16
Meaning of cylinder, cone,
frustum of a cone, sphere and
hemisphere.
Surface area and volume of
cylinder, cone, frustum of a
cone, sphere and hemisphere.
Combined figures, their area
and volume.
Scale drawing.
Mensuration
This unit facilitates you in,
defining cylinder, cone and sphere.
explaining the properties of cylinder, cone
and sphere and hemisphere.
deriving the formula to find the area and
volume of cylinder, cone, sphere and
hemisphere.
calculating the LSA and TSA of cylinder,
cone, sphere and hemisphere.
calculating the volume of cylinder, cone,
sphere and hemisphere.
solving application problems based on area
and volume of cylinder, cone and sphere.
explaining the meaning of frustum of a cone.
calculating the area and volume of frustum
of a cone.
identifying objects made up of two or more
solids.
finding the area and volume of combined
solid objects.
Archimedes
(287BC-212BC)
constructing scale drawing for irregular
shaped figures.
calcualting area of irregular shaped figures.
Archimedes of syracuse, Sicily, is
remembered as the greatest Greek
mathematician of the ancient era.
He contributed significantly in
geometry regarding the areas of
plane figures and the areas and
volumes of solid objects.
He proved that the volume of a
sphere is equal to two-thirds the
volume of a circumscribed cylinder.
Give me a place to stand,
and I will move the earth.
- Archimedes of Syracuse
378
UNIT-16
The study of measurement of objects is very essential because of its applications in
every day life, industry and engineering. We have already learnt to find the surface area
and volume of solid objects such as cube, cuboid, prism and pyramid.
Now let us consider some examples where measurement of objects are
applied.
• A village panchayat has built an over head water storage tank which
is cylindrical in shape. How to determine the size of the tank? What
is the cost incurred in painting the tank?
• How do you calculate the amount of diesel
that can be stored in a cylindrical shaped
fuel tank of a truck?
• What is the quantity of
colour paper required to
prepare a birthday cap?
• How many basin laddoos can be made from given weight of mixture
of basin laddoo?
The solution for all the above mentioned problems depends on finding the surface
area and volume of solid objects like cylinder, cone and sphere.
How to find the surface area and volume of these objects? Which are the formulae
used?
In this unit, let us learn about the surface area and volume of cylinder, cone and
sphere.
Cylinder
We have studied that when a number of rectangular sheets of paper are stacked up,
we get a cuboid or rectangular based prism.
What do we get when a number of congruent
circular sheets of paper are stacked up?
Observe the figures given.
The solid object obtained is called Cylinder.
 Cylinders are sort of prisms with circular bases.
Activity: Take a thick paper and cut a rectangle ABCD. Paste a long thick stick along one
side, say AB. Hold the stick with both your palms and rotate it fast (you can also paste a
long thick string, hold the string with hands, on either side and rotate it fast).
D
A
Do this activity in groups. Discuss and share
the ideas.
What shape can you recognise? From the above
activity, we can conclude that
If a rectangle revolves about one of its sides C
and completes a full rotation, the solid formed
is called a right circular cylinder.
B
Mensuration
379
In our daily life, we have
seen many objects which are
cylindrical in shape. Some of them
are given below.
The cylinders can be either
solid or hollow. Observe that the Milk can
Pipe
Gas cylinder
cylinder has two circular bases,
bottom base and top base which are congruent to each other.
Pillars
Activity: Place a cylinder on the table and hold a piece of cardboard
parallel to the top of the table and touching the top of the cylinder.
If we find the perpendicular distance between the top of the table
(i.e bottom of the cylinder) to the cardboard (i.e top of the cylinder), it is
the height of the cylinder.
Now let us consider a cylinder having the following properties.
• It has two congruent and parallel circular bases.
• It has a curved surface joining the edges of the two bases.
• The line segment joining the centres of the two bases is perpendicular to the base.
It is the height (h) of the cylinder and also called axis of the cylinder.
Such a cylinder is called right circular cylinder.
It is called so because, the cylinder has a circular base and its axis is at
right angle to the base.
h
In this unit, we will be dealing with only right circular cylinders. So unless stated
otherwise, the word cylinder would mean a right circular cylinder.
Surface area of a cylinder
We know that a cylinder can be either a hollow cylinder or a solid cylinder.
A hollow cylinder is formed by the curved surface only. Pipe, straw, tubes, flute, are
examples for hollow cylinders.
A solid cylinder is the object bounded by two circular plane surfaces and also the
curved surface. A cylindrical pillar, the wheels of road roller, a garden roller, pencil, filled
cylindrical tanks and filled cans are examples of solid cylinder.
The surface area of a cylinder refers to the area of the external surface of the
cylinder.
A hollow cylinder has only the curved surface, which is also called the lateral surface.
The area of this surface is the curved surface area (CSA) or the
Top circular
base
lateral surface area (LSA) of the hollow cylinder.
A solid cylinder has two circular surfaces and also the curved
surface. Hence, the area of the curved surface is the curved or
lateral surface area of the solid cylinder.
The total surface area (TSA) of the solid cylinder refers to
the sum of areas of the two circular regions and the curved surface
area.
Curved or
lateral surface
Bottom circular
base
380
UNIT-16
Now, let us derive the formulae used to find the curved surface area and total surface
area of cylinders.
i. Curved surface area of a cylinder: Take a rectangular sheet of paper whose length
is just enough to go round the given cylinder and whose breadth is equal to the
height of the cylinder. This activity can also be done by taking a cylinder made of
cardboard. Cut open the lateral surface along a line and unfold it, to get a rectangle.
Fold
b
h
b
Cut
l
2r
Observe that,
Area of the rectangular sheet = Curved surface area of the cylinder
2
Length of the rectangle = Circumference of the base of the cylinder  l = 2r
Breadth of the rectangle = Height of the cylinder b = h
We know that, Area of rectangle = l × b = 2r × h = 2rh
Curved surface area of the cylinder = 2rh sq. units
ii. Total surface area of a solid cylinder:
Total surface area
of the cylinder
=
Curved surface area
of the cylinder
+ 2 ×
=
2rh
= 2rh + 2r2 = 2r (h + r)
Area of the
circular bases
+ 2 ×
r2
Total surface area of a cylinder = 2r (h + r) sq. units.
Volume of a cylinder
You have already learnt that volume of a prism is the product of area of its
base and height. Since, cylinder is also a type of prism with circular base, its
volume should be calculated in the same manner.
A= r
h
Volume of a solid cylinder = Area of base × height = r2 × h = r2h
Volume of a cylinder = r2h cubic units
r
ILLUSTRATIVE EXAMPLES
Example 1 : Find the CSA of a right circular cylinder whose height and radius of base
are 30 cm and 3.5 cm respectively.
0.5
22
3.5 cm×30cm = 660cm2
7
 CSA of the cylinder = 660cm2
Sol. CSA of cylinder = 2rh = 2
Mensuration
381
Example 2 : The CSA of a right circular cylinder of height 14 cm is 88 cm2. Find the
radius of the base of the cylinder.
22
r 14 cm
7
1
2 4
88 cm2 7
r=
= 1 cm
2 22 14 2cm
 radius of the base of the cylinder = 1 cm
Example 3 : The radii of two right circular cylinders are in the ratio 2 : 3 and their
heights are in the ratio 5 : 4. Calculate the ratio of their curved surface areas.
Sol. Let the radii of 2 cylinders be 2r and 3r respectively and their heights be 5h and 4h
respectively.
Let S1 and S2 be the curved surface areas of two cylinders.
S1  CSA of cylinder of radius 2r and height 5h = S1 = 2 ×  × 2r × 5h = 20rh
S2 CSA of cylinder with radius 3r and height 4h= S2= 2 ×  × 3r × 4h = 24rh
Sol. CSA of a cylinder = 2rh
S1
S2 =
88 cm2= 2
5
20 rh 5
24 rh 6

S1 : S 2 = 5 : 6
Example 4 : The diameter of a roller 120cm long is 84 cm. If it takes 500 complete
revolutions to level a play ground, determine the cost of levelling it at the rate of
0.50 paise per square metre.
Sol. h = 120 cm, r = 42 cm.
Area covered by the roller in one revolution = CSA of the roller = 2rh

6
6
22
42 cm × 120cm = 31,680 cm2
7
 Area covered in 500 revolutions = (31,680 × 500) cm2 = 15,84,000 cm2
= 2
15840000 2
m = 1584 m2
100 100
Cost of levelling the playground per m2 = 0.50 ps
 Cost of levelling the playground= 1584 m2 × 0.50 ps/m2 = ` 792.
Example 5 : A metal pipe is 77 cm long. The inner diameter of cross section is 4 cm,
the outer diameter being 4.4cm. Find its (a) inner curved surface area (b) Outer curved
surface area (c) Total surface area.
Sol. Outer radius = R = 2.2 cm;
Inner radius = r = 2 cm;
h = 77cm
11
22
2cm 77 cm = 968 cm2
(a) Inner CSA = S1 = 2rh = 2
7
11
22
2.2cm 77 cm = 1064.8cm2
(b) Outer CSA = S2 = 2Rh = 2
7
(c) TSA = S1 + S2 + area of two bases = S1 + S2 + 2(R2 – r2)
22
= 968cm2 + 1064.8cm2 + 2
{(2.2)2 – 22}
7
0.12
44
cm2
0.84
= 968 1064.8
7
= (968 + 1064.8 + 5.28) cm2
Area covered inm 2
TSA = 2038.08 cm2
382
UNIT-16
Example 6 : Find the volume of a right circular cylinder if the radius of its base is 7cm
and height is 15 cm.
Sol. r = 7cm, h = 15 cm
22
7 cm × 7cm × 15cm = 2310 cm3
7
Example 7 : The circumference of the base of a cylinder is 132cm and its height is 25
cm. Find the volume of the cylinder.
Volume of a cylinder = r2h =
Sol. Let 'r' be the radius of the cylinder. Circumference = 132 cm
2r = 132 cm
2
22
7
6
3
132 cm 7
r = 132 cm  r =
 r = 21 cm
2 22
Volume of a cylinder = r2 h =
22
7
(21)2
25 =
22
7
21
3
21 25 = 34,650 cm3
Example 8 : A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it
is spread all around to a width of 5m to form an embankment. Find the height of the
embankment.
Sol. Volume of earth dug out = r2h =
22
7
52
5m
2
14 m3
3
5m
5m
3
= 22 × 25 × 2m = 1100 m
Area of the embankment (Shaded region) = (R2 – r2)
22
22
102 5 2
7
7
Volume of earth dugout = (r2) h = (Area of embankement) × h
=

height of the embankment =
75m2
Volume of earth dugout
Area of embankment
1100m3
7 1100
m = 4.66m
= 22
=
22 75
2
75m
7
 Height of the embankment = 4.66m
Example 9 : The difference between outside and inside surfaces of a cylindrical metallic
pipe 14 cm long is 44 cm2. If the pipe is made of 99cm3 of metal, find the outer and
inner radii of the pipe.
Sol. Let, R be the external radius and r be the inner radius of the metallic pipe. h=14 cm
Outer surface area – inner surface area = 44cm2
 2Rh – 2rh = 44 cm2  R – r=
2

R–r =
1
2
44
=
22
2
14
7
2
44
22
7
14 2
Mensuration
383
Volume of metal used = 99cm3  External volume – internal volume = 99cm3
R2h – r2h = 99 cm3h(R2 – r2) = 99 cm3

22
7
2
14 (R
22
r)(R
2 (R
R
r
9
R
r
1
2R
10
r) = 99 cm3
r)
1
2
[  a2 – b2 = (a + b)(a – b)] and  R
r
1
2
9
= 99 cm3 R + r =
2
R + r = 92
2
r + 2.5 = 4.5
99
9
=
2
22
r = 4.5 – 2.5 = 2 cm
2
2R = 5
R = 5 2 = 2.5 cm

External radius = 2.5 cm;
Internal radius = 2 cm
Example 10 : The circumference of the base of a cylindrical vessel is 132 cm and its
height is 25 cm. How many litres of water can it hold?
Sol. C= 132 cm, h = 25 cm Let the radius of the base = r
2r = 132cm  2
22
7
6
r = 132 cm
3
7
132
132
r=
= 2 22 = 21
22
2
7
Volume of the vessel = r2h =
22
7
 r = 21 cm
(21)2
25cm3 =
22
7
21
3
21 25 cm3 = 34650cm3
34650
[ 1000 cm3 = 1l ] = 34.65 litres
1000
 The vessel can hold 34.65 litres of water.
Volume in litres =
Example11 : If the radius of the base of a right circular cylinder is halved, keeping
the height same, what is the ratio of the volume of the new cylinder to that of the
original cylinder.
Sol. Let V1 be the volume of the original cylinder and V2 be the volume of the new cylinder.
Radius of original cylinder = r, radius of new cylinder =
V1= r2h and V2 =
 V2 : V1 = 1 : 4
r
2
2
V1
h 
V2 =
4
r2 h
=
2
1
r
h
4
V2
V1
1
4
r
2
384
UNIT-16
Example 12 : A solid cylinder has a TSA of 462 cm2. Its CSA is one - third of its TSA.
Find the volume of the cylinder.
1
1
1
TSA of the cylinder  2rh=
× 2r (h + r) =
3
3
3
2
 2rh = 154 cm
Sol. CSA of cylinder=
462
154
cm2
TSA of a cylinder = 2r (h + r)  462 cm2= 2rh + 2r2 = 154 cm2 + 2r2
22
7
462cm2 – 154 cm2 = 2r2  308 cm2= 2
14
7
308
7
r 
= r2
2 22
2
r2 = 49  r = 7 cm
CSA of cylinder = 154 cm2  2rh= 154 cm2
2
22
7
7 cm
h = 154 cm2
 h=
22
Volume of the cylinder = r h =
7
2
72
154
44
7
2
=
7
cm
2
7
cm3 =
2
11
22
7
49
7
cm3 = 539cm3
2
EXERCISE 16.1
1. The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm.
Find its (i) CSA (ii) TSA.
2. The CSA of a cylindrical pipe is 550 sq.cm. If the height of the pipe is 25 cm, find the
diameter of the base.
3. An iron pipe 20cm long has external radius equal to 12.5cm and internal radius
equal to 11.5cm. Find the TSA of the pipe.
4. The radii of two right circular cylinders are in the ratio 2 : 3 and the ratio of their
curved surface areas is 5 : 6. Find the ratio of their heights.
5. Find the ratio between TSA of a cylinder to its CSA given its height and radius are
7 cm and 3.5 cm respectively.
6. The inner diameter of a circular well is 2.8m. It is 10 m deep. Find its inner curved
surface area. Also find the cost of plastering this curved surface at the rate of `. 42
per m2?
7. Craft teacher of a school taught the students to prepare cylindrical pen holders out
of card board. In a class of strength 42, if each child prepared a pen holder of radius
5 cm and height 14 cm, how much cardboard was consumed?
8. A solid cylinder has a total surface area of 462cm2. If its curved surface area is one
third of its total surface area, find the radius and height of the cylinder.
9. A cylindrical vessel without lid has to be tin coated on its outside. If the radius of
the base is 70 cm and its height is 1.4m, calculate the cost of tin coating at the rate
of `. 3.50 per 1000 cm2.
10. The diameter of a garden roller is 1.4 m and is 2 m long. How much area will it cover
in 5 revolutions?
Mensuration
385
11. Find the volume of a right circular cylinder whose radius is 10.5 cm and height is
16 cm.
12. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is
28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1cm3 of wood has a mass
of 0.6 gm.
13. The lateral surface area of a cylinder of height 5 cm is 94.2 cm2. Find (i) radius of its
base (ii) Volume of the cylinder.
14. Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find
the ratio of their radii.
15. A rectangular sheet of paper, 44 cm × 20 cm is rolled along its length to form a
cylinder. Find the volume of the cylinder so formed.
16. The trunk of a tree is cylindrical. Its circumference is 176 cm. If the length of the
trunk is 3m, find the volume of the timber that can be obtained from the trunk.
17. A well of diameter 14 m is dug 8m deep. The earth taken out of it has been evenly
spread all around it to a width of 21m to form an embankement. Find the height of
the embankment.
18. In a village fair, a stall keeper has kept a large cylindrical vessel of base radius
15 cm filled upto a height of 32 cm with orange juice. He sells them at `. 3 per glass
in a cylindrical glass of radius 3 cm and height 8 cm. How much money does the
stall keeper earn by selling the
juice completely.
Cone
Observe the following objects.
They are conical in shape. They Ice-cream Birthday
can be hollow or solid.
cone
cap
Roof of the
hut
Heap of
sand
Observe that cones have a circular base and a curved surface which tapers from
the base to an end point.
Compare cones with pyramids. Cones are sort of pyramids with
circular bases.
A cone is a solid object that tapers smoothly from a flat circular
Pyramid
Cone
base to a point called vertex.
A right circular cone has the circular base and its axis passes through the centre of
the base and is perpendicular to the base.
Activity: Take a thick paper and cut a right angled triangle ABC, with B = 900. Paste a
long thick string along one of the perpendicular sides, say AB. Hold the string with the
B
hands on either side of the triangle and
B
D
C
C
C
B
D
rotate the triangle about the string.
Can you recognise the shape formed?
The shape formed is a right circular
A
A
cone. Therefore we can conclude that,
A
If a right angled triangle is revolved about one of the side containing the right angle,
the solid formed is called a right circular cone.
386
UNIT-16
Observe the figure of cone.
A
• The length AB or the distance from the centre of the circular base to
the vertex is called the height (h) of the cone. It is called the axis of
the cone.
• The length BC is the radius of the base (BC = r).
l
h
D
B
C
• The length AC or the distance from the vertex to any point on the
circular base is called the slant height (l) of the cone. (AC = AD = l)
Is there any relation between the height (h) and slant height (l) of a cone?
Consider the right angled triangle ABC in the figure
We have, AC2= AB2 + BC2 (Pythagoras theorem)  l2 = h2 + r2
where l is the slant height, h is the height, r is the radius of the base of the cone.

l=
h2
r 2 or h =
l2
r 2 or r =
l2
h2
Now, let us consider a cone having the following properties.
• It has a circular plane as its base.
• The axis of the cone and the slant height intersect at a point called vertex.
• It has a curved surface which connects the edge of circular base and the vertex.
• The line joining the vertex and the centre of the circular base is perpendicular to
the base.
Such a cone is called right circular cone. As in case of cylinder, since we will be
studying about only right circular cones in this unit, by "cone" we shall mean "right
circular cone".
Surface area of a cone
Observe the given below figures:
The surface of a cone consists of a
curved surface made by bending a sector
such that its arc touches with the edge of the circular base.
The area of the sector which forms the cone is called the
curved surface area or the lateral surface area of the cone.
Sector: is a part of
circular region bounded
by two radii and an arc.
If the base of the cone is closed with a circular piece of
cardboard, then we consider the Total surface area of a cone (TSA) by adding the curved
surface area (CSA) with the area of the circular base.
i. Curved surface area of cone
If a cone made up of cardboard is cut and spread out, we get a sector as shown below.
l
h
l
The area of the sector is the curved surface area of the cone.
Mensuration
387
How to find the area of the sector?
Let 'r' be the radius of the cone, 'l ' be the slant height of the cone and
'' be the central angle of the sector.
Here, radius 'r' of the sector is same as the slant height 'l' of the cone.  r = l
l2
360
Let 'L' denote the length of the arc of the sector.
 Area of the sector =
Then,
2 l
L
.....(1)
l
l
l
h
360
2 r
 Length of the arc L
360º
2 l
.....(2)
In the figure,
Length of the arc of the sector = Circumference of the circular base of the cone
 L = 2r
....(3)
From (2) and (3), we get 2r = 2 l
r
 =
l
360
360
Let 'A' be the area of the sector. Then,
....(4)
360
l2
=
A
r
2
= l
= rl
360
l
 The curved surface area of the cone = rl sq. units
2
 A= l
1
2 rl and 2r is the circumference of base of the
2
cone, l is the slant height. We have,
Since rl can be written as
Curved surface
1
area of the cone = 2
circumference of the
× Slant height
base of the cone
 Curved surface area (CSA) of the cone is half the product of the circumference
of its base (2r) and slant height (l)
(ii) Total surface area of the right circular cone
Total surface area
of the cone
=
Curved surface area
of the cone
=rl
+
Area of
the base
+ r2
rl
r2
=r (l + r)
 The total surface area of the cone = r (l + r) sq. units.
Volume of cone
Let us consider the experiment conducted to find the volume of a cone. Study the
experiment and repeat it in small groups.
388
UNIT-16
Take a hollow cylinder with one base closed and a hollow cone of same radius and
same height. It is useful, if they are made of transparent material. Let their thickness be
as less as possible and almost negligible.
Observe the given figure to easily check their
radius and height.
Cone is filled with sand or water and poured into
the cylinder. The cylinder is completely filled by
pouring sand or water three times by the cone.
Volume of a cylinder = r2h, where 'r' is the radius of its circular base and 'h' is the
height.
From the above experiment, we can conclude that,
1
Volume of cone = × Volume of cylinder
3
1
 Volume of cone = r2h cubic units
3
So far we have learnt about the surface area and volume of cylinder and cone.
It is interesting to note the change in the surface area and volume of cylinder and
cone when their radius and height are either doubled or halved. Study the following
table.
Sl.
Solids
No.
1.
2.
3.
4.
5.
6.
when
r=h
r
2r
h
r
h
h
r
2h
r
r
2
h
r
h
r
h
h
2
r
r
2
h
h
2
Cylinder
Cone
LSA
TSA
V
LSA
TSA
V
2r2
4r2
r3
–
–
1 3
r
3
4rh
4r(2r + h)
4r2h
-
-
4rh
2r(r + 2h)
2r2h
-
-
r
2
h
r 2h
4
-
-
r 2h
12
2 r r
h
2
r 2h
2
-
-
r 2h
6
r
2
h
2
r 2h
8
-
-
r 2h
24
rh
rh
rh
2
r
r
4 2
r h
3
2 2
r h
3
Mensuration
389
ILLUSTRATIVE EXAMPLES
Example 1 : The diameter of a cone is 14 cm and its slant height is 10 cm. Find its
curved surface area.
Sol. d = 14 cm  r =
14
= 7 cm, l = 10 cm
2
22
7 cm 10 cm = 220 cm2
7
Example 2 : Find the TSA of a cone, whose slant height is 9m and radius of the base is
14 m.
Curved surface area of a cone = rl =
Sol. l = 9 m, r = 14 m
22
7
 TSA of the cone= 1012 m2
TSA of a cone= r (r + l ) =
2
14 m (14
9) m = 44
23 m2 = 1012m2
Example 3 : How many metres of cloth 5m wide will be required to make a conical
tent, the radius of the base is 7m and whose height is 24 m?
Sol. r = 7m, h = 24m, l = ?
l2= r2 + h2 = 72 + 242 = 49 + 576 = 625  l =
Area of canvas used = CSA of conical tent
22
7
Area of canvas = 550m2
 CSA of a cone = rl =
length × width = 550m2
7m
625 = 25 m
25m = 550 m2
 length =
550m 2
550m 2
=
= 110 m
width
5
 Length of canvas used = 110 m
Example 4 : The CSA of a cone is 4070 cm2 and its diameter is 70 cm. What is its slant
height.
Sol. d = 70 cm  r= 35 cm CSA of a cone = rl.
37
5
407 0
22
35 cm l

=l
11 0
7
 Slant height of the cone = 37 cm
4070 cm2=
 l = 37 cm
Example 5 : The slant height and diameter of the base of a conical tomb are 25m and
14 m respectively. Find the cost of white washing its CSA at the rate of ` 210 per
100 m2.
14
Sol. l = 25m, d= 14 m  r =
= 7m
2
22
7 m 25m  CSA of conical tomb = 550 m2
7
Cost of white washing 100 m2 = ` 210
` 210
55 0 m2 = `1155
 Cost of white washing 550 m2 =
10 0 m2
CSA of a cone = rl =
390
UNIT-16
Example 6 : The diameters of 2 cones are equal. If their slant heights are in the ratio
5 : 4, find the ratios of their curved surfaces.
Sol. Let S1 be the CSA of 1st cone and S2 be the CSA of 2nd cone
Let their slant heights be 5l and 4l, respectively. CSA of a cone = rl
S1
 S =
2
r
r
5l
4l
 S1 : S 2 = 5 : 4
Example 7 : Find the volume of a right circular cone 90 cm high if the radius of the
base is 21 cm.
1
3
1 22
1 2
21
21 90cm3
Sol. r = 21 cm, h = 90 cm Volume of a cone= r h =
3
7
3
 Volume of the cone= 41,580 cm3
Example 8 : The volume of a cone is 18480 cm3. If the height of the cone is 40 cm, find
the radius of its base.
1 2
1 22
r h  18480=
r 2 40
Sol. Volume of a cone =
3
3 7
21
795
1848 0 3 7
= 441 cm2  r = 441cm2 = 21 cm
22 4 0
Example 9 : A cone of height 24 cm has a CSA of 550 cm? Find its volume.
2
r=
Sol. h = 24 cm, CSA = 550 cm2
We know that, l2 = r2 + h2 = r2 + 242 = r2 + 576
l=
r2
576
2
CSA of the cone = 550 cm rl = 550
22
r
r2 576 = 550  r r 2
7
Square on both sides
25
550
576 =
22
7
 r r 2
576 = 175
2
r r 2 576 = (175)2  r2 (r2 + 576) = 30625
r4 + 576r2– 30625 = 0 Factorising this equation we get,
(r2 + 625) (r2– 49) = 0
If r2 – 49 = 0, r2 = 49, r =
1
1 2
r h =
3
3
Volume of the cone = 1232 cm3
Volume of of a cone =

49  r = 7 cm
22
7
7
7
8
24 cm3
Example 10 : Find the weight of a solid cone whose base is of diameter 14 cm and
vertical height 51cm, if the material of which it is made weighs 10 gm/cm3.
14
Sol. d = 14 cm = r =
= 7 cm, h = 51 cm
2
17
1 22
1 2
7 7 51 cm3 = 2618 cm3
r h =
Volume of a cone=
3
7
3
 Volume of the cone = 2618 cm3
10
Weight= volume × density = 2618 cm3 ×
kg/cm3 = 26.18 kg
1000
Mensuration
391
Example 11 : A conical flask is full of water. The flask has base radius 3 cm and height
15 cm. The water is poured into cylindrical glass tube of uniform inner radius 1.5 cm,
placed vertically and closed at lower end. Find the height of water in the glass tube.
Sol. Volume of water in cone = Volume of water in cylinder 
1
3
22
7
3
3 15cm3 =
22
1.5 1.5
7
1 2
rc hc =
3
rcy2 hcy
hcy
1
3
2
2
5
22
3
3
15
7

hcy =
= 20 cm
22
1.5 1.5
7
 height of water in the glass tube = 20 cm
Example 12 : A solid metallic right circular cylinder 1.8m high with diameter of its
base 2 m is melted and recast into a right circular cone with base of diameter 3 m.
Find the height of the cone.
Sol. Cylinder: h= 1.8 cm, d = 2m  r =
2
3
= 1m Cone: d= 3m, r =
= 1.5m, h = ?
2
2
Volume of cylinder = Volume of cone  rcy2 hcy =
22
1
1 1 1.8m3 =
7
3
1 2
rc hc
3
22
1.5 1.5 hc m3
7
22
6
1 1 1.8
1.8
7

hc =
=
1 22
1.5 6
1.5 1.5
3
7
 height of the cone = 2.4m
3
2
1.5
=
12
= 2.4
5
EXERCISE 16.2
1. Find the curved surface area of a cone, if its slant height is 60 cm and the radius of
its base is 21 cm.
2. The radius of a cone is 7cm and area of curved surface is 176 cm2. Find its slant
height.
3. The area of the curved surface of a cone is 60cm2. If the slant height of the cone is
8 cm, find the radius of the base.
4. Curved surface area of a cone is 308cm2 and its slant height is 14cm. Find the
radius of the base and total surface area of the cone.
5. A clown's cap is in the form of a right circular cone of base radius 7cm and height
24 cm. Find the area of the sheet required to make 10 such caps.
6. Find the ratio of the curved surface areas of two cones if their diameters of the
bases are equal and slant heights are in the ratio 4 : 3.
392
UNIT-16
7. A cylinder and a cone have equal radii of their bases and equal heights. If their
curved surface areas are in the ratio 8 : 5, show that the ratio of the radius of each
to the height of each is 3 : 4.
8. Find the volume of a right circular cone with (i) radius 5 cm, height 7 cm (ii) radius
10.5 cm, height 20 cm (iii) height 21 cm, slant height 28 cm.
9. Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the
ratio 3 : 1. Find the ratio of their volumes.
10. A right angled triangle of which the sides containing the right angle are 6.3 cm and
10 cm in length, is made to turn around on the longer side. Find the volume of the
solid thus generated.
11. A right circular cone of height 81 cm and radius of base 16 cm is melted and recast
into a right circular cylinder of height 48 cm. Find the radius of the base of the
cylinder.
12 A right circular cone is of height 3.6cm and radius of its base is 1.6 cm. It is melted
and recast into a right circular cone with radius of its base 1.2cm. Find the height
of the cone so formed.
13. A conical flask is full of water. The flask has base radius 'a' and height '2a'. The
water is poured into a cylindrical flask of base radius
2a
. Find the height of water
3
in the cylindrical flask.
14. A tent is of the shape of a right circular cylinder upto a height of 3 m and then
becomes a right circular cone with a maximum height of 13.5m above the ground.
Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per sq m,
if the radius of the base is 14 m.
Frustum of a cone
Observe the shape of the objects that we
use in our daily life.
Tumbler
Bucket
These are neither cylinders nor cones.
What are these shapes called? How are they obtained?
Consider the following activity to answer these questions.
Cap
Activity: Take some clay or plasticine and form a right circular cone. Cut it with a knife
parallel to its base. You will get two solids as shown in the figure given below.
If we remove the smaller cone obtained
Know this!
at the top, we are left with a solid object at the
bottom. This part of the cone containing its 'Frustum' is a Latin word meaning "piece
cut off" and its plural form is 'Frusta'
base is called the frustum of the cone.
Mensuration
393
The frustum of a cone has two circular bases, one at the bottom and the other at the
top of it with different radii.
From the above activity, we can conclude that,
If a plane cuts a right circular cone parallel to its base and the upper smaller
cone is removed, then the remaining part of the cone containing the base is called
"Frustum of the cone".
Discuss :
Now, can you name the shape of tumbler,
Compare a frustum with cone and
bucket and cap?
cylinder. What similarities and
Why do you call it so? Discuss in groups.
differences do they have?
Is there any relationship between the dimensions
of the original cone and the smaller one cut off ? Observe the figures given below.
A
A
A
D
L
E
D H
l
F
h
B
E
O
C
B
R
O
r
C
D
F
Let the base radius, height and the slant height of the original cone be denoted by
R, H and L; and those of the smaller cone be r, h and l.
Observe that the two triangles ABO and AED are similar.
Hence, their corresponding sides are proportional.
i.e.
ED
AD AE
=
=
BO
AO AB

r
h
=
R
H
l
L
We can conclude that,
If from the top of a cone, a smaller cone is cut off by slicing parallel to the base,
then the base radius, height and the slant height of the two cones (smaller cone
obtained and the original cone) are proportional.
How is the slant height of the frustum related to its height and radii?
Let 'l ' be the slant height of the frustum, 'h' be the height, 'R' be the radius of the
bottom base, and r be the radius of the top base.
The relatonship is given by l2 = h2 + (R – r)2 or l =
h2
(R
C
r)2
l
Surface area and volume of frustum of a cone.
r
A
We know that frustum of a cone is a part of the cone ABC
obtained by cutting off the smaller cone A'B'C parallel to the base.
B
h
A
R
B
394
UNIT-16
Hence,
Lateral surface area
Lateral surface
Lateral surface
=
–
of frustum
area of cone ABC area of cone A'B'C
Total surface area of frustum = LSA of frustum + Area of the two circular bases
Volume of frustum = Volume of original Cone ABC – Volume of smaller Cone A'B'C
Surface area and volume of a frustum of a cone can be found by using the above
method only when the radius of the base and slant height of both the original (bigger)
cone and the smaller cone are given.
ar
2
Then, how to find the surface area and volume of a frustum when only
the heights and radii of the two bases of the frustum are given?
l
h
Let h be the height, l be the slant height and r1, and r2 be the radii of A
the two bases (r1> r2) of the frustum of a cone.
r
P
B
1
Then, we can directly find the surface area and volume of the frustum by using the
formulae given below:
* Lateral surface area of the frustum of the cone
= (r1+r2)l , where, l
=
h2 +(r1 r2 )2
* Total surface area of the frustum of the cone = (r1+r2)l +r12+r22
h2 +(r1 r2 )2
= {(r1+r2)l +r12+r22}, where, l =
1
h (r12+r22+r1r2)
3
[ Note : These formulae can be derived using the area of similarity of triangles but we
shall not be doing derivations here. We will use the formulae to solve problems.]
* Volume of the frustum of the cone =
ILLUSTRATIVE EXAMPLES
5cm
Example 1 : From the top a conical shaped jaggery of base diameter 10 cm and height
10 cm, a small cone is cut off by slicing parallel to the base, 4 cm from the vertex.
Find the surface area and volume of the frustum so obtained.
O
Height of the cone ABO = h1 = 10cm
I
r
A
I
10cm
Sol. Diameter of the cone ABO = 10cm
4cm
10
 radius r1 =
= 5cm
2
I
B
I
Height of the cone A B O = h2 = 4cm
First, let us find the other required data.
I
A
I
Radius of cone A B O = r2
I
I
Slant height of cones ABO and A B O , l1 and l2 respectively.
P
B
Mensuration
r1
r2
395
h1
5
h2  r2
10
4
5 4
= 2cm
10
r2 =
l12 = h12+ r12 = 102+ 52 = 100 + 25 = 125
 l1 = 125
l22 = h22+ r22 = 42 + 22 = 16 + 4 = 20
 l2 =
(i)
Lateral surface area of cone ABO =  r1l1 = 
5 5 cm
20
2 5 cm

5
5
5
25 5 

2
2
5
4 5
LSA of cone ABO = 25 5 sq.cm
I
I
Lateral surface area of cone A B O =  r2l2 = 
I
I
LSA of cone A B O = 4 5 sq.cm
I
I
Lateral surface area of frustum = LSA of cone ABO – LSA of cone A B O
= 25 5
3
4 5  21 5  21
5
22
= 66 5 sq. cm
71
LSA of frustum = 66 5 sq. cm. or 147.84 sq.cm
(ii) Total surface area of cone ABO =  r1 (r1+l1)
      5 5      5 
TSA of cone ABO = 25 5 sq.cm
I
I
Lateral surface area of cone A B O =  r2l2 = 
I

2
2
5
4 5
I
LSA of cone A B O = 4 5 sq.cm
I
I
Total surface area of frustum = TSA of cone ABO - LSA of cone A B O + r22

 5 - 4 5 + 4

 25
25 5
 29
21 5

22
7
4 5
4
75.95
TSA of frustum = 238.70 sq.cm
1
1
 r12 h1 =
3
3
(iii) Volume of cone ABO =
Volume of cone ABO =
I
I
  
5
 5  10
250
 cu.cm
3
Volume of cone A B O =
1
1
 r22 h2=
3
3
  
2
2 4
396
UNIT-16
16
 cu.cm
3
I I
Volume of the frustum = Volume of cone ABO Volume of cone A B O
I
I
Volume of cone A B O =
250
16
234
 –
 =

3
3
3
Volume of the frustum = 78  cu.cm or 245.14 cu.cm
=
We can also directly find the lateral surface area, total surface area and volume of
the frustum by using the formulae. Use the formulae and verify.
Example 2 : The radii of two circular ends of a frustum shaped dust bin are 15cm and
8cm. If its depth is 63cm, find volume of the dust bin.
15 cm
Sol. Given r1=15cm, r2 = 8cm, h = 63 cm
=
1
3
1
h (r12+r22+r1r2)
3
22
7
1
= 3
1
22
71
63 cm
Volume of the dustbin (frustum) =
63 (152+82+15×8)
8 cm
63
9
3
(225+64+120)
= 66 × 409 = 26,994
 Volume of the dustbin = 26,994 cu.cm
Example 3 : From the top of a cone of base radius 12cm and height 20cm, a small cone
of base radius 3cm is to be cut off. How far down the vertex is the cut to be made? Find
the volume of the frustum so obtained.
Sol. Given, r1 = 12cm, h1= 20cm, r2 = 3cm, h2 =?
r1
we know, r
2
h1
h2
12
3
20
h2
h2 =
3
20
12
5 cm
The cut to be made at 5cm from the vertex.
Volume of the frustum =
=
1
h (r12+r22+r1r2)
3
1
3
22
110
15 (122+32+12×3) =
×(144+9+36)
7
7
110
×189 = 2,970
7
Volume of the frustum = 2,970 cu.cm
=
Example 4 :The slant height of the frustum of a cone is 4cm, and the perimeter of its
circular bases are 18cm and 6cm respectively. Find the curved surface area and total
surface area of the frustum.
Sol. Given, 2r1 = 18cm,
2r2 = 6cm,
l = 4cm
r1 =
9
cm
r2 =
3
cm
Mensuration
397
Curved surface area of the frustum = (r1+r2)l
=
9
3
12
4=
4
CSA of the frustum = 48 sq.cm
Total surface area of the frustum =(r1+r2)l +r12+r22
9
= 48
= 48 +
81 7
22
9
3
3
48 +
81
+
9
9 7
= 48 + 25.8 + 2.9
22
TSA of the frustum = 76.7 sq.cm
EXERCISE 16.3
1. A flower vase is in the form of a frustum of a cone. The perimeter of the ends are 44
cm and 8.4cm. If the depth is 14 cm, find how much water it can hold?
2. A bucket is in the shape of a frustum with the top and bottom circles of radii
15cm and 10cm. Its depth is 12 cm. Find its curved surface area and total surface
area. (Express the answer in terms of 
3. From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height
17cm is cut off. What is the volume of the remaining frustum of the cone?
4. A vessel is in the form of a frustum of a cone . Its radius at one end is 8 cm and the
height is 14 cm. If its volume is
5676
cm3, find the radius of the other end.
3
5. A container, opened from the top and made up of a metal sheet, is in the form of a
frustum of a cone of height 16cm with radii of its lower and upper ends as 8 cm and
20 cm, respectively. Find the cost of the milk which can completely fill the container,
at the rate of ` 40 per litre. Also find the cost of metal sheet used to make the
container, if it costs ` 8 per 100 sq cm (take  = 3.14)
Sphere
Let us conduct an activity as done in the previous section. Take a
circular disc and paste a string along one of its diameter. Rotate it
and observe the new solid formed.
A
What does it look like? It is called a sphere.
If a circular disc is roated about one of its diameters, the
solid thus generated is called sphere.
B
The centre of the circle, when it forms a sphere on rotation becomes the centre of
the sphere. This centre point will be equidistant from all points in the space which have
formed the sphere.
398
UNIT-16
Therefore, a sphere is a three dimesnional figure or solid figure, which is made up
of all points in the space, which lie at a constant distance called the radius and from a
fixed point called the centre of the sphere.
The word solid sphere is used for the solid whose surface is a sphere.
It has both surface area and volume.
There are many things around us which are
sphere - shaped. Observe some examples.
How many surfaces do you see in the sphere - shaped
objects?
Ball
Marble
There is only one which is curved.
Now, let us assume that a solid sphere like a ball be sliced exactly
through the middle with a plane that passes through its centre.
The sphere gets divided into two equal parts. Each part of the sphere
is called a hemisphere.
Surface area of sphere and hemisphere
The formula used to find the surface area of a sphere can be obtained through an
interesting activity. You are familiar with the game played using a top, by winding a thick
thread around it. The following activity where a thread is wound around the spherical
object will help you to generate the formula for finding the surface area of a sphere. Study
the activity and try to conduct it in groups.
• Take a plastic ball.
• Fix a pin at the top of the ball.
• Wind a uniform thread over the ball so as to cover
the whole curved surface area. Use pins to fix the
thread in place.
• Mark the starting and finishing points on the thread.
• Slowly unwind the thread and measure the length of the thread used to wind around
the surface area of the sphere.
• Cut the thread into four equal parts.
• Measure the diameter of the ball and find its radius.
• Take a sheet of paper and draw four circles with radius equal to the radius of the
ball.
• Fill each circle with one piece of the thread as shown in the figure.
What do you observe and conclude from this activity?
The thread which had covered the surface area of the sphere (ball) has completely
filled the four regions of the circle, all of the same radius as of the sphere.
This suggests that surface area of the sphere is four times the surface area of
circles, where both the sphere and the circles have same radius (r).
Mensuration
399
 Curved surface area of the sphere = 4 × Area of the circle = 4 × r2 = 4r2
 The curved surface area of the sphere = 4r2 sq. units
 The surface area of the sphere = 4 r 2 sq.units
It is interesting to note that the surface area of a sphere is equal to the curved
surface area of a cylinder just containing it.
Observe the figure
A sphere with radius (r) is placed inside a cylinder whose radius is
also 'r' and height is '2r'.
r
Surface area of sphere = 4r2
Curved surface area of cylinder = 2rh = 2r × 2r = 4r2
 Surface area of the sphere = Curved surface area of the cylinder,
where, radius of the sphere and the cylinder are equal and height of the cylinder is
equal to the diameter of the sphere.
Now, let us find the surface area of hemisphere.
We know that hemisphere is exactly half of a sphere. Let us use
Solid Hemisphere
this idea and find surface area of hemisphere.
(i) Curved surface area of solid hemisphere
CSA of solid sphere 4 r2
C.S.A of solid hemisphere =
=
= 2r2
2
2
 CSA of solid hemisphere = 2r2 sq. units
(ii) Total surface area of solid hemisphere
TSA of solid hemisphere = CSA of hemisphere + surface area of base circle
= 2r2 + r2 = 3r2
 T.S.A of solid hemisphere = 3r2 sq. units
You might have observed the spherical object used to decorate
christmas trees, during birthdays etc.
This sphere is made up of several cones.
Imagine a sphere which is formed by using several tiny cones
whose radius is 'R' and height is 'h'
It can be considered that a sphere is made up miniature cones
whose height 'h' is equal to the radius 'r' of the sphere and each having a circular base.
Volume of each cone =
1
× Area of base × height
3
1
× B1 × r
Volume of cone 2 =
3
1
Volume of cone 3 =
× B3 × r
Volume of cone n =
3
 Volume of sphere = Sum of the volumes of all cones
Volume of cone 1 =
1
× B2 × r
3
1
× Bn × r
3
400
UNIT-16
1
1
1
1
1
B r + B2r + B3r+........+ Bnr = r(B1 + B2 + B3 + .......Bn)
3 1
3
3
3
3
=
We observe that the sum of base areas of all the cones which have formed the
sphere is equal to the surface area of the sphere. Since, the surface area of the sphere is
4r2 , we get
4 3
1
r
r(4r2) =
3
3
4 3
r cubic units
 Volume of sphere =
3
Volume of sphere =
There is an interesting relationship between volume of a sphere and a cone whose
radii (r) are same.
Study the relationship.
If a cone is inscribed within the sphere as shown in the figure,
Then volume of the sphere is four times the volume of the cone.
 Volume of sphere = 4 × volume of cone
=4×
1 2
1
1
4
r h = 4 × r2 × r = 4 × r2 = r3
3
3
3
3
4 3
r
3
Volume of a solid hemisphere
 Volume of sphere =
Volume of solid hemisphere =
1
1
× volume of solid sphere =
2
2
4 3
r
3
ILLUSTRATIVE EXAMPLES
Example 1 : Find the surface area of a sphere of radius 21 cm.
Sol. Surface area of a sphere = 4r2 = 4
22
7
21
3
21cm2
Surface area of the sphere = 5544cm2
Example 2 : Find the CSA and TSA of a solid hemisphere of radius 14 cm.
22
7
CSA of the hemisphere = 1232 cm2
Sol. CSA of a hemisphere = 2r2 = 2
22
7
TSA of the hemisphere = 1848 cm2
TSA of a hemisphere = 3r2 = 3
14
14
2
14cm2
2
14cm2
Example 3 : Find the volume of a sphere of radius 3 cm.
Sol. Volume of a sphere =
4
4 3
r =
3
3
22
7
3
3
3cm3 = 113.14 cm3
2 3
r
3
Mensuration
401
Example 4 : A hemispherical bowl has inner diameter 9 cm. Find the volume of milk
it can hold.
9
Sol. d = 9 cm r =
cm
2
3
2 22 9
2 3
r =
Volume of a hemisphere =
7
3
2
3
Volume of hemispherical bowl = 190.92 cm3
9
2
9
5346
cm3 =
2
28
190.92 cm3
 Volume of milk it can hold = 190.92 ml [1cm3 = 1ml]
Example 5 : A hemispherical bowl of internal radius 18 cm is full of fruit juice. The
juice is to be filled into cylinderical shaped bottles each of radius 3 cm and height 9
cm. How many bottles are required to empty the bowl?
Sol. Hemisphere: r = 18 cm, V = ? Volume of a hemisphere =
2 3
r
3
2 22
18 18 18cm3
3 7
Cylinder: r = 3, h = 9 Volume of a cylinder = r2h
 Volume of hemispherical bowl =
Volume of 1 bottle =
No. of bottles =
22
7
3
3 9cm3
Volume of fruit juice in hemispherical bowl
Volume of fruit juice in 1cylindrical bottle
=
2
3
2
6
6
22
18
18
7
22
3 3 9
7
18
2
= 48
 Number of bottles required = 48
Example 7 : The diameter of a metallic sphere is 4.2 cm. It is melted and recast into
a right circular cone of height 8.4 cm. Find the radius of the base of the cone.
4.2
Sol. Sphere: d = 4.2 cm r =
= 2.1 cm Cone: h = 8.4 cm
2
Volume of sphere = Volume of cone
4
rS3
3
4 2.1 2.1 2.1
4 3 1 2
rs =
rc hc  rC2 = 1
=
= 2.1 × 2.1
8.4 2.1
hC
3
3
3
rC= 2.1 2.1 = 2.1 cm
 radius of base of the cone = 2.1 cm
Example 8 : The internal and external diameters of a hollow hemispherical shell are
6cm and 10 cm respectively. It is melted and recast into a solid cone of base diameter
14 cm. Find the height of the cone so formed.
10
6
Sol. Hollow hemisphere: External radius, R =
= 5cm, Internal radius , r =
= 3 cm
2
2
402
UNIT-16
Cone: r =
14
= 7 cm, h = ?
2
Volume of hollow hemisphere = Volume of cone
2
1 2
(R 3 r3 ) =
r h
3
3
2
(R 3 r 3 )
2(R 3 r3 ) 2(53 33 ) 2(125 27)
3
h=
=
= 2
2
1
r2
2
7
49
r
3
 height of the cone of formed = 4cm
2
98
= 4 cm
49
Example 9 : The radii of two solid metallic spheres are r1 and r2. The spheres are
melted together and recast into a solid cone of height (r1 + r2). Show that the radius of
the cone is 2 r12
r22
r1r2 .
Sol. Volume of (sphere 1 + sphere 2) = Volume of cone
4 3 4
r1
3
3
4
r13
3
4
r13
3
rC= 1
r1
3
1
2
rC
(r1
3
1 2
rC (r1 r2 )
=
3
r23 =
r23
r23
r2
4 r1
=
rC=
 radius of the base of cone = 2 r12
r12
r2
r1
rC2 = 4 r12
4 r12
r1r2
r2 )
r1r2
r1r2
r1r2
r22
r2
r22
r22 = 2 r12
r1r2
r22
r22
Example 10 : A solid sphere of radius 1 cm is melted to stretch into a wire of length
100 cm. Find the radius of the wire.
Sol. Sphere  r = 1 cm
Wire is a cylinder
Cylinder  h = 100 cm, r = ?
Volume of sphere = Volume of wire (cylinder)
4 3
rS =
3
2
Cy
r =
rCy=
rcy2 hcy
4
3
rs3
hcy
1
75
4rs3
4 1 1 1cm3
1
cm2
= 3 h =
=
3 100 25cm
75
cy
1
5 3
 radius of the wire = 0.11 cm
cm = 0.11 cm
Mensuration
403
EXERCISE 16.4
1. Find the surface area of a sphere of radius (i) 14 cm
(ii)2.8 cm
(iii) 6.3cm
2. Find the TSA of a hemisphere of radius 5 cm.
3. A hemispherical bowl made of wood has inner diameter of 10.5 cm. Find the cost of
painting it on the inside at the rate of `12 per 100 sq. cm.
4. Calculate the surface area of the largest sphere that can be cut out of a cube of side
15 cm.
5. The surface area of a solid hemisphere is 432 cm2. Calculate its radius.
6. A hemispherical bowl made of steel is 0.25 cm thick. The inner radius of the bowl is
5 cm. Find the outer curved surface area of the bowl.
7. Find the volume of a sphere whose radius is (i) 7 cm
(ii) 10.5 cm
8. The diameter of a metal ball is 3.5cm. What is the mass of the ball, if the density of
the metal is 8.9g/cm3 [Hint: Mass = Volume × density)
9. Find the volume of a sphere whose surface area is 154 cm2.
10. The inner radius of a hollow sphere is 10 cm. Find its volume.
11. The volume of a solid hemisphere is 1152cm3. Find its curved surface area.
12. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much
medicine is needed to fill this capsule?
13. A solid hemisphere of wax of radius 12 cm is melted and made into a cone of base
radius 6 cm. Calculate the height of the cone.
14. A right circular metallic cone of height 20 cm and base radius 5 cm is melted and
recast into a sphere. Find the radius of the sphere.
15. The diameter of a metallic sphere is 18 cm. It is melted and drawn into a wire
having diameter of cross section 0.4 cm. Find the length of the wire.
16. The internal and external radii of metallic sherical shell are 4 cm and 8 cm
respectively. It is melted and recast into a solid right circular cylinder of height
9
1
cm. Find the diameter of the base of the cylinder.
3
17. A solid metallic sphere of diameter 28 cm is melted and recast into a number of
smaller cones each of diameter 4
2
cm and height 3cm. Find the number of cones
3
so formed.
Combination of solids
You are familiar with solid objects like cube, cuboid, prism, pyramid, cylinder, cone and
sphere. You know how to find their surface area and volume. It is common to identify
objects from our surroundings with these solid shapes.
404
UNIT-16
Now, observe the following objects and their shapes. Discuss in groups and identify
the shape of each one of them.
Toy
Oil tanker
Rocket model
Green-house shed
In each case, the object is made of two or more solid objects. Hence, each one of
them is a combination of solids.
Toy
Oil tank
Cone + Hemisphere
Rocket
Hemisphere + Cylinder
+ Hemisphere
Green-house
Half of cylinder +
Cuboid
Cone + Cylinder
How to find the surface area and volume of such solid objects? Let us learn to find
the surface area and volume of combination of solids.
Surface area of combination of solids
Consider the example of the toy. We observe that the total surface area
of the toy is sum of the curved surface area of the hemisphere and curved
surface area of the cone.
 TSA of the toy = CSA of the hemisphere + CSA of the cone
Toy
CSA of the toy = CSA of the hemisphere + CSA of the cone
Similarly, for the other examples we can find the total surface area and curved
surface area as follows.
TSA of oil tank = CSA of hemisphere + CSA of cylinder + CSA of hemisphere
CSA of oil tank = CSA of hemisphere + CSA of cylinder + CSA of hemisphere
TSA of rocket model = CSA of cylinder + Area of its base + CSA of cone
CSA of rocket model = CSA of cylinder + CSA of cone
Mensuration
405
TSA of shed = LSA of cuboid + Area of its base + Half of TSA of cylinder
LSA of shed = LSA of cuboid + Half of TSA of cylinder
From the above examples, we can conclude that, total surface
area of the combination of solids may or may not be the sum of
curved surface areas of the solids which are combined together.
We can note the following observations.
• The total surface area of a combiantion of solids is not simply the sum of total
surface area of solids combined together, because some part of the total surface
area disappears in the process of joining them.
• It is also not simply the sum of curved surface area of solids joined together because
some part other than the curved surface area has to be added to its total surface
area.
• The curved surface area of a combination of solids may be equal to the sum of
curved surface areas of the solids combined together.
• The curved surface area of a combination of solids may not be equal to the sum of
the curved surface areas of solids combined togehter, because some part other than
the curved surface area has to be added.
Now let us solve some problems based on the surface area of combination of solids.
ILLUSTRATIVE EXAMPLES
r
15 cm
Example 1 : A cylindrical container of radius 6 cm and height 15 cm is filled with ice
cream. The whole ice cream has to be distributed to 10 children in equal cones with
hemispherical tops. If the height of the conical portion is 4 times the radius of its
base, find the radius of the ice cream cone.
6 cm
r
4r
Sol. Volume of cone with hemispherical top
2
6
1 2
2 3
1 2
2 3
4 3 2 3
r h
r =
r
4r
r =
r
r =
=
3
3
3
3
3
3
3
 Volume of 10 cones with hemispherical tops = 10 × 2r3 = 20r3
r 3 = 2r3
Volume of ice-cream in cylindrical container = r2h   × 6cm × 6cm ×15cm = 540 cm3
Volume of 10 cones with hemispherical tops=Volume of cylinder containing ice-cream
3
20r = 540 
 r = 3 cm
540
r =
20
3
27
= 27  r =
3
27
406
UNIT-16
Example 2 : A solid is in the form of a cone mounted on a right circular cylinder, both
having same radii of their bases. Base of the cone is placed on the top base of the
cylinder. If the radius of the base and height of the cone be 7 cm and 10 cm respectivly,
and the total height of the solid be 30 cm, find the volume of the solid.
Sol. Cylindrical part: r = 7 cm, h = 30 – 10 = 20 cm
A
Conical part: r = 7 cm, h = 10 cm
Volume of the solid = 3593.33 cm
22
7
7
10 cm10 cm
C
30 cm
1 2
r h
= r2h +
3
22
1
7 7 20cm2 +
=
3
7
1540
cm3
= 3080 cm3 +
3
= 3080 + 513.33 cm3
10 cm
B
20 cm
Volume of the solid = Volume of cylinder + Volume of cone
7 10
D 7 cm 7 cm E
3
Example 3 : A toy is in the form of a cone mounted on a hemisphere with the same
radius. The diameter of the conical portion is 6 cm and its height is 4 cm. Determine
the surface area and volume of the solid.
C
Spherical portion
r = 3cm, h = 4 cm
4 cm
Sol. Conical portion
r = 3cm
l2 = r2 + h2 = 32 + 42 = 9 + 16
l2 = 25
l=
A
3 cm
3 cm
25 = 5 cm
B
Surface area of the toy = CSA of hemisphere + CSA of cone
22
22
3 3cm2
3 5cm2
= 2r2 + rl = 2
7
7
396
330
726
cm2
cm2 =
cm2 103.71cm2
=
7
7
7
Surface area of the toy = 103.71 cm2
Volume of the toy = Volume of hemisphere + Volume of cone
2 22
1
2 3 1 2
3 3 3cm3
r
r h =
=
7
3
3
3
3
22
7
3
3
4cm3
396
264
660
cm3
cm3
cm3
7
7
7
 Volume of the toy = 94.28 cm3
Example 4 : A circus tent is made of canvas and is in the form of a right circular cylinder
and a right circular cone above it. The diameter and height of the cylindrical part of the
tent are 126 m and 5 m respectively. The total height of the tent is 21m. Find the total
cost of the canvas used to make the tent when the cost per m2 of the canvas is ` 15.
=
Sol. Cylindrical part: r =
126
= 63m, h = 21 – 5 = 16 m
2
Mensuration
407
632
h2 =
3969
256 =
162
21 m
=
r2
16m
l=
4225
30 m
Total canvas used = CSA of cylindrical part + CSA of
conical part
9
22
22
63
5m2 +
7
7
= 1980 m2 + 12870 m2 = 14850 m2
= 2rh + rl = 2
63
9
O
30 m
5m
 l = 65 m
65m2
126 m
 Total canvas used = 14,850 m2
Cost of canvas at the rate of ` 16 per m2 = 14,850 × `15 = `2,22,750
= `2,22,750
Cost of canvas
Example 5 : A solid is composed of a cylinder with hemispherical ends. If the whole
length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find
the cost of polishing its surface at the rate of ` 4 per 100 cm2
Sol. Hemispherical part: r = 7 cm
7 cm
Cylindrical part: r = 7cm, h = 104 – (7 + 7) = 90 cm
7 cm
+ 2 × CSA of hemispherical part
= 2rh + 2 × 2r2 = 2r (h + 2r)
2(7) cm2
7 cm
= 44 × 104 cm2 = 4576 cm2
7 cm
=2
 TSA of the solid
22
7
90 cm
= CSA of cylindrical part
104 cm
Total surface area of the solid
7 90
2
= 4.576 cm
Cost of polishing at the rate of ` 4 per 100 cm2
`4
= `183.04
100
 Cost of polishing the solid = `183.04
= 4576 ×
EXERCISE 16.5
1. A petrol tank is in the shape of a cylinder with hemispheres of same radius attached
to both ends. If the total length of the tank is 6m and the radius is 1m, what is the
capacity of the tank in litres.
2. A rocket is in the shape of a cylinder with a cone attached to one end and a
hemisphere attached to the other. All of them are of the same radius of 1.5m. The
total length of the rocket is 7m and height of the cone is 2m. Calculate the volume
of the rocket.
408
UNIT-16
3. A cup is in the form of a hemisphere surmounted by a cylinder. The height of the
cylindrical portion is 8 cm and the total height of the cup is 11.5cm. Find the TSA of
the cup.
4. A storage tank consists of a circular cylinder with a hemisphere adjoined on either
ends. The external diameter of the cylinder is 1.4m and length is 8m, find the cost
of painting it on the outside at the rate of ` 10 per m2.
5. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of
the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the
toy at 7 per `100 cm2.
6. A circus tent is cylindrical upto a height of 3m and concial above it. If the diameter
of the base is 105 m and the slant height of the conical part is 53m, find the total
cost of canvas used to make the tent if the cost of the canvas per sqm is `10.
Scale drawing :
We are familiar with the formulae used to find area of plane geometrical figures. Observe
the figures given below and recall their formulae to find areas.
A
R
S
h
b
.
B
D
K
C
b
P
Triangle
1
A=
× base × height
2
A = 1/2 bh
N
A = length × breadth
M
b
Parallelogram
A = base × height
A = lb
A
E
A = bh
b
G
B
h2
h
Q
l
Rectangle
L
F
h1
h
d
F
C
D
Quadrilateral
a
Trapezium
D
E
A = 1/2 × diagonal × sum of heights
A = 1/2 × height × sum of parallel sides
A = 1/2 × d (h1+ h2)
A = 1/2 h (a + b)
Now, observe the following figures
B
E
C
A
E
D
A
D
B
C
C G
F
A
B
F
How are the areas of these figures found?
D
E
Mensuration
409
In our daily life we also come across situations where the figures formed have irregular
shapes. Observe the figures given below.
In all such cases, where the figures are not having triangle or quadrilateral shapes
and figures having irregular shapes, the simplest technique is to divide the given figure
into triangles and quadrilaterals and then find their areas.
Such figures can be divided into several familiar shapes as follows. study them.
Observe that each figure is divided into triangles and trapeziums.
How do you think that the total area of each figure is calculated by using the areas of
different triangles and trapeziums formed? Discuss in the class in groups.
Y
Do the following activity to understand this.
Activity 1:
5
Take a graph sheet.
Draw a pentagon A C B E D.
4
Join AB, which forms the base line.
3
Draw perpendiculars from C, D and E to AB.
Read the measure of the length of their
sides.
–2 –1
–3
–5
–4
Calculate the area of each triangle ADF, ACB
and EHB and area of the trapezium FHED.
D
F
C
G
2
Marks the points F, G and H
Identify the triangles and trapeziums formed.
A
E
1
O
H
B
1
2
3
4
5
X
Find the sum of these areas.
You have found the area of the pentagon. Repeat this activity for the other figures
discused earlier.
410
UNIT-16
Now consider the example of the piece of land. Suppose a farmer owns it and a land
surveyor wants to find its area. How is it done?
The procedure followed by surveyors while measuing and calculating areas of irregular
shaped lands are as follows.
Irregular shaped field is divided into known geometrical shaped fragments.
Measurements of the sides are made using Guntor’s chain.
Measurements are recorded in the surveyor’s field book .
The sketch or figure repesenting the irregular shaped field is drawn to scale.
Area of each fragment is calculated.
Total area of the field is found by finding the sum of the areas of the fragments.
How are the measurements taken and recorded in the field book?
How is scale drawing done?
Do the following activity to understand this .
Activity 2
Take a graph sheet.
Y
Draw an irregular shape as shown in the figure.
Mark the extreme end points, wherever
possible.
First mark the top and bottom points to obtain
the base line by joining them. AB is the base
line.
Now, mark the other end points on either side
of the base line, let the points be C,D, E and F.
Draw perpendiculars from C,D,E and F to the
base line AB.
5
A
4
3
D
J
C
G
2
E
1
O
K
F
B
1
2
3
4
5
X
Let the perpendicular lines be CG, DJ, EK,
and FB.
Join A and C, C and F, D and E, B and E and D and A.
Now you have obtained a regular shape ACFBED.
Read the measurements of lines formed and make entries in a table as given below.
Use the values in the table and find the area as discussed in activity 1.
The land surveyor’s method of recording the measurements of a land will be the same as
discussed above. But the measurements are usually in meters. Hence, while drawing
the rough sketch, a convenient scale is taken and all the original measurements are
converted according to the scale.
Now study the following examples.
Mensuration
411
ILLUSTRATIVE EXAMPLES
Example 1: Draw a plan and calculate the area of a level ground using the information
given below.
Sol.
to D (meter)
Take suitable scale, say 20m = 1cm.
Convert the given measurements
200 m =
200
1
× 200 = 10 cm
20
140
120 m =
120
To E 60
1
140 m =
×140 = 7 cm
20
40
1
× 120 = 6cm
20
40 m =
1
× 40 = 2 cm
20
60 m =
1
× 60 = 3cm
20
50 to C
30 to B
From A
D
R
1
50m =
× 50 = 2.5cm
20
C
Q
E
1
× 30 = 1.5 cm
20
Draw the vertical base line AD = 10cm (200 m)
30 m =
Mark points P, Q and R on AD such that AP = 2cm (40m),
AQ = 6cm (120 m) or PQ = 4cm (
AR = 7cm (140cm) or QR = 1cm (
P
6-2 is taken)
7-6 is taken)
Draw perpendiculars from P,Q and R with the given measurements.
PB = 1.5 cm (30m)
B
QE = 3 cm (60m)
A
RC = 2.5 cm (50m)
Join the points to get the figure ABCDE
Record the measurements.
Calculate the area of ABCDE.
Area of ABCDE = Area of ABP + Area of trap . PBCR+ Area of CRD+ Area of DEQ +
Area of EQA.
=
1
1
1
1
1
× AP × PB + ×PR(PB+RC) + × RC× RD+ × EQ × QD + × EQ × QA
2
2
2
2
2
=
1
1
1
1
1
× 40×30 + ×100(30 + 50) + × 50 × 60 + ×60×80+ × 60 × 120
2
2
2
2
2
= 600 + 4000 + 1500 + 2400 + 3600
Area of ABCD = 12,100 sq.m.
412
UNIT-16
Example 2: Plan out and find the area of the field from the following notes from the
field book.
Metre to D
150
100
To E 80
70 to C
D
80
30
P
From A
C
20
Sol. Scale 20 m = 1 cm
E
Observe that
N
80
50
AM = 30 m.
AN = 80 m.
40
M
ND = (150 – 80) = 70 m.
A
PD = (150 – 100) = 50 m.
1
bh
2
1
2
30
(2) Area of Trapezium MBCP =
B
30
MP = (100 – 30) = 70 m.
(1) Area of ABM =
50
70
40 to B
40
1
h a
2
600 sq.m.
b
1
2
(3) Area of DPC =
1
2
50
70
1750 sq.m.
(4) Area of DEN =
1
2
80
70
2800 sq.m.
(5) Area of NEA =
1
2
80
80
3200 sq.m.
70 70
40
3850 sq.m.
EXERCISE 16.6
1. Draw a plan and calculate the area of a level ground using the information given
below.
To D 120
To E 180
Metre to C
220
210
120
80
From A
200 to B
Mensuration
413
2. Plan out and find the area of the field from the data given from the Surveyor's field
book
Metre to E
350
300
250
150
50
From A
To D 100
To C 75
To B 50
150 to F
100 to G
3. Sketch a rough plan and calculate the area of the field ABCDEFG from the following
data
Metre to D
225
175
125
100
80
60
From A
To E 90
To F 60
To G 15
20 to C
70 to B
4. Calculate the area of the field shown in the diagram below:
[Measurements are in metre]
B
C
40
D
35 35
F25
30
E
30
G
70
A
414
2
UNIT-16
2
Statistics
Mensuration
Cylinder
Cone
Frustum of
a cone
LSA
2 rh
rl
(r1+r2)
TSA
2 r(h+r)
r (l+r)
{(r12+r22)l+
r1+r2 }
r2 h
1
3
Sphere
2 r2
1
3
h
(r1 +r2+r1+r2)
r2h
Hemisphere
4 r2
3 r2
4
3
2
3
r3
r3
Scale
drawing
Combination
of solids
ANSWERS
EXERCISE 16.1
1] (i) 176cm2 (ii) 201.14 cm2 2] 7cm 3] 3168cm2 4] 5:4 5] 3:2 6] 88m2, `3696
7] 21,780cm2
8] r = 7 cm, h = 3.5 cm
9] `269.50
10] 44 m2
11] 5,544 cm3
12] 3,432 g
3
3
15] 3,080 cm
EXERCISE 16.2
16] 7,39,200 cm
1] 3960 cm2
2] 8 cm
3] 7.5 cm 4] 7cm and 462 cm2
(ii) 2,310 cm3
3
12] 6.4cm
13] a
2
14]
2 :1
18] `300
17] 0.53m
8] (i) 183.33 cm3
11] 12 cm
(ii) 141.42 cm3
13] (i) 2.99cm`
(iii) 7,546 cm3
5] 5,500 cm2
6] 4:3
10] 415.8 cm3
9] 3:1
14] ` 2,068
EXERCISE 16.3
1] 1,408.58 cm3
4] 5 cm
2] CSA = 1,021.42 cm2, TSA = 1,335.71cm2
3] 26,148.57cm3
5] `418, `156.75
EXERCISE 16.4
1] (i) 2,464 cm2
2
4] 707.14 cm
8] 199.8g
13] 96 cm
EXERCISE 16.5
1] 16,761.9l
(ii) 98.56 cm2
2
5]12 cm
2] 36.53m3
3] 253 cm2
11] 905.14 cm2
16] 16 cm
2] 50,625m2
4] ` 413.6
3] 15,250m2
12] 22.45 mm3
17] 672
5] ` 58.08
EXERCISE 16.6
1] 49,300m2
(ii) 4851 cm3
7] (i) 1437.33 cm
10] 2095.24 cm3
15] 243 m
3] ` 20.79
2] 235.71cm2
3
6] 173.25 cm
9] 179.66 cm3
14] 5 cm
(iii) 498.96 cm2
4] 4,600m2
6] ` 97,350