Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Created by: Full Circle

Class 9 : Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

The document Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
 Solution:
Let ‘r ’ be the radius of the cylinder.
Here, height (h) = 14 cm and curved surface area = 88 cm2
∵ Curved surface area of a cylinder = 2πrh
∴ 2 πrh = 88
⇒ 2 x (22/7)r x 14 = 88

⇒      Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

⇒ Diameter = 2 x r = 2 x 1 cm = 2 cm


 Question 2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?
 Solution: 
Here, height (h) = 1 m
∵ Diameter of the base = 140 cm = 1.40 m
∴ Radius (r) = (1.40/2) = 0.70 m
Total surface area of the cylinder = 2πr (h + r)
= 2 x(22/7)x 0.70 (1 + 0.70) m
= 2 x 22 x 0.10 (1.70) m2

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Hence, the required sheet is 7.48 m


Question 3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure) Find its. (i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area.
 Solution: 
Length of the metal pipe = 77 cm

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

∵ It is in the form of a cylinder.
∴ Height (h) of the cylinder = 77 cm Inner diameter = 4 cm
⇒ Inner radius (r) =(4/2)= 2 cm
Outer diameter = 4.4 cm
⇒ Outer radius (R) = (4.4/2)= 2.2 cm

(i)  ∵ Curved surface area = 2πrh = 2 x (22/7) x 2 x 77 cm2

          = 2 x 22 x 2 11 cm2 = 968 cm2

(ii)  Outer curved surface area = 2πrh  

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

(iii) Total surface area

= [Inner curved surface] + [outer curved surface area] + [Two base circular lamina]

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

 

Question 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
 Solution:
The roller is in the form of a cylinder diameter of the roller = 84 cm
⇒ Radius of the cylinder =(84/2)
 = 42 cm
Length of the roller = 120 cm
⇒ Height of the cylinder (h) = 120 cm
∴ Curved surface area of the roller (cylinder) = 2 πrh
= 2 x (22/7)x 42 x 120 cm
= 2 x 22 x 6 x 120 cm= 31680 cm2
∴ Area of the playground levelled in one revolution of the roller = 31680 cm2

=   (31680 / 10000) m2

⇒ Area levelled in 500 revolutions = 500 x (31680 / 10000) m2

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev


Question 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹ 12.50 per m2.
 Solution:
Diameter of the pillar = 50 cm
⇒ Radius (r) = (50/2) = 25 cm = (1/4)m
Height (h) = 3.5 m = 3.50 m
∵ Curved surface area of a cylinder = 2πrh
∴ Curved surface area of a cylinder = 2 x (22/7) x(1/4)x 3.50 m2

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

⇒ Curved surface area to be painted = (11/2)m2
Cost of painting ∵ Rate of painting = ₹ 1250 per m2

∴ Cost of Painting 11/2 m2 

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

 


Question 6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
 Solution:
Radius (r) = 0.7 m Let height of the cylinder be ‘h’ metres.
∴ Curved surface area = 2πrh

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

But the curved surface area is 4.4 m2.

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the required height is 1 metre.


Question 7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find: (i) its inner curved surface area. (ii) the cost of plastering this curved surface at the rate of Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev40 per m2.
 Solution: 
Inner diameter of the well = 3.5 m
⇒ Radius of the well (cylinder) (r) =(3.5/2)
.
Depth of the well (= height of the cylinder) h = 10 m
∴ (i) Inner curved surface area = 2πrh

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

(ii) Cost of plastering
∵ Rate of plastering = ₹40 per m2
∴ Total cost of plastering 110 m2 = ₹110 x 40 = ₹4400


Question 8. 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.
 Solution: 
Length of the cylindrical pipe = 28 m
⇒ h = 28 m
Diameter of the pipe = 5 cm
∴ Radius (r) = (5/2) cm = (5/200) m

∴ Curved surface area = 2πrh

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

= (440/100)m2 = 4.40 m2

Thus, the total radiating surface is 4.40 m2


Question 9. Find: (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
 (ii) how much steel was actually used if (1/12) of the steel actually used was wasted in making the tank.
 Solution:
The storage tank is in the form of a cylinder, and diameter of the tank = 4.2 m
∴ The radius (r) = (42/2) = 2.1 m
Height (h) = 4.5 m

Now, (i) Lateral (or curved) surface area = 2πrh
 = 2 x (22/7) x 2.1 x 4.5 m2
= 2 x 22 x 0.3 x 4.5 m2
= 59.4 m2

(ii) Total surface area of the tank = 2πr(r + h)
= 2 x (22/7) x 2.1 (2.1 + 4.5) m2
= 44 x 0.3 x 6.6 m= 87.12 m

Let actual area of the steel used = x m2

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRevsteel was wasted.

∴ Area of steel that wasted = (1/12) x x
=  (x/12)m2

⇒ Area of steel used   Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

⇒ Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

 

Thus, the required area of the steel that was actually used is 95.04 m2


Question 10. In the figure you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.
 Solution:
The lampshade is in the form of a cylinder, where

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

r= (20/2)= 10 cm Height = 30 cm.
∵ A margin of 2.5 cm is to be added to top and bottom.
∴ Total height of the cylinder h = 30 cm + 2.5 cm + 2.5 cm = 35 cm
Now, curved surface area = 2 πrh = 2 x (22/7)x 10 x (35) cm
= 2 x 22 x 10 x 5 cm
= 2200 cm2
Thus, the required area of the cloth = 2200 cm2


Question 11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
 Solution: 
Here, the penholders are in the form of cylinders Radius of a cylinder (r) = 3 cm
Height of a cylinder (h) = 10.5 cm
Since, a penholder must be open from the top.
∴ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

⇒ Surface area of 35 penholders =  

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev                              [∵ There are 35 competitors]

Thus, 7920 cm2 of cardboard was required to be bought.


SURFACE AREA OF A RIGHT CIRCULAR CONE

If we rotate a right angled triangular lamina AOB about OA, a cone is generated, as shown in the adjoining figure. OA is its height (h) and OB is a radius of its base (r) such that OA ⊥ OB. Its base is a circle with centre O. The length AB = l is called its slant height.
We have:

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

(i) l (slant height) =Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev
(ii) Curved (= Lateral surface) area = πrl
(iii) Total surface area = Lateral surface area + Base area = πrl + πr2 = πr(l + r)

Note: In the following πroblems assume π =  (22/7)unless stated otherwise.

Complete Syllabus of Class 9

Dynamic Test

Content Category

Related Searches

Important questions

,

Viva Questions

,

practice quizzes

,

mock tests for examination

,

Sample Paper

,

past year papers

,

Objective type Questions

,

study material

,

shortcuts and tricks

,

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

,

Extra Questions

,

Exam

,

Summary

,

ppt

,

video lectures

,

pdf

,

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

,

MCQs

,

Semester Notes

,

Free

,

Ex 13.2 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

,

Previous Year Questions with Solutions

;