]
3
0.95 ×1.5 ×106 × 3
= 664
4π × 83
11. A spherical chemical storage tank has an internal diameter of 5.6 m. Calculate the storage
capacity of the tank, correct to the nearest cubic metre. If 1 litre = 1000 cm3 , determine the tank
capacity in litres.
3
4
4  5.6 
3
Volume of storage tank = π r 3 = π 
 = 91.95 = 92 m , correct to the nearest cubic metre
3
3  2 
Volume of tank = 92 ×106 cm3
If 1 litre = 1000 cm3 , then capacity of storage tank =
478
92 ×106
litres = 92 000 litres
1000
© 2014, John Bird
EXERCISE 126 Page 300
1. Find the total surface area of a hemisphere of diameter 50 mm.
Total surface area = π r 2 +
1
[ 4π r 2 ] =π r 2 + 2π r 2 =3π r 2
2
2
 50 
= 3π   = 5890 mm 2 or 58.90 cm 2
 2 
2. Find (a) the volume and (b) the total surface area of a hemisphere of diameter 6 cm.
Volume of hemisphere =
1
(volume of sphere)
2
3
2
2  6.0 
3
= πr3 = π 
 = 56.55 cm
3
3  2 
Total surface area = curved surface area + area of circle
=
1
(surface area of sphere) + πr2
2
=
1
 6.0 
(4πr 2 ) + πr2 = 2πr2 + πr2 = 3πr2 = 3π 

2
 2 
2
= 84.82 cm2
3. Determine the mass of a hemispherical copper container whose external and internal radii are
12 cm and 10 cm, assuming that 1 cm3 of copper weighs 8.9 g.
2 3 2
Volume of hemisphere = =
πr
π [123 − 103 ] cm3
3
3
Mass of copper = volume × density =
2
π [123 − 103 ] cm3 × 8.9 g/cm3 = 13570 g = 13.57 kg
3
4. A metal plumb bob comprises a hemisphere surmounted by a cone. If the diameter of the
hemisphere and cone are each 4 cm and the total length is 5 cm, find its total volume.
The plumb bob is shown sketched below.
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© 2014, John Bird
Volume of bob =
1 2
2 3 1
2
2
3
π r h +=
πr
π ( 2) (5 − 2) + π ( 2)
3
3
3
3
= 4π +
16
π = 29.32 cm3
3
5. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m and the
cylindrical portion has a height of 3.5 m, with a diameter of 15 m. Calculate the surface area of
material needed to make the marquee, assuming 12% of the material is wasted in the process
The marquee is shown sketched below.
Surface area of material for marquee = π rl + 2π rh
where l =
( 7.52 + 2.52 )
= 7.9057 m
Hence, surface area = π(7.5)(7.9057) + 2π(7.5)(3.5)
= 186.2735 + 164.9336 = 351.2071 m 2
If 12% of material is wasted then amount required = 1.12 × 351.2071 = 393.4 m 2
6. Determine (a) the volume and (b) the total surface area of the following solids:
(i) a cone of radius 8.0 cm and perpendicular height 10 cm
(ii) a sphere of diameter 7.0 cm
(iii) a hemisphere of radius 3.0 cm
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© 2014, John Bird
(iv) a 2.5 cm by 2.5 cm square pyramid of perpendicular height 5.0 cm
(v) a 4.0 cm by 6.0 cm rectangular pyramid of perpendicular height 12.0 cm
(vi) a 4.2 cm by 4.2 cm square pyramid whose sloping edges are each 15.0 cm
(vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 20 cm
(i) A sketch of the cone is shown below.
(a) Volume of cone =
1 2
1
2
π r h = π ( 8.0 ) (10 ) = 670 cm3
3
3
(b) Total surface area = π r 2 + π rl
where
l=
(102 + 8.02 )
= 12.80625 cm
= π ( 8.0 ) + π ( 8.0 )(12.80625 )
2
= 201.062 + 321.856 = 523 cm 2
3
4  7.0 
3
(ii) (a) Volume of sphere = π 
 = 180 cm
3  2 
2
 7.0 
2
(b) Surface area = 4π r = 4π 
 = 154 cm
2


2
(iii) (a) Volume of hemisphere =
(b) Surface area =
2 3 2
3
π r = π ( 3.0 ) = 56.5 cm3
3
3
1
2
(4π r 2 ) + π r 2 = 3π r 2 = 3π ( 3.0 ) = 84.8 cm 2
2
(iv) A sketch of the square pyramid is shown below, where AB = 5.0 cm
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© 2014, John Bird
1
2
( 2.5) ( 5.0 ) = 10.4 cm3
3
(a) Volume of pyramid =
( AB 2 + BC 2 ) = ( 5.02 + 1.252 )
(b) In the diagram, AC =
= 5.15388
2
1

Surface area = ( 2.5 ) + 4  × 2.5 × 5.15388 = 6.25 + 25.7694 = 32.0 cm 2
2

(v) A sketch of the rectangular pyramid is shown below.
(a) Volume of rectangular pyramid =
(b) In the diagram, AC =
and
AD =
1
( 6.0 × 4.0 )(12.0 ) = 96.0 cm3
3
(12.02 + 3.02 )
(12.02 + 2.02 )
= 12.3693 cm
= 12.1655 cm
1

1

Hence, surface area = ( 6.0 × 4.0 ) + 2  × 4.0 × 12.3696  + 2  × 6.0 × 12.1655
2

2

= 24 + 49.4784 + 72.993 = 146 cm 2
(vi) The square pyramid is shown sketched below.
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© 2014, John Bird
Diagonal on base =
5.9397
( 4.22 + 4.22 ) =
Hence, perpendicular height, h =
(a) Volume of pyramid =
(15.02 − 2.969852 )
1
× 5.9397 = 2.96985 cm
2
= 14.703 cm
1
2
( 4.2 ) (14.703) = 86.5 cm3
3
(14.7032 + 2.12 )
(b) AD =
cm hence, BC =
= 14.8522
2
1

Hence, surface area = ( 4.2 ) + 4  × 4.2 × 14.8522  = 17.64 + 124.75858
2

= 142 cm 2
(vii) A pyramid having an octagonal base is shown sketched below.
One sector is shown in diagram (p) below, where tan 22.5° =
from which,
x=
2.5
x
2.5
= 6.0355 cm
tan 22.5°
1

Hence, area of whole base = 8  × 5.0 × 6.0355 = 120.71 cm 2
2

(a) Volume of pyramid =
1
(120.71)( 20 ) = 805 cm3
3
(p)
(b) From diagram (q) above, y =
(q)
( 202 + 6.03552 )
483
= 20.891 cm
© 2014, John Bird
1

Total surface area = 120.71 + 8  × 5.0 × 20.891 = 120.71 + 417.817
2

= 539 cm 2
7. A metal sphere weighing 24 kg is melted down and recast into a solid cone of base radius 8.0 cm.
If the density of the metal is 8000 kg/m3 determine (a) the diameter of the metal sphere and (b) the
perpendicular height of the cone, assuming that 15% of the metal is lost in the process.
Volume of sphere =
mass
24 kg
3
=
= 0.003m
=
0.003 ×106 cm3 = 3000 cm3
density 8000 kg/m3
(a) Volume of sphere =
4 3
πr
3
and
i.e. 3000 =
radius, r =
4 3
πr
3
3
 3000 × 3 

 = 8.947 cm
 4π 
Hence, the diameter of the sphere, d = 2r = 2 × 8.947 = 17.9 cm
(b) Volume of cone = 0.85 × 3000 = 2550 cm3 =
from which, perpendicular height of cone, h =
1 2
1
2
π r h = π ( 8.0 ) h
3
3
2550 × 3
π ( 8.0 )
2
= 38.0 cm
8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the
side of base is 3.0 cm.
The hexagonal base is shown sketched below.
From the diagram, tan 30° =
1.5
1.5
from which, h =
= 2.598 cm
h
tan 30°
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© 2014, John Bird
1

Hence, area of hexagonal base = 6  × 3.0 × 2.598 = 23.3827 cm 2
2

and volume of hexagonal pyramid =
1
( 23.3827 )(16.0 ) = 125 cm3
3
9. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere
is 2.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the
buoy.
The buoy is shown in the sketch below.
( 4.02 − 1.252 )
Height of cone, h =
Volume of buoy =
= 3.80 m
2 3 1
2
1
3
2
π r + π r 2 h= π (1.25 ) + π (1.25 ) ( 3.80 )
3
3
3
3
= 4.0906 + 6.2177 = 10.3 m3
Surface area = π rl +
1
2
2
( 4π r=
) π (1.25)( 4.0 ) + 2π (1.25)
2
= 5π + 3.125π = 8.125π = 25.5 m 2
10. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical
section surmounted on each end. If the diameters of the hemisphere and cylinder are both 1.2 m
determine the capacity of the tank in litres (1 litre = 1000 cm3).
The petrol container is shown sketched below.
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© 2014, John Bird
4
3
2
2

Volume of container = 2  π r 3  + π r 2=
h
π ( 0.6 ) + π ( 0.6 ) ( 5.0 )
3
3

= 0.288π + 1.8π = 6.55965 m3
= 6.55965 ×106 cm3
and
tank capacity =
6.56 ×106 cm3
= 6560 litres
1000 cm3 /litre
11. The diagram below shows a metal rod section. Determine its volume and total surface area.
Volume of rod =
1 2
1
2
π r h=
+ (l × b × w)
π (1.0 ) (100 ) + (2.5 × 2.0 ×100)
2
2
= 50π +500 = 657.1 cm3
Surface area =
1
1
( 2π r h ) + 2  π r 2  + 2 ( 2.50 × 2.0 ) + 2 ( 2.5 ×100 ) + ( 2.0 ×100 )
2
2

= π(1.0)(100) + π (1.0 ) + 10 + 500 + 200
2
= 1027 cm 2
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© 2014, John Bird
12. Find the volume (in cm3 ) of the die-casting shown below. The dimensions are in millimetres.
Volume = 100 × 60 × 25 +
1
(π × 302 × 50 )
2
= 150 000 + 22 500π = 220 685.835 mm3
= 220 685.835 ×10−3 cm3 = 220.7 cm3
13. The cross-section of part of a circular ventilation shaft is shown below, ends AB and CD being
open. Calculate (a) the volume of the air, correct to the nearest litre, contained in the part of the
system shown, neglecting the sheet metal thickness, (given 1 litre = 1000 cm3 ),
(b) the cross-sectional area of the sheet metal used to make the system, in square metres, and
(c) the cost of the sheet metal if the material costs £11.50 per square metre, assuming that 25%
extra metal is required due to wastage.
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© 2014, John Bird
2
3
2
2
1  4  50  
 50 
 50 
 80 
(a) In cm , volume of air = π   ( 200 ) +  π    + π   (150 ) + π   (150 )
4  3  2  
 2 
 2 
 2 
3
= 125 000π + 5208.33π + 93 750π + 240 000π = 463 958.33π cm3
=
463 958.33π cm3
= 1458 litres, correct to the nearest litre
1000 cm3 /litre
(b) In m 2 , cross-sectional area of the sheet metal
1
2
= 2π ( 0.25 )( 2 ) +  4π ( 0.25 )  + 2π ( 0.25 )(1.5 ) + 2π ( 0.4 )(1.5 ) + π ( 0.42 − 0.252 )

4
= π + 0.0625π + 0.75π + 1.2π + 0.0975π
= 3.11π = 9.77035 m 2 = 9.77 m 2 correct to 3 significant figures
(c) Sheet metal required = 9.77035 × 1.25 m 2
Cost of sheet metal = 9.77035 × 1.25 × £11.50 = £140.45
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© 2014, John Bird
EXERCISE 127 Page 305
1. The radii of the faces of a frustum of a cone are 2.0 cm and 4.0 cm and the thickness of the
frustum is 5.0 cm. Determine its volume and total surface area.
A sketch of a side view of the frustum is shown below.
Volume of frustum =
1
π h ( R 2 + Rr + r 2 )
3
=
1
1
π ( 5.0 )( 4.02 + (4.0)(2.0) + 2.02 ) =
π ( 5.0 )( 28.0 ) = 147 cm3
3
3
From the diagram below, slant length, l =
29
( 5.02 + 2.02 ) =
Total surface area = π l ( R + r ) + π r 2 + π R 2
=π
(
)
29 ( 4.0 + 2.0 ) + π ( 2.0 ) + π ( 4.0 )
2
2
= 32.31π + 4π + 16π = 164 cm 2
2. A frustum of a pyramid has square ends, the squares having sides 9.0 cm and 5.0 cm, respectively.
Calculate the volume and total surface area of the frustum if the perpendicular distance between
its ends is 8.0 cm.
A side view of the frustum of the pyramid is shown below.
By similar triangles:
CG BH
=
BG AH
from which, height, CG = ( BG )
489
BH
 8.0 
= ( 2.5 ) 
 = 10.0 cm
AH
 2.0 
© 2014, John Bird
Height of complete pyramid = 10.0 + 8.0 = 18.0 cm
Volume of large pyramid =
1
2
( 9.0 ) (18.0 ) = 486 cm3
3
Volume of small triangle cut off =
1
2
( 5.0 ) (10.0 ) = 83.33 cm3
3
Hence, volume of frustum = 486 – 83.33 = 403 cm3
A cross-section of the frustum is shown below.
BC =
( 82 + 2 2 )
= 8.246 cm
1

230.888 cm 2
Area of four trapeziums = 4  ( 5.0 + 9.0 )( 8.246 )  =
2

Total surface area of frustum = 9.02 + 5.02 + 230.888 = 337 cm 2
3. A cooling tower is in the form of a frustum of a cone. The base has a diameter of 32.0 m, the top
has a diameter of 14.0 m and the vertical height is 24.0 m. Calculate the volume of the tower and
the curved surface area.
A sketch of the cooling tower is shown below.
490
© 2014, John Bird
Volume of frustum =
=
Slant length, l =
1
π h ( R 2 + Rr + r 2 )
3
1
π ( 24.0 )(16.02 + (16.0)(7.0) + 7.02 ) =
8π ( 417 ) = 10 480 m3
3
( AB 2 + BC 2 =)
( 24.0 + (16.0 − 7.0) =)
2
2
25.632 m
Curved surface area =
= π l ( R + r ) π ( 25.632 )(
=
16.0 + 7.0 ) 599.54π = 1852 m 2
4. A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 28.0 cm
and 6.00 cm and the vertical distance between the ends is 30.0 cm, find the area of material
needed to cover the curved surface of the speaker.
A sketch of the loudspeaker diaphragm is shown below.
Slant length, l =
( AC 2 + AB 2 =)
(30.0 + (14.0 − 3.0) =)
2
2
31.953cm
Curved surface area = πl (R + r) = π(31.953)(14.0 + 3.0) = 1707 cm 2
491
© 2014, John Bird
5. A rectangular prism of metal having dimensions 4.3 cm by 7.2 cm by 12.4 cm is melted down and
recast into a frustum of a square pyramid, 10% of the metal being lost in the process. If the ends of
the frustum are squares of side 3 cm and 8 cm respectively, find the thickness of the frustum.
Volume of frustum of pyramid = 90% of volume of rectangular prism
= 0.9(4.3 × 7.2 × 12.4) = 345.514 cm3 .
A cross-section of the frustum of the square pyramid is shown below (not to scale).
By similar triangles:
CG BH
=
BG AH
Volume of large pyramid =
BH
 h 
= (1.5 ) 
 = 0.6 h
AH
 2.5 
1 2
(8) ( h + 0.6h ) = 34.133h cm3
3
Volume of small triangle cut off =
Hence,
from which, height, CG = ( BG )
1 2
( 3) ( 0.6 h ) = 1.8 h cm3
3
345.514 = 34.133h – 1.8h = 32.333h
Thus, thickness of frustum, h =
345.514
= 10.69 cm
32.333
6. Determine the volume and total surface area of a bucket consisting of an inverted frustum of a
cone, of slant height 36.0 cm and end diameters 55.0 cm and 35.0 cm.
A sketch of the bucket is shown below.
492
© 2014, John Bird
Thickness of frustum, h =
Volume of frustum =
=
( 36.02 − (27.5 − 17.5)2 )
= 34.5832 cm
1
π h ( R 2 + Rr + r 2 )
3
1
π ( 34.5832 )( 27.52 + (27.5)(17.5) + 17.52 ) = 55 910 cm3 correct to 4
3
significant figures
Total surface area = π l ( R + r ) + π r 2
= π ( 36.0 )( 27.5 + 17.5 ) + π (17.5 )
2
= 1620π + 306.25π = 1926.25π = 6051 cm 2
7. A cylindrical tank of diameter 2.0 m and perpendicular height 3.0 m is to be replaced by a tank of
the same capacity but in the form of a frustum of a cone. If the diameters of the ends of the
frustum are 1.0 m and 2.0 m, respectively, determine the vertical height required.
Volume of cylinder
= π r 2 h π=
=
(1.0 ) ( 3.0 ) 3π m3
2
A sketch of the frustum of a cone is shown below.
493
© 2014, John Bird
Volume of frustum = 3π =
=
1
π h ( R 2 + Rr + r 2 )
3
1
1
π ( h )(1.02 + (1.0)(0.5) + 0.52 ) =
π h (1.75 )
3
3
from which, thickness of frustum = vertical height, h =
494
3π
1
π (1.75 )
3
=
9
= 5.14 m
1.75
© 2014, John Bird
EXERCISE 128 Page 307
1. Determine the volume and surface area of a frustum of a sphere of diameter 47.85 cm, if the radii
of the ends of the frustum are 14.0 cm and 22.0 cm and the height of the frustum is 10.0 cm.
Volume of frustum of sphere =
=
πh
6
( h2 + 3r12 + 3r2 2 )
π (10.0 )
6
(10.0 + 3 (14.0) + 3 ( 22.0) )
2
2
2
= 11205 cm3 = 11 210 cm3 correct to 4 significant figures
 47.85 
2
Surface area of frustum = 2πrh = 2π 
 (10.0 ) = 1503 cm
 2 
2. Determine the volume (in cm3) and the surface area (in cm2) of a frustum of a sphere if the
diameter of the ends are 80.0 mm and 120.0 mm and the thickness is 30.0 mm.
The frustum is shown shaded in the cross-section sketch below (in cm units).
Volume of frustum of sphere =
=
=
πh
6
( h2 + 3r12 + 3r2 2 )
π ( 3.0 ) 
6
π
2
2
2
 8.0 
 12.0  
 3.0 + 3 
 + 3
  in cm units
2
2



 

2
( 9 + 48 + 108) = 259.2 cm3
Surface area of frustum = 2πrh
From the above diagram,
2
r=
62 + OP 2
495
(1)
© 2014, John Bird
2
r=
42 + OQ 2
Now OQ = 3 + OP
r 2 =42 + ( 3.0 + OP )
Hence,
2
(2)
Equating equations (1) and (2) gives:
62 + OP 2 = 42 + ( 3.0 + OP )
2
36 + OP 2 = 16 + 9 + 6(OP) + OP 2
i.e.
Thus,
36 = 25 + 6(OP)
from which,
From equation (1),
OP =
 11 
r= 6 +  
6
2
2
36 − 25 11
=
6
6
2

 11  
2
6 +    = 6.274 cm
 6  

2
and
radius, r =
Surface area of frustum = 2πrh = 2π(6.274)(3.0) = 118.3 cm 2
3. A sphere has a radius of 6.50 cm. Determine its volume and surface area. A frustum of the sphere
is formed by two parallel planes, one through the diameter and the other at a distance h from the
diameter. If the curved surface area of the frustum is to be
1
of the surface area of the sphere, find
5
the height h and the volume of the frustum.
Volume of sphere =
4 3 4
3
π r = π ( 6.50 ) = 1150 cm3
3
3
Surface area = 4π r 2 = 4π ( 6.50 ) = 531 cm 2
2
The frustum is shown shaded in the sketch below.
496
© 2014, John Bird
Curved surface area = 2πrh =
2π(6.50)h =
i.e.
from which,
1
( 531) cm 2 in this case
5
1
( 531)
5
1
( 531)
5
height, h =
= 2.60 cm
2π ( 6.50 )
From the diagram,
=
r1
( 6.502 − 2.602 )
Volume of frustum of sphere =
=
πh
6
= 5.957 cm
( h2 + 3r12 + 3r2 2 )
π ( 2.60 )
6
( 2.60 + 3 ( 6.50) + 3 (5.957 ) )
2
2
2
= 326.7 cm3
4. A sphere has a diameter of 32.0 mm. Calculate the volume (in cm3) of the frustum of the sphere
contained between two parallel planes distances 12.0 mm and 10.00 mm from the centre and on
opposite sides of it.
A cross-section of the frustum is shown in the sketch below.
From the diagram,
=
r1
(16.02 − 12.02 )
= 10.583 mm
=
r2
(16.02 − 10.02 )
= 12.490 mm
and
Volume of frustum of sphere =
πh
6
( h2 + 3r12 + 3r2 2 )
497
© 2014, John Bird
=
π ( 22.0 )
6
( 22.0 + 3 (10.583) + 3 (12.490) )
2
2
2
= 14837 mm3 = 14.84 cm3
5. A spherical storage tank is filled with liquid to a depth of 30.0 cm. If the inner diameter of the
vessel is 45.0 cm, determine the number of litres of liquid in the container (1 litre = 1000 cm3).
A cross-section of the storage tank is shown sketched below.
Volume of water =
2 3
π r + volume of frustum
3
From the diagram,
=
r1
( 22.502 − 7.502 )
Volume of frustum of sphere =
=
πh
6
= 21.21 cm
( h2 + 3r12 + 3r2 2 )
π ( 7.5 )
6
( 7.5 + 3 ( 22.50) + 3 ( 21.21) )
2
2
2
= 11 485 cm3
Hence, total volume of water =
Number of litres of water =
2
3
π ( 22.50 ) + 11 845 = 35 341 cm3
3
35 341cm3
= 35.34 litres
1000 cm3 /litre
498
© 2014, John Bird
EXERCISE 129 Page 309
1. Use the prismoidal rule to find the volume of a frustum of a sphere contained between two parallel
planes on opposite sides of the centre, each of radius 7.0 cm and each 4.0 cm from the centre.
The frustum of the sphere is shown sketched in cross-section below.
Radius, r =
( 7.02 + 4.02 )
= 8.062 cm
Using the prismoidal rule, volume of frustum =
x
[ A1 + 4 A2 + A3 ]
6
8.0 
2
2
2
π ( 7.0 ) + 4π ( 8.062 ) + π ( 7.0 ) 

6 
=
= 1500 cm3
2. Determine the volume of a cone of perpendicular height 16.0 cm and base diameter 10.0 cm by
using the prismoidal rule.
Using the prismoidal rule:
Volume, V =
2
 10.0 
Area, =
A1 π=
r12 π 
=
 25π
 2 
Area, =
A3 π=
r32 π ( 0=
) 0
2
x
[ A1 + 4 A2 + A3 ]
6
2
 5.0 
Area, =
A2 π=
r2 2 π  =
 6.25π
 2 
and x = 16.0 cm
499
© 2014, John Bird
Hence, volume of cylinder, V =
16.0
x
A3 ]
[ A1 + 4 A2 +=
[ 25π + 4(6.25π ) + 0]
6
6
=
16.0
× 50π = 418.9 cm3
6
3. A bucket is in the form of a frustum of a cone. The diameter of the base is 28.0 cm and the
diameter of the top is 42.0 cm. If the length is 32.0 cm, determine the capacity of the bucket (in
litres) using the prismoidal rule (1 litre = 1000 cm3).
The bucket is shown in the sketch below.
The radius of the midpoint is
14 + 21
= 17.5 cm
2
Using the prismoidal rule, volume of frustum =
=
x
[ A1 + 4 A2 + A3 ]
6
32.0 
2
2
2
π ( 21.0 ) + 4π (17.5 ) + π (14.0 ) 

6 
= 31 200 cm3
Hence, capacity of bucket =
31 200 cm3
= 31.20 litres
1000 cm3 /litre
4. Determine the capacity of a water reservoir, in litres, the top being a 30.0 m by 12.0 m rectangle,
the bottom being a 20.0 m by 8.0 m rectangle and the depth being 5.0 m (1 litre = 1000 cm3).
The water reservoir is shown sketched below.
500
© 2014, John Bird
 30 + 20 
 12 + 8 
A mid-section will have dimensions of 
 = 25 m by 
 = 10 m
 2 
 2 
Using the prismoidal rule, volume of frustum =
=
x
[ A1 + 4 A2 + A3 ]
6
5.0
( 30 ×12 ) + 4 ( 25 ×10 ) + ( 20 × 8 ) 
6 
= 1266.7 m3 = 1266.7 ×106 cm3
Hence, capacity of water reservoir =
1266.7 ×106 cm3
= 1.267 ×106 litre
1000 cm3 /litre
501
© 2014, John Bird
EXERCISE 130 Page 310
1. The diameter of two spherical bearings are in the ratio 2:5. What is the ratio of their volumes?
Diameters are in the ratio 2:5
Hence, ratio of their volumes = 23 : 53 i.e. 8:125
2. An engineering component has a mass of 400 g. If each of its dimensions are reduced by 30%,
determine its new mass.
New mass = ( 0.7 ) × 400 = 0.343 × 400 = 137.2 g
3
502
© 2014, John Bird