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RD Sharma Solution for Class 9 Mathematics Chapter 20 - Surface Areas and Volume of A Right Circular Cone

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 20 - Surface Areas and Volume of A Right Circular Cone.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

RD Sharma Solution for Class 9 Mathematics Chapter 20 - Surface Areas and Volume of A Right Circular Cone Page/Excercise 20.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

(i)    Slant height of cone = 14 cm
    Let radius of circular end of cone be r.
    CSA of cone =
    
    Thus, the radius of circular end of the cone is 7 cm.

(ii)    Total surface area of cone = CSA of cone + Area of base
                     =
Thus, the total surface area of the cone is 462 .
    

                      


Solution 16

Solution 17

Solution 18

Solution 19

Height (h) of conical tent = 8 m
    Radius (r) of base of tent = 6 m
    Slant height (l) of tent =     
    CSA of conical tent =  = (3.14  10)  = 188.4

Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
    [(L - 0.2 m)  3] m = 188.4
    L - 0.2 m = 62.8 m
    L = 63 m
   
    Thus, the length of the tarpaulin sheet will be 63 m.

Solution 20

Solution 21

Solution 22

Solution 23

RD Sharma Solution for Class 9 Mathematics Chapter 20 - Surface Areas and Volume of A Right Circular Cone Page/Excercise 20.2

Solution 1

(i)    Radius (r) of cone = 6 cm
       Height (h) of cone = 7 cm
       Volume of cone   

(ii)   Radius (r) of cone = 3.5 cm
       Height (h) of cone = 12 cm
       Volume of cone   
(iii)

Solution 2

(i)    Radius (r) of cone = 7 cm
       Slant height (l) of cone = 25 cm
       Height (h) of cone   
       Volume of cone 
       Capacity of the conical vessel =  litres= 1.232 litres (ii)    Height (h) of cone = 12 cm
        Slant height (l) of cone = 13 cm         
        Radius (r) of cone
        Volume of cone   = 314.28 cm3
        Capacity of the conical vessel = litres =  litres.

Solution 3

Solution 4


Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

(i)    Radius of cone =   =14 cm
       Let height of cone be h.
       Volume of cone = 9856 cm3
               h = 48 cm        Thus, the height of the cone is 48 cm.   (ii)   Slant height (l) of cone              
       Thus, the slant height of the cone is 50 cm.   (iii)    CSA of cone = rl =

Solution 14

Radius (r) of pit = 
Depth (h) of pit = 12 m
Volume of pit = Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

Solution 15


RD Sharma Solution for Class 9 Mathematics Chapter 20 - Surface Areas and Volume of A Right Circular Cone Page/Excercise 20.24

Solution 1

A cone has two surfaces as follows: one curved surface and another bottom surface.

Hence, correct option is (b).

Solution 2

begin mathsize 12px style Curved space surface space area space of space straight a space cone space of space radius space apostrophe straight r apostrophe space and space slant space height space apostrophe straight l apostrophe space equals space πrl
Now comma space if space straight r apostrophe space equals space 2 straight r space and space straight l apostrophe space equals space straight l over 2 comma space then
straight C. straight S. straight A. space equals space straight pi open parentheses up diagonal strike 2 straight r close parentheses space open parentheses fraction numerator straight l over denominator up diagonal strike 2 end fraction close parentheses equals space πrl
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 3

begin mathsize 12px style Total space surface space Area space of space straight a space cone space equals space πR left parenthesis straight L space plus space straight R right parenthesis
Where comma space straight R space equals space Radius space and space straight L space equals space Slant space height
therefore space straight T. straight S. straight A. space equals space straight pi space open parentheses straight r over 2 close parentheses space open parentheses 2 straight l space plus space straight r over 2 close parentheses equals πr open parentheses straight l space plus space straight r over 4 close parentheses
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 4

begin mathsize 12px style Volume space of space cylinder space equals space Volume space of space cone
rightwards double arrow up diagonal strike straight pi up diagonal strike straight r squared end strike straight h subscript 1 equals 1 third up diagonal strike straight pi up diagonal strike straight r squared end strike straight h subscript 2 space space space space space space left parenthesis Let space Radius space be space straight r space for space both right parenthesis
rightwards double arrow fraction numerator straight h 1 over denominator straight h 2 end fraction equals 1 third
rightwards double arrow fraction numerator straight h subscript 2 open parentheses cone close parentheses over denominator straight h subscript 1 open parentheses cylinder close parentheses end fraction equals 3 over 1
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 5

begin mathsize 12px style Volume space of space straight a space cone space equals 1 third πR squared straight H
where comma space straight R space equals space Radius space of space Base comma space straight H space equals space height
If space straight R space equals space 3 straight r comma space straight H space equals space straight R space equals space 3 straight r comma
Then space space straight V space equals space fraction numerator 1 over denominator up diagonal strike 3 end fraction cross times space straight pi space cross times space open parentheses 3 straight r close parentheses squared space cross times space up diagonal strike 3 straight r space equals space 9 πr cubed
Heans comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 6

begin mathsize 12px style Let space the space volume space of space 1 to the power of st space cone space equals space 1 third πR subscript 1 superscript 2 straight H subscript 1 space equals space straight V subscript 1
Let space the space volume space of space 2 to the power of nd space cone space equals space 1 third πR subscript 2 superscript 2 straight H subscript 2 space equals space straight V subscript 2
straight V subscript 1 over straight V subscript 2 equals fraction numerator straight R subscript 1 superscript 2 straight H subscript 1 over denominator straight R subscript 2 superscript 2 straight H subscript 2 end fraction equals 1 fourth space space space space space and space space space space space straight d subscript 1 over straight d subscript 2 equals 4 over 5 rightwards double arrow fraction numerator 2 straight R subscript 1 over denominator 2 straight R subscript 2 end fraction equals 4 over 5 rightwards double arrow straight R subscript 1 over straight R subscript 2 equals 4 over 5
rightwards double arrow open parentheses 4 over 5 close parentheses squared space straight H subscript 1 over straight H subscript 2 equals 1 fourth space space space space space space space space space space space space space space space space space space space space space space space
rightwards double arrow straight H subscript 1 over straight H subscript 2 equals 25 over 64
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. space end style

Solution 7

begin mathsize 12px style Let space the space Curved space Surface space Area space of space one space cone space equals space πR subscript 1 open parentheses straight L subscript 1 close parentheses
where comma space straight L space equals space Slant space height comma space space straight R space equals space Radius
Curved space Surface space Area space of space other space cone space equals space πR subscript 2 open parentheses straight L subscript 2 close parentheses
Now comma space πR subscript 1 open parentheses straight L subscript 1 close parentheses space equals space 2 space πR subscript 2 open parentheses straight L subscript 2 close parentheses space and space space space straight L subscript 2 equals space 2 straight L subscript 1
rightwards double arrow up diagonal strike straight pi straight R subscript 1 straight L subscript 1 space equals space 2 up diagonal strike straight pi straight R subscript 2 space open parentheses 2 straight L subscript 1 close parentheses
rightwards double arrow straight R subscript 1 over straight R subscript 2 equals 4 over 1
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

RD Sharma Solution for Class 9 Mathematics Chapter 20 - Surface Areas and Volume of A Right Circular Cone Page/Excercise 20.25

Solution 8

begin mathsize 12px style straight V space equals 1 third πR squared straight H
If space straight R apostrophe space equals space 2 straight R space space and space straight H apostrophe space equals space 2 straight H comma space then
straight V apostrophe space equals space 1 third straight pi open parentheses 2 straight R close parentheses squared open parentheses 2 straight H close parentheses
space space space space space equals space 8 open parentheses 1 third πR squared straight H close parentheses space space space
space space space space space equals space 8 straight V
rightwards double arrow straight V apostrophe space equals space 8 straight V
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 9

begin mathsize 12px style Volume space of space straight a space right space circular space cylinder space of space Height space straight H space and space Radius space straight R space equals space πR squared straight H space equals space straight V subscript 1
Volume space of space straight a space cone space of space height space straight H space and space Radius space straight R space equals space 1 third πR squared straight H space equals space straight V subscript 2
straight V subscript 1 over straight V subscript 2 equals fraction numerator straight pi up diagonal strike straight R squared straight H end strike over denominator 1 third straight pi up diagonal strike straight R squared straight H end strike end fraction space equals 3 over 1 equals space 3 space colon space 1
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 10

begin mathsize 12px style Volume space of space straight a space Right space circular space cylinder space equals space πR subscript 1 superscript 2 straight H subscript 1 space equals space straight V subscript 1
Volume space of space straight a space Right space circular space cone space equals space 1 third πR subscript 2 superscript 2 straight H subscript 2 space equals space straight V subscript 2
If space straight V subscript 1 space equals space straight V subscript 2 space space and space straight R subscript 1 space space equals space straight R subscript 2 comma space then
up diagonal strike πR subscript 1 superscript 2 end strike straight H subscript 1 space equals space 1 third up diagonal strike straight pi open parentheses straight R subscript 1 close parentheses squared end strike space straight H subscript 2
rightwards double arrow straight H subscript 1 over straight H subscript 2 equals 1 third
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 13

begin mathsize 12px style straight C. straight S. straight A space of space straight a space cone space equals space πrl
If space straight l apostrophe space equals straight l space plus space 10 percent sign space of space straight l space equals space straight l space plus space 10 over 100 cross times straight l space equals space space straight l plus straight l over 10
And comma space straight r apostrophe space equals space straight r
straight C. straight S. straight A. equals space πr open parentheses straight l plus straight l over 10 close parentheses space equals space 11 over 10 πrl
So comma space increase space in space straight C. straight S. straight A. space equals space fraction numerator begin display style 11 over 10 πrl space minus space πrl space end style over denominator πrl end fraction cross times space 100 percent sign space equals space 10 percent sign
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 14

begin mathsize 12px style AB space equals space 12 space cm
Area space of space circular space Base space equals space up diagonal strike straight pi straight r squared space equals space 64 up diagonal strike straight pi
rightwards double arrow straight r space equals space 8 space cm
AD space equals space 9 space cm
Consider space triangle ADE space and space triangle ABC comma
angle DAE space equals space angle space BAC space space space space space open parentheses common close parentheses
angle ADE space equals space angle ABC space space space space left parenthesis each space 90 degree right parenthesis
angle AED space equals space angle ACB space space space space space space space space open parentheses third space angle space will space also space be space same close parentheses
Hence space triangle ADE space tilde space triangle ABC
So space AD over AB equals DE over BC
rightwards double arrow 9 over 12 equals DE over 8
rightwards double arrow DE space equals space 6 space cm
Radius space of space Base space of space new space cone space equals space 6 space cm
rightwards double arrow Area space equals space straight pi open parentheses 6 close parentheses squared space equals space 36 straight pi space cm squared
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 11

begin mathsize 12px style Curved space Surface space Area space of space cone space equals space πRL
where comma space straight R space equals space Radius space and space straight L space equals space Slant space height
Ratio space of space straight C. straight S. straight A. space of space two space cones comma
straight C. straight S. straight A subscript 1 space colon space straight C. straight S. straight A subscript 2 space equals space πR subscript 1 straight L subscript 1 space colon space πR subscript 2 straight L subscript 2
If space straight L subscript 1 over straight L subscript 2 equals 5 over 4 comma space then space fraction numerator 2 straight R subscript 1 over denominator 2 straight R subscript 2 end fraction equals 1 space space space space left parenthesis because space straight R subscript 1 equals straight R subscript 2 right parenthesis
rightwards double arrow fraction numerator straight C. straight S. straight A subscript 1 over denominator straight C. straight S. straight A subscript 2 end fraction space equals fraction numerator πR subscript 1 straight L subscript 1 over denominator πR subscript 2 straight L subscript 2 end fraction
rightwards double arrow fraction numerator straight C. straight S. straight A subscript 1 over denominator straight C. straight S. straight A subscript 2 end fraction equals straight R subscript 1 over straight R subscript 1 cross times space 5 over 4
rightwards double arrow fraction numerator straight C. straight S. straight A subscript 1 over denominator straight C. straight S. straight A subscript 2 end fraction equals 5 over 4
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 12

begin mathsize 12px style Let space the space volume space of space cone space 1 space equals 1 third πR subscript 1 superscript 2 straight H subscript 1 equals straight V subscript 1
Let space the space volume space of space cone space 2 space equals 1 third πR subscript 2 superscript 2 straight H subscript 2 equals straight V subscript 2
straight V subscript 1 over straight V subscript 2 equals fraction numerator up diagonal strike begin display style 1 third straight pi space end style end strike straight R subscript 1 superscript 2 straight H subscript 1 over denominator up diagonal strike 1 third straight pi space end strike straight R subscript 2 superscript 2 straight H subscript 2 end fraction equals fraction numerator straight R subscript 1 superscript 2 over denominator straight R subscript 2 superscript 2 end fraction space space straight H subscript 1 over straight H subscript 1 space space space space space space space open curly brackets straight H subscript 1 over straight H subscript 2 equals 1 fourth comma space space straight R subscript 1 over straight R subscript 2 equals 4 over 1 space space space space left parenthesis given right parenthesis close curly brackets space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open parentheses 4 over 1 close parentheses squared space open parentheses 1 fourth close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 over 1
rightwards double arrow straight V subscript 1 space colon space straight V subscript 2 space equals 4 space colon space 1
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 15

begin mathsize 12px style Let space the space radius space of space the space cone space equals space straight R space and space height space equals space straight H
Then comma space Volume space equals space 1 third πR squared straight H
Now comma space straight R apostrophe space equals space straight R space plus space 20 percent sign space of space straight R space equals straight R plus straight R over 5 equals fraction numerator 6 straight R over denominator 5 end fraction
space space space space space space space space space space straight H apostrophe space equals space straight H plus 20 percent sign space of space straight H space equals space straight H plus straight H over 5 equals fraction numerator 6 straight H over denominator 5 end fraction
New space volume comma space straight V apostrophe space equals space 1 third πR apostrophe squared straight H apostrophe
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 third straight pi open parentheses fraction numerator 6 straight R over denominator 5 end fraction close parentheses squared space open parentheses fraction numerator 6 straight H over denominator 5 end fraction close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 216 over 125 open parentheses 1 third πR squared straight H close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 216 over 125 straight V
percent sign space increase space in space Volume space equals space fraction numerator straight V apostrophe space minus space straight V over denominator straight V end fraction cross times 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator begin display style fraction numerator 216 straight v over denominator 125 end fraction minus straight V space end style over denominator straight v end fraction cross times 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 91 over 125 cross times 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 72.8 percent sign
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space almost equal to 73 percent sign
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 16

begin mathsize 12px style For space straight a space cone comma
straight V space equals space 1 third πR squared straight h space
straight S space equals space Curved space Surface space Area space equals space πRL
straight L space equals space square root of straight h squared space plus space straight R squared end root
3 πVh cubed space minus space straight S squared straight h squared space plus space 9 straight V squared space equals space 3 straight pi open parentheses 1 third πR squared straight h close parentheses straight h cubed space minus space straight pi squared straight R squared left parenthesis straight h squared space plus space straight R squared right parenthesis space straight h squared space plus space up diagonal strike 9 cross times fraction numerator 1 over denominator up diagonal strike 9 end fraction straight pi squared straight R to the power of 4 straight h squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space up diagonal strike straight pi squared straight R squared straight h to the power of 4 end strike space minus space up diagonal strike straight pi squared straight R squared straight h to the power of 4 end strike space minus space straight pi up diagonal strike blank squared straight R to the power of 4 straight h squared end strike space plus space up diagonal strike straight pi squared straight R to the power of 4 straight h squared end strike
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 17

begin mathsize 12px style AD over AB equals DF over BC
rightwards double arrow fraction numerator straight h divided by 2 over denominator straight h divided by 2 plus straight h divided by 2 end fraction equals DF over BC rightwards double arrow DF over BC equals 1 half rightwards double arrow DF equals BC over 2 equals straight r over 2
Volume space of space full space cone space equals space 1 third πr squared straight h
volume space of space small space cone space formed space equals 1 third straight pi open parentheses straight r over 2 close parentheses squared straight h over 2 equals 1 third straight pi straight r squared over 4 straight h over 2 space space equals 1 over 8 open parentheses fraction numerator πr squared straight h over denominator 3 end fraction close parentheses
Ratio space of space volume space of space two space parts
equals space fraction numerator Volume space of space small space cone over denominator Volume space of space full space cone space minus space Volume space of space small space cone end fraction
equals fraction numerator begin display style 1 over 8 open parentheses fraction numerator πr squared straight h over denominator 3 end fraction close parentheses end style over denominator begin display style fraction numerator πr squared straight h over denominator 3 end fraction minus fraction numerator πr squared straight h over denominator 8 cross times 3 end fraction end style end fraction equals 1 over 7
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 20 - Surface Areas and Volume of A Right Circular Cone for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 20 - Surface Areas and Volume of A Right Circular Cone.

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CBSE IX - Mathematics

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