Lesson: Surface areas and volumes

Exercise 13.1 (9)

Question: 1

2 cubes each of volume are joined end to

64 cm

end. Find the surface area of the resulting cuboid.

Solution:

Let be the edge of the cube.

So,

Volumeof each cube 64

 

a 64

a 4 cm

Fig. Exc_13.1_1

The dimensions of the cuboid formed are

4 cm,

and .

4 cm

8cm

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  

160cm

Thus, the surface area of cuboid is .

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

 

the cylindrical part is same (i.e., 7 cm).

Height of hemisphere = Radius = 7 cm

Height of cylindrical part = 13 − 7 = 6 cm

 

Now, find the inner surface area of the vessel.

Inner surface area CSA of CSA of hemi-

of the vessel cylindrical part spherical part

     

 

     

     

2πrh 2πr 

22 22

2 7 6 2 7 7

7 7

       

 

44 6 7 

572cm

Thus, the inner surface area of the vessel is 572 cm

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

 

Height of conical part = 15.5 − 3.5 = 12 cm

 

Now, find the slant height of the conical part.

2 2

Slant height r h 

   

2 2

3.5 12 

 

 

 

 

 

144

 

49 576





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

     

 

     

     

πrl 2πr 

22 7 25 22 7 7

7 2 2 7 2 2

      

.137 5 77 

.214 5cm

Thus, the total area of the toy is 214.5 cm

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

Radius of hemispherical part

 



Surface area CSA of hemi-

of cubical part spherical part

Total surface

area of solid

Area of base of

hemi-spherical part

 

   



 

   

 

   

 



 

 

 

 



 

 

 

2 2

6 edge 2πr πr  

 

6 edge πr 

 

22 7 7

6 7

7 2 2

   

294 38.5 

332.5cm

Thus, the total surface area of the solid is 332.5 cm

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

Radius of hemisphere



Surface area of CSA of hemi-

cubical part spherical part

Total surface

area of solid

Area of base of

hemi-spherical part

 

   



 

   

 

   

 



 

 

 

 



 

 

 

2 2

6 edge 2πr πr  

 

6 edge πr 

 

6 l π

 

  

 

 

 

πl

6 l

 

 

2 2

24 π l unit

 

Thus, the total surface area of solid is

 

2 2

24 π l unit



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

cylindrical part.

   

Radius r of Radius r of hemi-

cylindrical part spherical part

   



   

   

Diameter of thecapsule



 

Length of

Length of the

cylindrical part h entire capsule

 

 

 

 

14 5 

9cm

Now, find the surface area of the capsule.

Surface area CSA of hemi- CSA of

of capsule spherical part cylindrical part

     

  

     

     

2 2πr 2πrh  

 

5 5

4π 2π 9

2 2

   

 

   

   

25π 45π 

70π mm

70 mm

 

220mm

Thus, the surface area of the capsule is 220 mm

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

. (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

the slant height of conical part.

Height of the cylindrical part = 2.1 m

 

Diameter of the cylindrical part = 4 m

So, radius of the cylindrical part

Diameter 4

2 2

  

Slant height of conical part = 2.8 m

 

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 

2 7

  

44m

Cost of 1 m

canvas

Rs500

Cost of 44 m

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

Solution:

Fig. Exc_13.1_8

Let be the height of both the cylindrical and

conical part.

Height of the conical part = Height of the

 

cylindrical part = 2.4 cm

 

Diameter of the cylindrical part = 1.4 cm

So, radius of the cylindrical part

 

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

 

   



 

   

 

   

 



 

 

 

 



 

 

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  

17.60cm

Thus, the total surface of the remaining solid is 18

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:

Let be the radius.

Radius of cylindrical part = Radius of

 

hemispherical part = 3.5 cm

 

Height of cylindrical part = 10 cm

 

Total surface CSA of CSA of hemi-

area of article cylindrical part spherical part

     

  

     

     

2πrh 2 2πr  

2π 3.5 10 2 2π 3.5 3.5      

70π 49π 

119π

119

 

374 cm

Thus, the total surface area of the article is 374 cm

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 .

Solution:

Fig. Exc_13.2_1

Let be the radius and be the height.

Radius of conical part = Height of conical part

 

= 1 cm

 

Radius of conical part = Radius of hemispherical

 

part = 1 cm

 

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

 

π cm

Thus, the volume of solid is .

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

 

Height of

12 2

cylindrical part h conical part

 

 

  

 

 

 

 

12 2 2  

8cm

Radius of cylindrical part = Radius of conical

 

part



Volume of air contained Volume of Volume of

in the model cylinder cones

     

  

     

     

2 2

2 1

πr h 2 πr h

  

2 2

3 1 3

π 8 2 π 2

2 3 2

   

  

   

   

18π 3π 

 

66cm

Thus, the volume of air contained in the modal is 66

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,

Radius of cylindrical part = Radius of

 

hemispherical part

 

.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-

gulab jamun cylindrical part spherical part

     

  

     

     

2 3

πr h 2 πr

  

     

2 3

22 4 22

1.4 2.2 1.4

7 3 7

     

13.552 11.498 

25.05cm

Volume of 45 gulab jamuns

45 2 055. 

1127.25cm

Volume of sugar syrup

30%of volume

1127 25

100

. 

.1338 7cm

cm ap3 x3 8 pro

Thus, the volume of sugar syrup is 338 cm

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

conical depression.

Depth of each conical depression = 1.4 cm

 

Radius of each conical depression = 0.5 cm

 

Volume Volume of Volume of

of wood cuboid cones

     

  

     

     

lbh 4 πr h

  

1 22 1

15 10 3.5 4 1.4

3 7 2

 

       

 

 

525 1.47 

523.53cm

Thus, the volume of wood is .

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

the radius of conical vessel and be the radius of

each lead shot.

Height of conical vessel = 8 cm

 

Radius of conical vessel = 5 cm

 

Radius of lead shots = 0.5 cm

 

Let be the number of lead shots.

Volume of Volume of dropped

water spilled lead shots

   



   

   

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 

 

5 8 n 16 0.5   

25 8





 



 

 

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

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

 

Radius of larger cylinder

 

12cm

 

Height of smaller cylinder = 60 cm

 

Radius of smaller cylinder = 8 cm

 

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

1cm

8g

Mass of

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

Radius of hemispherical part = Radius of conical

 

part = 60 cm

 

Height of conical part of solid = 120 cm

 

Height of cylinder = 180 cm

 

Radius of cylinder = 60 cm

 

Volume of Volume of Volume

water left cylinder of solid

     

 

     

     

Volume of Volume Volume of

cylinder of cone hemisphere

 

     

  

 

   

 

     

 

2 2 3

1 2

πr h πr h πr

3 3

 

  

 

 

         

2 2 3

1 2

π 60 180 π 60 120 π 60

3 3

 

  

 

 

   

π 60 180 40 40  

 

 

   

π 60 100

360000π cm

1131428.571cm

1.131m

Thus, the volume of water left in the cylinder is

1.131 m

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

. 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

cylindrical part. Let be the radius of spherical

part.

Height of cylindrical part = 8 cm

 

Radius of cylindrical part = 1 cm

 

Radius spherical part = 4.25 cm

 

Volume Volume of Volume of

of vessel sphere cylinder

     

 

     

     

3 2

1 2

πr πr h

 

   

4 8.5

π π 1 8

3 2

 

 

 

 

3.14 76.765625 8 3.14

    

321.392 25.12 

346.512

346.51cm

Thus, the child is wrong, and the volume of the

vessel is 346.51 cm

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:

Let, radius of sphere be and radius of cylinder be

Let the height of the cylinder be .

Volume of Volume of

sphere cylinder

   



   

   

3 2

1 2

πr r h

 

Here,

r 4.2cm,

r 6cm

3 2

1 2

πr πr h



4 π 4.2

3 π 6

 



 

4 π 4.2 4.2 4.2

3 π 6 6

   



  

4 0.7 0.7 1.4

  



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:

Let, radius of sphere be , radius of sphere be

, and radius of sphere be .

Let the radius of the resulting sphere be .

Now,

r 6cm,

r 8cm,

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  

 

r 216 512 1000  

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

 

r m



Area of platform

l b 

22 14 

308m

Let, height of the platform be .

Volume of soil Volume of soil used to

dug from well make the platform

   



   

   

 

πr h Area of platform height H 

πr h

Area of platfrom



22 7 7 20

7 2 2 308

  



  



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

Inner radius of circular well

 

r m



Width of embankment

4m

Outer radius

 

2 1

r r width 

 



Height of circular well

 

h 14m

Let, be the height of the embankment.

Volume of soil Volume of soil used

dug from well to make embankment.

   



   

   

 

πr h Area of embankment height H 

 

2 2

2 1

Area of

embankmen

=π r

 







 

11 3

2 2

 

 

 

 

112

π m



Now, height of embankment

 

πr h

Area of embankment



π 14 4

112 π

 

  

 

 





3 3





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

 

Radius of end of container = 6 cm

 

Radius of end of ice-cream cone = 3 cm

 

Height of ice-cream cone = 12 cm

Let number of ice-cream cones to be filled.

Volume of ice-cream Volumeof cone

in cylinder Volumeof hemishpherical top



   

 

   

   

2 2 3

1 1 2 2 2

1 2

πr h n πr h πr

3 3

 

 

 

 

2 3

15 6

1 2

3 12 3

3 3





   

15 6 6

1 2

3 3 12 3 3 3

3 3

 



      

15 6 6

3 12 2 3 3

 



   

15 6 6

36 18

 





15 6 6

 



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.

Height of each coin

 

H 2mm

0.2cm

Radius of circular end of each coin

 

1.75

r 0.875cm

 

Let, , and be the dimensions of the cuboid.

l 5.5cm

b 10cm

h 3.5cm

Total volume Volume of

of coins melted cuboid formed

   



   

   

n πr H l b h   

 

5.5 10 3.5

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

 

Radius of end of bucket = 18 cm

 

Height of conical heap = 24 cm

 

Let the radius of the conical heap be .

Volume of sand in Volume of sand

the cylindrical bucket in conical heap

   



   

   

2 2

1 1 2 2

πr h πr h



3 18 32

 



3 18 18 32

  



r 18 18 4  

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 

9m

Speed

of water

10km / h

 

 

 

1000m

60min

 

10000

m/min



Volume of water through

lbh

canal in 1 minute

 



 

 

10000

 

1500m / min

Volume of water flowing through canal in 30

minutes

30 1500 

45000m

Volume of water Volume of water

irrigating the flowing through canal

required area in 30 minutes

   

   



   

   

A 8

45000

100





45000 100





A 562500m

Thus, the area is .

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)

Radius of circular pipe

 

0.2

r 0.1m

 

Area of cross-section of the pipe

πr

π 0.1 

0.01π m

Flow rate

of wat

3k / h

 

 



 

1000m

60min.

 

50m/min

Volume of water flowing through the pipe in 1

minute

50 0.01π 

0.5π m

Fig. Exc_13.3_9_(ii)

Radius of cylindrical tank = 5 m

 

Depth of cylindrical tank = 2 m

 

Volume of water flowing through Volume of water

pipein t minutes in tank

   



   

   

2 2

t 0.5π πr h 

5 2

0.5





5 5 2 10

  



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

Radius of upper circular end

 



2cm

Radius of lower circular end

 



1cm

Height of frustum

14 cm

Capacity Volume of

of glass frustum of cone

   



   

   

2 2

1 2 1 2

πh(r r rr )

  

 

1 22

14 4 1 2

3 7

    

22 14

  

308



102 cm



Thus, the capacity of the glass is .

102 cm

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

end.

Perimeter of upper end

2πr

18cm

Radius of upper end

 

2π



Perimeter of lower end

2πr

6cm

Radius of lower end

 

2π



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 

48cm

Thus, the curved surface area of frustum is .

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.)

Radius of upper circular end

 

r 4 cm

Radius of lower circular end

 

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

 

 

 

 

 

 

 

 

1 2 1

π r r l πr  

 

π 4 10 15 π 4    

210π 16π 

=226π

226

 

710 cm



Hence, the area of the material used for making the

cap is .

710 cm

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).

Solution:

Fig. Exc_13.4_4

Radius of upper end of container

 

r 20cm

Radius of lower end of container

 

r 8cm

Height of container

 

h 16cm

Volumeof milk in Volumeof

thecontainer frustum

   



   

   

2 2

1 2 1 2

πh(r r rr )

   

 

2 2

3.14 16 20 8 20 8

      

 

1 314

16 400 64 160

3 100

     

1 314

16 624

3 100

   

10449.92cm

10.45litres

Cost of milk per litre

Rs20

Therefore, cost of milk

10.45litres

10.45 20 

Rs209

Slant height of frustum

   

1 2

l r r h  

 

20 8 16  

144 256 

20cm

 

1 2 2

Area of sheet required

to make the

π r r l πr

container

 







 





 

π 20 8 20 π 8    

560π 64π 

624 3.14 

1959.36cm

Cost of the metal sheet

8Rs./ 100cm

Cost of metal sheet

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.

Solution:

Fig. Exc_13.4_5

Let, radius of the upper half be and that of lower

half be .

Height of cone

 

h 20cm

In ,

PTM

tan 30

 

TM PM

 

TM r



In ,

POR

tan30

 

OR r

 



So, radius of upper end of frustum

 

r cm



Radius of lower end of container

 

r cm



Height of container

 

h 10cm

Radius of wire

 

16 2







Let, be the length of the wire.

   

Volume of wire Area of cross-section l 

=πr l

Volumeof Volumeof

frustum wire

   



   

   

2 2 2

1 2 1 2

πh(r r rr ) πr l

    

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

 

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

Solution:

Fig. Exc_13.5_1

Height of the cylinder = 12 cm

Radius of the cylinder =

Diameter of cylinder



5cm

Observe that, 1 round of wire will cover 3 mm

height of the cylinder.

Height of cylinder

Number of rounds

Diameter of wire



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

 









 

400

 

8800



.1257 14 cm

12 m.57

Radius of wire

0 3

0 15

. 

12 m.57

 

π 0.15 1257.14 

0.15 0.15 1257.14

   

622.2843



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

(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 as

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

Area of ABC AB BC

   

1 1

AB BC AC BO

2 2

    

AB BC AC BO  

4 3 5 BO  

BO 24 cm

. 

Volume of Volume of Volume of

double cone cone 1 cone 2

     

 

     

     

2 2

1 2

1 1

πr h πr h

3 3

 

 

1 2

πr h h

 

 

πr OA OC

 

   

3.14 2.4 5

  

1 314 24 24

3 100 10 10

    

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  

52.75cm

Thus, the value of surface area is 52.75 cm

Question: 3

A cistern, internally measuring

, has 129600 cm

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

Volume of cistern

150cm 120cm 110cm  

1980000cm

Volume to be filled in cistern

1980000 129600 

1850400cm

Let be the number of bricks were placed in

cistern.

Volumeof nbrick .s n 225 6 5.7.5   

. n1096 875

Each brick absorbs one-seventeenth of its own

volume of water.

So,

   

1850400 1096 1096 n

.875 . 5

87 

 

16n

1850400 1096

.875

 

1850400 17

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

, 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

97280km

97280000000m

Rainfall in the river valley

10cm 

0.1m

Now, find the volume of water.

Volume of water

97280000000 0.1 

9728000000m

Now, find the volume of water of one river.

Volume of water of one river

1072000 75 3  

241200000m

So, volume of water of 3 rivers

3 241200000 

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

part and be the radius of lower circular end of

frustum part.

Radius 9 cm

 



Radius = Radius of circular end of cylindrical

 

part = 4 cm

Height of frustum part

 

22– 10 

12cm

Height of cylindrical part = 10 cm

 

   

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

       

 

169 80

 

22 249





782 cm



Thus, the area of tin sheet is .

782 cm

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

and be the height of the frustum of the cone.

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

 

 

1 1 1

r h l

1 1

r h l

   

That is,

1 1

r l

 

1 1

l r

 

1 2

1 1

l r r

l r





1 1

1 2

l r

l r r





1 2

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

   

  

   

 

   

1 1 1 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

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

and be the height of the frustum of the cone.

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

 

 

1 1 1

r h l

1 1

r h l

   

That is,

1 1

h r

 

1 1

h r

 

1 2

1 1

h r r

h r





1 2

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

r h r h h

 

   

 

2 2

1 2

1 1

1 2 1 2

rh r1

r r h

r r r r 

 

   

   

 

   

   

 

1 1 1

1 2 1 2

r h rh r1

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

r r

 

 













 

2 2

1 2 1 2 1 2

1 2

r r r r r

3 r r

r  

 

 







 

 

 

2 2

1 2 1 2

r r rr

  

Thus, the volume of frustum of cone is

 

2 2

1 2 1 2

r r

rr 