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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
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138 MATHEMATICS
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 =
× base of each triangle × l.
So, area of the entire piece of paper
Fig. 11.2
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SURFACE
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= sum of the areas of all the triangles
=
⋯
=
⋯
=
× 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 =
× l × 2
pp
pp
pr =
p
p
pp
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 =
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 =
pp
pp
prl +
pp
p
p
pr
2
=
pp
pp
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
=
× 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 =
cm = 20 cm
Fig. 11.4
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140 MATHEMATICS
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 =
=
cm
=
cm = 20.11 cm
Therefore, the curved surface area of the corn cob = prl
=
× 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 =
, 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
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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
= 1.02)
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
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142 MATHEMATICS
Can you guess what happens to the centre of the circle, when it forms a sphere on
rotation? Of course, it becomes the centre of the sphere. So, a sphere is a three
dimensional figure (solid figure), which is made up of all points in the space,
which lie at a constant distance called the radius, from a fixed point called the
centre of the sphere.
Note : A sphere is like the surface of a ball. The word solid sphere is used for the
solid whose surface is a sphere.
Activity : Have you ever played with a top or have you at least watched someone
play with one? You must be aware of how a string is wound around it. Now, let us take
a rubber ball and drive a nail into it. Taking support of the nail, let us wind a string
around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep
the string in place, and continue to wind the string around the remaining part of the ball,
till you have completely covered the ball [see Fig. 11.7(a)]. Mark the starting and
finishing points on the string, and slowly unwind the string from the surface of the ball.
Now, ask your teacher to help you in measuring the diameter of the ball, from which
you easily get its radius. Then on a sheet of paper, draw four circles with radius equal
to the radius of the ball. Start filling the circles one by one, with the string you had
wound around the ball [see Fig. 11.7(b)].
Fig. 11.7
What have you achieved in all this?
The string, which had completely covered the surface area of the sphere, has been
used to completely fill the regions of four circles, all of the same radius as of the sphere.
So, what does that mean? This suggests that the surface area of a sphere of radius r
= 4 times the area of a circle of radius r = 4 × (p r
2
)
So, Surface Area of a Sphere = 4
pp
pp
p r
2
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SURFACE
AREAS AND
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where r is the radius of the sphere.
How many faces do you see in the surface of a sphere? There is only one, which is
curved.
Now, let us take a solid sphere, and slice it exactly ‘through
the middle’ with a plane that passes through its centre. What
happens to the sphere?
Yes, it gets divided into two equal parts (see Fig. 11.8)! What
will each half be called? It is called a hemisphere. (Because
‘hemi’ also means ‘half’)
And what about the surface of a hemisphere? How many faces does it have?
Two! There is a curved face and a flat face (base).
The curved surface area of a hemisphere is half the surface area of the sphere, which
is
of 4pr
2
.
Therefore, Curved Surface Area of a Hemisphere = 2
pp
pp
pr
2
where r is the radius of the sphere of which the hemisphere is a part.
Now taking the two faces of a hemisphere, its surface area 2pr
2
+ pr
2
So, Total Surface Area of a Hemisphere = 3
pp
pp
pr
2
Example 4 : Find the surface area of a sphere of radius 7 cm.
Solution : The surface area of a sphere of radius 7 cm would be
4pr
2
= 4 ×
× 7 × 7 cm
2
= 616 cm
2
Example 5 : Find (i) the curved surface area and (ii) the total surface area of a
hemisphere of radius 21 cm.
Solution : The curved surface area of a hemisphere of radius 21 cm would be
= 2pr
2
= 2 ×
× 21 × 21 cm
2
= 2772 cm
2
Fig. 11.8
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144 MATHEMATICS
(ii) the total surface area of the hemisphere would be
3pr
2
= 3 ×
× 21 × 21 cm
2
= 4158 cm
2
Example 6 : The hollow sphere, in which the circus motorcyclist performs his stunts,
has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding
space available for the motorcyclist is the surface area of the ‘sphere’ which is
given by
4pr
2
= 4 ×
× 3.5 × 3.5 m
2
= 154 m
2
Example 7 : A hemispherical dome of a building needs to be painted
(see Fig. 11.9). If the circumference of the base of the dome is 17.6 m, find the cost of
painting it, given the cost of painting is ` 5 per 100 cm
2
.
Solution : Since only the rounded surface of the dome is to be painted, we would need
to find the curved surface area of the hemisphere to know the extent of painting that
needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2pr.
So, the radius of the dome = 17.6 ×
m = 2.8 m
The curved surface area of the dome = 2pr
2
= 2 ×
× 2.8 × 2.8 m
2
= 49.28 m
2
Now, cost of painting 100 cm
2
is ` 5.
So, cost of painting 1 m
2
= ` 500
Therefore, cost of painting the whole dome
= ` 500 × 49.28
= ` 24640
EXERCISE 11.2
Assume p =
, unless stated otherwise.
1. Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Fig. 11.9
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SURFACE
AREAS AND
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2. Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
3. Find the total surface area of a hemisphere of radius 10 cm. (Use p = 3.14)
4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped
into it. Find the ratio of surface areas of the balloon in the two cases.
5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of
tin-plating it on the inside at the rate of ` 16 per 100 cm
2
.
6. Find the radius of a sphere whose surface area is 154 cm
2
.
7. The diameter of the moon is approximately one fourth of the diameter of the earth.
Find the ratio of their surface areas.
8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is
5 cm. Find the outer curved surface area of the bowl.
9. A right circular cylinder just encloses a sphere of
radius r
(see Fig. 11.10). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
11.3 Volume of a Right Circular Cone
In earlier classes we have studied the volumes of
cube, cuboid and cylinder
In Fig 11.11, can you see that there is a right circular
cylinder and a right circular cone of the same base
radius and the same height?
Activity : Try to make a hollow cylinder and a hollow cone like this with the same
base radius and the same height (see Fig. 11.11). Then, we can try out an experiment
that will help us, to see practically what the volume of a right circular cone would be!
Fig. 11.12
Fig. 11.10
Fig. 11.11
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146 MATHEMATICS
So, let us start like this.
Fill the cone up to the brim with sand once, and empty it into the cylinder. We find
that it fills up only a part of the cylinder [see Fig. 11.12(a)].
When we fill up the cone again to the brim, and empty it into the cylinder, we see
that the cylinder is still not full [see Fig. 11.12(b)].
When the cone is filled up for the third time, and emptied into the cylinder, it can be
seen that the cylinder is also full to the brim [see Fig. 11.12(c)].
With this, we can safely come to the conclusion that three times the volume of a
cone, makes up the volume of a cylinder, which has the same base radius and the
same height as the cone, which means that the volume of the cone is one-third the
volume of the cylinder.
So, Volume of a Cone =
pp
pp
pr
2
h
where r is the base radius and h is the height of the cone.
Example 8 : The height and the slant height of a cone are 21 cm and 28 cm respectively.
Find the volume of the cone.
Solution : From l
2
= r
2
+ h
2
, we have
r =
So, volume of the cone =
pr
2
h =
×
cm
3
= 7546 cm
3
Example 9 : Monica has a piece of canvas whose area is 551 m
2
. She uses it to have
a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins
and the wastage incurred while cutting, amounts to approximately 1 m
2
, find the volume
of the tent that can be made with it.
Solution : Since the area of the canvas = 551 m
2
and area of the canvas lost in
wastage is 1 m
2
, therefore the area of canvas available for making the tent is
(551 – 1) m
2
= 550 m
2
.
Now, the surface area of the tent = 550 m
2
and the required base radius of the conical
tent = 7 m
Note that a tent has only a curved surface (the floor of a tent is not covered by
canvas!!).
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SURFACE
AREAS AND
VOLUMES 147
Therefore, curved surface area of tent = 550 m
2
.
That is, prl = 550
or,
× 7 × l = 550
or, l = 3
m = 25 m
Now, l
2
= r
2
+ h
2
Therefore, h =
=
= 24 m
So, the volume of the conical tent =
= 1232 m
3
.
EXERCISE 11.3
Assume p =
, unless stated otherwise.
1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm
2. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm
3. The height of a cone is 15 cm. If its volume is 1570 cm
3
, find the radius of the base.
(Use p = 3.14)
4. If the volume of a right circular cone of height 9 cm is 48 p cm
3
, find the diameter of its
base.
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
6. The volume of a right circular cone is 9856 cm
3
. If the diameter of the base is 28 cm,
find
(i) height of the cone (ii) slant height of the cone
(iii) curved surface area of the cone
7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
Find the volume of the solid so obtained.
8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find
the volume of the solid so obtained. Find also the ratio of the volumes of the two
solids obtained in Questions 7 and 8.
9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m.
Find its volume. The heap is to be covered by canvas to protect it from rain. Find the
area of the canvas required.
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148 MATHEMATICS
11.4 Volume of a Sphere
Now, let us see how to go about measuring the volume of a sphere. First, take two or
three spheres of different radii, and a container big enough to be able to put each of
the spheres into it, one at a time. Also, take a large trough in which you can place the
container. Then, fill the container up to the brim with water [see Fig. 11.13(a)].
Now, carefully place one of the spheres in the container. Some of the water from
the container will over flow into the trough in which it is kept [see Fig. 11.13(b)].
Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated
cylindrical jar) and measure the water over flowed [see Fig. 11.13(c)]. Suppose the
radius of the immersed sphere is r (you can find the radius by measuring the diameter
of the sphere). Then evaluate
pr
3
. Do you find this value almost equal to the
measure of the volume over flowed?
Fig. 11.13
Once again repeat the procedure done just now, with a different size of sphere.
Find the radius R of this sphere and then calculate the value of
Once again this
value is nearly equal to the measure of the volume of the water displaced (over flowed)
by the sphere. What does this tell us? We know that the volume of the sphere is the
same as the measure of the volume of the water displaced by it. By doing this experiment
repeatedly with spheres of varying radii, we are getting the same result, namely, the
volume of a sphere is equal to
times the cube of its radius. This gives us the idea
that
Volume of a Sphere =
where r is the radius of the sphere.
Later, in higher classes it can be proved also. But at this stage, we will just take it
as true.
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SURFACE
AREAS AND
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Since a hemisphere is half of a sphere, can you guess what the volume of a
hemisphere will be? Yes, it is
=
pr
3
.
So, Volume of a Hemisphere =
where r is the radius of the hemisphere.
Let us take some examples to illustrate the use of these formulae.
Example 10 : Find the volume of a sphere of radius 11.2 cm.
Solution : Required volume =
pr
3
=
cm
3
= 5887.32 cm
3
Example 11 : A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the
metal is 7.8 g per cm
3
, find the mass of the shot-putt.
Solution : Since the shot-putt is a solid sphere made of metal and its mass is equal to
the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere =
=
= 493 cm
3
(nearly)
Further, mass of 1 cm
3
of metal is 7.8 g.
Therefore, mass of the shot-putt = 7.8 × 493 g
= 3845.44 g = 3.85 kg (nearly)
Example 12 : A hemispherical bowl has a radius of 3.5 cm. What would be the
volume of water it would contain?
Solution : The volume of water the bowl can contain
=
=
cm
3
= 89.8 cm
3
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150 MATHEMATICS
EXERCISE 11.4
Assume p =
, unless stated otherwise.
1. Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of
the metal is 8.9 g per cm
3
?
4. The diameter of the moon is approximately one-fourth of the diameter of the earth.
What fraction of the volume of the earth is the volume of the moon?
5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m,
then find the volume of the iron used to make the tank.
7. Find the volume of a sphere whose surface area is 154 cm
2
.
8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed
at the cost of ` 4989.60. If the cost of white-washing is ` 20 per square metre, find the
(i) inside surface area of the dome, (ii) volume of the air inside the dome.
9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to
form a sphere with surface area S¢. Find the
(i) radius r¢ of the new sphere, (ii) ratio of S and S¢.
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much
medicine (in mm
3
) is needed to fill this capsule?
11.5 Summar
y
In this chapter, you have studied the following points:
1. Curved surface area of a cone =
pp
pp
prl
2. Total surface area of a right circular cone =
pp
p
p
prl +
pp
pp
pr
2
, i.e.,
p p
p p
pr (l + r)
3. Surface area of a sphere of radius r = 4
pp
pp
p r
2
4. Curved surface area of a hemisphere = 2
pp
pp
pr
2
5. Total surface area of a hemisphere = 3
pp
pp
pr
2
6. Volume of a cone =
pp
pp
pr
2
h
7. Volume of a sphere of radius r =
8. Volume of a hemisphere =
[Here, letters l, b, h, a, r, etc. have been used in their usual meaning, depending on the
context.]
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STATISTICS 151
CHAPTER 12
STATISTICS
12.1 Graphical Representation of Data
The representation of data by tables has already been discussed. Now let us turn our
attention to another representation of data, i.e., the graphical representation. It is well
said that one picture is better than a thousand words. Usually comparisons among the
individual items are best shown by means of graphs. The representation then becomes
easier to understand than the actual data. We shall study the following graphical
representations in this section.
(A) Bar graphs
(B) Histograms of uniform width, and of varying widths
(C) Frequency polygons
(A) Bar Graphs
In earlier classes, you have already studied and constructed bar graphs. Here we
shall discuss them through a more formal approach. Recall that a bar graph is a
pictorial representation of data in which usually bars of uniform width are drawn with
equal spacing between them on one axis (say, the x-axis), depicting the variable. The
values of the variable are shown on the other axis (say, the y-axis) and the heights of
the bars depend on the values of the variable.
Example 1 : In a particular section of Class IX, 40 students were asked about the
months of their birth and the following graph was prepared for the data so obtained:
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