CBSE Class 9 Mathematics Surface Areas And Volumes Worksheet Set C

Very Short Answer Type Questions:

Question: A solid cube is cut into two cuboids of equal volumes. Find the total surface area of one of the cuboids.
Answer: Let edge of the solid cube be a cm.
Then, dimensions of each of the cuboids will be a cm, a cm and a/2 cm.
∴ Total surface area of one of the cuboids
= 2( a × a + a × a/2 + a/2 × a) = 2( a² + a²/2 + a²/2) = 4a² cm²

Question: A metal sheet 27 cm long, 8 cm broad and 1 cm thick is melted and recast into a cube. Then find the volume of cube formed.
Answer: Volume of sheet = (27 × 8 × 1) cm³
= 216 cm³
Volume of cube formed = 216 cm³

Question: The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, find the ratio of their volumes.
Answer: Let the radii of the bases of a cylinder and a cone be 3x and 4x respectively and let their heights be 2y and 3y respectively.
∴ Ratio of their volumes = π × (3x)² × 2y / 1/3π × (4x)²× 3y = 9/8 or 9:8

Question: Area of base of a solid hemisphere is 81p sq. cm. Then find its volume.
Answer: Let r be the radius of the hemisphere.
∴ πr² = 81π ⇒ r² = 81 ⇒ r = 9 cm.
Volume of hemisphere = 2/3πr³ = 2/3π(9)³ = 486π cm³

Question: The dimensions of a cinema hall are 120 m, 50 m, 30 m. Then find the volume of hall.
Answer: Length(l) = 120 m, Breadth(b) = 50 m, Height (h) = 30 m
Since the shape of cinema hall is of cuboid
∴ Volume of hall = l × b × h
= 120 × 50 × 30 m³ = 180000 m³

Question. How many faces does a right circular cylinder have ? 
Answer: 3

Question. Find the capacity of a tank of dimensions 8 cm × 6 cm × 2.5 cm.
Answer: Capacity of the tank = 120 cm³ 
Detailed Solution :
Capacity of the tank = length × breadth × height
                              = 8 cm × 6 cm × 2.5 cm
                              = 120 cm³.

Question. Two cylinders have bases of same size. The diameter of each is 7 cm. If one of the cylinder is 10 cm high and the other is 20 cm high, then find the ratio of their volumes. 
Answer: Let r denotes the radius of both cylinders and h and h’ be their heights respectively.
Ratio of their volumes = 4πr²h/4πr²h’ = h/h’ = 10/20
                                 = 1 : 2.

Question. Find the volume of a right circular cone with radius 6 cm and height 7 cm.
Answer: Volume of right circular cone = 1/3πr²h
                                       = 1/3 × 22/7 × (6)² × 7 = 1/3× 22/7 × 36 × 7
                                       = 264 cm³.

Question. Calculate the volume of a cuboid whose dimensions are 3.6 cm, 8.2 cm and 11 cm. 
Answer: Volume of cuboid = length × breadth × height
                                 = 3.6 × 8.2 × 11
                                 = 324.72 cm³.

Short Answer Type Questions:

Question: Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes. 
Answer: Let the side of each cube be a units.
∴ TSA of 1 cube = 6a² sq. units
⇒ TSA of 3 cubes = 3 × 6a² = 18a² sq. units

Now length, breadth and height of the resulting cuboid is 3a units, a units and a units respectively.
∴ TSA of cuboid = 2(lb + bh + hl)
= 2(3a × a + a × a + a × 3a)
=2(3a² + a² + 3a²) = 2 × 7a² = 14a² sq. units.
So, required ratio = 14a²/18a² = 7/9 or : 7:9

Question: Find the diameter of the sphere, whose total surface area is 616 cm².
Answer: Let r be the radius of the sphere.
Total surface area of sphere = 4πr²
⇒ 616 = 4 × 22/7 × r² ⇒ r² = 616 × 7/4×22 = 49 ⇒ r = 7 cm
∴ Diameter = 2r = 2 × 7 = 14 cm

Question: A cuboidal oil tin box is 4 m by 2 m by 0.75 m. Find the cost of the tin sheet required for making 20 such tin boxes, if the cost of tin sheet is Rs 20 per square metre.
Answer: Length, l = 4 m, breadth, b = 2 m and height, h = 0.75 m
Surface area of one tin box = 2(lb + bh + hl)
= 2 (4 × 2 + 2 × 0.75 + 0.75 × 4)
= 2 (8 + 1.5 + 3) = 2 × 12.5 = 25 m²
∴ Surface area of 20 such tin boxes = (20 × 25) m²
= 500 m² 
Now, cost of 1 square metre of tin sheet = Rs 20
∴ Cost of 500 m² of tin sheet = Rs(20 × 500) = Rs 10000

Question: If volume and surface area of a sphere is numerically equal then find its radius (in units). 
Answer: Let r be the radius of the sphere.
∴ Volume of sphere = Surface area of sphere
∴ 4/3πr³ = 4πr² ⇒ r³/r² = 3 ⇒ r = units

Question: The total cost of making a solid spherical ball is Rs 67914 at the rate of Rs 14 per cubic metre. Find the radius of this ball. 
Answer: Volume of spherical ball = Total cost/Cost of 1m³
⇒ 4/3πr³ = 67914/14 ⇒ 4/3 × 22/7 × r³ = 67914/14
⇒ r³ = 101871/88 ⇒ r³ = 1157.625 ⇒ r = 10.5 m

Question: The external diameter of an iron pipe is 35 cm and its length is 30 cm. If the thickness of the pipe is 2.5 cm, find the curved surface area of the pipe. 
Answer: Length of the pipe, h = 30 cm
External radius of the pipe, R = 35/2 cm = 17.5 cm
∴ Thickness of the pipe = 2.5 cm
∴ Internal radius of the pipe, r = (17.5 – 2.5) cm
= 15 cm
Now, curved surface area of the pipe = External curved surface area + Internal curved surface area
= 2πRh + 2πrh = 2πh (R + r)
= 2 × 22/7 × 30(17.5 + 15) = 2 × 22/7 × 30 × 32.5 
= 42900/7 = 6128.57 cm²

Question: The length and breadth of a rectangular solid are respectively 35 cm and 20 cm. If its volume is 7000 cm³, then find its height (in cm). 
Answer: Let h be the height of the solid.
∴ Volume of cuboid = l × b × h
⇒ 7000 = 35 × 20 × h ⇒ h = 7000/700 = 10 cm

Question: A room is 16 m long, 9 m wide and 3 m high. It has two doors, each of dimensions (2 m × 2.5 m)and three windows, each of dimensions (1.6 m × 75 cm). Find the cost of distempering the walls of the room from inside at the rate of Rs 8 per sq. metre. 
Answer: Given, length (l) = 16 m, breadth (b) = 9 m and height (h) = 3 m
∴ Area of 4 walls of the room = 2(l + b) × h
= 2(16 + 9) × 3 = 150 m².
Area of 2 doors = 2 × (2 × 2.5) = 10 m²
Area of 3 windows = 3 × (1.6 × 75/100) = 3.6 m².
Area not to be distempered = 10 + 3.6 = 13.6 m²
Area to be distempered = 150 – 13.6 = 136.4 m²
Cost of distempering the walls = Rs (136.4 × 8) = Rs 1091.20

Question: Three cubes each of edge 5 cm are joined end to end. Find the surface area of the resulting cuboid.
Answer: When three cubes are joined end to end, we get a cuboid such that Length of the resulting cuboid, l = 5 cm + 5 cm + 5 cm = 15 cm
Breadth of resulting cuboid, b = 5 cm
Height of the resulting cuboid, h = 5 cm
Surface area of the cuboid = 2 (lb + bh + hl)
= 2 (15 × 5 + 5 × 5 + 5 × 15) cm²
= 2 (75 + 25 + 75) cm² = 350 cm²

Question: How many spherical balls of diameter 1 cm can be made from an iron ball of diameter 6 cm? 
Answer: Volume of iron ball having diameter, 6 cm
= 4/3π(6/2)³ =4/3π(3)³ 
Volume of small ball of diameter, 1 cm = 4/3π(1/2)³
∴ Number of balls = 4/3π(3)³/4/3π(1/2)³ = 27 × 8 = 216

Question: The length of a cold storage is three times its breadth. Its height is 5 m. The area of its four walls (including doors) is 256 m². Find its volume.
Answer: Let length, breadth and height of the cold storage be l, b and h respectively.
Then, l = 3b and h = 5 m.
Now, area of four walls = 256 m²
⇒ 2 (l + b)h = 256 ⇒ 2 (3b + b) × 5 = 256
⇒ 40b = 256 ⇒ b = 6.4 metres
∴ l = 3b = 3 × 6.4 = 19.2 m
Volume of the cold storage = l × b × h
= (19.2 × 6.4 × 5) m³ = 614.4 m³

Question: The curved surface area of a cone is 154 cm². If its radius is x cm and slant height is 7 cm. Find the value of 20x. 
Answer: We have, curved surface area = 154 cm²
⇒ πrl = 154 ⇒ r = 154 × 7/22 × 7 = 7
Now, r = x cm = 7 cm.
∴ x = 7 ⇒ 20x = 20 × 7 = 140

Question. A dome of a building is in the form of a hemisphere From inside, it was white washed at the cost of Rs 997.92. If the cost of white washing is 400 paise per square meter, find the volume of air inside the dome. (Take π = 22/7)
Answer: Cost of white washing hemispherical
                      dome = Rs 997.92
Cost of white washing per square meter
                               = 400 paise = Rs 4
∴  CSA = 997.92 ÷ 4 = 249.48 m²
                       2πr² = 249.48
          2 × 22/7 × r² = 249.48
or,                        r² = 39.69
or,                         r = 6.3 m
∴ Volume of air inside it
                              = 2/3πr³
                              = 2/3 × 22/7  × 6.3 × 6.3 × 6.3
                              = 523.90 m³.

Question. A solid piece of metal, cuboidal in shape, with dimensions 24 cm, 18 cm and 4 cm is recast into a cube. Calculate the lateral surface area of the cube.
Answer: Volume of cuboid = lbh
                                 = 24 × 18 × 4
                                 = 1728 cu.cm.
           Edge of a cube = ³√1728
                                 = 12 cm 
                           LSA = 4a²
                                  = 4 × 12 × 12 

Question. The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. If the cost of painting 1 cm² of the surface area is Rs 0.05, find the total cost of painting the vessel all over.
Answer: Internal radius (r) = 12 cm
      External radius (R) = 12.5 cm
T.S.A = 2πr²+ 2πR² + π(R²– r²)
        = 2π(r²+ R²) + π(R – r)(R + r)
        = 2π (144 + 156.25) + π (12.5 + 12)( 12.5 – 12)
        = (600.50 + 12.25) × 22/7
        = 1925.79 cm²
Cost of painting 1925.79 cm² at the rate of Rs 0.05/cm²
        = 1925.79 × 0.05
        = Rs 96.29.

Question. The length, breadth and height of a room are 5 m, 4 m and 3 m. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².
Answer: Given, l = 5 m, b = 4 m, h = 3 m.
Area to be painted = Area of walls + Area of ceiling
                            = 2h(l + b) + l × b
                            = 2 × 3(9) + 20
                            = 54 + 20 = 74 m²
    Cost of painting 1 m²= Rs 7.50
∴ Cost of painting 74 m²= 74 × 7.50
                                     = Rs 555

Question. The total surface area of a solid hemisphere is 5940 cm². Find the diameter of the hemisphere. 
Answer: Let the radius of hemisphere be r
∴         3πr² = 5940
or,          r² = 5940 × 7/3 × 22 = 630
or,             r = √630 or 3√70 cm.
So,            d = 2r = 6√70 cm.
Hence diameter of the hemisphere is 6√70 cm.

Question. Find the radius of the base of a right circular cylinder whose curved surface area is 2/3 of the sum of the surface areas of two circular faces. The height of the cylinder is given to be 15 cm.
Answer: Given, h = 15
          C.S.A. = 2/3 (Sum of circular faces)
             2πrh = 2/3(2πr²)
                15 = 2/3r
             45/2 = r
                  r = 22.5 cm.

Question. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold ?
Answer: Cuboidal water tank
                 length = 6 m
                  width = 5 m
                 height = 4.5 m
Volume of tank = length × width × height
                      = 6 × 5 × 4.5 m³
                      = 135 m³
                      = 135 × 1000 litre
                                                  (1 m³= 1000 litre)
∴  Volume of water = 135000 litre
or Capacity of tank = 135000 litre = 135 kl

Question. The surface area of a cuboid is 1372 cm². If its dimensions are in the ratio 4 : 2 : 1, find its length. 
Answer: 2(lb + bh + hl) = 1372
    l = 4x, b = 2x, h = x
2(4x × 2x + 2x × x + 4x × x) = 1372
or,            2(8x² + 2x² + 4x²) = 1372
or,                                 28x² = 1372
or,                                    x² = 49
or,                                     x = 7 cm
∴                                 Length = 4 × 7 = 28 cm.

Question. The total surface area of a solid right circular cylinder is 1540 cm². If the height is four times the radius of the base, then find the height of the cylinder.
Answer: Given, T.S.A. = 1540 cm²
∴         2πr(h + r) = 1540 cm²
Also,                 h = 4r
∴       2πr (4r + r) = 1540
or,         2π × 5r² = 1540
or,                  r² = 1540 × 7/2 × 5 × 22
or,                  r² = 49
or,                     r = 7 cm
Now                  h = 4r
or,                     h = 28 cm.

Long Answer Type Questions 

Question: A sector of a circle of radius 15 cm and central angle of 120°. It is rolled up and the two bounding radii are joined together to form a cone of radius 5 cm. Find :
(i) the volume of the cone.
(ii) the total surface area of the cone.

Answer: (i) The slant height of the cone = Radius of the given sector of a circle = 15 cm.
Now, let h be the height of the cone. Then,
h = √l² − r² [where l = slant height, r = radius of cone]
= √(15)² − (5)² = √225 − 25 = √200 = 10√2 cm
∴ Volume of the cone = 1/3πr²h
= 1/3π(5)² × 10√2 = 1/3 × 22/7 × 25 ×10 × √2 = 369.29 cm
(ii) Total surface area of the cone = πr(r + l)
= 22 × 5(5 + 15) = 22/7 × 5 × 20 = 314.29 cm²

Question: Coins of same size (say 10 rupee coin) are placed one above the other and a cylindrical block is obtained. The volume of this block is 67.76 cm³. Find the number of coins arranged in the block, if thickness of each coin is 2 mm and radius of each coin is 1.4 cm. 
Answer: Let h be the height of cylindrical block and n be the number of coins used to obtain it.
Volume of block = πr²h ⇒ 67.76 = 22/7× 1.4 × 1.4 × h
⇒ h = 67.76 × 7/22 × 1.4 × 1.4 =11cm = 110mm [∴ 1 cm = 10 mm]
Now, n × thickness of a coin = height of block
⇒ n × 2 = 110 ⇒ n = 55

Question: A plot of land is in the form of rectangle has dimension 240 m × 180 m. A drainlet 10 m wideis dug around it (on the outside) and the earth dug out is evenly sπread out over the plot increasing its surface level by 25 cm. Find the depth of the drainlet.
Answer: Volume of earth dug out = l × b × h
= 240 × 180 × 25/100 m³ = 10800 m³
Let the depth of the drainlet be x m.
∴ Volume of drainlet = 2[260 × 10 × x] + 2[180 × 10 × x]
= 8800 x m³

Now, volume of earth dug out = Volume of drainlet
⇒ 10800 = 8800x ⇒ x = 10800/8800 = 1.23 m (apπrox.)

Question. A pen stand is cylindrical in shape with the base radius 3.5 cm and height 10.5 cm. How much card board will be required to make 25 such pen stand ? Also, find volume of 1 pen stand.
Answer: Given, base radius of cylinder r = 3.5 cm
                        Height of cylinder h = 10.5 cm
Amount of card board required to make one pen stand
                                                     = πr(r + 2h)
                               One pen stand = 22/7 × 3.5(3.5 + 21)
                                                     = 22 × 0.5(24.5)
                                                     = 269.5 cm²
Amount of card board required to make 25 pen stand
                                                     = 269.5 × 25
                                                     = 6737.5 cm²
       1 pen stand, then for 25 stands = 25 × 308
                                                      = 7700 cm²
                     Volume of 1 pen stand = πr²h
                                                      = 22/7 × 3.5 × 3.5 × 10.5
                                                      = 404.25 cm³.

Question. Calculate the curved surface area of a cone whose radius of base and height are in the ratio 5 : 12 and its volume is 2512 cu. cm.
Answer: Let           r = 5x, h = 12x
Given :  1/3πr2h = 2512 cu. cm
∴         1/3π × 3.14(5x)² × 12x = 2512
or,                x³ = 2512 × 3 × 100/5 × 5 × 12 × 314
                        = 8
or,                  x = 2
∴                    r = 10, h = 24, l = 26
∴    CSA of cone = πrl
22/7 × 10 × 13 × 2 = 5720/7
                              = 817.14 cm².

Question. A cylindrical bowl of internal diameter 18 cm and height 15 cm is full of liquid. The whole of the liquid is to be filled in small cylindrical bottles of diameter 3 cm and height 4 cm, Each bottle is sold for Rs 5, then find the amount earned. 
Answer: Volume of liquid = πr²h
Given,                    r = 18/2 = 9 cm
and                       h = 15 cm
∴ Volume of liquid = π(9)² 15
                          = 1215 π cm³
Also, radius of small bottle
                    (r’) = 3/2 = 1.5 cm
and height of small bottle
                    (h’) = 4 cm
∴Volume of small bottle = π(r’)²h’
                                   = π(1.5)²4
                                   = 9π
Number of bottles = 1215π/9π = 135
∴  Amount earned = 135 × 5 = Rs 675

Question. The frame of a lampshade is cylindrical in shape. It has base diameter 28 cm and height 17 cm. It is to be covered with a decorative cloth. A margin of 2 cm is to be given for folding it over top and bottom of the frame. If 1/12 of cloth is wasted in cutting and pasting, find how much cloth is required to be purchased for covering the frame.
Answer: Base diameter = 28 m
           Base radius = 28/2 = 14 cm
Height of cloth required = 17 + 2 + 2 = 21 cm
Area of cloth required = Curved surface area of cylinder of radius 14 cm and height 21 cm
                                = 2πrh
                                = 2 × 22/7 × 14 × 21
                                = 1848 cm²
Let A sq. cm of cloth be purchased .
So, wastage of cloth for cutting and pasting
                                = A/12 cm²
Area of cloth actually used = A − A/12 = 11/12 A cm²
Area of cloth actually used = Area of cloth required
or,                        11/12A = 1848
or,                                A = 1848 × 12/11 = 2016 cm²

Question. An open box is made of wood 3 cm thick. Its external dimensions are 1.4 m, 1.1 m and 0.8 m. Find the cost of painting the outer surface of box at 75 paise per 100 cm².
Answer: l = 140 cm
       b = 110 cm
       h = 80 cm
Surface area of open box = lb + 2(bh + hl)
       Cost of painting box = Rs 75/100 × 100 [154 + 2(88 + 112)] × 100
                                     = Rs 3/4 [554] = Rs 3(138.5)
                                     = Rs 415.5