NCERT Textbook- Surface Areas and Volumes Class 10 Notes | EduRev

 Page 1


SURFACE AREAS AND VOLUMES 239
13
13.1 Introduction
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 13.1). Y ou have also learnt how to find their surface areas and volumes.
Fig. 13.1
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. 13.2),
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.
SURF ACE AREAS AND
VOLUMES
Fig. 13.2
2020-21
Page 2


SURFACE AREAS AND VOLUMES 239
13
13.1 Introduction
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 13.1). Y ou have also learnt how to find their surface areas and volumes.
Fig. 13.1
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. 13.2),
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.
SURF ACE AREAS AND
VOLUMES
Fig. 13.2
2020-21
240 MATHEMA TICS
Again, you may have seen an object like the
one in Fig. 13.3. 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.
13.2 Surface Area of a Combination of Solids
Let us consider the container seen in Fig. 13.2. 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. 13.4, after we put the pieces all together.
Fig. 13.4
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 ‘Total 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. 13.3
2020-21
Page 3


SURFACE AREAS AND VOLUMES 239
13
13.1 Introduction
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 13.1). Y ou have also learnt how to find their surface areas and volumes.
Fig. 13.1
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. 13.2),
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.
SURF ACE AREAS AND
VOLUMES
Fig. 13.2
2020-21
240 MATHEMA TICS
Again, you may have seen an object like the
one in Fig. 13.3. 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.
13.2 Surface Area of a Combination of Solids
Let us consider the container seen in Fig. 13.2. 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. 13.4, after we put the pieces all together.
Fig. 13.4
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 ‘Total 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. 13.3
2020-21
SURFACE AREAS AND VOLUMES 241
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. 13.5.
Fig. 13.5
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:
Total 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 13.6). 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 p = 
22
7
)
Solution : This top is exactly like the object we have discussed in Fig. 13.5. 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 =
2 2
1
(4 ) 2
2
r r p = p
=
2
22 3.5 3.5
2 cm
7 2 2
? ?
× × ×
? ?
? ?
Fig. 13.6
2020-21
Page 4


SURFACE AREAS AND VOLUMES 239
13
13.1 Introduction
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 13.1). Y ou have also learnt how to find their surface areas and volumes.
Fig. 13.1
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. 13.2),
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.
SURF ACE AREAS AND
VOLUMES
Fig. 13.2
2020-21
240 MATHEMA TICS
Again, you may have seen an object like the
one in Fig. 13.3. 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.
13.2 Surface Area of a Combination of Solids
Let us consider the container seen in Fig. 13.2. 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. 13.4, after we put the pieces all together.
Fig. 13.4
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 ‘Total 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. 13.3
2020-21
SURFACE AREAS AND VOLUMES 241
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. 13.5.
Fig. 13.5
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:
Total 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 13.6). 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 p = 
22
7
)
Solution : This top is exactly like the object we have discussed in Fig. 13.5. 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 =
2 2
1
(4 ) 2
2
r r p = p
=
2
22 3.5 3.5
2 cm
7 2 2
? ?
× × ×
? ?
? ?
Fig. 13.6
2020-21
242 MATHEMA TICS
Also, the height of the cone = height of the top – height (radius) of the hemispherical part
=
3.5
5 cm
2
? ?
-
? ?
? ?
 = 3.25 cm
So, the slant height of the cone (l ) = 
2
2 2 2
3.5
(3.25) cm
2
r h
? ?
+ = +
? ?
? ?
 = 3.7 cm (approx.)
Therefore, CSA of cone = prl = 
2
22 3.5
3.7 cm
7 2
? ?
× ×
? ?
? ?
This gives the surface area of the top as
=
2 2
22 3.5 3.5 22 3.5
2 cm 3.7 cm
7 2 2 7 2
? ? ? ?
× × × + × ×
? ? ? ?
? ? ? ?
= ( )
2
22 3.5
3.5 3.7 cm
7 2
× +
 = 
2 2
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. 13.7 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 p = 
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 – pr
2
 + 2 pr
2 
= (150 + pr
2
) cm
2
=
2 2
22 4.2 4.2
150 cm cm
7 2 2
? ?
+ × ×
? ?
? ?
= (150 + 13.86) cm
2
 = 163.86 cm
2
Fig. 13.7
.
2020-21
Page 5


SURFACE AREAS AND VOLUMES 239
13
13.1 Introduction
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 13.1). Y ou have also learnt how to find their surface areas and volumes.
Fig. 13.1
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. 13.2),
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.
SURF ACE AREAS AND
VOLUMES
Fig. 13.2
2020-21
240 MATHEMA TICS
Again, you may have seen an object like the
one in Fig. 13.3. 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.
13.2 Surface Area of a Combination of Solids
Let us consider the container seen in Fig. 13.2. 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. 13.4, after we put the pieces all together.
Fig. 13.4
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 ‘Total 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. 13.3
2020-21
SURFACE AREAS AND VOLUMES 241
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. 13.5.
Fig. 13.5
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:
Total 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 13.6). 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 p = 
22
7
)
Solution : This top is exactly like the object we have discussed in Fig. 13.5. 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 =
2 2
1
(4 ) 2
2
r r p = p
=
2
22 3.5 3.5
2 cm
7 2 2
? ?
× × ×
? ?
? ?
Fig. 13.6
2020-21
242 MATHEMA TICS
Also, the height of the cone = height of the top – height (radius) of the hemispherical part
=
3.5
5 cm
2
? ?
-
? ?
? ?
 = 3.25 cm
So, the slant height of the cone (l ) = 
2
2 2 2
3.5
(3.25) cm
2
r h
? ?
+ = +
? ?
? ?
 = 3.7 cm (approx.)
Therefore, CSA of cone = prl = 
2
22 3.5
3.7 cm
7 2
? ?
× ×
? ?
? ?
This gives the surface area of the top as
=
2 2
22 3.5 3.5 22 3.5
2 cm 3.7 cm
7 2 2 7 2
? ? ? ?
× × × + × ×
? ? ? ?
? ? ? ?
= ( )
2
22 3.5
3.5 3.7 cm
7 2
× +
 = 
2 2
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. 13.7 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 p = 
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 – pr
2
 + 2 pr
2 
= (150 + pr
2
) cm
2
=
2 2
22 4.2 4.2
150 cm cm
7 2 2
? ?
+ × ×
? ?
? ?
= (150 + 13.86) cm
2
 = 163.86 cm
2
Fig. 13.7
.
2020-21
SURFACE AREAS AND VOLUMES 243
Example 3 : A wooden toy rocket is in the
shape of a cone mounted on a cylinder, as
shown in Fig. 13.8. 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 p = 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 =
2 2
r h + = 
2 2
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
= prl + pr
2
 – p(r')
2
= p[(2.5 × 6.5) + (2.5)
2
  – (1.5)
2
] cm
2
= p[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
= 2pr'h' + p(r')
2
= pr' (2h' + r')
= (3.14 × 1.5) (2 × 20 + 1.5) cm
2
= 4.71 × 41.5 cm
2
= 195.465 cm
2
Fig. 13.8
2020-21