NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volumes

Get free online access to complete solutions for exercises from your NCERT Maths Class 9 textbook prepared by our experienced Maths teachers. With the NCERT solutions for Class 9 Maths Chapter 13, study the concepts of surface areas and volumes of numerous geometrical shapes. Utilise the model answers by experts for questions from your NCERT textbook while completing homework and when preparing for the exams. Through the Class 9 Maths NCERT solutions, our experts give you clear answers for questions based on shapes such as cuboid, cube, sphere, hemisphere and cone.

You can use TopperLearning’s NCERT Maths book Class 9 solutions to understand how to find the accurate area of a plastic sheet or to learn how to calculate the cost of painting a wall based on the given dimensions. Score more marks in CBSE Class 9 Maths by practising concepts with our detailed and step-wise NCERT solutions for Chapter 13. With conceptual clarity, you can obtain good marks in the Maths exam as well as find answers to real-life situations by applying the concepts you have learned.

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Chapter 13 - Surface Areas and Volumes Exercise Ex. 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 1.25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes0.65 + 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes1.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes0.65 + 1.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes1.25] m2
    = (1.625 + 1.95 + 1.875) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 5.45 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
(ii) Cost of sheet of area 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes= Rs 20
     Cost of sheet of area 5.45 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Rs (5.45 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes20) = 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 3 + 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 3 + 5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 4] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= (30 + 24 + 20) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= 74 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Cost of white washing 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs 7.50
Cost of white washing 74 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs (74 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes7.50) = 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
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Area of four walls = 2(l + b) h = 250h Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Cost of painting 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesarea = Rs 10
Cost of painting 250h Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs (250h Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 10) = Rs 2500h
It is given that the cost of paining the walls is Rs 15000.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes10 + 10 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 7.5 + 22.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 7.5)]Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 = 2(225 + 75 + 168.75) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
          = (2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 468.75) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 = 937.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes   
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes   
Area that can be painted by the container = 9.375 m2 = 93750 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Lateral surface area of cuboidal box = 2[lh + bh]
                            = [2(12.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 8 + 10 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 8)] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                                     = 360 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes - 360 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 40 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 600 cm2
Total surface area of cuboidal box = 2[lh + bh + lb]
= [2(12.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 8 + 10 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes8 + 12.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes10] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes - Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Thus, the total surface area of cubical box is smaller than that of cuboidal box by Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 25 + 30 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 25 + 25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 25)] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= [2(750 + 750 + 625)] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= (2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 2125) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= 4250 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Thus, the area of the glass is 4250 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.

(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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 20 + 25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 5 + 20 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 5)] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    = [2(500 + 125 + 100)] cm2 = 1450 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  

Extra area required for overlapping = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 72.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes           

Considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes =1522.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Area of cardboard sheet required for 250 such bigger box
= (1522.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes250) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 380625 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 

Total surface area of smaller box = [2(15 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes12 + 15 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 5 + 12 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 5] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes= [2(180 + 75 + 60)] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = (2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 315) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 630 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Extra area required for overlapping =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 31.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Considering all overlaps, total surface area of 1 smaller box
= (630 + 31.5) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 661.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Area of cardboard sheet required for 250 smaller box
= (250 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes661.5) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 165375 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Total cardboard sheet required = (380625 + 165375) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 546000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Cost of 1000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cardboard sheet = Rs 4
Cost of 546000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cardboard sheet = Rs Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 2.5 + 3 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 2.5) + 4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 3] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= [2(10 + 7.5) + 12] Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
= 47 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes


Chapter 13 - Surface Areas and Volumes Exercise Ex. 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Thus, the diameter of the base of the cylinder is 2 cm.    

Solution 2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 
    Height of frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
    Radius of the circular end of frame of lampshade = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumescm = 10cm
    Cloth required for covering the lampshade = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

                               Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                                 
                                = 2200 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
     Thus, for covering the lampshade 2200 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cloth will be required.

Solution 3
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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes + Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                  
Area of cardboard sheet used by 1 competitor = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
    Area of cardboard sheet used by 35 competitors

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 7920 cm2
    Thus, 7920 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cardboard sheet will be bought for the competition.


Solution 4
Height (h) of cylindrical tank = 1 m.
    Base radius (r) of cylindrical tank = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 70 cm = 0.7 m

Area of sheet required = total surface area of tank = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes   
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

So, it will require 7.48 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesarea of sheet.

Solution 5
Inner radius Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes of cylindrical pipe = 2 cm
 Outer radius Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes of cylindrical pipe = 2.2 cm
 Height (h) of cylindrical pipe = length of cylindrical pipe = 77 cm
    
(i) CSA of inner surface of pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

(ii) CSA of outer surface of pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes     
= 1064.8 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 
(iii) Total surface area of pipe =    
CSA of inner surface + CSA of outer surface+ area of both circular ends of pipe
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes      
   
 
 


Thus, the total surface area of cylindrical pipe is 2038.08 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.    

Solution 6
The roller is cylindrical.
Height of the roller = length of roller = 120 cm
Radius of the circular end of the roller =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
CSA of roller = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Area of field = 500 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes CSA of roller = (500 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes31680) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes= 15840000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    = 1584 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Solution 7
Height of the pillar = 3.5 m
Radius of the circular end of the pillar =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cm = 25 cm  = 0.25 m
CSA of pillar = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Cost of painting 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs 12.50
Cost of painting 5.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs (5.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes12.50) = Rs 68.75

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

Solution 8
    Let the height of the cylinder be h.
Radius of the base of the cylinder = 0.7 m
    CSA of cylinder = 4.4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 4.4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
    h = 1 m
    Thus, the height of the cylinder is 1 m.

Solution 9
Inner radius (r) of circular well
Depth (h) of circular well = 10 m 
(i) Inner curved surface area = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 

          Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                            
                            = (44 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 0.25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 10) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
            = 110 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
    (ii) Cost of plastering 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs 40                    
    Cost of plastering 110 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs (110 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 40) = Rs 4400

Solution 10
 Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
  Radius (r) of circular end of pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cm = 2.5 cm = 0.025 m
  CSA of cylindrical pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 4.4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Thus, the area of radiating surface of the system is 4.4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.    

Solution 11
Height (h) cylindrical tank = 4.5 m
    Radius (r) of circular end of cylindrical tank =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesm = 2.1m
    (i)    Lateral or curved surface area of tank = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

                                    = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                                          
                                    = 59.4 m2            
    
    (ii)    Total surface area of tank = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes (r + h)

              = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                           
              = 87.12 m2

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

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
    
Thus, 95.04 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes steel was used in actual while making the tank.    

Chapter 13 - Surface Areas and Volumes Exercise Ex. 13.3

Solution 1
Radius of base of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumescm = 5.25 cm
Slant height of cone = 10 cm
CSA of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Thus, the curved surface area of cone is 165 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.

Solution 2
Radius of base of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesm = 12 cm
Slant height of cone = 21 m
Total surface area of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes(r + l)

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 

Solution 3
(i)    Slant height of cone = 14 cm
    Let radius of circular end of cone be r.
    CSA of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
    Thus, the radius of circular end of the cone is 7 cm.

(ii)    Total surface area of cone = CSA of cone + Area of base
                     =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Thus, the total surface area of the cone is 462 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.
    

                      


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.
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
          Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesl = 26 m
.
        Thus, the slant height of the conical tent is 26 m.

    (ii)    CSA of tent = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
        Cost of 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes canvas = Rs 70
        Cost of Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes canvas = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
    CSA of conical tent = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = (3.14 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 10) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 188.4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

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) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 3] m = 188.4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    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 =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 7 m
    CSA of conical tomb = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
    Cost of white-washing 100 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs 210
    Cost of white-washing 550 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area =RsNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 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 =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
    CSA of 1 conical cap = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    CSA of 10 such conical caps = (10 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 550) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 5500 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
    Thus, 5500 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes sheet will be required to make the 10 caps.

Solution 8
Radius (r) of cone =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 0.2 m
    Height (h) of cone = 1 m

    Slant height (l) of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    CSA of each cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = (3.14 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 0.2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 1.02) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 0.64056 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
    CSA of 50 such cones = (50 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 0.64056) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 32.028 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
    
    Cost of painting 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs 12
    Cost of painting 32.028 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs (32.028 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 12) = Rs 384.336
     
    Thus, it will cost Rs 384.34 (approximately) in painting the 50 hollow cones.
 

Chapter 13 - Surface Areas and Volumes Exercise Ex. 13.4

Solution 1
(i)  Radius of sphere = 10.5 cm
     Surface area of sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

(ii) Radius of sphere = 5.6 cm
     Surface area of sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

(iii) Radius of sphere = 14 cm
     Surface area of sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Solution 2

(i)     Radius of sphere Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
        Surface area of sphere Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 

(ii)    Radius of sphere Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
        Surface area of sphere  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes     

(iii)    Radius of sphere Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
        Surface area of sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes   

Solution 3
Radius of hemisphere = 10 cm
Total surface area of hemisphere Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Solution 4
Radius Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes of spherical balloon = 7 cm
Radius Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesof spherical balloon, when air is pumped into it = 14 cm

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    



    

Solution 5
Inner radius (r) of hemispherical bowl = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Surface area of hemispherical bowl  = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes          
Cost of tin-plating 100 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area = Rs 16
Cost of tin-plating 173.25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes area Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes= 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
   Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 154 cm2    

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
Thus, the radius of the sphere is 3.5 cm.

Solution 7
Let diameter of earth be d. Then, diameter of moon will be Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.
    Radius of earth = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
    Radius of moon = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Surface area of moon = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
    Surface area of earth = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
    Required ratio = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
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
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesOuter radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                            
  Thus, the outer curved surface area of the bowl is 173.25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.

Solution 9
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
(i)    Surface area of sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

  (ii)  Height of cylinder = r + r = 2r
        Radius of cylinder = r
        CSA of cylinder =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
(iii)   Required ratio = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  

Chapter 13 - Surface Areas and Volumes Exercise Ex. 13.5

Solution 1
A matchbox is cuboidal in shape.
Volume of 1 match box = l Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes h = (4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 2.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 1.5) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 15 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Volume of the packet containing 12 such matchboxes = 12 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 15 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes =180 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  

Solution 2
Volume of tank = l Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes h = (6 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 4.5) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 135 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
  It is given that:
 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 1000 litres
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes h = 380 
       
 10 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes h = (8 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 3) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 144 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  

Cost of digging 1 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Rs 30
Cost of digging 144 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = Rs (144 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes30) = 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes h = (2.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 10) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 25b Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
Capacity of tank = 25b  = 25000 b litres
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 25000 b = 50000    (Given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 
  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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes bNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesh    = (20 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 15 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 6) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 1800 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 1800000 litres

Water consumed by people of village in 1 day = 4000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 600000 = 1800000
n = 3    
Thus, the water of tank will last for 3 days.

Solution 7
Length (l1) of the godown = 40 m

Breadth (b1) of the godown = 25 m

Height (h1) of the godown = 15 m

Volume of the godown = l1 × b1 × h1 = 40 × 25 × 15 = 15000m3

Length (l2) of a wooden crate = 40 m

Breadth (b2) of a wooden crate = 25 m

Height (h2) of a wooden crate = 15 m

Volume of a wooden crate = l2 × b2 × h2 = 1.5 × 1.25 × 0.5 = 0.9375m3

Let n wooden crates be stored in the godown

Volume of n wooden crates = volume of the godown

0.9375 × n = 15000

n = 16000

Thus, the number of wooden crates that can be stored in the godown is 16000.

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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes l = 6 cm
Thus, the side of each smaller cube is 6 cm.
Ratio between surface areas of the cubes =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
So, the required ratio between surface areas of the cubes is 4 : 1.    

Solution 9
Rate of water flow = 2 km per hour  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Thus, in 1 minute 4000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes water will fall into the sea.    

Chapter 13 - Surface Areas and Volumes Exercise Ex. 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
    2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr = 132 cm
          Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Volume of cylindrical vessel = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2h
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                                                            
Thus, the vessel can hold 34.65 litres of water.
Solution 2
Inner radius (r1) of cylindrical pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Outer radius (r2) of cylindrical pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = 5720 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes h = (5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 15) cm3 = 300 cm3
Radius (R) of circular end of plastic cylinder = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesR2H  =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes=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
        2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesrh = 94.2 cm2
        (2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 3.14 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 5) cm = 94.2 cm2
        r = 3 cm
 
(ii)    Volume of cylinder = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2h = (3.14 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes (3)2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 5) cm3 = 141.3 cm3

Solution 5
(i)    Cost of painting 1 m2 area = Rs 20
        So, Rs 2200 is cost of painting Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesarea , 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 = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesrh = 110 m2
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes        
(iii)    Capacity of vessel = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2h =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 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
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Total  Surface area of vessel = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes r(r+h)
                                         Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes   
Thus, 0.4708 m2 of metal sheet would be needed to make the cylindrical vessel.   
Solution 7
Radius (r1) of pencil = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 0.35 cm
Radius (r2) of graphite = 
                                 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Height (h) of pencil = 14 cm

Volume of wood in pencil = 
                                   Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                    
 
 
Volume of Graphite = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                            = 0.11 cm3
Solution 8
Radius (r) of cylindrical bowl = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumescm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2h= Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Volume of soup in 250 bowls = (250 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 154) cm3 = 38500 cm3 = 38.5 litres

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

Chapter 13 - Surface Areas and Volumes Exercise Ex. 13.7

Solution 1
(i)    Radius (r) of cone = 6 cm
       Height (h) of cone = 7 cm
       Volume of cone Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  

(ii)   Radius (r) of cone = 3.5 cm
       Height (h) of cone = 12 cm
       Volume of cone  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 

Solution 2
(i)    Radius (r) of cone = 7 cm
       Slant height (l) of cone = 25 cm
       Height (h) of cone  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
       Volume of cone  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       Capacity of the conical vessel = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes litres= 1.232 litres
(ii)    Height (h) of cone = 12 cm
        Slant height (l) of cone = 13 cm
        Radius (r) of cone Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
        Volume of cone   Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
        Capacity of the conical vessel = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumeslitres = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes litres.

Solution 3
Height (h) of cone = 15 cm
Let radius of cone be r.
Volume of cone = 1570 cm3
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
 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 = 48Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cm3
  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
 
  Thus, the diameter of the base of the cone is 2r = 8 cm.

Solution 5
Radius (r) of pit =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Depth (h) of pit = 12 m
Volume of pit =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 38.5 m3
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesCapacity of the pit = (38.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 1) kilolitres = 38.5 kilolitres

Solution 6
(i)    Radius of cone =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes =14 cm
       Let height of cone be h.
       Volume of cone = 9856 cm3
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
       h = 48 cm
       Thus, the height of the cone is 48 cm.
 
(ii)   Slant height (l) of cone  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes     
       Thus, the slant height of the cone is 50 cm.
 
(iii)    CSA of cone = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesrl= Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 2200 cm2
Solution 7
When the right angled Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesABC 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  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 100Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cm3
Thus, the volume of cone so formed by the triangle is 100Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes cm3.

Solution 8
When the right angled Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesABC 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 =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Required ratio  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Solution 9
Radius (r) of heapNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Height (h) of heap = 3 m
Volume of heap= Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
Slant height (l) =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Area of canvas required = CSA of cone
    
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Chapter 13 - Surface Areas and Volumes Exercise Ex. 13.8

Solution 1
(i)    Radius of sphere = 7 cm
        Volume of sphere =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
(ii)    Radius of sphere = 0.63 m
        Volume of sphere =Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
        m3(approximately)

Solution 2
Radius (r) of capsuleNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
Volume of spherical capsuleNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                        Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.

Solution 3
(i)    Radius (r) of ball = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       Volume of ball =   Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       Thus, the amount of water displaced is Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes.
(ii)    Radius (r) of ball = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes = 0.105 m
        Volume of ball =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 
        Thus, the amount of water displaced is 0.004851 m3.

Solution 4
    Radius (r) of metallic ball =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Volume of metallic ball =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 

    Mass = Density Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Volume = (8.9 * 38.808) g = 345.3912 g
    
    Thus, the mass of the ball is approximately 345.39 g.

Solution 5
    Let diameter of earth be d. So, radius earth will beNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  .
    Then, diameter of moon will be Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes . So, radius of moon will be Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes .
    Volume of moon =    Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Volume of earth =   Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    Thus, the volume of moon is Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes of volume of earth.

Solution 6
Radius (r) of hemispherical bowl = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 5.25 cm
Volume of hemispherical bowlNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                           Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes                      
                                            = 303.1875 cm3

Capacity of the bowl Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                              = 0.3031875 litre = 0.303 litre (approximately)

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

Solution 7
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  = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                                            Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Solution 8
Let the radius of the sphere be r.
Surface area of sphere = 154 cm2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2 = 154 cm2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes   
 Volume of sphereNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Solution 9
(i)    Cost of white washing the dome from inside = Rs 498.96
       Cost of white washing 1 m2 area = Rs 2
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesCSA of inner side of dome = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  = 249.48 m2
(ii)    Let inner radius of hemispherical dome be r.
        CSA of inner side of dome = 249.48 m2
        2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2 = 249.48 m2
         Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       
        Volume of air inside the dome = Volume of the hemispherical dome
                                                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                                    = 523.908 m3
        
        Thus, the volume of air inside the dome is approximately 523.9 m3.

Solution 10
(i)    Radius of 1 solid iron sphere = r
       Volume of 1 solid iron sphereNcert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes  
       Volume of 27 solid iron spheres  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       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  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
(ii)    Surface area of 1 solid iron sphere of radius r = 4Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2
        Surface area of iron sphere of radius r' = 4Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes (r')2    = 4 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes (3r)2 = 36 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes

Chapter 13 - Surface Areas and Volumes Exercise Ex. 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 110 + 2 (85 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 25 + 25 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 110)] cm2
                                                  = 19100 cm2
    Area of front face = [85 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 110 - 75 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 100 + 2 (75 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 0.20) = Rs 4340    
    
    Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and
    30cm Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes respectively.
    Area to be painted in 1 row = 2 (l + h) b + lh
                                           = [2 (75 + 30) Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 20 + 75 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 30] cm2
                                           = (4200 + 2250) cm2
                                           = 6450 cm2    
    Area to be painted in 3 rows = (3 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 6450) cm2 = 19350 cm2
    Cost of painting 1 cm2 area = Rs 0.10
    Cost of painting 19350 cm2 area = Rs (19350 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Surface area of a wooden sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes=Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
    
Radius (r') of cylindrical support = 1.5 cm
Height (h) of cylindrical support = 7 cm
CSA of cylindrical support = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr'h
                                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Area of circular end of cylindrical support = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumesr2 
                                                         Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                                           = 7.07 cm2
Area to be painted silver = [8 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes (1386 - 7.07)] cm2
                                       = (8 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 1378.93) cm2 = 11031.44 cm2
Cost occurred in painting silver colour = Rs (11031.44 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 0.25) = Rs 2757.86    

Area to painted black = (8 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 66) cm2 = 528 cm2
Cost occurred in painting black colour = Rs (528 Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes 0.05) = Rs 26.40
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And VolumesTotal 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 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
It is given that the diameter of the sphere is decreased by 25%.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes New radus (r2) of the sphere = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
CSA (S1) of the sphere =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes    
CSA (S2) of the new sphere =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Decrease in CSA of sphere = S1 - S2  Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
     
Percentage decrease in CSA of sphere= Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes%