mathlinE lesson
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Surface areas and volumes of some regular
composite solids
Preliminary: Consider a sphere of radius 2 3 units. It
is divided into a major dome and a minor dome such
that the minor dome subtends an angle of 120o at the
centre of the sphere.
(Suitable for year 8 to 10)
circle
Surface area formulae
radius 2 3
Sphere: SA  4r 2
(SA: surface area)
Cylinder:
Two ends + curved surface
TSA  2r 2  2rl
or  2r r  l 
(TSA: total surface area)
r
l
r
60o
The triangle in the above drawing is similar to the
special 30-60-90 triangle shown below. It is enlarged
by a scale factor of 3 .
3
30o
Cone: Base + curved surface
TSA  r 2  rs
2
s
1
60o
r
Do you know the following facts about a sphere of
radius r?
SA of minor dome
= r 2
To obtain the side lengths of the triangle, multiply
each side of the special 30-60-90 triangle by 3 .
3 3 3
30o
1 3  3
2 3  2 3
60o
120 0
SA of major dome
= 3r 2
Example 1 Find the area of the curved surface of the
frustum of the cone. (Measurements in cm)
60o
6
The above is true only when the minor dome subtends
an angle of 120o at the centre of the sphere.
The following examples involve the use of these two
particular domes, a cylinder and a cone to form
composite solids.
r=3
6
r=6
Area = rs (large cone) – rs (small cone)
= (6)(12) – (3)(6) = 170cm2
1
Example 2 Find the surface areas of the minor dome
and the major dome. (Measurements in cm)
Radius = 3
Minor dome
Volume formulae
4
Sphere: V  r 3
3
r
120 0
Cylinder: V  r 2 l
l
Major dome
1
Cone: V  r 2 h
3
The radius of the sphere is 2 3 cm (refer to
preliminary).
Area of major dome = 3r 2
h
   37.7 cm
 3 2 3   113.1 cm
Area of minor dome = r 2   2 3
2
r
r
2
2
2
Example 3 Find the TSA of the composite solid
formed by placing the minor dome in example 2 on
top of the frustum of the cone in example 1.
TSA = area of minor dome +
area of frustum of cone +
area of circular base
= 37.7 + 170 +  (6) 2
= 320.8cm2
Example 4 Find the TSA of the composite solid
formed by placing a major dome discussed in
example 2 on each end of a 10cm long cylinder.
Do you know the following facts about a sphere of
radius r?
Volume of minor dome
5 4 3
 r
32 3
120 0
Volume of major dome
27 4 3
 r
32 3
The above is true only when the minor dome subtends
an angle of 120o at the centre of the sphere.
Example 5 Find the volume of the frustum of the
cone. (Measurements in cm)
6 3
TSA = 2 major domes + cylindrical curved surface
= 2  113.1  2rl
= 2 113.1  2 (3)(10)
= 414.7cm2
60o
3 3
r=3
r=6
The heights of the cones were obtained by the method
outlined in the preliminary, using scale factors 3 and
6.
Volume = large cone – small cone
1
1
=  (6) 2 6 3   (3) 2 3 3 = 343cm3.
3
3
 
 
2
Example 6 Find the volumes of the minor dome and
the major dome. (Measurements in cm)
Exercise
Find the TSA and volume of each of the following
three-dimensional composite solids.
Radius = 3
(1)
120
Hemisphere
0
Cylinder
h  6cm
d  6cm
Minor dome:
3
5 4 3 5
V
 r 
  2 3  27.2 cm3
32 3
24
Major dome:
3
27 4 3 9
V
 r    2 3  146.9 cm3
32 3
8
 
 
60o
(2)
6
Cone
r = 3cm
120º
Example 7 Find the volume of material to be cut
from an appropriate cone to form the composite solid
discussed in example 3.
Major dome
60o
(3)
Volume of material to be removed
= volume of small cone – volume of minor dome
1
=  (3) 2 3 3  27.2  21.8 cm3
3
 
Cone-shape
hollow
inside the
cylinder
r  3cm
h  4cm
Example 8 Find the volume of material to be cut
from an appropriate cylinder to form the composite
solid discussed in example 4.
3 3
10
3 3
Radius of the large cylinder = 2 3 cm
Length of the large cylinder = 6 3 +10 = 20.4cm
Volume of material to be removed
= Large cylinder – small cylinder – 2 major domes
 
=  2 3 20.4   3 10  2146.9
= 192.5cm3
2
2
3