NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 13
Chapter Name : Surface Areas And Volumes4

Exercise 13.1

Q1 2 cubes each of volume 64cm 3
= are joined end to end. Find the surface area of the
resulting cuboid.

Answer. Given that ,
Volume of cubes= ( Edge ) 3
= 64
3
( Edge ) = 64

Edge = 4cm

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.
Surface area of cuboids
= 2(lb + bh + lh)

= 2(4 × 4 + 4 × 8 + 4 × 8)

= 2(16 + 32 + 32)

= 2(16 + 64)

2
= 2 × 80 = 160cm

Page : 244 , Block Name : Exercise 13.1

Q2 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.

It can be observed that radius (r) of the cylindrical pa t and the hemispherical part is the same
(i.e., 7 cm).
Height of hemispherical part = Radius = 7 cm
Height of cylindrical part (h) = 13-7 = 6 cm
Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part
2
= 2πrh + 2πr
22 22
= 2 × × 7 × 6 + 2 × × 7 × 7
7 7

Inner Surface area of vessel = 44(6 + 7) = 44 × 13
2
= 572cm

Page : 244 , Block Name : Exercise 13.1

Q3 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.

It can be observed that the radius of the conical part and the hemispherica part is same (i.e.,
3.5cm).
Height of hemispherical part=Radius(r) =3.5 =7/2cm
Height Of conical part (h) = 15.5 -3.5 = 12 cm

2 2
= √r + h

2
7 49 49+576

Slant Height = √(
of conical part + CSA of
2
) + (12) = √ + 144 = √
2 4 4

625 25
= √ =
4 2

hemispherical part
Total surface area of toy CSA of conica part + CSA of hemispherical part
2
= πrl + 2πr

22 7 25 22 7 7
= × × + 2 × × ×
7 2 2 7 2 2

2
= 137.5 + 77 = 214.5cm

Page : 244 , Block Name : Exercise 13.1

Q4 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 .

From the gure, it can be observed that the greatest diameter possible for such hemisphere is
equal to the cube's edge, i.e., 7cm.
Radius (r) of hemispherical part=7/2=3.5cm
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part -Area of
base of hemispherical part
2 2 2 2 2
= 6(Edge) + 2πr − πr = 6(Edge) + πv

2 22 7 7
= 6(7) + × ×
Total surface area of solid 7 2 2

2
= 294 + 38.5 = 332.5cm

Page : 244 , Block Name : Exercise 13.1

Q5 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.

Diameter of hemisphere =Edge Of cube = l
Radius of hemisphere = l/2
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of
base of hemispherical part
=6( Edge ) + 2πr − πr = 6( Edge ) + πr
2 2 2 2 2

2
2 l
= 6l + π × ( )
2

Total surface area of solid = 6l 2
+
πI
2

4

1 2 2
= (24 + π)l unit
4

Page : 244 , Block Name : Exercise 13.1

Q6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its
ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5
mm. Find its surface area.

It can be observed that
Radius (r) of cylindrical part = Radius (r) of hemispherical part
Diameter of the capsule 5
= =
2 2

Length of cylindrical part (h) = Length of the entire capsule - 2 x r
14-5= 9 cm
Surface area of capsule= 2 * CSA of hemispherical part + CSA of cylindrical part

2
= 2 × 2πr + 2πrh

2
5 5
= 4π( ) + 2π ( ) (9)
2 2

= 25π + 45π

2
= 70πmm

22
= 70 ×
7

2
= 220mm

Page : 244 , Block Name : Exercise 13.1

Q7 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,
nd the area of the canvas used for making the tent. Also, nd the cost of the canvas of the
tent at the rate of Rs 500 per m . (Note that the base of the tent will not be covered with
2

canvas.)

Given that,
Height (h) of the cylindrical part = 2.1 m
Diameter of the cylindrical part = 4 m
Radius of the cylindrical part =2 m
Slant height (l) of conical part =2.8 m
Area Of canvas used = CSA Of conical part + CSA Of cylindrical part
= πrl + 2πrh

= π × 2 × 2.8 + 2π × 2 × 2.1

22
= 2π[2.8 + 2 × 2.1] = 2π[2.8 + 4.2] = 2 × × 7
7

2
= 44m

Cost of 1m canvas = Rs 500
2

Cost of 44m canvas = 44*500=22000
2

Therefore, it will cost Rs 22000 for making such a tent .

Page : 245 , Block Name : Exercise 13.1

Q8 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

Given that,
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm
Diameter of the cylindrical part= 1.4 cm
Therefore, radius (r) of the cylindrical part = 0.7 cm
2 2
= √r + h

Slant height (l) of conical part = √(0.7) 2
+ (2.4)
2
= √0.49 + 5.76

= √6.25 = 2.5

Total surface area of the remaining solid will be
= CSA Of cylindrical part + CSA Of con cal part + 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

2
= 10.56 + 5.50 + 1.54 = 17.60cm

The total surface area of the remaining solid to the nearest cm 2
is 18cm
2

Page : 245 , Block Name : Exercise 13.1

Q9 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder,
as shown in Fig. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, nd the
total surface area of the article.

Answer. Given that,
Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm
Height of cylindrical part (h) = 10 cm
Surface area of article CSA of cylindrical part + 2 x CSA of hemispherical part
2
= 2πrh + 2 × 2πr

= 2π × 3.5 × 10 + 2 × 2π × 3.5 × 3.5

= 70π + 49π

= 119π

2
= 17 × 22 = 374cm

Page : 245 , Block Name : Exercise 13.1

Exercise 13.2

Q1 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.

Height (h) of conical part = Radius(r) of conical part = 1
Radius(r) of hemisphere cal part Radius of conical pat (r) = 1 cm
Volume of solid = Volume of conical part + Volume of hemispherical part
1 2 2 3
= πr h + πr
3 3

1 2 2π 2 1 3
= π(1) (1) + π(1) = + = πcm
3 3 3 3

Page : 247 , Block Name : Exercise 13.2

Q2 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 aluminum sheet. The diameter Of the model is 3
cm and its length is 12 cm. if each cone has a height of 2 cm, nd 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.) [Use π =
22
]
7

From the gure, it can be observed that
Height (h1) of each conica part = 2 cm
Height (h2) of cy indrical part =12 — 2 x Height of conical part =12-2*2 =8 cm
Radius (r) of cylindrical part = Radius of conical part = 3/2 cm
Volume of air present in the model = Volume of cylinder + 2 x Volume of cones
2 1 2
= πr h2 + 2 × πr h1
3

2 2
3 1 3 2
= π( ) (8) + 2 × π( ) (2) = 18π + 3π = 21π = 66cm
2 3 2

Page : 247 , Block Name : Exercise 13.2

Q3 A gulab jamun, contains sugar syrup up to about 30% Of its volume. Find approximately
how much syrup would be found in 45 gulab jarmuns, each shaped like a cylinder with two
hemispherical ends with length 5 cm and diameter 2.8 cm (see the given gure).

Answer. It can be observed that

Radius (r) of cylindrical part= Radius (r) of hemispherical part = 2.8/1.4cm
Length of each hemispherical part = Radius of hemispherical part 1 4 cm
Length (h) of cylindrical part = 5 — 2 x Length of hemispherical part
= 5 − 2 × 1.4 = 2.2cm

Volume of one gulab jamun=Vol of cylindrical part + 2* Vol. of hemispherical part

2 2 3 2 4 3
= πr h + 2 × πr = πr h + πr
3 3

2 4 3
= π × (1.4) × 2.2 + π(1.4)
3

22 4 22
= × 1.4 × 1.4 × 2.2 + × × 1.4 × 1.4 × 1.4
7 3 7

3
= 13.552 + 11.498 = 25.05cm

Volume of 45 gulab jamuns= 45 × 25.05 = 1, 127.25cm 3

Volume of sugar syrup = 30% of volume
30
= × 1, 127.25
100

3
= 338.17cm

3
= 338cm

Page : 248 , Block Name : Exercise 13.2

Q4 A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold
pens. The dimensions of the cuboids are 15 cm by 10 cm by 3.5 cm. The radius of each of the
depress ons is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see
the following gure).

Depth (h) of each conical depression = 1.4 cm
Radius (r) Of each conical depression =0.5 cm
Volume Of wood = Volume Of cuboid — 4 x Volume Of cones
1 2
= Ibh − 4 × πr h
3

2
1 22 1
= 15 × 10 × 3.5 − 4 × × × ( ) × 1.4
3 7 2

= 525 − 1.47

3
= 523.53cm

Page : 248 , Block Name : Exercise 13.2

Q5 A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top,
which is open, s 5 cm. It is lled 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 ows out. Find the
number Of lead shots dropped in the vessel.

Height (h)of conical vessel = 8cm
Radius (r ) of conical vessel =5cm
1

Radius (r ) of lead shots = 0.5cm
2

Let n number Of lead shots were dropped in the vessel.
Volume Of water spilled = Volume Of dropped lead shots
1 4 3
× Volume of coneπn × r
4 3 2

1 1 2 4 3
× πr h = n × πr
4 3 1 3 2

2 3
r h = n × 16r
1 2

2 3
5 × 8 = n × 16 × (0.5)

25×8
n = 3
= 100
1
16×( )
2

Hence the number of lead shots dropped in the vessel is 100.

Page : 248 , Block Name : Exercise 13.2

Q6 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 Of iron has approximately 8 g mass. [use π = 3.14]
3

From the gure, it can be observed that
Height (t ) of larger cylinder = 220 cm
1

Radius (r ) of larger cylinder 24 /2 —12 crn
1

Height (h ) Of smaller cylinder = 60 cm
2

Radius (r ) Of smaller cylinder = 8 cm
2

Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder
2 2
= πr h1 + πr h2
1 2

2 2
= π(12) × 220 + π(8) × 60

= π[144 × 220 + 64 × 60]

3
= 35520 × 3.14 = 1, 11, 532.8cm

Mass of 1cm iron = 8 g
3

Mass of 111532.8 cm iron = 111532.8 x 8 = 892262.4 g = 892.262 kg
3

Page : 248 , Block Name : Exercise 13.2

Q7 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 n a right circular cylinder full of water such that
it touches the bottom. Find the volume Of water left in the cylinder, Use = 22/ 7 if the radius of
the cylinder is 60 cm and its height is 180 cm.

Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm
Height (h ) of conical part of solid 120 cm
2

Height (h ) of cylinder =180 cm
1

Radius (r) Of cylinder = 60 cm
Volume of water left = Volume of cylinder — Volume Of solid
= Volume of cylinder − ( Volume of cone + Volume of hemisphere)

1 2
2 2 3
= πr h1 − ( πr h2 + πr )
3 3

1 2
2 2 3
= π(60) (180) − ( π(60) × 120 + π(60) )
3 3

2
= π(60) [(180) − (40 + 40)]

3 3 3
=π(3, 600)(100) = 3, 60, 000πcm = 1131428.57cm = 1.131m

Page : 248 , Block Name : Exercise 13.2

Q8 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 nds its
volume to be 345 3 . cm .Check whether she is correct, taking the above as the inside
3

measurements, and π = 3.14.

Height (h) Of cylindrical part = 8 cm
Radius (r ) of cylindrical part =2/2 =1 cm
2

Radius (r,) spherical part = 8.5/2 = 4.25 cm
Volume of vessel = Volume of sphere + Volume of cylinder
4 3 2
= πr + πr h
3 1 2

3
4 8.5 2
= π( ) + π(1) (8)
3 2

4
= × 3.14 × 76.765625 + 8 × 3.14
3

= 321.392 + 25.12

= 346.512

3
= 346.51cm

Hence ,she is wrong

Page : 248 , Block Name : Exercise 13.2

Exercise 13.3

Q1 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.

Radius (r1 ) of hemisphere = 4.2cm

Answer. Radius (r2 ) of cylinder = 6cm

Let the height of the cylinder be h .

The object formed by recasting the hemisphere will be the same in volume.

Volume of sphere = Volume of cylinder

4 3 2
πr = πr h
3 1 2

4 3 2
π(4.2) = π(6) h
3

4 4.2×4,2×4.2
× = h
3 36

3
h = (1.4) = 2.74cm

Hence, the height of the cylinder so formed will be 2.74 cm.

Page : 251 , Block Name : Exercise 13.3

Q2 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.[Use π = ] 22

7

r1 r2 , and r3 be the radius of metallic spheres, then r1 = 6cm, r2 = 8cm, r3 = 10cm

Let R cm be the radius of a single solid sphere. Since. three metallic spheres are formed from a
single solid sphere. so their volumes are equal.
Since, three metallic spheres are formed from a single solid sphere, so their volumes are equal.
4 3 4 3 4 3 4 3
πr + πr + πr = πR
3 1 3 2 3 3 3

4 3 3 3 4 3
π (r + r + r ) = π(R)
3 1 2 3 3

3 3 3 3
r + r + r = R
1 2 3

3 3 3 3
6 + 8 + 10 = R

3
216 + 512 + 1000 = R

3
R = 1728

R = 12

Page : 251 , Block Name : Exercise 13.3

Q3 A 20 m deep well with diameter 7m 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.
[Use π = ]
22

7

The shape of the well will be cylindrical.
Depth (h) of well =20 m
Area of platform = Length x Breadth = Let height of the platform = H
Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.
Volume of soil from well = Volume of soil used to make such platform
2
π × r × h = Area of platform × Height of platform

2
7
π × ( ) × 20 = 22 × 14 × H
2

22 49 20 5
∴ H = × × = m = 2.5m
7 4 22×14 2

Therefore, the height of such platform will be 2.5m

Page : 251 , Block Name : Exercise 13.3

Q4 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.

The shape of the well will be cylindrical.
Depth (h ) of well = 14 m
1

Radius (r ) of the circular end of well= 3/2 m
1

Width of embankment = 4 m
From the gure, it can be observed that our ernbankment will be in a cylindrical shape having
outer radius (r ) as 4 + = m and inner radius (r ) as 3/2 m
2
3

2
11

2
1

Let the height of ernbankment be h2.
Volume of soil dug from well = Volume of earth used to form embankment
2 2 2
π × r × h1 = π × (r − r ) × h2
1 2 1

2 2 2
3 11 3
π × ( ) × 14 = π × [( ) − ( ) ] × h
2 2 2

9 112
× 14 = × h
4 4

9
h = = 1.125m
8

Therefore, the height of the embankment will be 1.125 m

Page : 251 , Block Name : Exercise 13.3

Q5 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 lled 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 lled
with ice cream.

Answer. Height (h ) of cylindrical container =15 cm
1

Radius (r ) of circular end of container= 6 cm
1

Radius (r ) Of circular end Of ice-cream cone = 6/2=3 cm
2

Height (h ) Of conical part Of ice-cream cone = 12 cm
2

Let ry ice-cream cones be lled with ice-cream of the container.
Volume of ice-cream in cylinder = n x (Volume of 1 ice-cream cone + Volume of hemispherical
shape on the top)
2 1 2 2 3
πr h1 = n ( πr h2 + πr )
1 3 2 3 2

2
6 ×15 3
n = × (3)
6

36×15×3
n =
108+54

n = 10

Therefore, 10 ice cream cones can be lled with the ice-cream in the container.

Page : 251 , Block Name : Exercise 13.3

Q6 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to fo m
a cuboid of dimensions 5 cm x 10 cm x 3.5 cm?

Coins are cylindrical in shape.
Height (h ) of cy indrical coins = 2 mm = 0.2 cm
1

Radius (r) of circular end of coins = 1.75/2= 0.875 cm
Let ry coins be melted to form the required cuboids
Volume of n coins Volume of cuboids
2
n × π × r × h1 = l × b × h

2
n × π × (0.875) × 0.2 = 5.5 × 10 × 3.5

5.5×10×3.5×7
n = = 400
(0.875)2 ×0.2×22

Therefore, the number of coins melted to form such a cuboid is 400.

Page : 251 , Block Name : Exercise 13.3

Q7 A cylindrical bucket, 32 cm high and with radius of base 18 cm, is lled 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.

Height (h ) of cylindrical bucket = 32 cm
1

Radius (r ) of circular end of bucket = 18 cm
1

Height (h ) of conical heap = 24 cm
2

Let the radius of the circular end of conical heap be r 2

The volume of sand in the cylindrical bucket will be equal to the volume of sand in
the conical heap.
Volume Of sand in the cylindrical bucket = Volume Of sand in conical heap

2 1 2
π × r × h1 = π × r × h2
1 3 2

2 1 2
π × 18 × 32 = π × r × 24
3 2

2 1 2
π × 18 × 32 = π × r × 24
3 2

2
2 3×18 ×32 2
r = = 18 × 4
2 24

r2 = 18 × 2 = 36cm

Slant height= √36 2
+ 24
2
= √12
2
× (3
2 2
+ 2 ) = 12√13cm

Therefore the radius and slant height of the conical heap are 36 cm and 12√13cm respectively

Page : 252 , Block Name : Exercise 13.3

Q8 Water in canal, 6 m wide and 1.5 m deep, is owing with a speed of 10 km/h. how much
area will it irrigate in 30 minutes, if 8 of standing water is needed?

Consider an area of cross-section of canal as ABCD.
Area of cross-section = 6 x 1.5 = = 9m 2

Speed Of water = 10 km/h = metre /min 10000

60

Volume of water that ows in 1 minute from canal = 9 × 10000

60
= 1500m
3

Volume of water that ows in 30 minutes from canal 30 x 1500 = 45000m 3

Let the irrigated area be A. Volume of water irrigating the required area will be equal to the
volume Of water that owed n 30 minutes from the canal.
Vol. of water owing in 30 minutes from canal = Vol. of water irrigating the reqd. area
2
A = 562500m

Therefore, area irrigated in 30 minutes is 562500m 2

Page : 252 , Block Name : Exercise 13.3

Q9 A farmer connects a pipe of internal diameter 20 cm form a canal into a cylindrical tank in
her eld, which is 10 m in diameter and 2 m deep. If water ows through the pipe at the rate
Of 3 km/h, in how much time will the tank be lled?

Answer. Consider an area of cross-section of pipe as shown in the gure.

Radius (r ) of circular end of pipe=20/200 =0.1 m
1

Area of cross section = π × r 2

1
= π × (0.1)
2
= 0.01πm
2

Speed of water = 3km/h = 3000

60
= 50metre/min

Volume of water that ows in 1 minute from pipe = 50 × 0.01π = 0.5πm 3

Volume of water that ows in t minutes from pipe = t × 0.5πm 3

Radius (r ) of circular end of cylindrical tank =10/2=5
2

Depth (h ) of cylindrical tank = 2 m
2

Let the tank be lled completely in t minutes.
Volume of water lled in tank in t minutes is equal to the volume of water owed in t minutes
from the pipe.

Volume of water that ows in t minutes from pipe = Volume of water in tank
2
t × 0.5π = π × (r2 ) × h2

2
t × 0.5 = 5 × 2

t = 100

Therefore, the cylindrical tank will be lled in 100 minutes.

Page : 252 , Block Name : Exercise 13.3

Exercise 13.4

Q1 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

Radius (r ) of upper base of glass =4/2 =2 cm
1

Radius (r ) of upper base of glass =2/2 = 1 cm
2

Capacity of glass = Volume of frustum of Cone

1 2 2
= πh (r + r + r1 r2 )
3 1 2

1 2 2
= πh [(2) + (1) + (2)(1)]
3

1 22
= × × 14[4 + 1 + 2]
3 7

308 2 3
= = 102 cm
3 3

Therefore, the capacity of the glass is 102 2

3
cm
3

Page : 257 , Block Name : Exercise 13.4

Q2 The slant height of a frustum of a cone is 4 cm md the perimeters (circumference) of its
circular ends are 18 cm and 6 cm. nd the curved surface area of the frustum.

Perimeter of upper circular end of frustum = 18
2πr1 = 18

9
r1 =
π

Perimeter of lower end of frustum = 6
2πr2 = 6
3
r2 =
π

Slant height (l) of frustum =4
CSA of frustum = π (r + r ) 1 2

9 3
= π( + )4
π π

= 12 × 4

2
= 48cm

Therefore the curved surface area of the frustum is 48 cm 2

Page : 257 , Block Name : Exercise 13.4

Q3 A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the gure given
below). 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, nd the area of material use for making it.

Radius (r ) at upper circular end = 4 cm
2

Radius (r ) at lower circular end = 10 cm
1

Slant height (l) of frustum = 15 cm
Area of material used for making the fez = CSA of frustum + Area of upper circular end
2
= π (r1 + r2 ) l + πr
2

2
= π(10 + 4)15 + π(4)

= π(14)15 + 16π

226×22
= 210π + 16π =
7

2 2
= 710 cm
7

Therefore, the area of the material used for making it is 710 2

7
cm
2

Page : 257 , Block Name : Exercise 13.4

Q4 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 ll the container,at the rate of Rs. 20 per litre.
Also nd the cost of metal sheet used to make the container, if it costs Rs.8 per 100cm [Take
2

π = 3.14]

Radius (r ) of upper end of container = 20 cm
1

Radius (r ) Of lower end Of container =8 cm
2

Height (h) Of container = 16 cm
Slant height (l) of frustum = = √(r
2
2
1 − r2 ) + h

2 2
= √(20 − 8) + (16)

2 2
= √(12) + (16) = √144 + 256

= 20cm

Capacity of container = Volume of frustum
1 2 2
= πh (r + r + r1 r2 )
3 1 2

1 2 2
= × 3.14 × 16 × [(20) + (8) + (20)(8)]
3

1
= × 3.14 × 16(400 + 64 + 160)
3

1
= × 3.14 × 16 × 624
3

3
= 10449.92cm

= 10.45 litres.

Cost of 1 litre milk= Rs 20
Cost of 10.45 litre milk = 10.45 x 20 =Rs 209
Area of metal sheet used to make the container
2
= π (r1 + r2 ) l + πr
2

2
= π(20 + 8)20 + π(8)

2
= 560π + 64π = 624πcm

2
cost of 100cm metal sheet = Rs8
624×3.14×8
=
Cost of 624πcm metal sheet2 100

= 156.75

Therefore, the cost of the milk which can completely ll the container is Rs 209 and the cost of
metal sheet used to make the container is Rs 156.75.

Page : 257 , Block Name : Exercise 13.4

Q5 A metallic right circular cone 20 cm high and whose vertical angle s 600 is cut into two
parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is
drawn into a wire of diameter 1/16 cm nd the length of the wire .

In △AEG,

EG ∘
= tan 30
AG

10 10√3
EG = cm =
√3 3

In △ABD ,

BD ∘
= tan 30
AD

20 20√3
BD = = cm
√3 3

10√3
Radius (r ) of upper end of frustum =
1
3
cm

Radius (r ) of lower end of container =
20√3
2 cm
3

Height (h) of container=10 cm
Volume of frustum = πh (r + r 1

3
2

1
2

2
+ r1 r2 )

2 2

1 ⎡ 10√3 20√3 (10√3)(20√3) ⎤
= × π × 10 ( ) + ( ) +
3 ⎣ 3 3 3 × 3 ⎦

2
10 100 400 200 (10√3)(20√3)
= π[ + + ] + ]
3 3 3 3 3 × 3

10 22 700 22000
3
= × × = cm
3 7 3 9

Radius (r) of wire = 1

16
×
1

2
=
1

32
cm

Let the length of wire be l.
Volume of wire= Area of cross section * Length
2
= (πr ) (l)

2
1
= π( ) × l
32

Volume of frustum = Volume of wire

2
22000 22 1
= × ( ) × l
9 7 32

7000
× 1024 = l
9

l = 796444, 44cm

= 7964.44 metres

Page : 257 , Block Name : Exercise 13.4

Exercise 13.5

Q1 A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and
diameter 10 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

It can be observed that 1 round of wire will cover 3 mm height of cylinder.
Number of rounds =
Height of cylinder
=
Diameter of wire

12
= = 40 rounds
0.3

Length of wire required in 1 round = Circumference of base of cylinder
= 2πr = 2π × 5 = 10π

Length of wire required in 40 rounds = 40 × 10n
400×22 8800
= =
7 7

= 1257.14cm = 12.57m

Radius of wire = 0.3/2 = 0.15 cm
Volume of wire = Area of cross -section of wire * length of wire
2
= n(0.15) × 1257.14

3
= 88.898cm

Mass= Vol * Density

= 88.898 × 8.88

= 789.41gm

Page : 258 , Block Name : Exercise 13.5

Q2 A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve
about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose
value of n as found appropriate.)

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is
shown in the gure.
Hypotenuse AC = √3 2
+ 4
2

= √25 = 5cm

Area of △ABC = 1

2
× AB × AC

1 1
× AC × OB = × 4 × 3
2 2

1
× 5 × OB = 6
2

12
OB = = 2.4cm
5

Volume Of double cone Volume Of cone 1 + Volume Of cone 2
1 2 1 2
= πr h1 + πr h2
3 3

1 2 1 2
= πr (h1 + h2 ) = πr (OA + OC)
3 3

1 2
= × 3.14 × (2.4) (5)
3

3
= 30.14cm

Surface area of double cone Surface area of cone 1 + Surface area of cone 2

= πr/ + πr/
1 2

= πr[4 + 3] = 3.14 × 2.4 × 7

2
= 52.75cm

Page : 258 , Block Name : Exercise 13.5

Q3 A cistern, internally measuring 150 cm x 120 cm x 110 cm, has 129600 cm of water in it.
3

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
over owing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Answer. Volume of cistern 150 × 120 × 110
3
= 1980000cm

Volume to be lled in cistern = 1980000 − 129600
3
= 1850400cm

Let n numbers of porous bricks were placed in the cistern
= n × 22.5 × 7.5 × 6.5
3
= 1850400cm

Let n numbers of porous bricks were placed in the cistern.
Volume of n bricks = n × 22.5 × 7.5 × 6.5
= 1096.875n

As each brick absorbs one - seventeenth of its volume therefore volume absorbed by these
bricks = n
(1096.875)
17
n
1850400 + (1096.875) = (1096.875)n
17

16n
1850400 = (1096.875)
17

n = 1792.41

Therefore ,1792 bricks were placed in the cistern .

Page : 258 , Block Name : Exercise 13.5

Q4 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 7280 km , show that the total rainfall was approximately equivalent to the
2

addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer. Volume of rainfall
10
= 7280 ×
100×1000

3
= 0.7280km

Volume of three rivers
75 3 3
= (3 × 1072 × × ) km
1000 1000

1
= 0.7236km

Hence. the two are approximately equivalent.

Page : 258 , Block Name : Exercise 13.5

Q5 An oil funnel made of tn 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 crn, nd the area Of the tin sheet required to make
the funnel (see the given gure).

Radius (r ) of upper circular end of frustum part =
1
18

2
= 9cm

Radius (r ) of lower circular end of frustum part = Radius Of circular end Of cylindrica
2

part = 8 / 2=4 cm
Height (h ) of frustum part = 22 − 10 = 12cm
1

Height (h ) of cylindrical part =10 cm
2

Slant height (l) Of frustum part = √(r
2 2 2 2
1 − r2 ) + h = √(9 − 4) + (12) = 13cm

Area of tin sheet required = CSA Of frustum part + CSA Of cylindrical part
= π (r1 + r2 ) l + 2πr2 h2

22 22
= × (9 + 4) × 13 + 2 × × 4 × 10
7 7

22 22×249
= [169 + 80] =
7 7

4 2
= 782 cm
7

Page : 258 , Block Name : Exercise 13.5

Q6 Derive the formula for the curved surface area and total surface area of the frustum of cone.
given to you in Section 13.5, using the symbols as explained.

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r and r be the 1 2

radii of the ends of the frustum of the cone and h be the height of the frustum of the cone,
In △ABG and △ADF, DFI IBG
∴ △ABG ∼ △ADF

DF AF AD
= =
BG AG AB

r2 h1 −h l1 −1
= =
r1 h1 l1

r2 h 1
= 1 − = 1 −
r1 h1 l1

l r2
1 − =
l1 r1

l1 r1
= 1 −
l1 r1 −r2

l1 r1
=
l r1 −r2

r1
l1 =
r1 −r2

CSA of frustum DECB= CSA of cone ABC - CSA cone ADE
= πr1 l1 − πr2 (l1 − l)

lr1 r1 l
= πr1 ( ) − πr2 [ − l]
r1 − r2 r1 − r2

2
πr l r1 l − r1 l + r2 l
1
= − πr2 ( )
r1 − r2 r1 − r2

2 2
πr l πr l
1 2
= −
r1 − r2 r1 − r2

2 2
r − r
1 2
= πl [ ]
r1 − r2

CSA of frustum = π (r 1 + r2 ) l

Total surface area of frustum = CSA of frustum + Area of upper circular end

+ Area of lower circular end

2 2
= π (r1 + r2 ) I + πr + πv
2 1

2 2
= π [(r1 + r2 ) l + r + r ]
1 2

Page : 258 , Block Name : Exercise 13.5

Q7 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5,
using the symbols as explained.

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let rt and r2 be the
radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.
In △ABG and △ADF , DF ∥BG

∴ △ABG ∼ ΔADF

DF AF AD
= =
BG AG AB

r2 h1 −h t1 −l
= =
r1 h1 l1

r2 h I
= 1 − = 1 −
r1 h1 l1

h r2
1 − =
h1 r1

h r2 r1 −r2
= 1 − =
h1 r1 r1

h1 r1
=
h r1 −r2

r1 h
h1 =
r1 −r2

Volume of frustum of cone — Volume of cone ABC — Volume of cone ADE

1 1
2 2
= πr h1 − πr (h1 − h)
1 2
3 3
π
2 2
= [r h1 − r (h1 − h)]
1 2
3

π hr1 hr1
2 2
= [r ( ) − r ( − h)]
1 2
3 r1 − r2 r1 − r2

3
π hr hr1 − hr1 + hr2
1 2
= [( ) − r ( )]
2
3 r1 − r2 r1 − r2

3 3
π hr hr
1 2
= [ − ]
3 r1 − r2 r1 − r2

3 3
π r − r
1 2
= h[ ]
3 r1 − r2

2 2
π (r1 − r2 ) (r + r + r1 r2 )]
1 2
= h[ ]
3 r1 − r2

1
2 2
= πh [r + r + r1 r2 ]
1 2
3

Page : 258 , Block Name : Exercise 13.5