Class 9 maths Surface areas and Volume Important Questions
Q 1 – If the radius of a cylinder is 4cm and height is 10cm, then the total surface area of a cylinder is:
a. 440 sq.cm
b. 352 sq.cm
c. 400 sq.cm
d. 412 sq.cm
352 sq.cm
Q 2 – If slant height of the cone is 21cm and the diameter of the base is 24 cm. The total surface area of a cone is
1200.77 sq.cm
b. 1177 sq.cm
1222.77 sq.cm
d. 1244.57 sq.cm
d. 1244.57 sq.cm
Q 3 – The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit that is 16 m long, 12m wide and 4 m deep is
a. 1900
b. 1920
c.1800
d. 1840
1920
Q 4 – Match the column:
Curved surface area of Cone | 3π r2 |
Curved surface of Hemisphere | 2π rh |
Curved surface area of Cylinder | 2π r2 |
Total surface area of Hemisphere | πr2 +πrl |
Total surface area of cone | πrl |
Curved surface area of Cone | πrl |
Curved surface of Hemisphere | 2π r2 |
Curved surface area of Cylinder | 2π rh |
Total surface area of Hemisphere | 3π r2 |
Total surface area of cone | π r2 +πrl |
Q 5 – Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see Fig.). Find how much he would spend for the tiles, if the cost of the tiles is Rs 360 per dozen.
The Length of the each side of water tank = 1.5 m = 150 cm.
As it is given that base is not covered with tiles so, five sides of the tank are covered with tiles.
Therefore,
Surface area of water tank =5a2 = 5 (150)2 = 112500 cm2
Side of tile = 25 cm
Area of tile = 25 2 = 625 cm2
Number of tiles to cover five sides of the tank = 112500 / 625 = 180
Now,
Cost of one dozen tiles = Rs.360
Hence, total cost of 180 = 360/12 × 180 = Rs.5400
Q 6 – The paint in a certain container is sufficient to paint an area equal to 9.375 sq.m. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Volume of paint = 9.375 m2 = 93750 cm2
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm2
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm2
= 2(225 + 75 + 168.75) cm2
= 2×468.75 cm2 = 937.5 cm2
Number of bricks can be painted = 93750/937.5 = 100
Q 7 – A circus tent consists of cylindrical base surmounted by a conical roof. The radius of the cylinder is 20 m. The height of the tent is 63 m and that of the cone is 21 m. find the area of the canvas used for making it.
V = 879203
A = 97342
Given : A circus tent consists of a cylindrical base surmounted by a conical roof. The radius of the base is 20m. If the height of the tent is 63m and that of the conical part is 21m.
To find : The volume of the tent also find the amount of canvas required to make this tent?
Solution : Radius of the base is r=20 m
Height of the tent h=63 m
Height of the conical part is H=21 m
Volume of the tent = Volume of cylinder + Volume of the cone
Now, the slant height is
Curved surface area of the tent = area of cylinder + Area of cone
Q 8 – A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
(i) Lateral surface area of cube = 4 edge2
=4 (10)2
= 400
lateral surface area of cuboid =2h (l+b)
2 × 8 (12.5 + 10)
=16 × 22.5
= 360
So, the lateral surface area of the cube is larger by (400−360=40) cm2
(ii) Total surface area of cube = 6edge 2
= 6 (10) 2
= 600
lateral surface area of cuboid =2 (lb + bh + hl)
2 (12.5 × 10 + 10 × 8 + 8 × 12.5)
=2 (125+80+100)=610
So, the total surface area of cuboid is larger by (610 – 600 = 10) cm2
Q 9 – Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope.(see Fig).
What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take 
Radius of the base of the cylindrical kaleidoscope (r) = 3.5 cm.
Height (length) of kaleidoscope (h) = 25 cm.
Area of chart paper required = curved surface area of the kaleidoscope
Q 10 – A 20 m deep well with diameter 14 m is dug up and the earth from digging is evenly spread to form a platform 22 m 14 m. Find the height of the platform.
Diameter = 7m
so, radius = 3.5 m
Earth dug out from (cylindrical) well
= 770m3
to from platform …… (cuboidal)
So, Height = 
× 14 = 2.5 m
Q 11 – Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Given,
Slant height of cone is l = 21 m.
Diameter of its base is d = 24 m.
hence, radius of its base is 
Now, the total surface area of cone is given by
S = πrl + πr2
S = π (rl + r2)
S = π (12 × 21 + 122)
S = π (252 + 144)
hence ,S = 1244.57m2
Q 12 – The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Inner diameter of the hollow sphere, 2r =7m
Available area to the motorcyclist for riding = Inner surface area of the sphere
=4πr2=π(2r)2
Available area to the motorcyclist for riding =154sq.m