 Lesson: Surface areas and volumes
Exercise 13.1 (9)
Question: 1
2 cubes each of volume are joined end to
3
64 cm
end. Find the surface area of the resulting cuboid.
Solution:
Let be the edge of the cube.
a
So,
Volumeof each cube 64
3
a 64
3
a 64
a 4 cm
Fig. Exc_13.1_1 The dimensions of the cuboid formed are
4 cm,
and .
4 cm
Now, calculate the surface area of the cuboid.
Surfacearea of cuboid 2 lb bh lh
2 4 4 4 8 4 8
2 16 32 32
2
160cm
Thus, the surface area of cuboid is .
2
160cm
Question: 2
A vessel is in the form of a hollow hemisphere
mounted by a hollow cylinder. The diameter of the
hemisphere is 14 cm and the total height of the
vessel is 13 cm. Find the inner surface area of the
vessel.
Solution:
Fig. Exc_13.1_2 Observe that, the radius of the hemisphere and
r
the cylindrical part is same (i.e., 7 cm).
Height of hemisphere = Radius = 7 cm
Height of cylindrical part = 13 7 = 6 cm
h
Now, find the inner surface area of the vessel.
Inner surface area CSA of CSA of hemi-
of the vessel cylindrical part spherical part
2
2πrh 2πr
22 22
2 7 6 2 7 7
7 7
44 6 7
2
572cm
Thus, the inner surface area of the vessel is 572 cm
2
.
Question: 3
A toy is in the form of a cone of radius 3.5 cm
mounted on a hemisphere of same radius. The total
height of the toy is 15.5 cm. Find the total surface
area of the toy.
Solution: Fig. Exc_13.1_3
Observe that, the radius of the hemispherical part =
the conical part = 3.5 cm
Height of hemispherical part = Radius = 3.5 cm
r
Height of conical part = 15.5 3.5 = 12 cm
h
Now, find the slant height of the conical part.
2 2
Slant height r h
2 2
3.5 12
2
2
7
12
2
49
144
4
49 576
4 625 25
4 2
Now, find the total surface area of the toy.
Total surface CSA of CSA of hemi-
area of toy conical part spherical part
2
πrl 2πr
22 7 25 22 7 7
2
7 2 2 7 2 2
.137 5 77
2
.214 5cm
Thus, the total area of the toy is 214.5 cm
2
.
Question: 4
A cubical block of side 7 cm is surmounted by a
hemisphere. What is the greatest diameter the
hemisphere can have? Find the surface area of the
solid.
Solution: Fig. Exc_13.1_4
From the figure, observe that the greatest diameter
of hemisphere = the cube’s edge = 7 cm
r
7
cm
2
Surface area CSA of hemi-
of cubical part spherical part
Total surface
area of solid
Area of base of
hemi-spherical part
2
2 2
6 edge 2πr πr
2
2
6 edge πr
2
22 7 7
6 7
7 2 2
294 38.5
2
332.5cm Thus, the total surface area of the solid is 332.5 cm
2
.
Question: 5
A hemispherical depression is cut out from one face
of a cubical wooden block such that the diameter l
of the hemisphere is equal to the edge of the cube.
Determine the surface area of the remaining solid.
Solution:
Fig. Exc_13.1_5
Edge of cube = Diameter of hemisphere = l
l
2
Surface area of CSA of hemi-
cubical part spherical part
Total surface
area of solid
Area of base of
hemi-spherical part
2
2 2
6 edge 2πr πr 2
2
6 edge πr
2
2
l
6 l π
2
2
2
πl
6 l
4
2 2
1
24 π l unit
4
Thus, the total surface area of solid is
.
2 2
1
24 π l unit
4
Question: 6
A medicine capsule is in the shape of a cylinder
with two hemispheres stuck to each of its ends (see
the given figure). The length of the entire capsule is
14 mm and the diameter of the capsule is 5 mm.
Find its surface area.
Fig. Exc_13.1_6 (Ques.) Solution:
Fig. Exc_13.1_6 (Sol.)
Let be the radius and be the length of the
r
h
cylindrical part.
cylindrical part spherical part
Diameter of thecapsule
2
5
2
Length of
Length of the
2r
cylindrical part h entire capsule
14 5
9cm
Now, find the surface area of the capsule.
Surface area CSA of hemi- CSA of
2
of capsule spherical part cylindrical part
2
2 2πr 2πrh 2
5 5
4π 2π 9
2 2
25π 45π
2
70π mm
2
22
70 mm
7
2
220mm
Thus, the surface area of the capsule is 220 mm
2
.
Question: 7
A tent is in the shape of a cylinder surmounted by a
conical top. If the height and diameter of the
cylindrical part are 2.1 m and 4 m respectively, and
the slant height of the top is 2.8 m, find the area of
the canvas used for making the tent. Also, find the
cost of the canvas of the tent at the rate of Rs 500
per m
2
. (Note that the base of the tent will not be
covered with canvas.)
Solution: Fig. Exc_13.1_7
Let be the height of the cylindrical part and be
h
l
the slant height of conical part.
Height of the cylindrical part = 2.1 m
h
Diameter of the cylindrical part = 4 m
So, radius of the cylindrical part
Diameter 4
2m
2 2
Slant height of conical part = 2.8 m
l
Area of canvas used CSA of CSA of
for making tent conical part cylindrical part
πrl 2πrh
π 2 2.8 2π 2 2.1
2π 2.8 4.2 22
2 7
7
2
44m
Cost of 1 m
2
canvas
Rs500
Cost of 44 m
2
canvas
44 500
22000
Thus, the total cost for making a tent is Rs 22000.
Question: 8
From a solid cylinder, whose height is 2.4 cm and
diameter 1.4 cm, a conical cavity of the same height
and same diameter is hollowed out. Find the total
surface area of the remaining solid to the nearest
cm
2
.
Solution:
Fig. Exc_13.1_8 Let be the height of both the cylindrical and
h
conical part.
Height of the conical part = Height of the
h
cylindrical part = 2.4 cm
h
Diameter of the cylindrical part = 1.4 cm
So, radius of the cylindrical part
r
Diameter 1 4
0 7cm
2 2
.
.
Slant height
2 2
l r h
2 2
0.7 42.
70.49 5. 6
6 25. cm2.5
CSA of CSA of
cylindrical part conical part
Total surface area
of remaining solid
Area of
cylindrical base
2
2πrh πrl πr
22 22 22
2 0.7 2.4 0.7 2.5 0.7 0.7
7 7 7
4.4 2.4 2.2 2.5 2.2 0.7
10.56 5.50 1.54
2
17.60cm Thus, the total surface of the remaining solid is 18
cm
2
.
Question: 9
A wooden article was made by scooping out a
hemisphere from each end of a solid cylinder, as
shown in the given figure. If the height of the
cylinder is 10 cm, and its base is of radius 3.5 cm,
find the total surface area of the article.
Fig. Exc_13.1_9
Solution:
r
r
hemispherical part = 3.5 cm
r
Height of cylindrical part = 10 cm
h Total surface CSA of CSA of hemi-
2
area of article cylindrical part spherical part
2
2πrh 2 2πr
2π 3.5 10 2 2π 3.5 3.5
70π 49π
119π
22
119
7
2
374 cm
Thus, the total surface area of the article is 374 cm
2
.
Exercise 13.2 (8)
Question: 1
A solid is in the shape of a cone standing on a
hemisphere with both their radii being equal to 1
cm and the height of the cone is equal to its radius.
Find the volume of the solid in terms of .
π
Solution: Fig. Exc_13.2_1
Let be the radius and be the height.
r
h
Radius of conical part = Height of conical part
r
= 1 cm
h
r
part = 1 cm
r
Volume Volume of Volume of hemi-
of solid conical part spherical part
2 3
1 2
πr h πr
3 3
2 3
1 2
π 1 1 π 1
3 3
3
π cm
Thus, the volume of solid is .
3
π cm
Question: 2
Rachel, an engineering student, was asked to make
a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium
sheet. The diameter of the model is 3 cm and its
length is 12 cm. If each cone has a height of 2 cm,
find the volume of air contained in the model that
Rachel made. (Assume the outer and inner
dimensions of the model to be nearly the same.)
Solution:
Fig. Exc_13.2_2
From the figure, observe that,
Height of each conical part = 2 cm
1
h
 
2
Height of
Height of
12 2
cylindrical part h conical part
12 2 2
8cm
r
part
3
cm
2 Volume of air contained Volume of Volume of
2
in the model cylinder cones
2 2
2 1
1
πr h 2 πr h
3
2 2
3 1 3
π 8 2 π 2
2 3 2
18π 3π
22
21
7
3
66cm
Thus, the volume of air contained in the modal is 66
cm
3
.
Question: 3
A gulab jamun, contains sugar syrup up to about
30% of its volume. Find approximately how much
syrup would be found in 45 gulab jamuns, each
shaped like a cylinder with two hemispherical ends
with length 5 cm and diameter 2.8 cm. Fig. Exc_13.2_3 (Ques.)
Solution:
Fig. Exc_13.2_3 (Sol.)
Observe that,
r
hemispherical part
r
28
1
2
.
.4 cm
Length of each hemispherical part = Radius of
hemispherical part = 1.4 cm Length of
Length of hemi-
5 2
cylindrical part spherical parth
5 2 1.4 2.2 cm
Volume of one Volume of Volume of hemi-
2
gulab jamun cylindrical part spherical part
2 3
2
πr h 2 πr
3
2 3
22 4 22
1.4 2.2 1.4
7 3 7
13.552 11.498
3
25.05cm
Volume of 45 gulab jamuns
45 2 055.
3
1127.25cm
Volume of sugar syrup
30%of volume
30
1127 25
100
.
3
.1338 7cm
3
cm ap3 x3 8 pro
Thus, the volume of sugar syrup is 338 cm
3
approx.
Question: 4
A pen stand made of wood is in the shape of a
cuboid with four conical depressions to hold pens.
The dimensions of the cuboid are 15 cm by 10 cm
by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of
wood in the entire stand.
Fig. Exc_13.2_4 (Ques.)
Solution:
Fig. Exc_13.2_4 (Sol.)
Let be the depth and be the radius of the
h
r
conical depression.
Depth of each conical depression = 1.4 cm
h
Radius of each conical depression = 0.5 cm
r Volume Volume of Volume of
4
of wood cuboid cones
2
1
lbh 4 πr h
3
2
1 22 1
15 10 3.5 4 1.4
3 7 2
525 1.47
3
523.53cm
Thus, the volume of wood is .
3
523.53cm
Question: 5
A vessel is in the form of an inverted cone. Its
height is 8 cm and the radius of its top, which is
open, is 5 cm. It is filled with water up to the brim.
When lead shots, each of which is a sphere of
radius 0.5 cm are dropped into the vessel, one-
fourth of the water flows out. Find the number of
lead shots dropped in the vessel.
Solution: Fig. Exc_13.2_5
Let be the height of the conical vessel and be
h
1
r
2
r
Height of conical vessel = 8 cm
h
Radius of conical vessel = 5 cm
1
r
2
r
Let be the number of lead shots.
n
Volume of Volume of dropped
3
2
Volume
1 4
n πr
of cone
4 3
2 3
1 2
1 1 4
πr h n πr
4 3 3
2 3
1 2
r h n 16r
3
2
5 8 n 16 0.5 3
25 8
n
1
16
2
100
Thus, the number of lead shots dropped are 100.
Question: 6
A solid iron pole consists of a cylinder of height 220
cm and base diameter 24 cm, which is surmounted
by another cylinder of height 60 cm and radius 8
cm. Find the mass of the pole, given that 1 cm
3
of
iron has approximately 8 g mass.
Solution:
Fig. Exc_13.2_6 Let be the height and radius of the larger
1 1
h andr
cylinder and be the height and radius of
2 2
h andr
the smaller circle respectively.
Observe that,
Height of larger cylinder = 220 cm
1
h
1
r
24
12cm
2
Height of smaller cylinder = 60 cm
 
2
h
Radius of smaller cylinder = 8 cm
2
r
Total volume Volumeof Volumeof
of pole larger cylinder smaller cylinder
2 2
1 1 2 2
πr h πr h
2 2
π 12 220 π 8 60
π 144 220 64 60
π 31680 3840
35520 3 14.
1115328.
Mass of iron
3
1cm
8g
Mass of
3
1115328cm 1115328 8. .
892262 g.4
8922. 62kg
Thus, the mass of pole is 892.262 kg. Question: 7
A solid consisting of a right circular cone of height
120 cm and radius 60 cm standing on a hemisphere
of radius 60 cm is placed upright in a right circular
cylinder full of water such that it touches the
bottom. Find the volume of water left in the
cylinder, if the radius of the cylinder is 60 cm and
its height is 180 cm.
Solution:
Fig. Exc_13.2_7
r
part = 60 cm
r Height of conical part of solid = 120 cm
 
2
h
Height of cylinder = 180 cm
1
h
Radius of cylinder = 60 cm
r
Volume of Volume of Volume
water left cylinder of solid
Volume of Volume Volume of
cylinder of cone hemisphere
2 2 3
1 2
1 2
πr h πr h πr
3 3
2 2 3
1 2
π 60 180 π 60 120 π 60
3 3
2
π 60 180 40 40
2
π 60 100
3
360000π cm
3
1131428.571cm
3
1.131m
Thus, the volume of water left in the cylinder is
1.131 m
3
.
Question: 8
A spherical glass vessel has a cylindrical neck 8 cm
long, 2 cm in diameter; the diameter of the
spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345
cm
3
. Check whether she is correct, taking the above
as the inside measurements, and .
π 3 4.1
Solution:
Fig. Exc_13.2_8
Let and be the height and radius of the
h
2
r
cylindrical part. Let be the radius of spherical
1
r
part.
Height of cylindrical part = 8 cm
h
Radius of cylindrical part = 1 cm
2
r Radius spherical part = 4.25 cm
1
r
Volume Volume of Volume of
of vessel sphere cylinder
3 2
1 2
4
πr πr h
3
3
2
4 8.5
π π 1 8
3 2
4
3.14 76.765625 8 3.14
3
321.392 25.12
346.512
3
346.51cm
Thus, the child is wrong, and the volume of the
vessel is 346.51 cm
3
.
Exercise 13.3 (9)
Question: 1
A metallic sphere of radius 4.2 cm is melted and
recast into the shape of a cylinder of radius 6 cm.
Find the height of the cylinder.
Solution: 1
r
.
2
r
Let the height of the cylinder be .
h
Volume of Volume of
sphere cylinder
3 2
1 2
4
πr r h
3
Here,
1
r 4.2cm,
2
r 6cm
3 2
1 2
4
πr πr h
3
3
2
4 π 4.2
h
3 π 6
4 π 4.2 4.2 4.2
3 π 6 6
4 0.7 0.7 1.4
3
2.74 cm
Thus, the height of the cylinder is .
2.74 cm
Question: 2
Metallic spheres of radii 6 cm, 8 cm and 10 cm,
respectively, are melted to form a single solid
sphere. Find the radius of the resulting sphere. Solution:
st
1
1
r
nd
2
, and radius of sphere be .
2
r
rd
3
3
r
Let the radius of the resulting sphere be .
r
Now,
1
r 6cm,
2
r 8cm,
3
r 10cm
Volume of Sum of Volumes
resulting sphere of the 3 spheres
3 3 3 3
1 2 3
4 4
π r r r πr
3 3
3 3 3 3
1 2 3
r r r r
3 3 3 3
r 6 8 10
3
r 216 512 1000
3
r 1728
r 12cm
Thus, the radius of the resulting sphere is .
12cm
Question: 3
A 20 m deep well with diameter 7 m is dug and the
earth from digging is evenly spread out to form a
platform 22 m by 14 m. Find the height of the
platform.
Solution: Fig. Exc_13.3_3
Depth of well
h 20m
Radius of circular end of well
7
r m
2
Area of platform
l b
22 14
2
308m
Let, height of the platform be .
H
Volume of soil Volume of soil used to
dug from well make the platform
2
πr h Area of platform height H
2
πr h
H
Area of platfrom 22 7 7 20
7 2 2 308
5
2
2.5m
Thus, the height of the platform is .
2.5m
Question: 4
A well of diameter 3 m is dug 14 m deep. The earth
taken out of it has been spread evenly all around it
in the shape of a circular ring of width 4 m to form
an embankment. Find the height of the
embankment.
Solution: Fig. Exc_13.3_4
3
r m
2
Width of embankment
4m
2 1
r r width
3
4
2
11
m
2
Height of circular well
h 14m
Let, be the height of the embankment.
H Volume of soil Volume of soil used
dug from well to make embankment.
2
πr h Area of embankment height H
2 2
2 1
Area of
embankmen
r
t
=π r
11 3
π
2 2
2
112
π m
4
Now, height of embankment
2
πr h
H
Area of embankment
2
3
π 14 4
2
112 π
3 3
8
1.125m
Hence, the height of the embankment is .
1.125m
Question: 5
A container shaped like a right circular cylinder
having diameter 12 cm and height 15 cm is full of
ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a
hemispherical shape on the top. Find the number of
such cones which can be filled with ice cream.
Solution:
Height of cylindrical container = 15 cm
1
h
Radius of end of container = 6 cm
1
r
Radius of end of ice-cream cone = 3 cm
2
r
Height of ice-cream cone = 12 cm
Let number of ice-cream cones to be filled.
n
Volume of ice-cream Volumeof cone
n
in cylinder Volumeof hemishpherical top
2 2 3
1 1 2 2 2
1 2
πr h n πr h πr
3 3
2
2 3
15 6
n
1 2
3 12 3
3 3
15 6 6
n
1 2
3 3 12 3 3 3
3 3
15 6 6
n
3 12 2 3 3
15 6 6
n
36 18 15 6 6
n
54
n 10
Hence, 10 ice-cream cones can be filled.
Question: 6
How many silver coins, 1.75 cm in diameter and of
thickness 2 mm, must be melted to form a cuboid of
dimensions ?
5.5cm 10cm 3.5cm
Solution:
Fig. Exc_13.3_6
Let, number of coins are melted.
n
Height of each coin
H 2mm
0.2cm
Radius of circular end of each coin
1.75
r 0.875cm
2
Let, , and be the dimensions of the cuboid.
l
b
h l 5.5cm
b 10cm
h 3.5cm
Total volume Volume of
of coins melted cuboid formed
2
n πr H l b h
2
5.5 10 3.5
n
π 0.875 0.2
7 55 10 35 10 1000 1000
22 875 875 2 10 10
7 55 10 1000
11 875
7 10 1000
175
400
Hence, the number of coins melted to form the
desired cuboid is 400.
Question: 7
A cylindrical bucket, 32 cm high and with radius of
base 18 cm, is filled with sand. This bucket is
emptied on the ground and a conical heap of sand
is formed. If the height of the conical heap is 24 cm,
find the radius and slant height of the heap.
Solution: Fig. Exc_13.3_7
Height of cylindrical bucket = 32 cm
1
h
Radius of end of bucket = 18 cm
1
r
Height of conical heap = 24 cm
 
2
h
Let the radius of the conical heap be .
2
r
Volume of sand in Volume of sand
the cylindrical bucket in conical heap
2 2
1 1 2 2
1
πr h πr h
3
2
2
2
3 18 32
r
24
2
2
3 18 18 32
r
24
2
2
r 18 18 4
2
r 1296
36cm Hence, radius of the conical heap is .
36cm
Slant height
2 2
r h
2 2
36 24
12 13 cm
Thus, the radius and slant height of the conical
heap are and , respectively.
36cm
12 13 cm
Question: 8
Water in a canal, 6 m wide and 1.5 m deep, is
flowing with a speed of 10 km/h. How much area
will it irrigate in 30 minutes, if 8 cm of standing
water is needed?
Solution:
Fig. Exc_13.3_8
Area of cross-section of canal ABCD
l b
6 1.5
2
9m Speed
of water
10km / h
1000m
10
60min
10000
m/min
60
Volume of water through
lbh
canal in 1 minute
10000
9
60
3
1500m / min
Volume of water flowing through canal in 30
minutes
30 1500
3
45000m
Volume of water Volume of water
irrigating the flowing through canal
required area in 30 minutes
A 8
45000
100
45000 100
A
8
2
A 562500m
Thus, the area is .
2
562500m Question: 9
A farmer connects a pipe of internal diameter 20 cm
from a canal into a cylindrical tank in her field,
which is 10 m in diameter and 2 m deep. If water
flows through the pipe at the rate of 3 km/h, in how
much time will the tank be filled?
Solution:
Fig. Exc_13.3_9_(i)
1
0.2
r 0.1m
2
Area of cross-section of the pipe
2
1
πr
2
π 0.1
2
0.01π m
Flow rate
of wat
m
er
3k / h
1000m
3
60min.
50m/min Volume of water flowing through the pipe in 1
minute
50 0.01π
3
0.5π m
Fig. Exc_13.3_9_(ii)
Radius of cylindrical tank = 5 m
2
r
Depth of cylindrical tank = 2 m
 
2
h
Volume of water flowing through Volume of water
pipein t minutes in tank
2
2 2
t 0.5π πr h
2
5 2
t
0.5
5 5 2 10
5
100min
Thus, tank will take 100 minutes to fill completely. Exercise 13.4 (5)
Question: 1
A drinking glass is in the shape of a frustum of a
cone of height 14 cm. The diameters of its two
circular ends are 4 cm and 2 cm. Find the capacity
of the glass.
Solution:
Fig. Exc_13.4_1 1
4
r
2
2cm
2
2
r
2
1cm
Height of frustum
14 cm
Capacity Volume of
of glass frustum of cone
2 2
1 2 1 2
1
πh(r r rr )
3
1 22
14 4 1 2
3 7
1
22 14
3
308
3
3
2
102 cm
3
Thus, the capacity of the glass is .
3
2
102 cm
3
Question: 2
The slant height of a frustum of a cone is 4 cm and
the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of
the frustum.
Solution:
Fig. Exc_13.4_2
Let, and be the radius of the upper and lower
1
r
2
r
end.
Perimeter of upper end
1
2πr
18cm
1
18
r
2π
9
cm
π
Perimeter of lower end
2
2πr
6cm
2
6
r
2π 3
cm
π
Slant height of frustum
l 4 cm
Curved surface area of frustum
1 2
π r r l
9 3
π 4
π π
12 4
2
48cm
Thus, the curved surface area of frustum is .
2
48cm
Question: 3
A fez, the cap used by the Turks, is shaped like the
frustum of a cone (see the given figure). If its radius
on the open side is 10 cm, radius at the upper base
is 4 cm and its slant height is 15 cm, find the area of
material used for making it.
Fig. Exc_13.4_3 (Ques.) Solution:
Fig. Exc_13.4_3 (Sol.)
1
r 4 cm
2
r 10cm
Slant height of frustum
l 15cm
Area of material
Area of upper
used for making CSA of frustum
circular end
the fez cap
2
1 2 1
π r r l πr
2
π 4 10 15 π 4
210π 16π
=226π
22
226
7 2
2
710 cm
7
Hence, the area of the material used for making the
cap is .
2
2
710 cm
7
Question: 4
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 16 cm 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 Rs. 20 per litre. Also find the cost of metal
sheet used to make the container, if it costs Rs. 8
per 100 . (Take π = 3.14).
2
cm
Solution: Fig. Exc_13.4_4
Radius of upper end of container
1
r 20cm
Radius of lower end of container
2
r 8cm
Height of container
h 16cm
Volumeof milk in Volumeof
thecontainer frustum
2 2
1 2 1 2
1
πh(r r rr )
3
2 2
1
3.14 16 20 8 20 8
3
1 314
16 400 64 160
3 100
1 314
16 624
3 100
3
10449.92cm
10.45litres
Cost of milk per litre
Rs20
Therefore, cost of milk
10.45litres
10.45 20
Rs209
Slant height of frustum
2
2
1 2
l r r h
2
2
20 8 16
144 256
20cm 2
1 2 2
Area of sheet required
to make the
π r r l πr
container
2
π 20 8 20 π 8
560π 64π
624 3.14
2
1959.36cm
Cost of the metal sheet
2
8Rs./ 100cm
Cost of metal sheet
2
1959.36cm
1959.36 8
100
Rs156.75
Hence, the cost of milk required to fill the container
is and the cost of metal sheet required to
Rs209
make the container is
Rs156.75.
Question: 5
A metallic right circular cone 20 cm high and whose
vertical angle is is cut into two parts at the
60
middle of its height by a plane parallel to its base. If
the frustum so obtained be drawn into a wire of
diameter cm, find the length of the wire.
1
16
Solution: Fig. Exc_13.4_5
Let, radius of the upper half be and that of lower
1
r
half be .
2
r
Height of cone
h 20cm
In ,
PTM
TM
tan 30
PM
1
TM PM
3
1
TM r
10
cm
3
In ,
POR
OR
tan30
OP 2
OR r
1
OP
3
1
20
3
20
cm
3
So, radius of upper end of frustum
1
10
r cm
3
Radius of lower end of container
2
20
r cm
3
Height of container
h 10cm
1
r
16 2
1
cm
32
Let, be the length of the wire.
l
Volume of wire Area of cross-section l
2
=πr l
Volumeof Volumeof
frustum wire
2 2 2
1 2 1 2
1
πh(r r rr ) πr l
3 2 2
2
1 20 10 20 10 1
10 l
3 32
3 3 3 3
1 400 100 200 1
10 l
3 3 3 3 1024
7000
l 1024
9
796444.44 cm
7964.44m
Thus, the length of the wire is .
7964.44m
Exercise 13.5 (7)
Question: 1
A copper wire, 3 mm in diameter, is wound about a
cylinder whose length is 12 cm, and diameter 10 cm,
so as to cover the curved surface of the cylinder.
Find the length and mass of the wire, assuming the
density of copper to be 8.88 g per cm
3
.
Solution: Fig. Exc_13.5_1
Height of the cylinder = 12 cm
Diameter of cylinder
2
10
2
5cm
Observe that, 1 round of wire will cover 3 mm
height of the cylinder.
Height of cylinder
Number of rounds
Diameter of wire
12
0 3.
40rounds
Now, find the length of wire required. Length of wire Circumferenceof
required in 1 round baseof cylinder
2πr
2π 5
10π
Length of wire
in 40 rounds
40 10π
22
400
7
8800
7
.1257 14 cm
12 m.57
0 3
0 15
2
.
.
12 m.57
2
π 0.15 1257.14
22
0.15 0.15 1257.14
7
622.2843
7
3
88.898cm
Now, calculate the mass of the wire.
Mass Volume Density .89888 8.88
789.41gm
Thus, the mass of the wire is 789.41 gm.
Question: 2
A right triangle, whose sides are 3 cm and 4 cm
its hypotenuse. Find the volume and surface area of
the double cone so formed. (Choose value of as
π
found appropriate.)
Solution:
Fig. Exc_13.5_2
Sides of right-angled triangle are 3 cm and 4 cm.
Now, find the value of hypotenuse.
2 2
Hypotenuse AC AB BC
2 2
3 4 25
5cm
1
Area of ABC AB BC
2
1 1
AB BC AC BO
2 2
AB BC AC BO
4 3 5 BO
12
BO 24 cm
5
.
Volume of Volume of Volume of
double cone cone 1 cone 2
2 2
1 2
1 1
πr h πr h
3 3
2
1 2
1
πr h h
3
2
1
πr OA OC
3
2
1
3.14 2.4 5
3
1 314 24 24
5
3 100 10 10
3
30.14 cm
Now, find the surface area of the double cone. Surfacearea of Surfacearea Surfacearea
doublecone of cone1 of cone2
1 2
πrl πrl
πr 4 3
3.14 2.4 7
2
52.75cm
Thus, the value of surface area is 52.75 cm
2
.
Question: 3
A cistern, internally measuring
, has 129600 cm
3
of water in
150cm 120cm 110cm
it. Porous bricks are placed in the water until the
cistern is full to the brim. Each brick absorbs one-
seventeenth of its own volume of water. How many
bricks can be put in without overflowing the water,
each brick being ?
22.5cm 7.5cm 6.5cm
Solution:
Internal measurement of water = 129600 cm
3
Volume of cistern
150cm 120cm 110cm
3
1980000cm
Volume to be filled in cistern
1980000 129600
3
1850400cm Let be the number of bricks were placed in
n
cistern.
Volumeof nbrick .s n 225 6 5.7.5
. n1096 875
Each brick absorbs one-seventeenth of its own
volume of water.
So,
n
1850400 1096 1096 n
1
.875 . 5
7
87
16n
1850400 1096
17
.875
1850400 17
.
n
1096 16875
1792 1.4
Thus, there are 1792 bricks placed in the cistern.
Question: 4
In one fortnight of a given month, there was a
rainfall of 10 cm in a river valley. If the area of the
valley is 97280 km
2
, show that the total rainfall was
approximately equivalent to the addition to the
normal water of three rivers each 1072 km long, 75
m wide and 3 m deep. Solution:
Area of the valley
2
97280km
2
97280000000m
Rainfall in the river valley
10cm
0.1m
Now, find the volume of water.
Volume of water
97280000000 0.1
3
9728000000m
Now, find the volume of water of one river.
Volume of water of one river
1072000 75 3
3
241200000m
So, volume of water of 3 rivers
3 241200000
3
723600000m
Thus, .
3 3
9 7728 236000000m 00000m
Question: 5
An oil funnel made of tin sheet consists of a 10 cm
long cylindrical portion attached to a frustum of a
cone. If the total height is 22 cm, diameter of the
cylindrical portion is 8 cm and the diameter of the
top of the funnel is 18 cm, find the area of the tin
sheet required to make the funnel. Fig. Exc_13.5_5 (Ques.)
Solution:
Fig. Exc_13.5_5 (Ans.)
Let be the radius of upper circular end of frustum
1
r
part and be the radius of lower circular end of
2
r
frustum part.
1
r 2
r
part = 4 cm
Height of frustum part
1
h
22 10
12cm
Height of cylindrical part = 10 cm
 
2
h
2
2
1 2
Slant height l r r h
2 2
9 4 12
2 2
5 12
25 144
12cm
Area of tin CSA of CSA of
sheet required frustum part cylindrical part
1 2 2 2
π r r l 2πr h
22 22
9 4 13 2 4 10
7 7
22
169 80
7
22 249
7
2
4
782 cm
7 Thus, the area of tin sheet is .
2
4
782 cm
7
Question: 6
Derive the formula for the curved surface area and
total surface area of the frustum of a cone, given to
you in Section 13.5, using the symbols as explained.
Solution:
Fig. Exc_13.5_6
Let and be the radii of the frustum of the cone
1
r
2
r
and be the height of the frustum of the cone.
h
In and
PDF
PQG,DF||QG
So, .
PDF|| PQG DF PF PD
QG PG PQ
2 1 1
1 1 1
r h h l l
r h l
2
1 1 1
r h l
1 1
r h l
That is,
2
1 1
r l
1
r l
2
1 1
l r
1
l r
1 2
1 1
l r r
l r
1 1
1 2
l r
l r r
1
1
1 2
lr
l
r r
CSA of CSA of CSA of
frustum DEQR cone PQR cone PDE
1 1 2 1
πrl πr l l
1 1
1 2
1 2 1 2
lr lr
πr πr l
r r r r
2
1 1 1 2
2
1 2 1 2
πlr lr rl r l
πr
r r r r 2 2
1 2
1 2 1 2
πlr πr l
r r r r
2 2
1 2
1 2 1 2
r r
πl
r r r r
2 2
1 2
1 2
r r
πl
r r
CSA of frustum
1 2
πl r r
Now, find the total surface area of frustum.
CSA of Area of upper
frustum circular end
Total surface
area of frustum
Area of lower
circular end
2 2
1 2 2 1
π r r l πr πr
2 2
1 2 2 1
π r r l r r
Thus, the total surface area of frustum is
.
2 2
1 2 2 1
π r r l r r
Question: 7
Derive the formula for the volume of the frustum of
a cone, given to you in Section 13.5, using the
symbols as explained.
Solution: Fig. Exc_13.5_7
Let and be the radii of the frustum of the cone
1
r
2
r
and be the height of the frustum of the cone.
h
In and
PDF
PQG,DF||QG
So, .
PDF|| PQG
DF PF PD
QG PG PQ
2 1 1
1 1 1
r h h l l
r h l
2
1 1 1
r h l
1 1
r h l
That is,
2
1 1
h r
1
h r 2
1 1
h r
1
h r
1 2
1 1
h r r
h r
1
1
1 2
rh
h
r r
Volume of Volume of Volume of
frustum of cone cone ABC cone ADE
2 2
1 1 2 1
1 1
r h r h h
3 3
2 2
1 1 2 1
1
r h r h h
3
2 2
1 2
1 1
1 2 1 2
rh r1
r r h
3
h
r r r r
3
1 1 1
2
2
2
1 2 1 2
r h rh r1
r
h hr
r r r r3
3 3
1 2
1 2 1 2
r h r
3 r
1 h
r r r
3 3
1 2
1 2
r r
r
1
h
r3
2 2
1 2 1 2 1 2
1 2
r r r r r
1
3 r r
h
r
2 2
1 2 1 2
r r rr
1
h
3 Thus, the volume of frustum of cone is
.
2 2
1 2 1 2
r r
1
h
3
rr