RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

Other Exercises

Mark correct alternative in each of the following:
Question 1.
In a cylinder, if radius is doubled and height is halved, curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
Let radius of the first cylinder (r1) = r
and height (h1) = h
Surface area = 2πrh
If radius is doubled and height is halved
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q1.1
∴ Their surface area remain same (c)

Question 2.
Two cylindrical jars have their diameters in the ratio 3:1, but height 1:3. Then the ratio of their volumes is
(a) 1 : 4
(b) 1 : 3
(c) 3 : 1
(d) 2 : 5
Solution:
Sol. Ratio in the diameters of two cylinder = 3:1
and ratio in their heights = 1:3
Let radius of the first cylinder (r1) = 3x
and radius of second (r2) = x
and height of the first (h1) = y
and height of the second (h2) = 3y
Now volume of the first cylinder = πr2h
= π(3x)2 x y = 9πx2y
and of second cylinder = π(x2) (3y)
∴ Ratio between then = 9πx2y : 3πx2y
= 3 : 1 (c)

Question 3.
The number of surfaces in right cylinder is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The number of surfaces of a right cylinder is three. (c)

Question 4.
Vertical cross-section of a right circular cylinder is always a
(a) square
(b) rectangle
(c) rhombus
(d) trapezium
Solution:
The vertical cross-section of a right circular cylinder is always a rectangle. (b)

Question 5.
If r is the radius and h is height of the cylinder the volume will be
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q5.1
Solution:
Volume of a cylinder = πr2h (b)

Question 6.
The number of surfaces of a hollow cylindrical object is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The number of surfaces of a hollow cylindrical object is 4. (d)

Question 7.
If the radius of a cylinder is doubled and the height remains same, the volume will be
(a) doubled
(b) halved
(c) same
(d) four times
Solution:
If r be the radius and h be the height, then volume = πr2h
If radius is doubled and height remain same,
the volume will be
= π(2r)2h = π x 4r2h
= 4πr2h = 4 x Volume
The volume is four times (d)

Question 8.
If the height of a cylinder is doubled and radius remains the same, then volume will be
(a) doubled
(b) halved
(c) same
(d) four times
Solution:
If r be the radius and h be the height, then volume of a cylinder = πr2h
If height is doubled and radius remain same, then volume = πr2(2h) = 2πr2h
∴ Its doubled (a)

Question 9.
In a cylinder, if radius is halved and height is doubled, the volume will be
(a) same
(b) doubled
(c) halved
(d) four times
Solution:
Let r be radius and h be height, then Volume = πr2h
If radius is halved and height is doubled
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q9.1

Question 10.
If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q10.1
Solution:
Let diameter of the base of a cylinder (r) = h
Then its height (h) = h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q10.2

Question 11.
A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m2, is
(a) 40π
(b) 80π
(c) 160π
(d) 200π
Solution:
Diameter of a cylindrical tunnel = 2 m
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1m
and length (h) = 40 m
Curved surfae area = 2πrh = 2 x π x 1 x 40 = 80π (b)

Question 12.
Two circular cylinders of equal volume have their heights in the ratio 1 : 2. Ratio of their radii is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q12.1
Solution:
Let r1 and h1 be the radius and height of the
first cylinder, then
Volume = πr12h1
Similarly r1 and h2 are the radius and height of the second cylinder
∴ Volume = πr2h2
But their volumes are equal,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q12.2

Question 13.
The radius of a wire is decreased to one- third. If volume remains the same, the length will become
(a) 3 times
(b) 6 times
(c) 9 times
(d) 27 times
Solution:
In the first case, r and h1, be the radius and height of the cylindrical wire
∴ Volume = πr2h1 …(i)
In second case, radius is decreased to one third
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q13.1
∴ In second case height is 9 times (c)

Question 14.
If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q14.1
Solution:
Let r be the radius and h be the height then volume = πr2h
If height is doubled and volume is same and let x be radius then πr2h = π(x)2 x 2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q14.2

Question 15.
The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q15.1
Solution:
Let r be the radius and h be the height, then volume = πr2h
This volume is \(\frac { 1 }{ 4 }\) of the volume of a rectangular box
∴ Volume of box = 4πr2h
Let side of base of box = x and height h,
then volume = x2h
∴ 4πr2h = x2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q15.2

Question 16.
The height ft of a cylinder equals the circumference of the cylinder. In terms of ft, what is the volume of the cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q16.1
Solution:
In a cylinder,
h = circumference of the cylinder
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q16.2

Question 17.
A cylinder with radius r and height ft is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?
(a) 2πr(r + h)
(b) πr(r + 2h)
(c) πr(2r + h)
(d) 2πr2 + h
Solution:
r is the radius of the base and ft is the height of a closed cylinder
Then total surface area = 2πr(r + h ) (a)

Question 18.
The height of sand in a cylindrical-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q18.1
Solution:
Let h be the height and d be the diameter of a cylinder, then
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q18.2

Question 19.
Two steel sheets each of length a1 and breadth a2 are used to prepare the surfaces of two right circular cylinders – one having volume v1 and height a2 and other having volume v2 and height a1. Then,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.1
Solution:
Length of each sheet = a1
and breadth = a2
Volume of cylinder = πr2h
In first case,
v1 is volume and a2 is the height
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.3

Question 20.
The altitude of a circualr cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q20.1
Solution:
In first case,
Let r be the radius and h be the height of the cylinder. Then,
∴ Lateral surface area = 2πrh
In second case,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q20.2

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS are helpful to complete your math homework.

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NCERT Solutions for Class 9 Science Chapter 12 Sound

NCERT Solutions for Class 9 Science Chapter 12 Sound

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 12 Sound. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Physics) Chapter 12 – Sound solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 12 – Sound Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
Explain how sound is produced by your school bell.

                                                            Or

Sound is produced when your school bell is struck with a hammer. Why ? (CBSE 2011)
Answer:
When school bell is struck by a hammer, it starts vibrating. Since the vibrating bodies produce sound, so the vibrating school bell produces the sound.

Question 2.
Why are sound waves called mechanical waves ? (CBSE 2012)
Answer:
Sound waves are characterised by the motion of particles of a medium. Hence sound waves are called mechanical waves.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend ?
Answer:
Sound waves need material medium like air to move from one place to another place. Since there is no air on the moon, so sound cannot travel from one place to another place. Hence, we cannot hear sound on the moon.

Question 4.
Which wave property determines
(a) loudness,
(b) pitch ? (CBSE 2011, 2012)
Answer:
(a) Amplitude of the wave determines loudness.
(b) Frequency of the wave determines pitch.

Question 5.
Guess which sound has a higher pitch : guitar or car horn ?
Answer:
Guitar, because frequency of sound produced by guitar is higher than the sound produced by car horn.

Question 6.
What are wavelength, frequency, time period and amplitude of a sound wave ? (CBSE 2012)
Answer:
Wavelength (or length of a wave): The distance between two successive regions of high pressure or high density {or compressions) or the distance between two successive regions of low pressure or low density (or rarefactions) is known as wavelength of a sound wave. It is denoted by λ (read as lambda).
In S.I., unit of wavelength is metre (m).
Frequency: The number of compressions or rarefactions crossing a point per unit time is known as the frequency of a sound wave. It is denoted by μ (read as Neu). In S.I., unit of frequency is hertz (Hz).
1 hertz = one oscillation completed by a vibrating body or a vibrating particle in one second.
Time period: Time taken by two consecutive compressions or rarefactions to cross a fixed point. Amplitude. The maximum displacement of a vibrating body from its rest position or mean position.

Question 7.
How are the wavelength and frequency of a sound wave related to its speed ? (CBSE 2011, 2012)
Answer:
V = vλ.

Question 8.
Calculate the wavelength of a sound wave whose frequency is 200 Hz and speed is 440 m/s in a given medium.
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 1

Question 9.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of sound. What is the time interval between successive compressions from the source ?
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 2

Question 10.
Distinguish between loudness and intensity of sound.
Answer:

Loudness Intensity of a sound
1.Loudness is a subjective quantity. It depends upon the sensitivity of the human ear. A sound may be loud for a person but the same sound may be feeble for another person who is hard of hearing even when both are sitting at the same distance from the source of sound.

 

 Intensity of a sound is an objective physical quantity. It does not depend on the sensitivity of a human ear.
 2. Loudness cannot be measured as a physical quantity because it is just sensation which can be felt only.

 

 Intensity of a sound can be measured as a physical quantity.

Question 11.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature ?
Answer:
Sound travels the fastest in iron.

Question 12.
An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m s-1 ? (CBSE 2011)
Answer:
Time taken by sound to travel from the source to the reflecting surface, t = 3/2 = 1.5 s
Speed, v = 342 m s
Distance of reflecting surface from the source, S = vt = 342 x 1.5 = 513 m.

Question 13.
Why are the ceilings of concert halls curved ? (CBSE 2011, 2012)
Answer:
So that the sound after reflection from the ceiling reaches all the corners of the hall.

Question 14.
What is the audible range of the average human ear ? (CBSE, 2011, 2012, 2015)
Answer:
20 Hz to 20,000 Hz.

Question 15.
What is the range of frequencies associated with :
(a) infra sound ?
(b) ultra sound ? (CBSE 2011, 2013)
Answer:
(a) Frequencies less than 20 Hz and greater than zero.
(b) Frequencies greater than 20,000 Hz and equal to 107 Hz.

Question 16.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff ?
Answer:
Time taken by the pulse to go from submarine to the cliff, t = 1.02/2 =0.51 s
Speed of sound, v = 1531 m/s
Distance of cliff from the submarine, S = vt = 1531 x 0.51 = 780.81 m.

NCERT CHAPTER END EXERCISE

Question 1.
What is sound and how is it produced ? (CBSE2012)
Answer:
Sound is a form of energy which produces the sensation of hearing in our ears. Sound is produced by forcing an object to vibrate. In other words, sound is produced by a vibrating object.

Question 2.
Why is sound wave called a longitudinal wave ? (CBSE 2012)
Answer:
When sound waves travel in medium, the particles of the medium vibrate about their equilibrium positions along the direction of the propagation of the waves.

Question 3.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room ? (CBSE 2011)
Answer:
Timber or quality of sound.

Question 4.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why ?

          Or

There is some time interval between observing a flash and hearing a thunder. Explain. (CBSE 2013)
Answer:
Thunder is heared after some time interval the flash is seen because speed of sound is less than the speed of light.

Question 5.
A person has a hearing range of 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies ?
Answer:
Take the speed of sound.in air as 344 m s-1.
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 3

Question 6.
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound waves in air and in the aluminium to reach the second child. (Speed of sound in aluminium is 6420 m s-1 and in air is 346 m s-1). (CBSE 2013)
Answer:
Let l = length of the rod
Time taken by sound to travel distance / in aluminium rod,
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 4
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 5

Question 7.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute ?
(CBSE 2011, 2012)
Answer:
Frequency of source = 100 Hz.
∴ Number of times the source of sound vibrates in 1 s = 100
∴ Number of times the source vibrates in a minute or 60 s = 100 x 60 = 6000.

Question 8.
Does sound follow the same laws of reflection as light does ? Explain.
Answer:
Yes. Sound waves are reflected just like light waves.

Question 9.
When a sound is reflected from a distant object, an echo is produced. Let the distance of the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day ?
Answer:
Let d = distance between the reflecting surface and the source of sound
v = speed of sound in air.
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 6
On a hotter day, speed of sound increases with increase in temperature. Hence, the time after which echo is heard decreases. If the time taken by the reflected sound is less than 0-1 s after the production of original sound, then echo is not heard. However, if this time is greater than 0-1 s, then echo will be heard.

Question 10.
Give two practical applications of multiple reflection of sound waves. (CBSE 2011, 2012)
Answer:

  1. Megaphone
  2. Hearing aid.

Question 11.
A stone is dropped from the top of a tower 500 m high into a pond,of water at the base of the tower. When is the splash heard at the top ? Given g = 10 m s-2 and speed of sound = 340 m s-1 .
(CBSE 2011, 2012)
Answer:
Time, after which splash is heard = time taken by the stone to reach the surface of water in a pond + time taken by the sound of splash to reach the top of tower.
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 7
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 8

Question 12.
A sound wave travels at a speed of 399 m s-1. If its wavelength is 1.5 m, what is the frequency of the wave ? Will it be audible ?
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 9
Since audible range of frequencies is 20 Hz to 20,000 Hz. Hence, the given frequency will not be audible.

Question 13.
What is reverberation ? How can it be reduced ? (CBSE 2012, 2014)
Answer:
The phenomenon of prolongation of original sound due to the multiple reflection of sound waves even after the source of sound stops producing sound is called reverberation.
Reverberation can be reduced by covering the roof and walls of a hall by sound absorbing materials.

Question 14.
What is loudness of sound ? What factors does it depend on ? (CBSE 2011)
Answer:
Loudness of a sound is a subjective quantity which causes unpleasant effect in our ear.
Loudness depends upon the amplitude of the vibrating body and the sensitivity of human ear.

Question 15.
Explain how bats use ultrasound to catch a prey. (CBSE 2011, 2012)

                                                  Or

Bats have no eyes, yet they can ascertain distances. (CBSE 2013)
Answer:
Bats can produce ultrasonic waves by flapping their wings. They can also detect these waves. The ultrasonic waves produced by the bats after reflection from the obstacles like buildings guide them to remain away from the obstacles during their flights. Hence, they can fly during night without hitting the obstacles. Bats also catch their prey during night with the help of ultrasonic waves. The ultrasonic waves produced by a bat spread out. These waves after reflecting from a prey say an insect reach the bat. Hence, the bat can easily locate its prey (Figure 24).
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 10

Question 16.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m. (CBSE 2011)
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 11

NCERT Solutions for Class 9 Science Chapter 12 Sound

Hope given NCERT Solutions for Class 9 Science Chapter 12 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B.

Other Exercises

Question 1.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 1.1
(i) Draw a line segment AB = 5.2 cm.
(ii) With centre A and radius 7.6 cm. and with centre B and radius 4.7 cm. draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 4.7 cm and with centre C and radius 5.2 cm draw arcs which intersect each other at D.
(v) Join AD and CD.
ABCD is the required parallelogram.

Question 2.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 4.3 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 2.1
(ii) With centre A and radius 4 cm. and with centre B and radius 6.8 cm., draw arcs which intersect each other at D.
(iii) Join AD and BD.
(iv) Again with centre B and radius 4 cm. and with centre D and radius 4.3 cm., draw arcs intersecting each other at C.
(v) Join DC and BC. ABCD is the required parallelogram.

Question 3.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At Q, draw a ray making an angle of 60° and cut off QR = 6 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 3.1
(iii) With centre P and radius 6 cm. and with centre R and radius 4 cm draw arcs intersecting each other at S.
(iv) Join RS and PS.
PQRS is the required parallelogram. Q.

Question 4.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 4.1
(ii) At C, draw a ray making an angle of 120° and cut off CD = 4.8 cm.
(iii) With centre B and radius 4.8 cm. with centre D and radius 5 cm, draw arcs intersecting each other at A.
(iv) Join AB and AD
ABCD is the required parallelogram.

Question 5.
Solution:
Steps of Construction :
We know that diagonals of a parallelogram bisect each other.
(i) Draw a line segment AB = 4.4 cm.
(ii) With centre A and radius \(\\ \frac { 5.6 }{ 2 } \) cm and with centre B and radius \(\\ \frac { 7 }{ 2 } \) = 3.5 cm. draw arcs
intersecting each other at O.
(iii) Join AO and BO and produce them to C and D respectively such that OC = 2.8 cm and OD = 3.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 5.1
(iv) Join AD, CD and BC
ABCD is the required parallelogram

Question 6.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 6.1
(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a perpendicular AX and cut off AL = 2.5 cm.
(iii) Through L, draw a line PQ parallel to AB.
(iv) From A, draw an arc of radius 3 -4 cm which intersects the line PQ at C.
(v) Join AC. BC
(vi) From PQ, cut off CD = AB.
(vii) Join AD
(viii) From C, draw a perpendicular CM to AB.
ABCD is the required parallelogram.

Question 7.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 7.1
(i) Draw a line segment AC = 3.8 cm.
(ii) Bisect it at O.
(iii) At O, draw a ray making an angle of 60° and produce it both sides.
(iv) From O cut off OB = OD = 2.3 cm.
(v) Join AB, BC, CD and AD.
ABCD is the required parallelogram.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 11 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 8.1
(ii) At B, draw a perpendicular and cut off BC = 8.5 cm.
(iii) With centre A and radius 8.5 cm and with centre C and radius 11 cm, draw arcs intersecting each other at D.
(iv) Join AD and CD.
ABCD is the required rectangle.

Question 9.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 9.1
(ii) At A and B draw perpendiculars and
cut off AD = BC = AB = 6.4 Cm.
(iii) Join CD.
ABCD is the required square.

Question 10.
Solution:
Steps of Construction :
(i) Draw a line segment AC = 5.8 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 10.1
(ii) Draw its perpendicular bisector intersecting AC at O.
(iii) From O, cut off OD = OB = 2.9 cm.
\(\qquad =\left( \frac { 1 }{ 2 } BD \right) \)
(iv) Join AB, BC, CD and DA. ABCD is the required square.

Question 11.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 11.1
(i) Draw a line segment QR = 3.6 cm.
(ii) At Q, draw a ray QX making an angle of 90°.
(iii) With centre R and radius 6 cm. draw an arc which intersects QX at P.
(iv) Join PR.
(v) With centre P and radius equal to QR and with centre R and radius equal to QP, draw arcs intersecting each other at S.
(vi) Join PS and RS.
PQRS is the required rectangle.
The length of other side PQ = 4.8 cm.

Question 12.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 12.1
(i)Draw a line segment AC = 8 cm.
(ii)Draw its perpendicular bisector intersecting it at O.
(iii)From O, cut off OB = OD = 3 cm.
(iv)Join AB, BC, CD and DA.
ABCD is the required rhombus.

Question 13.
Solution:
Steps of Construction :
(i)Draw a line segment AC = 6.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 13.1
(ii) With centres A and C and radius equal to 4 cm., draw arcs which intersect each other on both sides of line segment AC at B and D respectively.
(iii) Join AB, BC, CD and DA.
ABCD is the required rhombus.

Question 14.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 14.1
(i)Draw a line segmentAB = 7.2 cm.
(ii)At A draw a ray AX making an angle of 60° and cut off AD = 7.2 cm.
(iii)With centres D and B, and radius 7.2 cm., draw arcs intersecting each other at C.
(iv)Join CD and CB.
ABCD is the required rhombus.

Question 15.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6 cm
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 15.1
(ii)At B, draw a ray BX making an angle of 75° and cut off BC = 4 cm.
(iii) At C, draw a ray CY making an angle of 180° – 75° = 105°
So that CY may be parallel to AB.
(iv) From CY, Cut off CD = 3.2 cm.
(v) Join DA.
ABCD is the required trapezium.

Question 16.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 7 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 16.1
(ii) At B, draw a ray BX making an angle of 60° and cut off BC = 5 cm.
(iii) At C, draw a ray CY making an angle of (180° – 60°) = 120° so that CY || AB.
(iv) With centre A and radius 6.5 cm. draw an arc intersecting CY at D.
(v) Join AD,
ABCD is the required trapezium.

Hope given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A.

Other Exercises

Question 1.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 4.2 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 1.1
(ii) With centre A and radius 8 cm and with centre B and radius 6 cm., draw arcs intersecting each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 5 cm. and with centre C, radius 5 2 cm. draw arcs intersecting each other at D.
(v) Join AD and CD. ABCD is the required quadrilateral.

Question 2.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 5.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 2.1
(ii) With Centre P and radius 4 cm. and with centre Q and radius 4.6 cm., draw arcs intersecting each other at R.
(iii) Join PR and QR.
(iv) Again with centre P and radius 3.5 cm. and with centre R and radius 4.3 cm. draw arcs intersecting each other at S.
(v) Join PS and RS. PQRS is the required quadrilateral.

Question 3.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 3.1
(ii) With centre A and radius 4.5 cm. and with centre B and radius 5.6 cm. draw arcs intersecting each other at D.
(iii) Join AD and BD.
(iv) With centre B and radius 3.8 cm. and with centre D and radius 4.5 cm., draw arcs intersecting each other at C.
(v) Join BC and DC. ABCD is the required quadrilateral.

Question 4.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.6 cm.
(ii) With centre A and radius 4.6 cm. and with centre B and radius 3.3 cm. draw arcs intersecting each other at C.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 4.1
(iii) Join AC and BC.
(iv) Again with centre A and radius 2.7 cm. and centre B and radius 4 cm., draw arcs intersecting each other at D.
(v) Join BD, AD and CD. ABCD is the required quadrilateral.

Question 5.
Solution:
Steps of Construction :
(i) Draw a line segment RS = 5 cm.
(ii) With centre R and S, radius 6 cm. each, draw arcs intersecting each other at R
(iii) Join PR and PS.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 5.1
(iv) With centre R and radius 7.5 cm. and with centre S and radius 10 cm, draw arcs intersecting each other at Q.
(v) Join RQ, SQ and PQ. PQRS is the required quadrilateral. Measuring the fourth sides PQ, it is 4.7 cm. (approx.)

Question 6.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 6.1
(ii) With centre A and radius 5.7 cm. and with centre B and radius 4 cm., draw arcs intersecting each other at D.
(iii) Join BD and AD.
(iv) Again with centre A and radius 8 cm and with centre D and radius 3 cm., draw arcs intersecting each other at C.
(v) Join AC, BC and DC. ABCD is the required quadrilateral

Question 7.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 7.1
(ii) At B, draw a ray BX making an angle of 120° using protractor and cut off BC = 3.5 cm
(iii) With centres A and C and radius 5.2 cm, draw arcs intersecting each other at D.
(iv) Join CD and AD. ABCD is the required quadrilateral.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line AB = 2.9 cm.
(ii) At A, draw a ray AX making an angle of 70° with AB. Using protractor and cut off AD = 3.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 8.1
(iii) With centre B and radius 3.2 cm and with centre D and radius 2.7 cm., draw arcs intersecting each other at C.
(iv) Join BC and DC. ABCD is the required quadrilateral.

Question 9.
Solution:
Steps of Construction
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 9.1
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray BX making an angle of 125° and cut off BA = 3.5 cm.
(iii) At C, draw a ray CY making an angle of 60° and cut off CD = 4.6 cm
(iv) Join AD. ABCD is the required quadrilateral.

Question 10.
Solution:
Steps of Construction :
(i) Draw a line segment QR = 5.6 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 10.1
(ii) At Q, draw a ray QX making an angle of 45° and cut off QP = 6 cm.
(iii) At R, draw a ray RY making an angle of 90° and cut off RS = 2.7 cm.
(iv) Join SP PQRS is the required quadrilateral.

Question 11.
Solution:
Steps of Construction :
∠A = 50°, ∠B = 105° and ∠D = 80°
and ∠A + ∠B + ∠C + ∠D = 360°
=> 50° + 105° + ∠C + 80° = 360°
=> ∠C + 235° = 360°
=> ∠C = 360° – 235°
=> ∠C = 125°
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 11.1
(i) Draw a line segment AB = 5.6 cm.
(ii) At B, draw a ray BY making an angle of 105° and cut off BC = 4 cm.
(iii) At C, draw a ray CZ making an. angle of 125° and at A, a ray AX making an angle of 50° intersecting each other at D.
then ∠D = 80°
ABCD is the required quadrilateral.

Question 12.
Solution:
∠P + ∠Q + ∠R + ∠S = 360°
100° + ∠Q + 100° + 75° = 360°
=> ∠Q + 275° = 360°
=> ∠Q = 360° – 275°
∠Q = 85°
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 12.1
(i) Draw a line segment PQ = 5 cm.
(it) At Q, draw a ray QX making an angle of 85° and cut off QR = 6.5 cm.
(iii) At R, draw a ray making an angle of 100° and at P, another ray making an angle of 100° which intersect each other at S. then ∠S = 75°
PQRS is the required quadrilateral.

Question 13.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 13.1
(i) Draw a line segment AB = 4 cm.
(ii) At B, draw a ray BX making an angle of 90°.
(iii) From A, draw an arc of 5 cm. radius intersecting BX at C.
(iv) Join AC.
(v) At C, draw a ray CY making an angle of 90°.
(vi) From A, draw an arc of radius 5.5 cm. which intersects CY at D.
(vii) Join AD.
ABCD is the required quadrilateral.

 

Hope given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

Other Exercises

Question 1.
Write the number of surface of a right circular cylinder.
Solution:
Three, two circular and one curved.

Question 2.
Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.
Solution:
∵ Radius = r
and height = h
∴ Curved surface area = 2πrh
and total surface area = 2πr(h + r)
∴ Ratio = 2πr(h + r) : 2πrh
= h + r : h

Question 3.
The ratio between the radius of the base and height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.
Solution:
Ratio in radius and height of the cylinder = 2 : 3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Q3.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Q3.2

Question 4.
If the radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then find the ratio of their volumes.
Solution:
Ratio of radii of two cylinder = 2:3
Let radius of first cylinder (r1) = 2x
and second cylinder (r2) = 3x
and ratio in their heights = 5:3
Let height of first cylinder (h1) = 5y
and height of second (h2) = 3y
∴ Volume of the first cylinder =πr2h
= π x (2x)2 x 5y = 20πx2y
and volume of second cylinder = π(3x)2 x 3y = 27πx2y
Now ratio between then,
= 20πx2y: 21πx2y
= 20 : 27

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS are helpful to complete your math homework.

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NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy

NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy

MULTIPLE CHOICE QUESTIONS

Question 1.
When a body falls freely towards the earth, then its total energy
(a) increases
(b) decreases
(c) remains constant
(d) first increases and then decreases,
Answer:
(c) Total energy = K.E. + P.E.
When a body falls freely, its K.E. increases and P.E. decreases but the sum of K.E and P.E. remains the same.

More Resources

Question 2.
A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process, the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial.
Answer:
(a). Potential energy does not depend on the velocity of a body.

Question 3.
In case of negative work, the angle between the force and displacement is
(a) 0°
(b) 45°
(c) 90°
(d) 180°.
Answer:
(d) Explanation : W = FS cos θ. When θ = 180°, cos 180° = – 1 and W = – FS

Question 4.
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower.
When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy.
Answer:
(a). Freely falling bodies moves with constant acceleration.

Question 5.
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g = 10 m s-2)
(a) 6 x 103 J
(b) 6 J
(c) 0.6 J
(d) zero.
Answer:
(d) Explanation : W = FS cos 90° = 0.

Question 6.
Which one of the following is not the unit of energy ?
(a) joule
(b) newton metre
(c) kiiowatt
(d) kolwatt hour.
Answer:
(c) It is the unit of power.

Question 7.
The work done on an object does not depend upon the
(a) displacement
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object.
Answer:
(d) W = FS cos θ.

Question 8.
Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy
Answer:
(d). The energy possessed by a body by virtue of its position is called potential energy.

Question 9.
A body is falling from a height h. After it has fallen a height h/2, it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy.
Answer:
(c).

SHORT ANSWER QUESTIONS

Question 10.
A rocket is moving up with a velocity u. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 1

Question 11.
Avinash can run with a speed of 8 m s-1 against the frictional force of 10 N, and Kapil can move with a speed of 3 m s-1 against the frictional force of 25 N. Who is more powerful and why ?
Answer:
P = Fu
Power of Avinash = 10 x 8 = 80 W
Power of Kapil = 25 x 3 = 75 W
So, Avinash is more powerful.

Question 12.
A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round about (Fig. 1) of radius 100 m.
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 2
Answer:
However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.
Here, F = 5N, Distance travelled S = 1500 + 3 π r + 2000 = 4442.86 m
∴ W = F x S = 5 x 4442.86 = 22214.3 J.

Question 13.
Can any object have mechanical energy even if its momentum is zero ? Explain.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 3

Question 14.
Can any object have momentum even if its mechanical energy is zero ? Explain.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 4

Question 15.
The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m ? (Given g = 10 m s-2).
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 5

Question 16.
The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 6

Question 17.
The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
Answer:
Consider a body or an object of mass m moving with velocity u. Let a force F be applied on the body so that the velocity attained by the body after travelling a distance S is v (Figure 5).
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 7
Work done by the force on the body is given by
W =      FS                                                                               …(i)
Since velocity of the body changes so the body is accelerated. Let a be the acceleration of the body. Therefore, according to Newton’s second law of motion,
F =     ma                                                                             …(ii)
Using eqn. (ii) in eqn. (i), we get
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 8
Thus, work done by a force on a body is equal to the change in kinetic energy of the body This is known as work-energy theorem.

Question 18.
Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example.
Answer:
Yes. When body moves in a circular path.

Question 19.
A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back ? (g = 10 m s-2). (CBSE 2013)
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 9

Question 20.
If an electric iron of 1200 W is used for 30 minutes everyday, find electric energy consumed in the month of April.
Answer:
Energy consumed in one day = P x t= 1200 W x ½ h = 600 Wh
Energy consumed in 30 days = 600 Wh x 30 = 1800 Wh =18 kWh.

LONG ANSWER QUESTIONS

Question 21.
A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 10

Question 22.
An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h-1. Find power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stops at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of car (B) just after the collision.
Answer:
Force = 100 N, v = 36 km h-1 = 36 x (5/18) = 10 m s-1
∴ Power, P = Force x velocity = 100 x 10 = 1000 W
According to law of conservation of linear momentum
Momentum of car A + Mementum of car B before collision = momentum of car A + momentum of car B after collision
i.e. 1000 x 10 + 1000 x 0 = 1000 x 0 + 1000 x v
∴ v = 10 m s-1
Thus, speed of car B just after collision =10 ms-1

Question 23.
A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms-1 by applying a force. The trolley comes to rest after travelling a distance of 16m.
(a) How much work is done on the trolley ?
(b) How much work is done by the girl ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 11

Question 24.
Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.
(a) How much work is done by the men in lifting the box ?
(b) How much work do they do in just holding it ?
(c) Why do they get tired while holding it ? (g = 10m s-2)
Answer:

Question 25.
(a) W = F x S = mgS= 250 x 10 x 1 =25.0 J.
(b) Zero. This is because displacement of box is zero.
(c) They get tired because muscular force applied by them is needed to balance the weight of the box.
Answer:
The Jog Falls in Karnataka State are nearly 200m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized ? (g = 10 ms-2)
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 12

Question 26.
How is the power related to the speed at which a body can be lifted ? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 ms-1 vertically ? (g = 10 ms-2)
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 13

Question 27.
Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s-1 ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy image - 14

Question 28.
Compare the power at which each of the following is moving upwards against the force of gravity ? (given g = 10 ms-2)

  1. a butterfly of mass 1.0 g that flies upward at a rate of 0.5 m s-1 .
  2. a 250 g squirrel climbing up on a tree at a rate of 0.5 m s-1.

Answer:

  1. P = Fv = mgv = 1 x 10-3 x 10 x 0.5 = 5 x 10-3 W
  2. P = Fv = mg= 250 x 10-3 x 10 x 0.5 = 1.25 W.

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work, Power and Energy are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B

RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 8. RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16B.

Other Exercises

Questions Tick the correct answer in each of the following.

Question 1.
Solution:
Answer = (c)
The diagonals of a rhombus are not necessarily equal but the diagonals in rectangle, square and isosceles trapezium are always equal.

Question 2.
Solution:
Answer = (c)
Each side of a rhombus
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B 2.1

Question 3.
Solution:
Answer = (b)
The sum of adjacent angles of a || gm = 180°
2x + 25° + 3x – 5° = 180°
=> 5x + 20° = 180°
=> 5x = 180° – 20° = 160°
=> x = \(\\ \frac { 160 }{ 5 } \)
= 32°

Question 4.
Solution:
Answer = (a)
The diagonals in rhombus, kite intersect each other at right angles.
But the diagonals of parallelogram do not necessarily intersect at right angles.

Question 5.
Solution:
Answer = (c)
Let l = 4x, b = 3x,
Then (diagonal)² = l² + b²
=> (25)² = 16x² + 9x²
=> 25x² = 625
=> x² = 25
=> x = 5
=> l = 4x = 4 x 5 = 20cm
b = 3x = 3 x 5 = 15cm
Perimeter = 2(l + b) = 2 (20 + 15)
= 2 x 35 = 70 cm

Question 6.
Solution:
Answer = (d)
AP and BP are the bisector of ∠A and ∠B
Sum of two adjacent angles of a ||gm = 180°
or ∠A + ∠B = 180°
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B 6.1
But ∠1 = \(\\ \frac { 1 }{ 2 } \) ∠A and ∠2 =\(\\ \frac { 1 }{ 2 } \) ∠B
∠1 + ∠2 = \(\\ \frac { 1 }{ 2 } \) ∠A + \(\\ \frac { 1 }{ 2 } \) ∠B
= \(\\ \frac { 1 }{ 2 } \) (∠A + ∠B)
= 180° x \(\\ \frac { 1 }{ 2 } \) = 90°
∠P = 180° – (∠1 + ∠2)
= 180° – 90° = 90°

Question 7.
Solution:
Answer = (b)
Let one adjacent angle = x
Then second angle (smallest) = \(\frac { 2 }{ 3 } x \)
x + \(\frac { 2 }{ 3 } x \) = 180°
= \(\frac { 5 }{ 3 } x \) = 180°
=> x = 180° x \(\\ \frac { 3 }{ 5 }\) = 108°
=> Smallest angle = 108° x \(\\ \frac { 2 }{ 3 }\) = 72°

Question 8.
Solution:
Answer = (a)
The diagonals of square, rhombus bisect the interior angle but the diagonals of a rectangle do not.

Question 9.
Solution:
Answer = (d)
Sides of a square are equal
2x + 3 = 3x – 5
=> 3x – 2x = 3 + 5
=> x = 8

Question 10.
Solution:
Answer = (c)
Let smallest angle = x
then largest angle = 2x – 24°
But x + 2x – 24° = 180°
=> 3x – 24° = 180°
=> 3x = 180° + 24 = 204°
=> x = \(\\ \frac { 204 }{ 3 }\) = 68°
largest angle = 180° – 68° = 112°

Hope given RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2

Other Exercises

Question 1.
A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much? [NCERT]
Solution:
In first case, in a rectangular container of soft drink, the length of base = 5 cm
and Width = 4 cm
Height = 15 cm
∴ Volume of soft drink = lbh = 5 x 4 x 15 = 300 cm3
and in second case, in a cylindrical container, diameter of base = 7 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q1.1
∴ The soft drink in second container is greater and how much greater = 385 cm – 380 cm2 = 85 cm2

Question 2.
The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars? [NCERT]
Solution:
Radius of each pillar (r) = 20 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q2.1

Question 3.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm. [NCERT]
Solution:
Inner diameter of a cylindrical wooden pipe = 24 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q3.1

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, find:
(i) radius of its base
(ii) volume of the cylinder [Use π = 3.14] [NCERT]
Solution:
Lateral surface area of a cylinder = 94.2 cm2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q4.1

Question 5.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it? [NCERT]
Solution:
The capacity of a closed cylindrical vessel = 15.4 l
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q5.1

Question 6.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? [NCERT]
Solution:
Diameter of the cylindrical bowl = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\)cm
Level of soup in it = 4 cm
∴ Volume of soup in one bowl for one patient
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q6.1

Question 7.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.
Solution:
Width of hollow cylinder (A) = 63 cm
Girth = 440 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q7.1

Question 8.
The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is ₹ 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.
Solution:
Rate of painting = 50 paise per dm2
Total cost = ₹198
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q8.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q8.2

Question 9.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:3. Calculate the ratio of their volumes and the ratio of their curved surfaces.
Solution:
Ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3
Let radius of the first cylinder (r1) = 2x
and radius of second cylinder (r2) = 3x
and height of first cylinders (h1) = 5y
and height of second cylinder (h2) = 3y
(i) Now volume of the first cylinder = πr2h = π(2x)2 x 5y = 20πx22y
and volume of tlie second cylinder = π(3x)2 x 3y = π x 9×2 x 3y = 27πx2y
Now ratio in their volume
= 20πx2y : 21πx2y = 20 : 27
(ii) Curved surface area of first cylinder = 2πrh = 2π x 2x x 5y =20πxy
and curved surface area of second cylinder = 2π x 3x x = 1 8πxy
∴ Ratio in their curved surface area
= 20πxy : 18πxy = 10 : 9

Question 10.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2.
Solution:
Ratio in curved surface area and total surface area of a cylinder =1:2
Total surface area = 616 cm2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q10.1

Question 11.
The curved surface area of a cylinder is 1320 cm2 and its base had diameter 21 cm. Find the height and the volume of the cylinder. [Use π = 22/7]
Solution:
Curved surface area of a cylinder = 1320 cm2
Diameter of the base = 21 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q11.1

Question 12.
The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm3.
Solution:
Ratio between radius and height of a cylinder = 2:3
Volume =1617 cm3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q12.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q12.2

Question 13.
A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form a cylinder. Find the volume of the cylinder so formed.
Solution:
Length of sheet = 44 cm
Breadth = 20 cm
By rolling along length, the height of cylinder (h) = 20cm
and circumference of the base = 44cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q13.1

Question 14.
The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.
Solution:
Curved surface area of a pillar = 264 m2
and volume = 924 m3
Let r be the radius and It be height, then 2πrh = 264
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q14.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q14.2

Question 15.
Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.
Solution:
Volumes of two cylinders are equal Ratio in their height h1 :h2 = 1: 2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q15.1

Question 16.
The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the” area of the curved surface. Find the volume of the cylinder.
Solution:
Height of a right circular cylinder = 10.5 m
3 x sum of areas of two circular faces
= 2 x area of curved surface
Let r be that radius,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q16.1

Question 17.
How many cubic metres of earth must be dugout to sink a well 21 m deep and 6 m diameter? Find the cost of plastering the inner surface of the well at ₹9.50 per m2.
Solution:
Diameter of a well = 6 m
∴ Radius (r) = \(\frac { 6 }{ 2 }\) = 3 m
Depth (h) = 21 m
∴ Volume of earth dugout = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q17.1

Question 18.
The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m. Find the volume of the timber that can be obtained from the trunk.
Solution:
Circumference of a cylindrical trunk of a tree = 176 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q18.1

Question 19.
A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.
Solution:
Diameter of cylindrical container = 56 cm
∴ Radius (r) = \(\frac { 56 }{ 2 }\) = 28 cm
Dimensions of a rectangular solid are = 32 cm x 22 cm x 14 cm
∴ Volume of solid = lbh
= 32 x 22 x 14 = 9856 cm3
∴ Volume of water in the container = 9856 cm3
Let h be the level of water, then
πr2h = 9856
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q19.1

Question 20.
A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.
Solution:
Length of metallic tube = 25 cm
Inner diameter = 10.4 cm
∴ Radius (r) = \(\frac { 10.4 }{ 2 }\) = 5.2 cm
Thickness of metal = 8 mm
∴ Outer radius (R) = 5.2 + 0.8 = 6.0 cm
Volume of metal used = π(R2 – r2) x h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q20.1

Question 21.
From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.
Solution:
Inner radius of a tap = 0.75 cm
Speed of flow of water in it = 7 m/s
Time = 1 hour
∴ Length of flow of water (h)
= 7 x 60 x 60 m = 25200 m
∴ Volume of water = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q21.1

Question 22.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Solution:
Size of rectangular sheet = 30 cm x 18 cm
∴ Length of sheet = 30 cm
and breadth = 18 cm
By folding length wise,
Height = 18 cm
and circumference = 30 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q22.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q22.2

Question 23.
How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec?
Solution:
Area of the cross-section of the pipe = 5 cm2
Speed of water flow = 30 cm/sec
Period = 1 minute
∴ Flow of water in 1 minute = 30 x 60 cm = 1800 cm
Area of mouth of pipe = 5 cm2
∴ Volume = 1800 x 5 = 9000 cm3
Volume of water in litres = 9000 ml
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q23.1

Question 24.
Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of ₹3.60 per cubic metre. Find also the cost of cementing its inner curved surface at ₹2.50 per square metre.
Solution:
Depth of well (h) = 280 m
Diameter = 3 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q24.1

Question 25.
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Weights of copper wire = 13.2 kg
Diamter = 4 mm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q25.1

Question 26.
A solid cylinder has a total surface area of 231 cm2. Its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.
Solution:
Surface area of solid cylinder = 231 cm2
and curved surface area = \(\frac { 2 }{ 3 }\) of 231 cm2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q26.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q26.2

Question 27.
A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of a well = 14 m
∴ Radius (r) = y = 7 m
Depth (h) = 8 m
∴ Volume of the earth dugout = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q27.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q27.2

Question 28.
The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.
Solution:
Length of cylindrical tube = 14 cm
Difference betveen the outer surface and inner surface = 88 cm2
and volume of the tube = 176 cm3
Let R and r be the outer and inner radius of the tube
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q28.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q28.2

Question 29.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank. The radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?
Solution:
Internal diameter of the pipe = 2 cm
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm
Speed of water flow = 6m per second Water in 30 minutes (h) = 6 x 60 x 30 m = 10800 m
Volume of water = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q29.1

Question 30.
A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled?
Solution:
Diameter of cylindrical tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 m
and height (h) = 2.1 m
∴ Volume of water in the tank = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q30.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q30.2

Question 31.
The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m2. Find the volume of the cylinder.
Solution:
Sum of radius and height of a cylinder = 37 m
Let r be the radius and h be the height, then r + h = 37m …(i)
Total surface area of a solid cylinder = 1628m3
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q31.1

Question 32.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 10 m 10
∴ Radius (r) = \(\frac { 10 }{ 2 }\) = 5 m
Depth (h) = 8.4 m
∴ Volume of earth dugout = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 Q32.1

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.2 are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A

RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A

These Solutions are part of RS Aggarwal Solutions Class 8. RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16A.

Other Exercises

Question 1.
Solution:
In ||gm ABCD,
∠A = 110°
But ∠ C = ∠ A {Opposite angles of a ||gm are equal}
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 1.1
∴ ∠C = 110°
But ∠A + ∠B = 180°
(Sum of adjacent angles)
=> 110° + ∠B = 180°
=> ∠B – 180° – 110° = 70°
But ∠ D = ∠ B (opposite angles)
∴ ∠ D = 70°
Hence ∠B = 70°, ∠C = 110° and ∠D = 70° Ans.

Question 2.
Solution:
In a parallelogram, sum of two adjacent angles is 180°
But these are equal to each other
∴ Each angle will be \(\frac { 180^{ o } }{ 2 } \)
= 90° Ans.

Question 3.
Solution:
The ratio between two adjacent angles of a ||gm ABCD are in the ratio 4 : 5
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 3.1
Let ∠ A = 4x and ∠ B = 5x
But ∠A + ∠B = 180°
=> 4x + 5x = 180°
=> 9x = 180°
∴ x = \(\frac { 180^{ o } }{ 9 } \)
= 20°
∴ ∠A = Ax = 4 x 20° = 80°
∠B = 5x = 5 x 20 = 100° Ans.

Question 4.
Solution:
In || gm ABCD, ∠ A and ∠ B are two adjacent angles
Let ∠ A = (3x – 4)° and ∠ B = (3x + 16)°
But ∠A + ∠B = 180°
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 4.1
=> (3x – 4)° + (3x + 16) = 180°
=> 3x – 4° + 3x + 16° = 180°
=> 6x + 12° = 180°
=> 6x= 180° – 12°
=> 6x = 168
=> x = \(\\ \frac { 168 }{ 6 } \) = 28°
∴ x = 28°
Now ∠A = 3x – 4 = 3 x 28° – 4° = 84° – 4° = 80°
∠B = 3x + 16
= 3 x 28 + 16
= 84°+ 16° = 100°
But ∠C = ∠A (opposite angles of ||gm)
∴ ∠ C = 80°
Similarly ∠ D = ∠ B = 100°
Hence ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠D= 100° Ans.

Question 5.
Solution:
In ||gm ABCD, ∠A and ∠C are opposite angles.
∴ ∠A = ∠C= 130°
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 5.1
But ∠A = ∠C (opposite angles)
∴ ∠A = ∠C
= \(\frac { 130^{ o } }{ 2 } \)
= 65°
But ∠A + ∠B = 180°
(sum of adjacent angles)
=> 65° + ∠B = 180°
=> ∠B = 180° – 65° = 115°
But ∠ D = ∠ B (opposite angles)
∴ ∠D = 115°
Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠ D = 115° Ans.

Question 6.
Solution:
Let ABCD is a parallelogram in which AB : BC = 5:3
Let AB = 5x: and BC = 3x.
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 6.1
But perimeter = 64 cm.
∴ 2(5x + 3x) = 64
=> 2 x 8x = 64
=> 16x = 64
x = \(\\ \frac { 64 }{ 16 } \)
= 4
∴ AB = 5x = 5 x 4 = 20 cm
BC = 3x = 3 x 4=12 cm
But CD = AB and AD = BC
(opposite sides of ||gm)
∴ CD = 20 cm and AD = 12 cm Ans.

Question 7.
Solution:
Perimeter of parallelogram ABCD = 140 cm.
=> ∴ 2 (AB + BC) = 140 cm.
=> AB + BC = \(\\ \frac { 140 }{ 2 } \) = 70 cm.
Let BC = x
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 7.1
then AB = x + 10
∴ x + x + 10 = 70
=> 2x + 10 = 70
=> 2x = 70 – 10 = 60
=>x = \(\\ \frac { 60 }{ 2 } \) = 30
∴ BC = 30 cm. and
AB = 30 + 10 = 40 cm.
But AD = BC and CD = AB
(Opposite sides of parallelogram)
∴ AD = 30 cm. and CD = 40 cm.

Question 8.
Solution:
In rectangle ABCD, AC is diagonal BM ⊥ AC and DN ⊥ AC.
Now, we have to prove that
∆BMC ≅ ∆DNA
In ∆BMC and ∆DNA,
BC = AD (opposite sides of the rectangle)
∠M = ∠N (each = 90°)
∠BCM = ∠D AN (Alternate angles)
∴ ∆BMC ≅ ∆DNA
(S.A.A. axiom of congruency)
∴ BM = DN (c.p .c.t.)

Question 9.
Solution:
ABCD is a parallelogram.
AE and CF are the bisectors of ∠A and ∠C respectively.
In ∆ADE and ∆CBF,
AD = BC
(Opposite sides of the parallelogram)
∠D = ∠B
(Opposite angles of the parallelogram)
∠DAE = ∠FCB (\(\\ \frac { 1 }{ 2 } \) of equal angles)
∴ ∆ADE ≅ ∆CBF
(S.A.A. axiom of congruency)
∴ DE = BF (c.p.c.t.)
But CD = AB
(Opposite sides of the parallelogram)
∴ CD – DE = AB – BF
=> EC = AF
and AB || CD
∴ AFCE is a parallelogram
∴ AE || CF.

Question 10.
Solution:
Let ABCD is a rhombus AC and BD are its diagonals which bisect each other at right angles at O.
AC = 16cm and BD = 12cm
∴ AO = \(\\ \frac { 16 }{ 2 } \) = 8cm
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 10.1
BO = \(\\ \frac { 12 }{ 2 } \) = 6 cm
Now, in right ∆AOB
AB² = AO² + BO²
(Pythagorus Theorem)
= (8)² + (6)²
= 64 + 36 = 100 = (10)²
∴ AB = 10 cm
But all the sides of a rhombus are equal
∴ Each side will be 10 cm Ans.

Question 11.
Solution:
In square ABCD, AC is its diagonal
∴ Diagonals of a square bisect each angle at the vertex
∴ ∠ CAD = ∠ CAB
But ∠ DAB = 90° (Angle of a square)
∴ ∠ CAD = ∠ CAB = \(\\ \frac { 1 }{ 2 } \) ∠ DAB
= \(\\ \frac { 1 }{ 2 } \) x 90° = 45°
Hence ∠ CAD = 45° Ans.

Question 12.
Solution:
Let ABCD is a rectangle
AB : BC = 5 : 4
Let AB = 5x and BC = 4x.
But perimeter = 90cm
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A 12.1
2(AB + BC) = 90
=> 2(5x + 4x) = 90
=> 2 x 9x = 90
=> 18x = 90
x = \(\\ \frac { 90 }{ 18 } \) = 5
∴ Length (AB) = 5x = 5 x 5 = 25 cm
Breadth (BC) = 4x = 4 x 5 = 20 cm Ans.

Question 13.
Solution:
(i) It is a rectangle
(ii) Square
(iii) rhombus
(iv) rhombus
(v) square
(vi) rectangle.

Question 14.
Solution:
(i) False
(ii) False
(iii) False
(iv) False
(v) False
(vi) True
(vii) True
(viii) True
(ix) False
(x) True

 

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RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15

RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15.

Question 1.
Solution:
(i) Four
(ii) Four
(iii) 4, collinear
(iv) two
(v) opposite
(vi) 360°

Question 2.
Solution:
(i) There are four pairs of adjacent sides which are (AB, BC), (BC, CD), (CD, DA) and (DA, AB)
(ii) There are two pairs of opposite sides which are (AB, CD) and (BC, AD)
(iii) There are four pairs of adjacent angles which are (∠ A, ∠ B), (∠ B, ∠ C), (∠ C, ∠ D) and (∠ D, ∠ A)
(iv) There are two pairs of opposite angles which are (∠A, ∠C) and (∠B, ∠D)
(v) There are two diagonals which are AC and BD.

Question 3.
Solution:
Given : ABCD is a quadrilateral
RS Aggarwal Class 8 Solutions Chapter 15 Quadrilaterals Ex 15 3.1
To prove : ∠A + ∠B + ∠C + ∠D = 360°
Construction : Join BD
Proof : In ∆ ABD,
∠ A + ∠1 + ∠ 4 = 180° (sum of angles of a triangle)
Similarly ∠2 + ∠C + ∠ 3 = 180° Adding we get,
∠ A + ∠1 + ∠4 + ∠2 + ∠C + ∠ 3
= 180° + 180°
=> ∠A + ∠1 + ∠2 + ∠C + ∠3 + ∠4 = 360°
=> ∠A + ∠B + ∠C + ∠D = 360° Hence proved.

Question 4.
Solution:
We know that
Sum of 4 angles of a quadrilateral = 360°
But sum of 3 angles = 76° + 54° + 108°
= 238°
4th angle = 360 – 238°
= 122°
Hence, measure of fourth angle = 122° Ans

Question 5.
Solution:
Ratio of four angles of a quadrilateral = 3 : 5 : 7 : 9
Let these angles be 3x, 5x, 7x and 9x
then 3x + 5x + 7x + 9x = 360° (sum of angles)
=> 24x = 360°
First angle = 3x = 3 x 15° = 45°
Second angle = 5x = 5 x 15° = 75°
Third angle = 7x = 7 x 15° = 105°
Fourth angle = 9x = 9 x 15° = 135° Ans.

Question 6.
Solution:
Three acute angles of a quadrilateral are 75° each
Sum of three angles = 3 x 75° = 225°
But sum of 4 angles = 360°
Fourth angle = 360° – 225°
= 135° Ans.

Question 7.
Solution:
Sum of 4 angles of a quadrilateral 360°
One angles = 120°
Sum of other three angles = 360° – 120° = 240°
But each of these 3 angles are equal
Each of equal angles = \(\frac { 240^{ o } }{ 3 } \)
= 80°

Question 8.
Solution:
Sum of 4 angles of a quadrilateral = 360°
Sum of two angles = 85° + 75° = 160°
Sum of other two angles = 360° – 160° = 200°
But each of these two angles are equal
Measure of each equal angle = \(\frac { 200^{ o } }{ 2 } \)
= 100° Ans.

Question 9.
Solution:
In quadrilateral ABCD
∠C = 100°, ∠D = 60°
and ∠A + ∠B + ∠C + ∠D = 360°
(sum of angles of a quadrilateral)
∴ ∠ A + ∠ B = 360° – (100° + 60°)
= 360° – 160° = 200°
But AP and BP are the bisectors of ∠ A and ∠ B
∴ \(\\ \frac { 1 }{ 2 } \) – (∠ A + ∠B) = 200° x \(\\ \frac { 1 }{ 2 } \) = 100°
i.e. ∠ 1 + ∠2 = 100°
But in ∆ APB,
∠1 + ∠2 + ∠P = 180°
=> 100° + ∠P = 180°
=> ∠P = 180° – 100° = 80°
or ∠APB = 80° Ans.

 

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RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1

Other Exercises

Question 1.
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. [NCERT]
Solution:
Curved surface area of a cylinder = 4.4 m2
Radius (r) = 0.7 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q1.1

Question 2.
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. [NCERT]
Solution:
Diameter of the pipe = 5 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q2.1

Question 3.
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. [NCERT]
Solution:
Diameter of cylindrical pillar = 50 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q3.1

Question 4.
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? [NCERT]
Solution:
Height of cylinder (h) = 1 m = 100 cm
Diameter of box = 140 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q4.1

Question 5.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder open from both sides = 4620 cm2
Area of base of ring = 115.5 cm2
Height (h) = 7 cm
Let outer radius (R) = R
and inner radius = r
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q5.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q5.2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q5.3

Question 6.
Find the ratio between the total surface area of a cylinder to its curved surface area, given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius of the cylinder (r) = 3.5 cm
and height (h) = 7.5 cm
Total surface area = 2πr (h + r)
and curved surface area = 2πrh
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q6.1

Question 7.
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of ₹3.50 per 1000 cm2.
Solution:
Radius of the base of a cylindrical vessel (r) = 70 cm
and height (h) = 1.4 m = 140 cm
Total surface area (excluding upper lid) on both sides = 2πrh x 2 + πr2 x 2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q7.1

Question 8.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:
(i) inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of ₹40 per m2. [NCERT]
Solution:
Inner diameter of a well = 3.5 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q8.1

Question 9.
The students of a Vidyalaya were asked to participate s a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder 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? [NCERT]
Solution:
Radius of cylinderical pen holder (r) = 3 cm
Height (h) = 10.5 cm
∴ Surface area of the pen holder
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q9.1

Question 10.
The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a play¬ground, find the cost of levelling this ground at the rate of 50 paise per square metre.
Solution:
Diameter of a roller = 1.5 m
∴ Radius = \(\frac { 1.5 }{ 2 }\) = 0.75 m = 75 cm
and length (h) = 84 cm
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q10.1

Question 11.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m. What will be the cost of cleaning them at the rate of ₹2.50 per square metre? [NCERT]
Solution:
Number of pillars = 20
Diameter of one pillar = 0.50 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q11.1

Question 12.
A solid cylinder has total surface area of 462 cm2. Its curved surface area is one- third of its total surface area. Find the radius and height of the cylinder.
Solution:
Total surface of solid cylinder = 462 cm2
Curved surface area = \(\frac { 1 }{ 3 }\) of total surface area
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q12.1

Question 13.
The total surface area of a hollow metal cylinder, open at both ends of external radius 8 cm and height 10 cm is 338 π cm2. Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.
Solution:
Total surface area of a hollow metal cylinder = 338π cm2
Let R be the outer radius, r be inner radius and h be the height of the cylinder of the cylinder
∴ 2πRh + 2πrh + 2πR2 – 2πr2 = 338π
R = 8 cm, h = 10 cm
⇒ 2πh (R + r) + 2π(R2 – r2) = 338π
⇒ Dividing by 2π , we get
⇒ h(R + r) + (R2 – r2) = 169
⇒ 10(8 + r) + (8 + r) (8 – r) = 169
⇒ 80 + 10r + 64 – r2 = 169
⇒ 10r – r2 + 144 – 169 = 0
⇒ r2 – 10r + 25 = 0
⇒ (r-5)2 = 0
⇒ r = 5
∴ Thickness of the metal = R – r = 8 – 5 = 3 cm

Question 14.
Find the lateral curved surface area of a cylinderical petrol storage tank that is 4.2m in diameter and 4.5 m high. How much steel was actually used, if \(\frac { 1 }{ 12 }\) of steel actually used was wasted in making the closed tank? [NCERT]
Solution:
Diameter of a cylinderical tank = 4.2 m
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q14.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 19.1 Q14.2

 

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