Chapter 13 : Surface Areas and Volumes - Ncert Solutions for Class 9 Maths CBSE

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Chapter 13 - Surface Areas and Volumes Excercise 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  1.25 0.65 + 2 1.5 0.65 + 1.5 1.25] m2
    = (1.625 + 1.95 + 1.875)  = 5.45   
(ii) Cost of sheet of area 1 = Rs 20
     Cost of sheet of area 5.45  = Rs (5.45 20) = Rs 109

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

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

Solution 3
Let length, breadth and height of rectangular hall be l, b and h respectively.
Area of four walls = 2lh + 2bh = 2(l + b) h
Perimeter of floor of hall = 2(l + b) = 250 m
 Area of four walls = 2(l + b) h = 250h
Cost of painting 1 area = Rs 10
Cost of painting 250h  area = Rs (250h  10) = Rs 2500h
It is given that the cost of paining the walls is Rs 15000.
 15000 = 2500h
h = 6
Thus, the height of hall is 6 m.

Solution 4
Total surface area of one brick = 2(lb + bh + lh)
 = [2(22.5 10 + 10  7.5 + 22.5  7.5)]
 = 2(225 + 75 + 168.75)
          = (2  468.75)
 = 937.5    
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n    
Area that can be painted by the container = 9.375 m2 = 93750
 93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

 

Solution 5
Edge of the cubical box = 10 cm
    
Length of the cuboidal box = 12.5 cm
Breadth of the cuboidal box = 10 cm
Height of the cuboidal box = 8 cm

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

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

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

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

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

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

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

Solution 6
(i)    Length of green house = 30 cm
Breadth of green house = 25 cm
Height of green house = 25 cm

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

Thus, the area of the glass is 4250 .

(ii)    Total length of tape = 4(l + b + h)
        = [4(30 + 25 + 25)] cm
        = 320 cm
        
        Therefore, 320 cm tape is needed for all the 12 edges.

Solution 7
Length of bigger box = 25 cm
Breadth of bigger box = 20 cm
Height of bigger box = 5 cm

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

Extra area required for overlapping = = 72.5            

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

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

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

Extra area required for overlapping =  = 31.5

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

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

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

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

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

Solution 8
Length of shelter = 4 m
Breadth of shelter = 3 m
Height of shelter = 2.5 m

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


Chapter 13 - Surface Areas and Volumes Excercise 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
    
   
    
     
Thus, the diameter of the base of the cylinder is 2 cm.    

Solution 2
Height (h) of cylindrical tank = 1 m.
    Base radius (r) of cylindrical tank =  = 70 cm = 0.7 m

Area of sheet required = total surface area of tank = 
   

So, it will require 7.48 area of sheet.

Solution 3
Inner radius  of cylindrical pipe = 2 cm
 Outer radius  of cylindrical pipe = 2.2 cm
 Height (h) of cylindrical pipe = length of cylindrical pipe = 77 cm
    
(i) CSA of inner surface of pipe =

(ii) CSA of outer surface of pipe =

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


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

Solution 4
The roller is cylindrical.
Height of the roller = length of roller = 120 cm
Radius of the circular end of the roller =  
CSA of roller =
Area of field = 500  CSA of roller = (500 31680) = 15840000
    = 1584

Solution 5
Height of the pillar = 3.5 m
Radius of the circular end of the pillar = cm = 25 cm  = 0.25 m
CSA of pillar =  =
Cost of painting 1  area = Rs 12.50
Cost of painting 5.5  area = Rs (5.5 12.50) = Rs 68.75

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

Solution 6
    Let the height of the cylinder be h.
Radius of the base of the cylinder = 0.7 m
    CSA of cylinder = 4.4
     = 4.4
    
    h = 1 m
    Thus, the height of the cylinder is 1 m.

Solution 7
Inner radius (r) of circular well
Depth (h) of circular well = 10 m 
(i) Inner curved surface area =
 

                                     
                            = (44  0.25  10)
            = 110 
    
    (ii) Cost of plastering 1  area = Rs 40                    
    Cost of plastering 110  area = Rs (110  40) = Rs 4400

Solution 8
 Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
  Radius (r) of circular end of pipe =  cm = 2.5 cm = 0.025 m
  CSA of cylindrical pipe =   = 4.4
    Thus, the area of radiating surface of the system is 4.4 .    

Solution 9
Height (h) cylindrical tank = 4.5 m
    Radius (r) of circular end of cylindrical tank =m = 2.1m
    (i)    Lateral or curved surface area of tank =

                                    =                                           
                                    = 59.4 m2            
    
    (ii)    Total surface area of tank = 2 (r + h)

              =                           
              = 87.12 m2

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

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

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

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

Solution 11
Radius of circular end of cylindrical penholder = 3 cm
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of     
penholder =  +
                  
Area of cardboard sheet used by 1 competitor =  
    Area of cardboard sheet used by 35 competitors

 = 7920 cm2
    Thus, 7920  cardboard sheet will be bought for the competition.


Chapter 13 - Surface Areas and Volumes Excercise Ex. 13.3

Solution 1
Radius of base of cone = cm = 5.25 cm
Slant height of cone = 10 cm
CSA of cone =  =
Thus, the curved surface area of cone is 165 .

Solution 2
Radius of base of cone = m = 12 cm
Slant height of cone = 21 m
Total surface area of cone = (r + l)

 

Solution 3
(i)    Slant height of cone = 14 cm
    Let radius of circular end of cone be r.
    CSA of cone =

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

(ii)    Total surface area of cone = CSA of cone + Area of base
                     =

Thus, the total surface area of the cone is 462 .
    

                      


Solution 4
(i)    Height (h) of conical tent = 10 m
        Radius (r) of conical tent = 24 m
        Let slant height of conical tent be l.
          
          l = 26 m
.
        Thus, the slant height of the conical tent is 26 m.

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

Solution 5
Height (h) of conical tent = 8 m
    Radius (r) of base of tent = 6 m
    Slant height (l) of tent =     
    CSA of conical tent =  = (3.14  10)  = 188.4

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

Solution 6
Slant height (l) of conical tomb = 25 m
    Base radius (r) of tomb = = 7 m
    CSA of conical tomb =  = 
    Cost of white-washing 100  area = Rs 210
    Cost of white-washing 550  area =Rs = Rs 1155
    Thus, the cost of white washing the conical tomb is Rs 1155.

Solution 7
Radius (r) of conical cap = 7 cm
    Height (h) of conical cap = 24 cm
    Slant height (l) of conical cap =  
    CSA of 1 conical cap = 
    CSA of 10 such conical caps = (10  550)  = 5500
    
    Thus, 5500  sheet will be required to make the 10 caps.

Solution 8
Radius (r) of cone =  = 0.2 m
    Height (h) of cone = 1 m

    Slant height (l) of cone = 
    CSA of each cone =  = (3.14  0.2  1.02)  = 0.64056   
    CSA of 50 such cones = (50  0.64056)  = 32.028   
    
    Cost of painting 1  area = Rs 12
    Cost of painting 32.028  area = Rs (32.028  12) = Rs 384.336
     
    Thus, it will cost Rs 384.34 (approximately) in painting the 50 hollow cones.
 

Chapter 13 - Surface Areas and Volumes Excercise Ex. 13.4

Solution 1
(i)  Radius of sphere = 10.5 cm
     Surface area of sphere =

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

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

Solution 2

(i)     Radius of sphere  
        Surface area of sphere  

(ii)    Radius of sphere
        Surface area of sphere       

(iii)    Radius of sphere
        Surface area of sphere =    

Solution 3
Radius of hemisphere = 10 cm
Total surface area of hemisphere

Solution 4
Radius  of spherical balloon = 7 cm
Radius of spherical balloon, when air is pumped into it = 14 cm

    



    

Solution 5
Inner radius (r) of hemispherical bowl =
Surface area of hemispherical bowl  =

          
Cost of tin-plating 100  area = Rs 16
Cost of tin-plating 173.25  area = Rs 27.72
Thus, the cost of tin-plating the inner side of hemispherical bowl is Rs 27.72
Solution 6
Let radius of the sphere be r.
 Surface area of the sphere = 154
   = 154 cm2    

    
Thus, the radius of the sphere is 3.5 cm.

Solution 7
Let diameter of earth be d. Then, diameter of moon will be .
    Radius of earth =  
    Radius of moon =
    Surface area of moon =  
    Surface area of earth =  
    Required ratio = 
    
Thus, the required ratio of the surface areas is 1:16.

Solution 8
Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl =

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

Solution 9
(i)    Surface area of sphere =

  (ii)  Height of cylinder = r + r = 2r
        Radius of cylinder = r
        CSA of cylinder =
    
(iii)   Required ratio =   

Chapter 13 - Surface Areas and Volumes Excercise Ex. 13.5

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

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

Solution 2
Volume of tank = l  h = (6  4.5)  = 135
  It is given that:
 1  = 1000 litres

    
 Thus, the tank can hold 135000 litres of water.

Solution 3
Let height of cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380
  h = 380 
       
 10  h = 380
 h = 4.75
    
 Thus, the height of the vessel should be 4.75 m.    

Solution 4
Length (l) of the cuboidal pit = 8 m
Width (b) of the cuboidal pit = 6 m
 Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l  h = (8  3)  = 144   

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

Solution 5
Let the breadth of the tank be 'b' m.
Length (l) of the tank = 2.5 m
Depth (h) of the tank = 10 m
Volume of tank = l  h = (2.5  10)  = 25b 
    
Capacity of tank = 25b  = 25000 b litres
25000 b = 50000    (Given)
 
  Thus, the breadth of the tank is 2 m.

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

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

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

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

Solution 7
Length (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  

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

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

Solution 9
Rate of water flow = 2 km per hour  
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min  
Thus, in 1 minute 4000  water will fall into the sea.    

Chapter 13 - Surface Areas and Volumes Excercise 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
    2r = 132 cm
         
    Volume of cylindrical vessel = r2h
                                                            
Thus, the vessel can hold 34.65 litres of water.
Solution 2
Inner radius (r1) of cylindrical pipe =
Outer radius (r2) of cylindrical pipe =
Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe =
Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = 5720  0.6 g = 3432 g = 3.432 kg

Solution 3
The tin can will be cuboidal in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l  h = (5  15) cm3 = 300 cm3
Radius (R) of circular end of plastic cylinder =
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = R2H  ==385 cm3
Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 - 300) cm3 = 85 cm3

Solution 4
(i)    Height (h) of cylinder = 5 cm
        Let radius of cylinder be r.
        CSA of cylinder = 94.2 cm2
        2rh = 94.2 cm2
        (2  3.14  5) cm = 94.2 cm2
        r = 3 cm
 
(ii)    Volume of cylinder = r2h = (3.14  (3)2  5) cm3 = 141.3 cm3

Solution 5
(i)    Cost of painting 1 m2 area = Rs 20
        So, Rs 2200 is cost of painting area , i.e, 110 m2 area.
        Thus, the inner surface area of the vessel is 110 m2.
    
(ii)    Let radius of base of vessel be r.
        Height (h) of vessel = 10 m
        Surface area = 2rh = 110 m2
                
(iii)    Capacity of vessel = r2h =  = 96.25 m3
Solution 6
Let radius of the circular ends of the cylinder be r.
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
Total  Surface area of vessel = 2 r(r+h)
                                            
Thus, 0.4708 m2 of metal sheet would be needed to make the cylindrical vessel.   
Solution 7
Radius (r1) of pencil =  = 0.35 cm
Radius (r2) of graphite = 
                                 
Height (h) of pencil = 14 cm

Volume of wood in pencil = 
                                   
                                    
 
 
Volume of Graphite =
                            = 0.11 cm3
Solution 8
Radius (r) of cylindrical bowl = cm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = r2h=  
Volume of soup in 250 bowls = (250  154) cm3 = 38500 cm3 = 38.5 litres

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

Chapter 13 - Surface Areas and Volumes Excercise Ex. 13.7

Solution 1
(i)    Radius (r) of cone = 6 cm
       Height (h) of cone = 7 cm
       Volume of cone   

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

Solution 2
(i)    Radius (r) of cone = 7 cm
       Slant height (l) of cone = 25 cm
       Height (h) of cone   
       Volume of cone 
       Capacity of the conical vessel =  litres= 1.232 litres
(ii)    Height (h) of cone = 12 cm
        Slant height (l) of cone = 13 cm
        Radius (r) of cone
        Volume of cone   
        Capacity of the conical vessel = litres =  litres.

Solution 3
Height (h) of cone = 15 cm
Let radius of cone be r.
Volume of cone = 1570 cm3
 
 
 r = 10 cm

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

Solution 4
   Height (h) of cone = 9 cm
   Let radius of cone be r.
   Volume of cone = 48 cm3
   
 
  Thus, the diameter of the base of the cone is 2r = 8 cm.

Solution 5
Radius (r) of pit = 
Depth (h) of pit = 12 m
Volume of pit =  = 38.5 m3
Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

Solution 6
(i)    Radius of cone =   =14 cm
       Let height of cone be h.
       Volume of cone = 9856 cm3
       
       h = 48 cm
       Thus, the height of the cone is 48 cm.
 
(ii)   Slant height (l) of cone  
           
       Thus, the slant height of the cone is 50 cm.
 
(iii)    CSA of cone = rl=   = 2200 cm2
Solution 7
When the right angled ABC is revolved about its side 12 cm, a cone of height (h) 12 cm, radius (r) 5 cm, and slant height (l) 13 cm will be formed.    
Volume of cone    = 100 cm3
Thus, the volume of cone so formed by the triangle is 100 cm3.

Solution 8
When the right angled ABC is revolved about its side 5 cm, a cone of radius (r) 12 cm, height (h) 5 cm, and slant height (l) 13 cm will be formed.
    
Volume of cone = 
Required ratio  

Solution 9
Radius (r) of heap
Height (h) of heap = 3 m
Volume of heap=  
Slant height (l) =

Area of canvas required = CSA of cone
    

Chapter 13 - Surface Areas and Volumes Excercise Ex. 13.8

Solution 1
(i)    Radius of sphere = 7 cm
        Volume of sphere = 
(ii)    Radius of sphere = 0.63 m
        Volume of sphere =
        m3(approximately)

Solution 2
(i)    Radius (r) of ball =
       Volume of ball =   
       Thus, the amount of water displaced is .
(ii)    Radius (r) of ball =  = 0.105 m
        Volume of ball =   
        Thus, the amount of water displaced is 0.004851 m3.

Solution 3
    Radius (r) of metallic ball =  
    Volume of metallic ball =   

    Mass = Density  Volume = (8.9 * 38.808) g = 345.3912 g
    
    Thus, the mass of the ball is approximately 345.39 g.

Solution 4
    Let diameter of earth be d. So, radius earth will be  .
    Then, diameter of moon will be  . So, radius of moon will be  .
    Volume of moon =    
    Volume of earth =   
      
    Thus, the volume of moon is  of volume of earth.

Solution 5
Radius (r) of hemispherical bowl =   = 5.25 cm
Volume of hemispherical bowl
                                                                
                                            = 303.1875 cm3

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

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

Solution 6
Inner radius (r1) of hemispherical tank  = 1 m
     Thickness of hemispherical tank       = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Volume of iron used to make the tank  =
                                                           
Solution 7
Let the radius of the sphere be r.
Surface area of sphere = 154 cm2
r2 = 154 cm2
   
 Volume of sphere

Solution 8
(i)    Cost of white washing the dome from inside = Rs 498.96
       Cost of white washing 1 m2 area = Rs 2
       CSA of inner side of dome =   = 249.48 m2
(ii)    Let inner radius of hemispherical dome be r.
        CSA of inner side of dome = 249.48 m2
        2r2 = 249.48 m2
        
       
        Volume of air inside the dome = Volume of the hemispherical dome
                                                   
                                                    = 523.908 m3
        
        Thus, the volume of air inside the dome is approximately 523.9 m3.

Solution 9
(i)    Radius of 1 solid iron sphere = r
       Volume of 1 solid iron sphere  
       Volume of 27 solid iron spheres  
       It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of
       this iron sphere will be equal to volume of 27 solid iron spheres.
       Radius of the new sphere = r'.
       Volume of new sphere  
      
(ii)    Surface area of 1 solid iron sphere of radius r = 4r2
        Surface area of iron sphere of radius r' = 4 (r')2    = 4 (3r)2 = 36 r2
        

Solution 10
Radius (r) of capsule  
Volume of spherical capsule
                                       
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.

Chapter 13 - Surface Areas and Volumes Excercise 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  110 + 2 (85  25 + 25  110)] cm2
                                                  = 19100 cm2
    Area of front face = [85  110 - 75  100 + 2 (75  5)] cm2
                                                  = 1850 + 750 cm2
                                                  = 2600 cm2
    Area to be polished = (19100 + 2600) cm2 = 21700 cm2    
    Cost of polishing 1 cm2 area = Rs 0.20
    Cost of polishing 21700 cm2 area = Rs (21700  0.20) = Rs 4340    
    
    Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and
    30cm  respectively.
    Area to be painted in 1 row = 2 (l + h) b + lh
                                           = [2 (75 + 30)  20 + 75  30] cm2
                                           = (4200 + 2250) cm2
                                           = 6450 cm2    
    Area to be painted in 3 rows = (3  6450) cm2 = 19350 cm2
    Cost of painting 1 cm2 area = Rs 0.10
    Cost of painting 19350 cm2 area = Rs (19350  0.10) = Rs 1935    
    Total expense required for polishing and painting the surface of the bookshelf 
                                            = Rs(4340 + 1935) = Rs 6275

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

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

Solution 3
Let the diameter of the sphere be d.
Radius (r1) of the sphere =
It is given that the diameter of the sphere is decreased by 25%.
New radus (r2) of the sphere =
CSA (S1) of the sphere =      
CSA (S2) of the new sphere =  
Decrease in CSA of sphere = S1 - S2 
     
Percentage decrease in CSA of sphere=
                                                      %

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