NCERT Textbook: Surface Area & Volumes

# NCERT Textbook: Surface Area & Volumes | Mathematics (Maths) Class 9 PDF Download

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``` Page 1

SURFACE AREAS AND VOLUMES 137
CHAPTER 11
SURFACE AREAS AND VOLUMES
11.1 Surface Area of a Right Circular Cone
We have already studied the surface areas of cube, cuboid and cylinder. We will now
study the surface area of cone.
So far, we have been generating solids by stacking up congruent figures. Incidentally,
such figures are called prisms. Now let us look at another kind of solid which is not a
prism (These kinds of solids are called pyramids.). Let us see how we can generate
them.
Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick
string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)].
Hold the string with your hands on either sides of the triangle and rotate the triangle
about the string a number of times. What happens? Do you recognize the shape that
the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind
you of the time you had eaten an ice-cream heaped into a container of that shape [see
Fig. 11.1 (c) and (d)]?
Fig. 11.1
Rationalised 2023-24
Page 2

SURFACE AREAS AND VOLUMES 137
CHAPTER 11
SURFACE AREAS AND VOLUMES
11.1 Surface Area of a Right Circular Cone
We have already studied the surface areas of cube, cuboid and cylinder. We will now
study the surface area of cone.
So far, we have been generating solids by stacking up congruent figures. Incidentally,
such figures are called prisms. Now let us look at another kind of solid which is not a
prism (These kinds of solids are called pyramids.). Let us see how we can generate
them.
Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick
string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)].
Hold the string with your hands on either sides of the triangle and rotate the triangle
about the string a number of times. What happens? Do you recognize the shape that
the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind
you of the time you had eaten an ice-cream heaped into a container of that shape [see
Fig. 11.1 (c) and (d)]?
Fig. 11.1
Rationalised 2023-24
138 MATHEMA TICS
This is called a right circular cone. In Fig. 11.1(c)
of the right circular cone, the point A is called the
vertex, AB is called the height, BC is called the radius
and AC is called the slant height of the cone. Here B
will be the centre of circular base of the cone. The
height, radius and slant height of the cone are usually
denoted by h, r and l respectively. Once again, let us
see what kind of cone we can not call a right circular
cone. Here, you are (see Fig. 11.2)! What you see in
these figures are not right circular cones; because in
(a), the line joining its vertex to the centre of its base
is not at right angle to the base, and in (b) the base is
not circular.
As in the case of cylinder, since we will be studying only about right circular cones,
remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’
Activity : (i) Cut out a neatly made paper cone that does not have any overlapped
paper, straight along its side, and opening it out, to see the shape of paper that forms
the surface of the cone. (The line along which you cut the cone is the slant height of
the cone which is represented by l). It looks like a part of a round cake.
(ii) If you now bring the sides marked A and B at the tips together, you can see that
the curved portion of Fig. 11.3 (c) will form the circular base of the cone.
Fig. 11.3
(iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces,
along the lines drawn from the point O, each cut portion is almost a small triangle,
whose height is the slant height l of the cone.
(iv) Now the area of each triangle =
1
2
× base of each triangle × l.
So, area of the entire piece of paper
Fig. 11.2
Rationalised 2023-24
Page 3

SURFACE AREAS AND VOLUMES 137
CHAPTER 11
SURFACE AREAS AND VOLUMES
11.1 Surface Area of a Right Circular Cone
We have already studied the surface areas of cube, cuboid and cylinder. We will now
study the surface area of cone.
So far, we have been generating solids by stacking up congruent figures. Incidentally,
such figures are called prisms. Now let us look at another kind of solid which is not a
prism (These kinds of solids are called pyramids.). Let us see how we can generate
them.
Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick
string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)].
Hold the string with your hands on either sides of the triangle and rotate the triangle
about the string a number of times. What happens? Do you recognize the shape that
the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind
you of the time you had eaten an ice-cream heaped into a container of that shape [see
Fig. 11.1 (c) and (d)]?
Fig. 11.1
Rationalised 2023-24
138 MATHEMA TICS
This is called a right circular cone. In Fig. 11.1(c)
of the right circular cone, the point A is called the
vertex, AB is called the height, BC is called the radius
and AC is called the slant height of the cone. Here B
will be the centre of circular base of the cone. The
height, radius and slant height of the cone are usually
denoted by h, r and l respectively. Once again, let us
see what kind of cone we can not call a right circular
cone. Here, you are (see Fig. 11.2)! What you see in
these figures are not right circular cones; because in
(a), the line joining its vertex to the centre of its base
is not at right angle to the base, and in (b) the base is
not circular.
As in the case of cylinder, since we will be studying only about right circular cones,
remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’
Activity : (i) Cut out a neatly made paper cone that does not have any overlapped
paper, straight along its side, and opening it out, to see the shape of paper that forms
the surface of the cone. (The line along which you cut the cone is the slant height of
the cone which is represented by l). It looks like a part of a round cake.
(ii) If you now bring the sides marked A and B at the tips together, you can see that
the curved portion of Fig. 11.3 (c) will form the circular base of the cone.
Fig. 11.3
(iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces,
along the lines drawn from the point O, each cut portion is almost a small triangle,
whose height is the slant height l of the cone.
(iv) Now the area of each triangle =
1
2
× base of each triangle × l.
So, area of the entire piece of paper
Fig. 11.2
Rationalised 2023-24
SURFACE AREAS AND VOLUMES 139
= sum of the areas of all the triangles
=
1 2 3
1 1 1
2 2 2
b l b l b l + + +? = ( )
1 2 3
1
2
l b b b + + +?
=
1
2
× l × length of entire curved boundary of Fig. 11.3(c)
(as b
1
+ b
2
+ b
3
+ . . . makes up the curved portion of the figure)
But the curved portion of the figure makes up the perimeter of the base of the cone
and the circumference of the base of the cone = 2pr, where r is the base radius of the
cone.
So, Curved Surface Area of a Cone =
1
2
× l × 2p p p p pr = p p p p prl
where r is its base radius and l its slant height.
Note that l
2
= r
2
+ h
2
(as can be seen from Fig. 11.4), by
applying Pythagoras Theorem. Here h is the height of the
cone.
Therefore, l =
2 2
r h +
Now if the base of the cone is to be closed, then a circular piece of paper of radius r
is also required whose area is pr
2
.
So, Total Surface Area of a Cone = p p p p prl + p p p p pr
2
= p p p p pr(l + r)
Example 1 : Find the curved surface area of a right circular cone whose slant height
is 10 cm and base radius is 7 cm.
Solution : Curved surface area = prl
=
22
7
× 7 × 10 cm
2
= 220 cm
2
Example 2 : The height of a cone is 16 cm and its base radius is 12 cm. Find the
curved surface area and the total surface area of the cone (Use p = 3.14).
Solution : Here, h = 16 cm and r = 12 cm.
So, from l
2
= h
2
+ r
2
, we have
l =
2 2
16 12 + cm = 20 cm
Fig. 11.4
Rationalised 2023-24
Page 4

SURFACE AREAS AND VOLUMES 137
CHAPTER 11
SURFACE AREAS AND VOLUMES
11.1 Surface Area of a Right Circular Cone
We have already studied the surface areas of cube, cuboid and cylinder. We will now
study the surface area of cone.
So far, we have been generating solids by stacking up congruent figures. Incidentally,
such figures are called prisms. Now let us look at another kind of solid which is not a
prism (These kinds of solids are called pyramids.). Let us see how we can generate
them.
Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick
string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)].
Hold the string with your hands on either sides of the triangle and rotate the triangle
about the string a number of times. What happens? Do you recognize the shape that
the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind
you of the time you had eaten an ice-cream heaped into a container of that shape [see
Fig. 11.1 (c) and (d)]?
Fig. 11.1
Rationalised 2023-24
138 MATHEMA TICS
This is called a right circular cone. In Fig. 11.1(c)
of the right circular cone, the point A is called the
vertex, AB is called the height, BC is called the radius
and AC is called the slant height of the cone. Here B
will be the centre of circular base of the cone. The
height, radius and slant height of the cone are usually
denoted by h, r and l respectively. Once again, let us
see what kind of cone we can not call a right circular
cone. Here, you are (see Fig. 11.2)! What you see in
these figures are not right circular cones; because in
(a), the line joining its vertex to the centre of its base
is not at right angle to the base, and in (b) the base is
not circular.
As in the case of cylinder, since we will be studying only about right circular cones,
remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’
Activity : (i) Cut out a neatly made paper cone that does not have any overlapped
paper, straight along its side, and opening it out, to see the shape of paper that forms
the surface of the cone. (The line along which you cut the cone is the slant height of
the cone which is represented by l). It looks like a part of a round cake.
(ii) If you now bring the sides marked A and B at the tips together, you can see that
the curved portion of Fig. 11.3 (c) will form the circular base of the cone.
Fig. 11.3
(iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces,
along the lines drawn from the point O, each cut portion is almost a small triangle,
whose height is the slant height l of the cone.
(iv) Now the area of each triangle =
1
2
× base of each triangle × l.
So, area of the entire piece of paper
Fig. 11.2
Rationalised 2023-24
SURFACE AREAS AND VOLUMES 139
= sum of the areas of all the triangles
=
1 2 3
1 1 1
2 2 2
b l b l b l + + +? = ( )
1 2 3
1
2
l b b b + + +?
=
1
2
× l × length of entire curved boundary of Fig. 11.3(c)
(as b
1
+ b
2
+ b
3
+ . . . makes up the curved portion of the figure)
But the curved portion of the figure makes up the perimeter of the base of the cone
and the circumference of the base of the cone = 2pr, where r is the base radius of the
cone.
So, Curved Surface Area of a Cone =
1
2
× l × 2p p p p pr = p p p p prl
where r is its base radius and l its slant height.
Note that l
2
= r
2
+ h
2
(as can be seen from Fig. 11.4), by
applying Pythagoras Theorem. Here h is the height of the
cone.
Therefore, l =
2 2
r h +
Now if the base of the cone is to be closed, then a circular piece of paper of radius r
is also required whose area is pr
2
.
So, Total Surface Area of a Cone = p p p p prl + p p p p pr
2
= p p p p pr(l + r)
Example 1 : Find the curved surface area of a right circular cone whose slant height
is 10 cm and base radius is 7 cm.
Solution : Curved surface area = prl
=
22
7
× 7 × 10 cm
2
= 220 cm
2
Example 2 : The height of a cone is 16 cm and its base radius is 12 cm. Find the
curved surface area and the total surface area of the cone (Use p = 3.14).
Solution : Here, h = 16 cm and r = 12 cm.
So, from l
2
= h
2
+ r
2
, we have
l =
2 2
16 12 + cm = 20 cm
Fig. 11.4
Rationalised 2023-24
140 MATHEMA TICS
So, curved surface area = prl
= 3.14 × 12 × 20 cm
2
= 753.6 cm
2
Further, total surface area = prl + pr
2
= (753.6 + 3.14 × 12 × 12) cm
2
= (753.6 + 452.16) cm
2
= 1205.76 cm
2
Example 3 : A corn cob (see Fig. 11.5), shaped somewhat
like a cone, has the radius of its broadest end as 2.1 cm and
length (height) as 20 cm. If each 1 cm
2
of the surface of the
cob carries an average of four grains, find how many grains
you would find on the entire cob.
Solution : Since the grains of corn are found only on the curved surface of the corn
cob, we would need to know the curved surface area of the corn cob to find the total
number of grains on it. In this question, we are given the height of the cone, so we
need to find its slant height.
Here, l =
2 2
r h + =
2 2
(2.1) 20 + cm
=
404.41
cm = 20.11 cm
Therefore, the curved surface area of the corn cob = prl
=
22
7
× 2.1 × 20.11 cm
2
= 132.726 cm
2
= 132.73 cm
2
(approx.)
Number of grains of corn on 1 cm
2
of the surface of the corn cob = 4
Therefore, number of grains on the entire curved surface of the cob
= 132.73 × 4 = 530.92 = 531 (approx.)
So, there would be approximately 531 grains of corn on the cob.
EXERCISE 11.1
Assume p =
22
7
, unless stated otherwise.
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved
surface area.
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base
is 24 m.
Fig. 11.5
Rationalised 2023-24
Page 5

SURFACE AREAS AND VOLUMES 137
CHAPTER 11
SURFACE AREAS AND VOLUMES
11.1 Surface Area of a Right Circular Cone
We have already studied the surface areas of cube, cuboid and cylinder. We will now
study the surface area of cone.
So far, we have been generating solids by stacking up congruent figures. Incidentally,
such figures are called prisms. Now let us look at another kind of solid which is not a
prism (These kinds of solids are called pyramids.). Let us see how we can generate
them.
Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick
string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)].
Hold the string with your hands on either sides of the triangle and rotate the triangle
about the string a number of times. What happens? Do you recognize the shape that
the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind
you of the time you had eaten an ice-cream heaped into a container of that shape [see
Fig. 11.1 (c) and (d)]?
Fig. 11.1
Rationalised 2023-24
138 MATHEMA TICS
This is called a right circular cone. In Fig. 11.1(c)
of the right circular cone, the point A is called the
vertex, AB is called the height, BC is called the radius
and AC is called the slant height of the cone. Here B
will be the centre of circular base of the cone. The
height, radius and slant height of the cone are usually
denoted by h, r and l respectively. Once again, let us
see what kind of cone we can not call a right circular
cone. Here, you are (see Fig. 11.2)! What you see in
these figures are not right circular cones; because in
(a), the line joining its vertex to the centre of its base
is not at right angle to the base, and in (b) the base is
not circular.
As in the case of cylinder, since we will be studying only about right circular cones,
remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’
Activity : (i) Cut out a neatly made paper cone that does not have any overlapped
paper, straight along its side, and opening it out, to see the shape of paper that forms
the surface of the cone. (The line along which you cut the cone is the slant height of
the cone which is represented by l). It looks like a part of a round cake.
(ii) If you now bring the sides marked A and B at the tips together, you can see that
the curved portion of Fig. 11.3 (c) will form the circular base of the cone.
Fig. 11.3
(iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces,
along the lines drawn from the point O, each cut portion is almost a small triangle,
whose height is the slant height l of the cone.
(iv) Now the area of each triangle =
1
2
× base of each triangle × l.
So, area of the entire piece of paper
Fig. 11.2
Rationalised 2023-24
SURFACE AREAS AND VOLUMES 139
= sum of the areas of all the triangles
=
1 2 3
1 1 1
2 2 2
b l b l b l + + +? = ( )
1 2 3
1
2
l b b b + + +?
=
1
2
× l × length of entire curved boundary of Fig. 11.3(c)
(as b
1
+ b
2
+ b
3
+ . . . makes up the curved portion of the figure)
But the curved portion of the figure makes up the perimeter of the base of the cone
and the circumference of the base of the cone = 2pr, where r is the base radius of the
cone.
So, Curved Surface Area of a Cone =
1
2
× l × 2p p p p pr = p p p p prl
where r is its base radius and l its slant height.
Note that l
2
= r
2
+ h
2
(as can be seen from Fig. 11.4), by
applying Pythagoras Theorem. Here h is the height of the
cone.
Therefore, l =
2 2
r h +
Now if the base of the cone is to be closed, then a circular piece of paper of radius r
is also required whose area is pr
2
.
So, Total Surface Area of a Cone = p p p p prl + p p p p pr
2
= p p p p pr(l + r)
Example 1 : Find the curved surface area of a right circular cone whose slant height
is 10 cm and base radius is 7 cm.
Solution : Curved surface area = prl
=
22
7
× 7 × 10 cm
2
= 220 cm
2
Example 2 : The height of a cone is 16 cm and its base radius is 12 cm. Find the
curved surface area and the total surface area of the cone (Use p = 3.14).
Solution : Here, h = 16 cm and r = 12 cm.
So, from l
2
= h
2
+ r
2
, we have
l =
2 2
16 12 + cm = 20 cm
Fig. 11.4
Rationalised 2023-24
140 MATHEMA TICS
So, curved surface area = prl
= 3.14 × 12 × 20 cm
2
= 753.6 cm
2
Further, total surface area = prl + pr
2
= (753.6 + 3.14 × 12 × 12) cm
2
= (753.6 + 452.16) cm
2
= 1205.76 cm
2
Example 3 : A corn cob (see Fig. 11.5), shaped somewhat
like a cone, has the radius of its broadest end as 2.1 cm and
length (height) as 20 cm. If each 1 cm
2
of the surface of the
cob carries an average of four grains, find how many grains
you would find on the entire cob.
Solution : Since the grains of corn are found only on the curved surface of the corn
cob, we would need to know the curved surface area of the corn cob to find the total
number of grains on it. In this question, we are given the height of the cone, so we
need to find its slant height.
Here, l =
2 2
r h + =
2 2
(2.1) 20 + cm
=
404.41
cm = 20.11 cm
Therefore, the curved surface area of the corn cob = prl
=
22
7
× 2.1 × 20.11 cm
2
= 132.726 cm
2
= 132.73 cm
2
(approx.)
Number of grains of corn on 1 cm
2
of the surface of the corn cob = 4
Therefore, number of grains on the entire curved surface of the cob
= 132.73 × 4 = 530.92 = 531 (approx.)
So, there would be approximately 531 grains of corn on the cob.
EXERCISE 11.1
Assume p =
22
7
, unless stated otherwise.
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved
surface area.
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base
is 24 m.
Fig. 11.5
Rationalised 2023-24
SURFACE AREAS AND VOLUMES 141
3. Curved surface area of a cone is 308 cm
2
and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m
2
canvas is ` 70.
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m
and base radius 6 m? Assume that the extra length of material that will be required for
stitching margins and wastage in cutting is approximately 20 cm (Use p = 3.14).
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively.
Find the cost of white-washing its curved surface at the rate of ` 210 per 100 m
2
.
7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height
24 cm. Find the area of the sheet required to make 10 such caps.
8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow
cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height
1 m. If the outer side of each of the cones is to be painted and the cost of painting is
` 12 per m
2
, what will be the cost of painting all these cones? (Use p = 3.14 and take
1.04
= 1.02)
11.2 Surface Area of a Sphere
What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes,
you can, because a circle is a plane closed figure whose every point lies at a constant
distance (called radius) from a fixed point, which is called the centre of the circle.
Now if you paste a string along a diameter of a circular disc and rotate it as you had
rotated the triangle in the previous section, you see a new solid (see Fig 11.6). What
does it resemble? A ball? Yes. It is called a sphere.
Fig. 11.6
Rationalised 2023-24
```

## Mathematics (Maths) Class 9

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## FAQs on NCERT Textbook: Surface Area & Volumes - Mathematics (Maths) Class 9

 1. What is surface area?
Ans. Surface area is the total area that covers the outer part of a three-dimensional object. It includes the sum of all the areas of its faces or surfaces.
 2. How can I find the surface area of a cube?
Ans. To find the surface area of a cube, you need to know the length of one side. The surface area can be calculated by using the formula: Surface Area = 6 × (side)^2.
 3. What is the difference between lateral surface area and total surface area?
Ans. Lateral surface area refers to the sum of the areas of all the sides of a three-dimensional object, excluding the top and bottom faces. Total surface area, on the other hand, includes the areas of all the faces, including the top and bottom.
 4. How can I find the volume of a sphere?
Ans. The volume of a sphere can be calculated using the formula: Volume = (4/3) × π × (radius)^3, where π is a constant approximately equal to 3.14.
 5. Can you provide an example of finding the surface area of a cylinder?
Ans. Sure! Let's say we have a cylinder with radius 5 cm and height 10 cm. To find its surface area, we can use the formula: Surface Area = 2πr(r+h), where r is the radius and h is the height. Plugging in the values, we get: Surface Area = 2 × 3.14 × 5(5+10) = 2 × 3.14 × 5 × 15 = 471 cm².

## Mathematics (Maths) Class 9

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