Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Short Answer Type Questions: Surface Areas & Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

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 m= 42 m2
∴ The required lateral surface of the room is 42 m2.
 

Question 2. 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

⇒    Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Now, Base area =   Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes


Question 3. 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 = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes


Question 4. 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 =  Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Question 5. If the radius of a sphere is such that πr2 = 6cm2 then find its total surface area.
 Solution: 
∵ πr2 = 6cm
∴ Curved S.A. of the hemisphere Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes
= 2 × 6 cm2 = 12 cm
Also, plane S.A. of the hemisphere = π r2 = 6 cm
⇒ Total S.A. = C.S.A. + plane S.A. = 12 cm2 + 6 cm2 = 18 cm2

 

 

Question 6. 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,    Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ 4 × π × 5 × 5 = 5 × π × 4 × ℓ
⇒              Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ Volume of the cone = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes


Question 7. 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 = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Increased radius = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ Increased volume = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

= (4/3) x π x 1.331r3

Thus, Increase in volume Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ Percentage increase in volume Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

= 0.331 × 100%
= 33.1%.


Question 8. 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,

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

= 25 cm
∴ Slant height = 25 cm.
 

Question 9. 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 Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes2 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 cm
= 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

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ Cost of levelling the playground = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes2 x 15840 = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes31680.


Question 10. 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 Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes50 per metre.
 Solution:
Radius of the base of the tent (r) = 7 m
Height (h) = 24 m
Slant height (ℓ)=Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Now, curved surface area of the conical tent = πrℓ
= (22/7) x 7 x 25 m2 = 22 x 25 m2 = 550 m
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 = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes50 x 110 = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes5500.


Question 11. 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 = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes x 1 x 1 x 1 cm3

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

∴ Volume of a sphere = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumesx 8 x 8 x 8 cm3

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

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Thus, the required number of balls is 512.

The document Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

1. What is the formula for finding the surface area of a rectangular prism?
Ans. The formula for finding the surface area of a rectangular prism is 2(length × width + width × height + height × length).
2. How do you find the volume of a cone?
Ans. The volume of a cone can be found using the formula 1/3 × π × radius^2 × height, where π is approximately 3.14.
3. What is the surface area of a sphere?
Ans. The surface area of a sphere can be found using the formula 4 × π × radius^2, where π is approximately 3.14.
4. How do you calculate the total surface area of a cylinder?
Ans. The total surface area of a cylinder can be calculated using the formula 2πr(r+h), where r is the radius of the base and h is the height of the cylinder.
5. How do you find the volume of a pyramid?
Ans. The volume of a pyramid can be found using the formula 1/3 × base area × height, where the base area is the area of the base of the pyramid and height is the perpendicular height from the base to the apex.
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