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
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
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 5 3 + 2 4 3 + 5 4]
= (30 + 24 + 20)
Cost of white washing 1 area = Rs 7.50
Cost of white washing 74 area = Rs (74 7.50) = Rs 555
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.
= [2(22.5 10 + 10 7.5 + 22.5 7.5)]
= 2(225 + 75 + 168.75)
= (2 468.75)
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.
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)]
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
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)
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.
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.
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]
Chapter 13 - Surface Areas and Volumes Excercise Ex. 13.2
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.
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.
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 =
Thus, the total surface area of cylindrical pipe is 2038.08 .
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
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.
Radius of the base of the cylinder = 0.7 m
CSA of cylinder = 4.4
h = 1 m
Thus, the height of the cylinder is 1 m.
Depth (h) of circular well = 10 m
= (44 0.25 10)
(ii) Cost of plastering 1 area = Rs 40
Cost of plastering 110 area = Rs (110 40) = Rs 4400
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 .
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.
Radius of the circular end of frame of lampshade = cm = 10cm
Cloth required for covering the lampshade =
Thus, for covering the lampshade 2200 cloth will be required.
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of
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
Slant height of cone = 10 cm
CSA of cone = =
Thus, the curved surface area of cone is 165 .
Slant height of cone = 21 m
Total surface area of cone = (r + l)
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
Radius (r) of conical tent = 24 m
Let slant height of conical tent be l.
(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.
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent = = (3.14 6 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.
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.
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.
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
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 = =
(i) Radius of sphere
Surface area of sphere
(ii) Radius of sphere
Surface area of sphere
(iii) Radius of sphere
Surface area of sphere =
Total surface area of hemisphere
Radius of spherical balloon, when air is pumped into it = 14 cm
Surface area of hemispherical bowl =
Cost of tin-plating 100 area = Rs 16
Cost of tin-plating 173.25 area = Rs 27.72
Surface area of the sphere = 154
= 154 cm2
Thus, the radius of the sphere is 3.5 cm.
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.
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 .
(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
Volume of 1 match box = l b h = (4 2.5 1.5) = 15
Volume of the packet containing 12 such matchboxes = 12 15 =180
1 = 1000 litres
Thus, the tank can hold 135000 litres of water.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380
l b h = 380
10 8 h = 380
h = 4.75
Thus, the height of the vessel should be 4.75 m.
Width (b) of the cuboidal pit = 6 m
Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l b h = (8 6 3) = 144
Cost of digging 1 = Rs 30
Cost of digging 144 = Rs (144 30) = Rs 4320
Length (l) of the tank = 2.5 m
Depth (h) of the tank = 10 m
Capacity of tank = 25b = 25000 b litres
25000 b = 50000 (Given)
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
n 600000 = 1800000
n = 3
Thus, the water of tank will last for 3 days.
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.
Volume of the cube = a3= (12 cm)3 = 1728
Let the side of each smaller cube be l.
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.
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
Height (h) of the vessel = 25 cm
Circumference of the vessel = 132 cm
2r = 132 cm
Volume of pipe =
Mass of 5720 cm3 wood = 5720 0.6 g = 3432 g = 3.432 kg
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l b h = (5 4 15) cm3 = 300 cm3
Difference in capacity = (385 - 300) cm3 = 85 cm3
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm2
2rh = 94.2 cm2
(2 3.14 r 5) cm = 94.2 cm2
r = 3 cm
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
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
Radius (r2) of graphite =
Height (h) of pencil = 14 cm
Volume of wood in pencil =
Height (h) up to which the bowl is filled with soup = 4 cm
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
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
Slant height (l) of cone = 25 cm
Height (h) of cone
Volume of cone
Capacity of the conical vessel = litres= 1.232 litres
Slant height (l) of cone = 13 cm
Radius (r) of cone
Volume of cone
Capacity of the conical vessel = litres = litres.
Let radius of cone be r.
Volume of cone = 1570 cm3
Thus, the radius of the base of the cone is 10 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.
Depth (h) of pit = 12 m
Volume of pit = = 38.5 m3
Let height of cone be h.
Volume of cone = 9856 cm3
Thus, the slant height of the cone is 50 cm.
Volume of cone = 100 cm3
Thus, the volume of cone so formed by the triangle is 100 cm3.
Volume of cone =
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
Volume of sphere =
Volume of sphere =
Volume of ball =
Thus, the amount of water displaced is .
Volume of ball =
Thus, the amount of water displaced is 0.004851 m3.
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.
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.
= 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.
Thickness of hemispherical tank = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Surface area of sphere = 154 cm2
r2 = 154 cm2
Volume of sphere
Cost of white washing 1 m2 area = Rs 2
CSA of inner side of dome = = 249.48 m2
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.
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
Radius of the new sphere = r'.
Volume of new sphere
Surface area of iron sphere of radius r' = 4 (r')2 = 4 (3r)2 = 36 r2
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
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
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
Radius (r') of cylindrical support = 1.5 cm
Height (h) of cylindrical support = 7 cm
Area of circular end of cylindrical support = r2
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
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 (S2) of the new sphere =
Decrease in CSA of sphere = S1 - S2
Percentage decrease in CSA of sphere=
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