SURFACE
AREAS AND
VOLUMES 161
12
12.1 Introduction
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 12.1). You have also learnt how to find their surface areas and volumes.
Fig. 12.1
In our day-to-day life, we come across a number of solids made up of combinations
of two or more of the basic solids as shown above.
You must have seen a truck with a
container fitted on its back (see Fig. 12.2),
carrying oil or water from one place to
another. Is it in the shape of any of the four
basic solids mentioned above? You may
guess that it is made of a cylinder with two
hemispheres as its ends.
SURFACE AREAS AND
VOLUMES
Fig. 12.2
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162 MATHEMATICS
Again, you may have seen an object like the
one in Fig. 12.3. Can you name it? A test tube, right!
You would have used one in your science laboratory.
This tube is also a combination of a cylinder and a
hemisphere. Similarly, while travelling, you may have
seen some big and beautiful buildings or monuments
made up of a combination of solids mentioned above.
If for some reason you wanted to find the
surface areas, or volumes, or capacities of such
objects, how would you do it? We cannot classify
these under any of the solids you have already studied.
In this chapter, you will see how to find surface areas and volumes of such
objects.
12.2 Surface Area of a Combination of Solids
Let us consider the container seen in Fig. 12.2. How do we find the surface area of
such a solid? Now, whenever we come across a new problem, we first try to see, if
we can break it down into smaller problems, we have earlier solved. We can see that
this solid is made up of a cylinder with two hemispheres stuck at either end. It would
look like what we have in Fig. 12.4, after we put the pieces all together.
Fig. 12.4
If we consider the surface of the newly formed object, we would be able to see
only the curved surfaces of the two hemispheres and the curved surface of the cylinder
.
So, the total surface area of the new solid is the sum of the curved surface
areas of each of the individual parts. This gives,
TSA of new solid = CSA of one hemisphere + CSA of cylinder
+ CSA of other hemisphere
where TSA, CSA stand for ‘T
otal Surface Area’ and ‘Curved Surface Area’
respectively.
Let us now consider another situation. Suppose we are making a toy by putting
together a hemisphere and a cone. Let us see the steps that we would be going
through.
Fig. 12.3
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SURFACE AREAS AND VOLUMES 163
First, we would take a cone and a hemisphere and bring their flat faces together.
Here, of course, we would take the base radius of the cone equal to the radius of the
hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown
in Fig. 12.5.
Fig. 12.5
At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if
we want to find how much paint we would require to colour the surface of this toy,
what would we need to know? We would need to know the surface area of the toy,
which consists of the CSA of the hemisphere and the CSA of the cone.
So, we can say:
T
otal surface area of the toy = CSA of hemisphere + CSA of cone
Now, let us consider some examples.
Example 1 : Rasheed got a playing top (lattu) as his
birthday present, which surprisingly had no colour on
it. He wanted to colour it with his crayons. The top is
shaped like a cone surmounted by a hemisphere
(see Fig 12.6). The entire top is 5 cm in height and
the diameter of the top is 3.5 cm. Find the area he
has to colour. (Take =
22
7
)
Solution : This top is exactly like the object we have discussed in Fig. 12.5. So, we
can conveniently use the result we have arrived at there. That is :
TSA of the toy = CSA of hemisphere + CSA of cone
Now, the curved surface area of the hemisphere =
22
1
(4 ) 2
2
rr
=
2
22 3.5 3.5
2cm
722
Fig. 12.6
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164 MATHEMATICS
Also, the height of the cone = height of the top – height (radius) of the hemispherical part
=
3.5
5cm
2
= 3.25 cm
So, the slant height of the cone (l ) =
2
22 2
3.5
(3.25) cm
2
rh
= 3.7 cm (approx.)
Therefore, CSA of cone = rl =
2
22 3.5
3.7 cm
72
This gives the surface area of the top as
=
22
22 3.5 3.5 22 3.5
2cm3.7cm
722 72
=
2
22 3.5
3.5 3.7 cm
72
=
22
11
(3.5 3.7) cm 39.6 cm (approx.)
2
You may note that ‘total surface area of the top’ is not the sum of the total
surface areas of the cone and hemisphere.
Example 2 : The decorative block shown
in Fig. 12.7 is made of two solids — a cube
and a hemisphere. The base of the block is a
cube with edge 5 cm, and the hemisphere
fixed on the top has a diameter of 4.2 cm.
Find the total surface area of the block.
(Take =
22
7
)
Solution : The total surface area of the cube = 6 × (edge)
2
= 6 × 5 × 5 cm
2
= 150 cm
2
.
Note that the part of the cube where the hemisphere is attached is not included in the
surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere
+ CSA of hemisphere
= 150 – r
2
+ 2 r
2
= (150 + r
2
) cm
2
=
22
22 4.2 4.2
150 cm cm
722
= (150 + 13.86) cm
2
= 163.86 cm
2
.
Fig. 12.7
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SURFACE AREAS AND VOLUMES 165
Example 3 : A wooden toy rocket is in the
shape of a cone mounted on a cylinder, as
shown in Fig. 12.8. The height of the entire
rocket is 26 cm, while the height of the conical
part is 6 cm. The base of the conical portion
has a diameter of 5 cm, while the base
diameter of the cylindrical portion is 3 cm. If
the conical portion is to be painted orange
and the cylindrical portion yellow, find the
area of the rocket painted with each of these
colours. (Take = 3.14)
Solution : Denote radius of cone by r, slant
height of cone by l, height of cone by h, radius
of cylinder by r and height of cylinder by h.
Then r = 2.5 cm, h = 6 cm, r = 1.5 cm,
h = 26 – 6 = 20 cm and
l =
22
rh
=
22
2.5 6 cm
= 6.5 cm
Here, the conical portion has its circular base resting on the base of the cylinder, but
the base of the cone is larger than the base of the cylinder. So, a part of the base of the
cone (a ring) is to be painted.
So, the area to be painted orange = CSA of the cone + base area of the cone
– base area of the cylinder
= rl + r
2
– r
2
= [(2.5 × 6.5) + (2.5)
2
– (1.5)
2
] cm
2
= [20.25] cm
2
= 3.14 × 20.25 cm
2
= 63.585 cm
2
Now, the area to be painted yellow = CSA of the cylinder
+ area of one base of the cylinder
=2rh + (r)
2
= r (2h + r)
= (3.14 × 1.5) (2 × 20 + 1.5) cm
2
= 4.71 × 41.5 cm
2
= 195.465 cm
2
Fig. 12.8
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166 MATHEMATICS
Example 4 : Mayank made a bird-bath for his garden
in the shape of a cylinder with a hemispherical
depression at one end (see Fig. 12.9). The height of
the cylinder is 1.45 m and its radius is 30 cm. Find the
total surface area of the bird-bath. (Take =
22
7
)
Solution : Let h be height of the cylinder, and r the
common radius of the cylinder and hemisphere. Then,
the total surface area of the bird-bath = CSA of cylinder
+ CSA of hemisphere
=2rh + 2r
2
= 2 r(h + r)
=
2
22
2 30(145 30) cm
7
= 33000 cm
2
= 3.3 m
2
EXERCISE 12.1
Unless stated otherwise, take =
22
7
1. 2 cubes each of volume 64 cm
3
are joined end to end. Find the surface area of the
resulting cuboid.
2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The
diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the
inner surface area of the vessel.
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.
The total height of the toy is 15.5 cm. Find the total surface area of the toy.
4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest
diameter the hemisphere can have? Find the surface area of the solid.
5. A hemispherical depression is cut out from one face of a cubical wooden block such
that the diameter l of the hemisphere is equal to the edge of the cube. Determine the
surface area of the remaining solid.
6. A medicine capsule is in the shape of a
cylinder with two hemispheres stuck to each
of its ends (see Fig. 12.10). The length of
the entire capsule is 14 mm and the diameter
of the capsule is 5 mm. Find its surface area.
Fig. 12.10
Fig. 12.9
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SURFACE AREAS AND VOLUMES 167
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and
diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the
top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of
the canvas of the tent at the rate of
` 500 per m
2
. (Note that the base of the tent will not
be covered with canvas.)
8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm
2
.
9. A wooden article was made by scooping
out a hemisphere from each end of a solid
cylinder, as shown in Fig. 12.11. If the
height of the cylinder is 10 cm, and its
base is of radius 3.5 cm, find the total
surface area of the article.
12.3 Volume of a Combination of Solids
In the previous section, we have discussed how to find the surface area of solids made
up of a combination of two basic solids. Here, we shall see how to calculate their
volumes. It may be noted that in calculating the surface area, we have not added the
surface areas of the two constituents, because some part of the surface area disappeared
in the process of joining them. However, this will not be the case when we calculate
the volume. The volume of the solid formed by joining two basic solids will actually be
the sum of the volumes of the constituents, as we see in the examples below.
Example 5 : Shanta runs an industry in
a shed which is in the shape of a cuboid
surmounted by a half cylinder (see Fig.
12.12). If the base of the shed is of
dimension 7 m × 15 m, and the height of
the cuboidal portion is 8 m, find the volume
of air that the shed can hold. Further,
suppose the machinery in the shed
occupies a total space of 300 m
3
, and
there are 20 workers, each of whom
occupy about 0.08 m
3
space on an
average. Then, how much air is in the
shed? (Take =
22
7
)
Fig. 12.12
Fig. 12.11
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168 MATHEMATICS
Solution : The volume of air inside the shed (when there are no people or machinery)
is given by the volume of air inside the cuboid and inside the half cylinder, taken
together.
Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively.
Also, the diameter of the half cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid +
1
2
volume of the cylinder
=
3
12277
15 7 8 15 m
2722
= 1128.75 m
3
Next, the total space occupied by the machinery = 300 m
3
And the total space occupied by the workers = 20 × 0.08 m
3
= 1.6 m
3
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 – (300.00 + 1.60) = 827.15 m
3
Example 6 : A juice seller was serving his
customers using glasses as shown in Fig. 12.13.
The inner diameter of the cylindrical glass was
5 cm, but the bottom of the glass had a
hemispherical raised portion which reduced the
capacity of the glass. If the height of a glass
was 10 cm, find the apparent capacity of the
glass and its actual capacity. (Use = 3.14.)
Solution : Since the inner diameter of the glass = 5 cm and height = 10 cm,
the apparent capacity of the glass = r
2
h
= 3.14 × 2.5 × 2.5 × 10 cm
3
= 196.25 cm
3
But the actual capacity of the glass is less by the volume of the hemisphere at the
base of the glass.
i.e., it is less by
2
3
r
3
=
3
2
3.14 2.5 2.5 2.5 cm
3
= 32.71 cm
3
So, the actual capacity of the glass = apparent capacity of glass – volume of the
hemisphere
= (196.25 – 32.71) cm
3
= 163.54 cm
3
Fig. 12.13
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SURFACE AREAS AND VOLUMES 169
Example 7 : A solid toy is in the form of a
hemisphere surmounted by a right circular cone. The
height of the cone is 2 cm and the diameter of the
base is 4 cm. Determine the volume of the toy
. If a
right circular cylinder circumscribes the toy, find the
difference of the volumes of the cylinder and the toy.
(Take = 3.14)
Solution : Let BPC be the hemisphere and
ABC be the cone standing on the base
of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as
of the cone) =
1
2
× 4 cm = 2 cm.
So, volume of the toy =
32
21
33
rrh
=
323
21
3.14 (2) 3.14 (2) 2 cm
33
= 25.12 cm
3
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of
the base of the right circular cylinder = HP = BO = 2 cm, and its height is
EH = AO + OP = (2 + 2) cm = 4 cm
So, the volume required = volume of the right circular cylinder – volume of the toy
= (3.14 × 2
2
× 4 – 25.12) cm
3
= 25.12 cm
3
Hence, the required difference of the two volumes = 25.12 cm
3
.
EXERCISE 12.2
Unless stated otherwise, take =
22
7
.
A solid is in the shape of a cone standing on a hemisphere with both their radii being
equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid
in terms of .
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with
two cones attached at its two ends by using a thin aluminium sheet. The diameter of the
model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume
of air contained in the model that Rachel made. (Assume the outer and inner dimensions
of the model to be nearly the same.)
Fig. 12.14
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170 MATHEMATICS
3. A gulab jamun, contains sugar syrup up to about
30% of its volume. Find approximately how much
syrup would be found in 45 gulab jamuns, each
shaped like a cylinder with two hemispherical ends
with length 5 cm and diameter 2.8 cm (see Fig. 12.15).
4. A pen stand made of wood is in the shape of a
cuboid with four conical depressions to hold pens.
The dimensions of the cuboid are 15 cm by 10 cm by
3.5 cm. The radius of each of the depressions is 0.5
cm and the depth is 1.4 cm. Find the volume of
wood in the entire stand (see Fig. 12.16).
5. A vessel is in the form of an inverted cone. Its
height is 8 cm and the radius of its top, which is
open, is 5 cm. It is filled with water up to the brim.
When lead shots, each of which is a sphere of radius
0.5 cm are dropped into the vessel, one-fourth of
the water flows out. Find the number of lead shots
dropped in the vessel.
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which
is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the
pole, given that 1 cm
3
of iron has approximately 8g mass. (Use = 3.14)
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on
a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water
such that it touches the bottom. Find the volume of water left in the cylinder, if the radius
of the cylinder is 60 cm and its height is 180 cm.
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter
of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its
volume to be 345 cm
3
. Check whether she is correct, taking the above as the inside
measurements, and = 3.14.
12.4 Summary
In this chapter, you have studied the following points:
1. To determine the surface area of an object formed by combining any two of the basic
solids, namely, cuboid, cone, cylinder, sphere and hemisphere.
2. To find the volume of objects formed by combining any two of a cuboid, cone, cylinder,
sphere and hemisphere.
Fig. 12.16
Fig. 12.15
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