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# NCERT Textbook- Surface Area & Volumes Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Indu Gupta

## Class 9 : NCERT Textbook- Surface Area & Volumes Class 9 Notes | EduRev

``` Page 1

208 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
CHAPTER 13
SURFACE AREAS AND VOLUMES
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our study, we have been dealing
with figures that can be easily drawn on our notebooks or blackboards. These are
called plane figures. We have understood what rectangles, squares and circles are,
what we mean by their perimeters and areas, and how we can find them. We have
learnt these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheet and
stack them up in a vertical pile. By this process, we shall obtain some solid figures
(briefly called solids) such as a cuboid, a cylinder, etc. In the earlier classes, you have
also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We
shall now learn to find the surface areas and volumes of cuboids and cylinders in
details and extend this study to some other solids such as cones and spheres.
13.2 Surface Area of a Cuboid and a Cube
Have you looked at a bundle of many sheets of paper? How does it look? Does it look
like what you see in Fig. 13.1?
Fig. 13.1
That makes up a cuboid. How much of brown paper would you need, if you want
to cover this cuboid? Let us see:
Page 2

208 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
CHAPTER 13
SURFACE AREAS AND VOLUMES
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our study, we have been dealing
with figures that can be easily drawn on our notebooks or blackboards. These are
called plane figures. We have understood what rectangles, squares and circles are,
what we mean by their perimeters and areas, and how we can find them. We have
learnt these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheet and
stack them up in a vertical pile. By this process, we shall obtain some solid figures
(briefly called solids) such as a cuboid, a cylinder, etc. In the earlier classes, you have
also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We
shall now learn to find the surface areas and volumes of cuboids and cylinders in
details and extend this study to some other solids such as cones and spheres.
13.2 Surface Area of a Cuboid and a Cube
Have you looked at a bundle of many sheets of paper? How does it look? Does it look
like what you see in Fig. 13.1?
Fig. 13.1
That makes up a cuboid. How much of brown paper would you need, if you want
to cover this cuboid? Let us see:
SURFACE AREAS AND VOLUMES 209
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
First we would need a rectangular piece to cover
the bottom of the bundle. That would be as shown in
Fig. 13.2 (a)
Then we would need two long rectangular pieces
to cover the two side ends. Now, it would look like
Fig. 13.2 (b).
Now to cover the front and back ends, we would
need two more rectangular pieces of a different size.
With them, we would now have a figure as shown in
Fig. 13.2(c).
This figure, when opened out, would look like
Fig. 13.2 (d).
Finally, to cover the top of the bundle, we would
require another rectangular piece exactly like the one
at the bottom, which if we attach on the right side, it
would look like Fig. 13.2(e).
So we have used six rectangular pieces to cover
the complete outer surface of the cuboid.
Fig. 13.2
Page 3

208 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
CHAPTER 13
SURFACE AREAS AND VOLUMES
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our study, we have been dealing
with figures that can be easily drawn on our notebooks or blackboards. These are
called plane figures. We have understood what rectangles, squares and circles are,
what we mean by their perimeters and areas, and how we can find them. We have
learnt these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheet and
stack them up in a vertical pile. By this process, we shall obtain some solid figures
(briefly called solids) such as a cuboid, a cylinder, etc. In the earlier classes, you have
also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We
shall now learn to find the surface areas and volumes of cuboids and cylinders in
details and extend this study to some other solids such as cones and spheres.
13.2 Surface Area of a Cuboid and a Cube
Have you looked at a bundle of many sheets of paper? How does it look? Does it look
like what you see in Fig. 13.1?
Fig. 13.1
That makes up a cuboid. How much of brown paper would you need, if you want
to cover this cuboid? Let us see:
SURFACE AREAS AND VOLUMES 209
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
First we would need a rectangular piece to cover
the bottom of the bundle. That would be as shown in
Fig. 13.2 (a)
Then we would need two long rectangular pieces
to cover the two side ends. Now, it would look like
Fig. 13.2 (b).
Now to cover the front and back ends, we would
need two more rectangular pieces of a different size.
With them, we would now have a figure as shown in
Fig. 13.2(c).
This figure, when opened out, would look like
Fig. 13.2 (d).
Finally, to cover the top of the bundle, we would
require another rectangular piece exactly like the one
at the bottom, which if we attach on the right side, it
would look like Fig. 13.2(e).
So we have used six rectangular pieces to cover
the complete outer surface of the cuboid.
Fig. 13.2
210 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
This shows us that the outer surface of a cuboid is made up of six rectangles (in
fact, rectangular regions, called the faces of the cuboid), whose areas can be found by
multiplying length by breadth for each of them separately  and then adding the six
areas together.
Now, if we take the length of the cuboid as l, breadth as b and the height as h, then
the figure with these dimensions would be like the shape you see in Fig. 13.2(f).
So, the sum of the areas of the six rectangles is:
Area of rectangle 1 (= l × h)
+
Area of rectangle 2 (= l × b)
+
Area of rectangle 3 (= l × h)
+
Area of rectangle 4 (= l × b)
+
Area of rectangle 5 (= b × h)
+
Area of rectangle 6 (= b × h)
= 2(l × b) + 2(b × h) + 2(l × h)
= 2(lb + bh + hl)
This gives us:
Surface Area of a Cuboid = 2(lb + bh + hl)
where l, b and h are respectively the three edges of the cuboid.
Note : The unit of area is taken as the square unit, because we measure the magnitude
of a region by filling it with squares of side of unit length.
For example, if we have a cuboid whose length, breadth and height are 15 cm,
10 cm and 20 cm respectively, then its surface area would be:
2[(15 × 10) + (10 × 20) + (20 × 15)] cm
2
= 2(150 + 200 + 300) cm
2
= 2 × 650 cm
2
= 1300 cm
2
Page 4

208 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
CHAPTER 13
SURFACE AREAS AND VOLUMES
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our study, we have been dealing
with figures that can be easily drawn on our notebooks or blackboards. These are
called plane figures. We have understood what rectangles, squares and circles are,
what we mean by their perimeters and areas, and how we can find them. We have
learnt these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheet and
stack them up in a vertical pile. By this process, we shall obtain some solid figures
(briefly called solids) such as a cuboid, a cylinder, etc. In the earlier classes, you have
also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We
shall now learn to find the surface areas and volumes of cuboids and cylinders in
details and extend this study to some other solids such as cones and spheres.
13.2 Surface Area of a Cuboid and a Cube
Have you looked at a bundle of many sheets of paper? How does it look? Does it look
like what you see in Fig. 13.1?
Fig. 13.1
That makes up a cuboid. How much of brown paper would you need, if you want
to cover this cuboid? Let us see:
SURFACE AREAS AND VOLUMES 209
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
First we would need a rectangular piece to cover
the bottom of the bundle. That would be as shown in
Fig. 13.2 (a)
Then we would need two long rectangular pieces
to cover the two side ends. Now, it would look like
Fig. 13.2 (b).
Now to cover the front and back ends, we would
need two more rectangular pieces of a different size.
With them, we would now have a figure as shown in
Fig. 13.2(c).
This figure, when opened out, would look like
Fig. 13.2 (d).
Finally, to cover the top of the bundle, we would
require another rectangular piece exactly like the one
at the bottom, which if we attach on the right side, it
would look like Fig. 13.2(e).
So we have used six rectangular pieces to cover
the complete outer surface of the cuboid.
Fig. 13.2
210 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
This shows us that the outer surface of a cuboid is made up of six rectangles (in
fact, rectangular regions, called the faces of the cuboid), whose areas can be found by
multiplying length by breadth for each of them separately  and then adding the six
areas together.
Now, if we take the length of the cuboid as l, breadth as b and the height as h, then
the figure with these dimensions would be like the shape you see in Fig. 13.2(f).
So, the sum of the areas of the six rectangles is:
Area of rectangle 1 (= l × h)
+
Area of rectangle 2 (= l × b)
+
Area of rectangle 3 (= l × h)
+
Area of rectangle 4 (= l × b)
+
Area of rectangle 5 (= b × h)
+
Area of rectangle 6 (= b × h)
= 2(l × b) + 2(b × h) + 2(l × h)
= 2(lb + bh + hl)
This gives us:
Surface Area of a Cuboid = 2(lb + bh + hl)
where l, b and h are respectively the three edges of the cuboid.
Note : The unit of area is taken as the square unit, because we measure the magnitude
of a region by filling it with squares of side of unit length.
For example, if we have a cuboid whose length, breadth and height are 15 cm,
10 cm and 20 cm respectively, then its surface area would be:
2[(15 × 10) + (10 × 20) + (20 × 15)] cm
2
= 2(150 + 200 + 300) cm
2
= 2 × 650 cm
2
= 1300 cm
2
SURFACE AREAS AND VOLUMES 211
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Recall that a cuboid, whose length, breadth and height are all equal, is called a
cube. If each edge of the cube is a, then the surface area of this cube would be
2(a × a + a × a + a × a) i.e., 6a
2
(see Fig. 13.3), giving us
Surface Area of a Cube = 6a
2
where a is the edge of the cube.
Fig. 13.3
Suppose, out of the six faces of a cuboid, we only find the area of the four faces,
leaving the bottom and top faces. In such a case, the area of these four faces is called
the lateral surface area of the cuboid. So, lateral surface area of a cuboid of
length l, breadth b and height h is equal to 2lh + 2bh or 2(l + b)h. Similarly,
lateral surface area of a cube of side a is equal to 4a
2
.
Keeping in view of the above, the surface area of a cuboid (or a cube) is sometimes
also referred to as the total surface area. Let us now solve some examples.
Example 1 : Mary wants to decorate her Christmas
tree. She wants to place the tree on a wooden box
covered with coloured paper with picture of Santa
Claus on it (see Fig. 13.4). She must know the exact
quantity of paper to buy for this purpose. If the box
has length, breadth and height as 80 cm, 40 cm and
20 cm respectively how many square sheets of paper
of side 40 cm would she require?
Solution : Since Mary wants to paste the paper on
the outer surface of the box; the quantity of paper
required would be equal to the surface area of the
box which is of the shape of a cuboid. The dimensions
of the box are:
Fig. 13.4
Page 5

208 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
CHAPTER 13
SURFACE AREAS AND VOLUMES
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our study, we have been dealing
with figures that can be easily drawn on our notebooks or blackboards. These are
called plane figures. We have understood what rectangles, squares and circles are,
what we mean by their perimeters and areas, and how we can find them. We have
learnt these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheet and
stack them up in a vertical pile. By this process, we shall obtain some solid figures
(briefly called solids) such as a cuboid, a cylinder, etc. In the earlier classes, you have
also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We
shall now learn to find the surface areas and volumes of cuboids and cylinders in
details and extend this study to some other solids such as cones and spheres.
13.2 Surface Area of a Cuboid and a Cube
Have you looked at a bundle of many sheets of paper? How does it look? Does it look
like what you see in Fig. 13.1?
Fig. 13.1
That makes up a cuboid. How much of brown paper would you need, if you want
to cover this cuboid? Let us see:
SURFACE AREAS AND VOLUMES 209
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
First we would need a rectangular piece to cover
the bottom of the bundle. That would be as shown in
Fig. 13.2 (a)
Then we would need two long rectangular pieces
to cover the two side ends. Now, it would look like
Fig. 13.2 (b).
Now to cover the front and back ends, we would
need two more rectangular pieces of a different size.
With them, we would now have a figure as shown in
Fig. 13.2(c).
This figure, when opened out, would look like
Fig. 13.2 (d).
Finally, to cover the top of the bundle, we would
require another rectangular piece exactly like the one
at the bottom, which if we attach on the right side, it
would look like Fig. 13.2(e).
So we have used six rectangular pieces to cover
the complete outer surface of the cuboid.
Fig. 13.2
210 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
This shows us that the outer surface of a cuboid is made up of six rectangles (in
fact, rectangular regions, called the faces of the cuboid), whose areas can be found by
multiplying length by breadth for each of them separately  and then adding the six
areas together.
Now, if we take the length of the cuboid as l, breadth as b and the height as h, then
the figure with these dimensions would be like the shape you see in Fig. 13.2(f).
So, the sum of the areas of the six rectangles is:
Area of rectangle 1 (= l × h)
+
Area of rectangle 2 (= l × b)
+
Area of rectangle 3 (= l × h)
+
Area of rectangle 4 (= l × b)
+
Area of rectangle 5 (= b × h)
+
Area of rectangle 6 (= b × h)
= 2(l × b) + 2(b × h) + 2(l × h)
= 2(lb + bh + hl)
This gives us:
Surface Area of a Cuboid = 2(lb + bh + hl)
where l, b and h are respectively the three edges of the cuboid.
Note : The unit of area is taken as the square unit, because we measure the magnitude
of a region by filling it with squares of side of unit length.
For example, if we have a cuboid whose length, breadth and height are 15 cm,
10 cm and 20 cm respectively, then its surface area would be:
2[(15 × 10) + (10 × 20) + (20 × 15)] cm
2
= 2(150 + 200 + 300) cm
2
= 2 × 650 cm
2
= 1300 cm
2
SURFACE AREAS AND VOLUMES 211
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Recall that a cuboid, whose length, breadth and height are all equal, is called a
cube. If each edge of the cube is a, then the surface area of this cube would be
2(a × a + a × a + a × a) i.e., 6a
2
(see Fig. 13.3), giving us
Surface Area of a Cube = 6a
2
where a is the edge of the cube.
Fig. 13.3
Suppose, out of the six faces of a cuboid, we only find the area of the four faces,
leaving the bottom and top faces. In such a case, the area of these four faces is called
the lateral surface area of the cuboid. So, lateral surface area of a cuboid of
length l, breadth b and height h is equal to 2lh + 2bh or 2(l + b)h. Similarly,
lateral surface area of a cube of side a is equal to 4a
2
.
Keeping in view of the above, the surface area of a cuboid (or a cube) is sometimes
also referred to as the total surface area. Let us now solve some examples.
Example 1 : Mary wants to decorate her Christmas
tree. She wants to place the tree on a wooden box
covered with coloured paper with picture of Santa
Claus on it (see Fig. 13.4). She must know the exact
quantity of paper to buy for this purpose. If the box
has length, breadth and height as 80 cm, 40 cm and
20 cm respectively how many square sheets of paper
of side 40 cm would she require?
Solution : Since Mary wants to paste the paper on
the outer surface of the box; the quantity of paper
required would be equal to the surface area of the
box which is of the shape of a cuboid. The dimensions
of the box are:
Fig. 13.4
212 MATHEMA TICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Length =80 cm, Breadth = 40 cm, Height = 20 cm.
The surface area of the box = 2(lb + bh + hl)
= 2[(80 × 40) + (40 × 20) + (20 × 80)] cm
2
= 2[3200 + 800 + 1600] cm
2
= 2 × 5600 cm
2
= 11200 cm
2
The area of each sheet of the paper = 40 × 40 cm
2
= 1600 cm
2
Therefore, number of sheets required =
surface area of box
area of one sheet of paper
=
11200
1600
= 7
So, she would require 7 sheets.
Example 2 : Hameed has built a cubical water tank with lid for his house, with each
outer edge 1.5 m long. He gets the outer surface of the tank excluding the base,
covered with square tiles of side 25 cm (see Fig. 13.5). Find how much he would
spend for the tiles, if the cost of the tiles is Rs 360 per dozen.
Solution : Since Hameed is getting the five outer faces of the tank covered with tiles,
he would need to know the surface area of the tank, to decide on the number of tiles
required.
Edge of the cubical tank = 1.5 m = 150 cm (= a)
So, surface area of the tank = 5 × 150 × 150 cm
2
Area of each square tile = side × side = 25 × 25 cm
2
So, the number of tiles required =
surface area of the tank
area of each tile
=
5 × 150 × 150
25 × 25
= 180
Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs 360
Therefore, cost of one tile = Rs
360
12
= Rs 30
So, the cost of 180 tiles = 180 × Rs 30 = Rs 5400
Fig. 13.5
```
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## Mathematics (Maths) Class 9

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