Class 9 Maths Chapter 12 Question Answers - Surface Area & Volumes

Q1: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Sol:
Given,
Length of the cylindrical pipe = h = 28 m
Diameter of the pipe = 5 cm
Now, the radius of piper (r) = 5/ 2 cm = 2.5 cm = 0.025 m
Total radiating surface in the system = Total surface area of the cylinder  
= 2πr(h + r)  
= 2 × (22/7) × 0.025 (28 + 0.025) m² 
= (44 x 0.025 x 28.025)/7 m²
= 4.4 m² (approx)

Q2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Sol:
Given,
Diameter of the cone = 24 m
Radius of the cone (r) = 24/2 = 12 m
Slant height of the cone (l) = 21 m
Total surface area of a cone = πr(l + r)
= (22/7) × 12 × (21 + 12)
= (22/7) × 12 × 33
= 1244.57 m²

Q3: 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.
Sol:
Given,
Diameter of the sphere = 7 m
Radius (r) = 7/2 = 3.5 m
Now, the riding space available for the motorcyclist = Surface area of the sphere
= 4πr²
= 4 × (22/7) × 3.5 × 3.5
= 154 m²  

Q4: Hameed has built a cubical water tank with a 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 in the figure below). Find how much he would spend on the tiles if the cost of the tiles is Rs.360 per dozen.

Sol:
Given,
Edge of the cubical tank (a) = 1.5 m = 150 cm
So, surface area of the tank = 5 × 150 × 150 cm²
The measure of side of a square tile = 25 cm
Area of each square tile = side × side = 25 × 25 cm²
Required number of tiles = (Surface area of the tank)/(area of each tile)
= (5 × 150 × 150)/(25 × 25)
= 180
Also, given that the cost of the tiles is Rs. 360 per dozen.
Thus, the cost of each tile = Rs. 360/12 = Rs. 30
Hence, the total cost of 180 tiles = 180 × Rs. 30 = Rs. 5400

Q5: The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of  Rs.7.50 per sq.m.
Sol:
Given,
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area of walls of the room = Lateral surface area of cuboid
= 2h(l + b)
= 2 × 3(5 + 4)
= 6 × 9
= 54 sq.m
Area of ceiling = Area of base of the cuboid = lb
= 5 × 4
= 20 sq.m
Area to be white washed = (54 + 20) sq.m = 74 sq.m
Given that, the cost of white washing 1 sq.m = Rs. 7.50
Therefore, the total cost of white washing the walls and ceiling of the room = 74 × Rs. 7.50 = Rs. 555

Q6: Curved surface area of a right circular cylinder is 4.4 sq.m. If the radius of the base of the cylinder is 0.7 m, find its height.
Sol:
Let h be the height of the cylinder.
Given,
Radius of the base of the cylinder (r) = 0.7 m
Curved surface area of cylinder = 4.4 m²
Thus,
2πrh = 4.4  
2 × 3.14 × 0.7 × h = 4.4
4.4 × h = 4.4
h = 4.4/4.4
h = 1  
Therefore, the height of the cylinder is 1 m.

Q7: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. (Take π = 3.14)
Sol:
Given
Height of a cone (h) = 16 cm
Radius of the base (r) = 12 cm
Now,
Slant height of cone (l) = √(r² + h²)
= √(256 + 144)
= √400
= 20 cm
Curved surface area of cone = πrl
= 3.14 × 12 × 20 cm²
= 753.6 cm²
Total surface area = πrl + πr²
= (753.6 + 3.14 × 12 × 12) cm²
= (753.6 + 452.16) cm²
= 1205.76 cm²

Q8: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs.210 per 100 sq.m.
Sol:
Given,
Slant height of a cone (l) = 25 m  
Diameter of the base of cone = 2r = 14 m  
∴ Radius = r = 7 m  
Curved Surface Area = πrl
= (22/7) x 7 x 25
= 22 × 25  
= 550 sq.m  
Also, given that the cost of white-washing 100 sq.m = Rs. 210
Hence, the total cost of white-washing for 550 sq.m = (Rs. 210 × 550)/100 = Rs. 1155

Q9: 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?
Sol:
Given,
Dimensions of the brick = 22.5 cm × 10 cm × 7.5 cm
Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm
Surface area of 1 brick  = 2(lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2(225 + 75 + 168.75) cm² 
= 2 x 468.75 cm² 
= 937.5 cm² 
Area that can be painted by the container = 9.375 m² (given)
= 9.375 × 10000 cm²
= 93750 cm²
Thus, the required number of bricks = (Area that can be painted by the container)/(Surface area of 1 brick)
= 93750/937.5
= 937500/9375
= 100

Q10: The curved surface area of a right circular cylinder of height 14 cm is 88 sq.cm. Find the diameter of the base of the cylinder.
Sol:
Let d be the diameter and r be the radius of a right circular cylinder.
Given,
Height of cylinder (h) = 14 cm
Curved surface area of right circular cylinder = 88 cm2
⇒ 2πrh = 88 cm²
⇒ πdh = 88 cm² (since d = 2r)
⇒ 22/7 x d x 14 cm = 88 cm²
⇒ d = 2 cm
Hence, the diameter of the base of the cylinder is 2 cm.