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# NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes

NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 9 Mathematics exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 13 - Surface Areas and Volumes.

All our solutions for Chapter 13 - Surface Areas and Volumes are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

Exercise/Page

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.1

Solution 1

Length of box = 1.5 m
Breadth of box = 1.25 m
Depth of box = 65 cm = 0.65 m

(i) The box is open at the top.
Area of sheet required = 2bh + 2lh + lb
= [2  1.25 0.65 + 2 1.5 0.65 + 1.5 1.25] m2
= (1.625 + 1.95 + 1.875)  = 5.45
(ii) Cost of sheet of area 1 = Rs 20      Cost of sheet of area 5.45  = Rs (5.45 20) = Rs 109

Solution 2

Length of room = 5 m
Breadth of room = 4 m
Height of room = 3 m
Area to be white washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
= [2  3 + 2  3 + 5  4]
= (30 + 24 + 20)
= 74

Cost of white washing 1  area = Rs 7.50
Cost of white washing 74  area = Rs (74 7.50) = Rs 555

Solution 3

Let length, breadth and height of rectangular hall be l, b and h respectively.
Area of four walls = 2lh + 2bh = 2(l + b) h
Perimeter of floor of hall = 2(l + b) = 250 m
Area of four walls = 2(l + b) h = 250h
Cost of painting 1 area = Rs 10
Cost of painting 250h  area = Rs (250h  10) = Rs 2500h
It is given that the cost of paining the walls is Rs 15000.
15000 = 2500h
h = 6
Thus, the height of hall is 6 m.

Solution 4

Total surface area of one brick = 2(lb + bh + lh)
= [2(22.5 10 + 10  7.5 + 22.5  7.5)]
= 2(225 + 75 + 168.75)
= (2  468.75)
= 937.5
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n
Area that can be painted by the container = 9.375 m2 = 93750
93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

Solution 5

Edge of the cubical box = 10 cm

Length of the cuboidal box = 12.5 cm
Breadth of the cuboidal box = 10 cm
Height of the cuboidal box = 8 cm

Lateral surface area of cubical box =  =
Lateral surface area of cuboidal box = 2[lh + bh]
= [2(12.5  8 + 10  8)]
= 360
The lateral surface area of cubical box is greater than lateral surface area of cuboidal box.

Lateral surface area of cubical box - lateral surface area of cuboidal box = 400  - 360  = 40

Thus, the lateral surface area of cubical box is greater than that of cuboidal box by 40 cm2.

(ii)    Total surface area of cubical box =  = 600 cm2
Total surface area of cuboidal box = 2[lh + bh + lb]
= [2(12.5  8 + 10 8 + 12.5 10]

The total surface area of cubical box is smaller than that of cuboidal box

Total surface area of cuboidal box - total surface area of cubical box =  - =

Thus, the total surface area of cubical box is smaller than that of cuboidal box by

Solution 6

(i)    Length of green house = 30 cm
Breadth of green house = 25 cm
Height of green house = 25 cm

Total surface area of green house = 2[lb + lh + bh]
= [2(30  25 + 30  25 + 25  25)]
= [2(750 + 750 + 625)]
= (2  2125)
= 4250

Thus, the area of the glass is 4250 .

(ii)    Total length of tape = 4(l + b + h)
= [4(30 + 25 + 25)] cm
= 320 cm

Therefore, 320 cm tape is needed for all the 12 edges.

Solution 7

Length of bigger box = 25 cm
Breadth of bigger box = 20 cm
Height of bigger box = 5 cm

Total surface area of bigger box = 2(lb + lh + bh)
= [2(25  20 + 25  5 + 20  5)]
= [2(500 + 125 + 100)] cm2 = 1450

Extra area required for overlapping = = 72.5

Considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5)  =1522.5

Area of cardboard sheet required for 250 such bigger box
= (1522.5 250)  = 380625

Total surface area of smaller box = [2(15 12 + 15  5 + 12  5] = [2(180 + 75 + 60)]  = (2  315)  = 630

Extra area required for overlapping =  = 31.5

Considering all overlaps, total surface area of 1 smaller box
= (630 + 31.5)  = 661.5

Area of cardboard sheet required for 250 smaller box
= (250 661.5)  = 165375

Total cardboard sheet required = (380625 + 165375)  = 546000

Cost of 1000  cardboard sheet = Rs 4
Cost of 546000  cardboard sheet = Rs = Rs 2184

So, cost of cardboard sheet required for 250 boxes of each kind will be Rs 2184.

Solution 8

Length of shelter = 4 m
Breadth of shelter = 3 m
Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4  2.5 + 3  2.5) + 4  3]
= [2(10 + 7.5) + 12]
= 47

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.2

Solution 1

Height of the cylinder = 14 cm
Let diameter of cylinder be d and the radius of its base be r.
Curved surface area of cylinder = 88

Thus, the diameter of the base of the cylinder is 2 cm.

Solution 2

Height (h) of cylindrical tank = 1 m.
Base radius (r) of cylindrical tank =  = 70 cm = 0.7 m
Area of sheet required = total surface area of tank =
So, it will require 7.48 area of sheet.

Solution 3

Inner radius  of cylindrical pipe = 2 cm
Outer radius  of cylindrical pipe = 2.2 cm
Height (h) of cylindrical pipe = length of cylindrical pipe = 77 cm

(i) CSA of inner surface of pipe =

(ii) CSA of outer surface of pipe =
=       = 1064.8   (iii) Total surface area of pipe =     CSA of inner surface + CSA of outer surface+ area of both circular ends of pipe

Thus, the total surface area of cylindrical pipe is 2038.08 .

Solution 4

The roller is cylindrical.
Height of the roller = length of roller = 120 cm
Radius of the circular end of the roller =
CSA of roller =
Area of field = 500  CSA of roller = (500 31680) = 15840000
= 1584

Solution 5

Height of the pillar = 3.5 m
Radius of the circular end of the pillar = cm = 25 cm  = 0.25 m
CSA of pillar =  =
Cost of painting 1  area = Rs 12.50
Cost of painting 5.5  area = Rs (5.5 12.50) = Rs 68.75

Thus, the cost of painting the CSA of pillar is Rs 68.75.

Solution 6

Let the height of the cylinder be h.
Radius of the base of the cylinder = 0.7 m
CSA of cylinder = 4.4
= 4.4
h = 1 m
Thus, the height of the cylinder is 1 m.

Solution 7

Inner radius (r) of circular well
Depth (h) of circular well = 10 m  (i) Inner curved surface area =

= (44  0.25  10)
= 110

(ii) Cost of plastering 1  area = Rs 40
Cost of plastering 110  area = Rs (110  40) = Rs 4400

Solution 8

Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe =  cm = 2.5 cm = 0.025 m
CSA of cylindrical pipe =   = 4.4
Thus, the area of radiating surface of the system is 4.4 .

Solution 9

Height (h) cylindrical tank = 4.5 m
Radius (r) of circular end of cylindrical tank =m = 2.1m
(i)    Lateral or curved surface area of tank =
=
= 59.4 m2

(ii)    Total surface area of tank = 2 (r + h)
=
= 87.12 m2

Let A m2 steel sheet be actually used in making the tank.

Thus, 95.04  steel was used in actual while making the tank.

Solution 10

Height of frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
Radius of the circular end of frame of lampshade = cm = 10cm
Cloth required for covering the lampshade =

= 2200

Thus, for covering the lampshade 2200  cloth will be required.

Solution 11

Radius of circular end of cylindrical penholder = 3 cm
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of      penholder =  +
Area of cardboard sheet used by 1 competitor =
Area of cardboard sheet used by 35 competitors
= 7920 cm2
Thus, 7920  cardboard sheet will be bought for the competition.

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.3

Solution 1

Radius of base of cone = cm = 5.25 cm
Slant height of cone = 10 cm
CSA of cone =  =
Thus, the curved surface area of cone is 165 .

Solution 2

Radius of base of cone = m = 12 cm
Slant height of cone = 21 m
Total surface area of cone = (r + l)

Solution 3

(i)    Slant height of cone = 14 cm
Let radius of circular end of cone be r.
CSA of cone =

Thus, the radius of circular end of the cone is 7 cm.

(ii)    Total surface area of cone = CSA of cone + Area of base
=
Thus, the total surface area of the cone is 462 .

Solution 4

(i)    Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let slant height of conical tent be l.
l = 26 m .         Thus, the slant height of the conical tent is 26 m.

(ii)    CSA of tent =  =
Cost of 1  canvas = Rs 70
Cost of  canvas = = Rs 137280
Thus, the cost of canvas required to make the tent is Rs 137280.

Solution 5

Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent =  = (3.14  10)  = 188.4

Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(L - 0.2 m)  3] m = 188.4
L - 0.2 m = 62.8 m
L = 63 m

Thus, the length of the tarpaulin sheet will be 63 m.

Solution 6

Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb = = 7 m
CSA of conical tomb =  =
Cost of white-washing 100  area = Rs 210
Cost of white-washing 550  area =Rs = Rs 1155
Thus, the cost of white washing the conical tomb is Rs 1155.

Solution 7

Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 cm
Slant height (l) of conical cap =
CSA of 1 conical cap =
CSA of 10 such conical caps = (10  550)  = 5500

Thus, 5500  sheet will be required to make the 10 caps.

Solution 8

Radius (r) of cone =  = 0.2 m
Height (h) of cone = 1 m
Slant height (l) of cone =
CSA of each cone =  = (3.14  0.2  1.02)  = 0.64056
CSA of 50 such cones = (50  0.64056)  = 32.028

Cost of painting 1  area = Rs 12
Cost of painting 32.028  area = Rs (32.028  12) = Rs 384.336

Thus, it will cost Rs 384.34 (approximately) in painting the 50 hollow cones.

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.4

Solution 1

(i)  Radius of sphere = 10.5 cm
Surface area of sphere =

(ii) Radius of sphere = 5.6 cm
Surface area of sphere =  =

(iii) Radius of sphere = 14 cm
Surface area of sphere =

Solution 2

Surface area of sphere

Surface area of sphere

Surface area of sphere =

Solution 3

Radius of hemisphere = 10 cm
Total surface area of hemisphere

Solution 4

Radius  of spherical balloon = 7 cm
Radius of spherical balloon, when air is pumped into it = 14 cm

Solution 5

Inner radius (r) of hemispherical bowl =
Surface area of hemispherical bowl  =

Cost of tin-plating 100  area = Rs 16
Cost of tin-plating 173.25  area = Rs 27.72 Thus, the cost of tin-plating the inner side of hemispherical bowl is Rs 27.72

Solution 6

Let radius of the sphere be r.
Surface area of the sphere = 154
= 154 cm2

Thus, the radius of the sphere is 3.5 cm.

Solution 7

Let diameter of earth be d. Then, diameter of moon will be .
Surface area of moon =
Surface area of earth =
Required ratio =

Thus, the required ratio of the surface areas is 1:16.

Solution 8

Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl =

Thus, the outer curved surface area of the bowl is 173.25 .

Solution 9

(i)    Surface area of sphere =

(ii)  Height of cylinder = r + r = 2r
CSA of cylinder =

(iii)   Required ratio =

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.5

Solution 1

A matchbox is cuboidal in shape.
Volume of 1 match box = l  h = (4  2.5  1.5)  = 15

Volume of the packet containing 12 such matchboxes = 12 15  =180

Solution 2

Volume of tank = l  h = (6  4.5)  = 135   It is given that:
1  = 1000 litres

Thus, the tank can hold 135000 litres of water.

Solution 3

Let height of cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380
h = 380
10  h = 380
h = 4.75

Thus, the height of the vessel should be 4.75 m.

Solution 4

Length (l) of the cuboidal pit = 8 m
Width (b) of the cuboidal pit = 6 m
Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l  h = (8  3)  = 144

Cost of digging 1  = Rs 30
Cost of digging 144  = Rs (144 30) = Rs 4320

Solution 5

Let the breadth of the tank be 'b' m.
Length (l) of the tank = 2.5 m
Depth (h) of the tank = 10 m Volume of tank = l  h = (2.5  10)  = 25b

Capacity of tank = 25b  = 25000 b litres
25000 b = 50000    (Given)     Thus, the breadth of the tank is 2 m.

Solution 6

Length (l) of the cuboidal tank = 20 m
Breadth (b) of the cuboidal tank = 15 m
Height (h) of the cuboidal tank = 6 m

Capacity of tank = l  bh    = (20  15  6)  = 1800  = 1800000 litres

Water consumed by people of village in 1 day = 4000  150 litres = 600000 litres

Let water of this tank lasts for n days.
Water consumed by all people of village in n days = capacity of tank
600000 = 1800000
n = 3
Thus, the water of tank will last for 3 days.

Solution 7

Length  of the godown = 40 m
Breadth  of the godown = 25 m
Height  of the godown = 10 m

Volume of godown = l1 b1 h1 = (40  25  10)  = 10000

Length  of a wooden crate = 1.5 m
Breadth  of a wooden crate = 1.25 m
Height  of a wooden crate = 0.5 m

Volume of a wooden crate =      = (1.5  1.25  0.5) m3 = 0.9375

Let n wooden crates be stored in the godown.
Volume of n wooden crates = volume of godown
0.9375  n = 10000

Thus, 10666 number of wooden crates can be stored in godown.

Solution 8

Side (a) of the cube = 12 cm
Volume of the cube = a3= (12 cm)3 = 1728

Let the side of each smaller cube be l. Volume of each smaller cube

l = 6 cm
Thus, the side of each smaller cube is 6 cm.
Ratio between surface areas of the cubes =

So, the required ratio between surface areas of the cubes is 4 : 1.

Solution 9

Rate of water flow = 2 km per hour
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min
Thus, in 1 minute 4000  water will fall into the sea.

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.6

Solution 1

Let the radius of the cylindrical vessel be r.
Height (h) of the vessel = 25 cm
Circumference of the vessel = 132 cm
2r = 132 cm               Volume of cylindrical vessel = r2h                                                              Thus, the vessel can hold 34.65 litres of water.

Solution 2

Inner radius (r1) of cylindrical pipe = Outer radius (r2) of cylindrical pipe = Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe = Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = 5720  0.6 g = 3432 g = 3.432 kg

Solution 3

The tin can will be cuboidal in shape. Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l  h = (5  15) cm3 = 300 cm3 Radius (R) of circular end of plastic cylinder = Height (H) of plastic cylinder = 10 cm Capacity of plastic cylinder = R2H  ==385 cm3 Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 - 300) cm3 = 85 cm3

Solution 4

(i)    Height (h) of cylinder = 5 cm
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm2
2rh = 94.2 cm2
(2  3.14  5) cm = 94.2 cm2
r = 3 cm   (ii)    Volume of cylinder = r2h = (3.14  (3)2  5) cm3 = 141.3 cm3

Solution 5

(i)    Cost of painting 1 m2 area = Rs 20
So, Rs 2200 is cost of painting area , i.e, 110 m2 area.
Thus, the inner surface area of the vessel is 110 m2.

(ii)    Let radius of base of vessel be r.
Height (h) of vessel = 10 m
Surface area = 2rh = 110 m2
(iii)    Capacity of vessel = r2h =  = 96.25 m3

Solution 6

Let radius of the circular ends of the cylinder be r.
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3 Total  Surface area of vessel = 2 r(r+h)                                              Thus, 0.4708 m2 of metal sheet would be needed to make the cylindrical vessel.

Solution 7

Radius (r1) of pencil =  = 0.35 cm
Height (h) of pencil = 14 cm
Volume of wood in pencil =                                                                               Volume of Graphite =                             = 0.11 cm3

Solution 8

Radius (r) of cylindrical bowl = cm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cm Volume of soup in 1 bowl = r2h=
Volume of soup in 250 bowls = (250  154) cm3 = 38500 cm3 = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.7

Solution 1

(i)    Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone

(ii)   Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone

Solution 2

(i)    Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height (h) of cone
Volume of cone
Capacity of the conical vessel =  litres= 1.232 litres (ii)    Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Volume of cone
Capacity of the conical vessel = litres =  litres.

Solution 3

Height (h) of cone = 15 cm
Let radius of cone be r.
Volume of cone = 1570 cm3
r = 10 cm

Thus, the radius of the base of the cone is 10 cm.

Solution 4

Height (h) of cone = 9 cm
Let radius of cone be r.
Volume of cone = 48 cm3

Thus, the diameter of the base of the cone is 2r = 8 cm.

Solution 5

Depth (h) of pit = 12 m
Volume of pit =  = 38.5 m3 Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

Solution 6

(i)    Radius of cone =   =14 cm
Let height of cone be h.
Volume of cone = 9856 cm3
h = 48 cm        Thus, the height of the cone is 48 cm.   (ii)   Slant height (l) of cone
Thus, the slant height of the cone is 50 cm.   (iii)    CSA of cone = rl=   = 2200 cm2

Solution 7

When the right angled ABC is revolved about its side 12 cm, a cone of height (h) 12 cm, radius (r) 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone    = 100 cm3
Thus, the volume of cone so formed by the triangle is 100 cm3.

Solution 8

When the right angled ABC is revolved about its side 5 cm, a cone of radius (r) 12 cm, height (h) 5 cm, and slant height (l) 13 cm will be formed.

Volume of cone =  Required ratio

Solution 9

Height (h) of heap = 3 m
Volume of heap=
Slant height (l) =

Area of canvas required = CSA of cone

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.8

Solution 1

(i)    Radius of sphere = 7 cm
Volume of sphere =  (ii)    Radius of sphere = 0.63 m
Volume of sphere =         m3(approximately)

Solution 2

(i)    Radius (r) of ball =
Volume of ball =
Thus, the amount of water displaced is . (ii)    Radius (r) of ball =  = 0.105 m
Volume of ball =
Thus, the amount of water displaced is 0.004851 m3.

Solution 3

Radius (r) of metallic ball =
Volume of metallic ball =

Mass = Density  Volume = (8.9 * 38.808) g = 345.3912 g

Thus, the mass of the ball is approximately 345.39 g.

Solution 4

Let diameter of earth be d. So, radius earth will be  .
Then, diameter of moon will be  . So, radius of moon will be  .
Volume of moon =
Volume of earth =

Thus, the volume of moon is  of volume of earth.

Solution 5

Radius (r) of hemispherical bowl =   = 5.25 cm Volume of hemispherical bowl
= 303.1875 cm3

Capacity of the bowl
= 0.3031875 litre = 0.303 litre (approximately)

Thus, the hemispherical bowl can hold 0.303 litre of milk.

Solution 6

Inner radius (r1) of hemispherical tank  = 1 m
Thickness of hemispherical tank       = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m Volume of iron used to make the tank  =

Solution 7

Let the radius of the sphere be r.
Surface area of sphere = 154 cm2
r2 = 154 cm2
Volume of sphere

Solution 8

(i)    Cost of white washing the dome from inside = Rs 498.96
Cost of white washing 1 m2 area = Rs 2
CSA of inner side of dome =   = 249.48 m2 (ii)    Let inner radius of hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
2r2 = 249.48 m2

Volume of air inside the dome = Volume of the hemispherical dome

= 523.908 m3

Thus, the volume of air inside the dome is approximately 523.9 m3.

Solution 9

(i)    Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere
Volume of 27 solid iron spheres
It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of        this iron sphere will be equal to volume of 27 solid iron spheres.
Radius of the new sphere = r'.
Volume of new sphere
(ii)    Surface area of 1 solid iron sphere of radius r = 4r2
Surface area of iron sphere of radius r' = 4 (r')2    = 4 (3r)2 = 36 r2

Solution 10

Volume of spherical capsule

Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.

## NCERT Solution for Class 9 Mathematics Chapter 13 - Surface Areas and Volumes Page/Excercise 13.9

Solution 1

External length (l) of bookshelf = 85 cm
External breadth (b) of bookshelf = 25 cm
External height (h) of bookshelf = 110 cm
External surface area of shelf while leaving front face of shelf
= lh + 2 (lb + bh)
= [85  110 + 2 (85  25 + 25  110)] cm2
= 19100 cm2
Area of front face = [85  110 - 75  100 + 2 (75  5)] cm2
= 1850 + 750 cm2
= 2600 cm2
Area to be polished = (19100 + 2600) cm2 = 21700 cm2
Cost of polishing 1 cm2 area = Rs 0.20
Cost of polishing 21700 cm2 area = Rs (21700  0.20) = Rs 4340

Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and     30cm  respectively.
Area to be painted in 1 row = 2 (l + h) b + lh
= [2 (75 + 30)  20 + 75  30] cm2
= (4200 + 2250) cm2
= 6450 cm2
Area to be painted in 3 rows = (3  6450) cm2 = 19350 cm2
Cost of painting 1 cm2 area = Rs 0.10
Cost of painting 19350 cm2 area = Rs (19350  0.10) = Rs 1935
Total expense required for polishing and painting the surface of the bookshelf                                              = Rs(4340 + 1935) = Rs 6275

Solution 2

Radius (r) of a wooden sphere = Surface area of a wooden sphere = =

Radius (r') of cylindrical support = 1.5 cm
Height (h) of cylindrical support = 7 cm CSA of cylindrical support = 2r'h
Area of circular end of cylindrical support = r2                                                                                                                       = 7.07 cm2
Area to be painted silver = [8  (1386 - 7.07)] cm2
= (8 1378.93) cm2 = 11031.44 cm2
Cost occurred in painting silver colour = Rs (11031.44 0.25) = Rs 2757.86

Area to painted black = (8 66) cm2 = 528 cm2
Cost occurred in painting black colour = Rs (528 0.05) = Rs 26.40
Total cost occurred in painting = Rs (2757.86 + 26.40) = Rs 2784.26

Solution 3

Let the diameter of the sphere be d.
Radius (r1) of the sphere =
It is given that the diameter of the sphere is decreased by 25%.
New radus (r2) of the sphere = CSA (S1) of the sphere =
CSA (S2) of the new sphere =
Decrease in CSA of sphere = S1 - S2

Percentage decrease in CSA of sphere=                                                       %

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Our NCERT Solutions for Class 9 Mathematics are by our subject matter experts. These NCERT Textbook Solutions will help you to revise the whole chapter, and you can increase your knowledge of Mathematics. If you would like to know more, please get in touch with our counsellor today!

# Text Book Solutions

CBSE IX - Mathematics

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