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

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
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 3
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 4
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 5
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 6
    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 7
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 8
 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 9
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.    

Solution 10
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 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 = 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.


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
(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 3
    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 4
    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 5
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 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  = Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
                                                            Ncert Solutions Cbse Class 9 Mathematics Chapter - Surface Areas And Volumes
Solution 7
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 8
(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 9
(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

Solution 10
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.

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%