Short Answers Type Questions- Surface Areas and Volumes Class 9 Notes | EduRev

Question 1. The length, breadth and height of a room are 4 m, 3 m and 3 m respectively. Find the lateral surface area of the room.
 Solution: Here, l = 4 m, b = 3 m and h = 3 m.
Lateral surface area = Area of four walls = 2(l + b)h = 2(4 + 3) x 3 m² = 2 x 7 x 3 m² = 42 m²
∴ The required lateral surface of the room is 42 m².
 

Question 2. The floor area of a room is 100 m² and its height is 8 m. Find its volume.
 Solution: ∵ Volume of a cuboid = [Base area] x Height
∴ Volume of the room = [Area of the floor] x height = 100 m² x 8 m = 800 m³
Thus, the volume of the room = 800 m³.
 

Question 3. If the total surface area of a cube is 216 cm², then find its volume.
 Solution: Let each side of the cube be x.
∴ Total surface area = 6x2
⇒ 6x² = 216
⇒   x² =(216/6) = 36
⇒ x = √36 = 6 cm

∴ Volume = (side)³ = 6³ = 6 x 6 x 6
= 216 cm³

Question 4. If the circumference of the base of a right circular cylinder is 110 cm, then find its base area.
 Solution: Let r be the radius of the base of the cylinder.

∴ Circumference = 2πr = 2 x(22/7)x r
Now, 2 x(22/7) x r= 110

⇒    

Now, Base area =   

Question 5. The curved surface area of a cylinder is 4400 cm². If the circumference of its base is 110 cm, then find its height.
 Solution: Curved surface area of the cylinder = 2πrh
Circumference = 2πr

∴          

Thus, the height of the cylinder is 40 cm.
 

Question 6. The radii of two cylinders are in the ratio of 2 : 3 and heights are in the ratio of 5 : 3.
 Find the ratio of their volumes.
 Solution: Ratio of the radii = 2 : 3 Let the radii be 2r and 3r Also, their heights are in the ratio of 5:3 Let the height be 5h and 3h
∴ Ratio of their volumes = 

Question 7. If the radius of a sphere is doubled, then find the ratio of their volumes.
 Solution: Let the radius of the original sphere = r
∴ Radius of new sphere = 2r

∴  Ratio of their volumes = 

Question 8. If the radius of a sphere is such that πr² = 6cm² then find its total surface area.
 Solution: ∵ πr² = 6cm² 
∴ Curved S.A. of the hemisphere
= 2 × 6 cm² = 12 cm² 
Also, plane S.A. of the hemisphere = π r² = 6 cm² 
⇒ Total S.A. = C.S.A. + plane S.A. = 12 cm² + 6 cm² = 18 cm²

Question 1. The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone (taking π = (22/7).
  Solution: Surface area of the sphere = 4πr² = 4 × π × 5 × 5 cm²
Curved surface area of the cone (with slant height as ‘ℓ’) = πrℓ = π × 4 × ℓ cm²

Since,    

∴ 4 × π × 5 × 5 = 5 × π × 4 × ℓ
⇒              

∴ Volume of the cone = 

Question 2. The radius of a sphere is increased by 10%. πrove that the volume will be increased by 33.1% apπroximately.
 Solution: The volume of a sphere = 

Increased radius = 

∴ Increased volume = 

= (4/3) x π x 1.331r³

Thus, Increase in volume

∴ Percentage increase in volume

= 0.331 × 100%
= 33.1%.

Question 3. Find the slant height of a cone whose radius is 7 cm and height is 24 cm.
 Solution: Here, h = 24 cm and r = 7 cm

Since,

= 25 cm
∴ Slant height = 25 cm.
 

Question 4. The radius of a cylinder is 7 cm. If its volume is 2002 cm³, then find its height and total surface area.
 Solution: Here, radius (r) = 7 cm
∴ Volume of the cylinder = = πr²h = (22/7) x 7 x 7 x h
Now,(22/7)x 7 x 7 x h = 2002

∴  

Now, total surface area of the cylinder = 2πr² + 2πrh = 2πr(r + h)
= 2 x(22/7) x 7 x (7 + 13) cm² 
= 44 × 20 cm² = 880 cm² 

Question 5. The diameter of a road roller, 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, find the cost of levelling it at 2 per square metre.
 Solution: Here, radius (r) = 42 cm
Length of the roller = Height of the cylinder
⇒ h = 120 cm
∴ Curved surface area of the roller = 2πrh =2 x(22/7) x 42 x 120 cm² 
= 2 x 22 x 6 x 120 cm² = 31680 cm²
∴ Area levelled in one revolution = 31680 cm²

⇒ Area levelled in 500 revolutions = 31680 x 500 cm²

∴ Cost of levelling the playground = 2 x 15840 = 31680.

Question 6. A conical tent of radius 7 m and height 24 m is to be made. Find the cost of the 5 m wide cloth required at the rate of 50 per metre.
 Solution: Radius of the base of the tent (r) = 7 m
Height (h) = 24 m
Slant height (ℓ)=

Now, curved surface area of the conical tent = πrℓ
= (22/7) x 7 x 25 m² = 22 x 25 m² = 550 m² 
Let ‘ℓ’ be the length of the cloth.

∴ ℓ  x b = 550
⇒ ℓ  x 5 = 550
⇒ ℓ = (550/5) m = 110 m
∴ Cost of the cloth = 50 x 110 = 5500.

Question 7. How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?
 Solution: Radius of the lead ball (r) = 1 cm

∴ Volume of a lead ball =  x 1 x 1 x 1 cm³

= (4/3) x (22/7) cm³
Radius of the sphere (r) = 8 cm

∴ Volume of a sphere = x 8 x 8 x 8 cm³

Let the required number of balls = n
∴ [Volume of n-lead balls] = [Volume of the sphere]

Thus, the required number of balls is 512.