RD Sharma Solutions Class 9 Maths Chapter 19 – Surface Area And Volume of A Right Circular Cylinder (Updated for 2024)

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RD Sharma Class 9 Solutions Chapter 19

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RD Sharma class 9 chapter 19 exercise 19a

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

Question 1.
The 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:
The curved surface area of a cylinder = 4.4 m²
Radius (r) = 0.7 m

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

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:
The diameter of the cylindrical pillar = is 50 cm

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 meters of the sheet is required for the same? [NCERT]
Solution:
Height of cylinder (h) = 1 m = 100 cm
Diameter of box = 140 cm

Question 5.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, the area of the base ring is 115.5 sq. cm and the height is 7 cm. Find the thickness of the cylinder.
Solution:
The 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

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:
The 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

Question 7.
A cylindrical vessel, without a lid, has to be tin-coated on 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:
The 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

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

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:
The radius of the cylindrical pen holder (r) = 3 cm
Height (h) = 10.5 cm
∴ The surface area of the pen holder

Question 10.
The diameter of the roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a play¬ground, find the cost of leveling this ground at the rate of 50 paise per square meter.
Solution:
The diameter of a roller = 1.5 m
∴ Radius = 1.52 = 0.75 m = 75 cm
and length (h) = 84 cm

Question 11.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and the height is 4 m. What will be the cost of cleaning them at the rate of ₹2.50 per square meter? [NCERT]
Solution:
Number of pillars = 20
The diameter of one pillar = 0.50 m

Question 12.
A solid cylinder has a 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:
The total surface of the solid cylinder = 462 cm²
Curved surface area = 13 of the total surface area

Question 13.
The total surface area of a hollow metal cylinder, open at both ends of an external radius of 8 cm and height of 10 cm is 338 π cm². Taking r to be the inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.
Solution:
The total surface area of a hollow metal cylinder = 338π cm²
Let R be the outer radius, r be the inner radius and h be the height of the cylinder 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 cylindrical petrol storage tank that is 4.2m in diameter and 4.5 m high. How much steel was actually used, if 112 of steel actually used was wasted in making the closed tank? [NCERT]
Solution:
The diameter of a cylindrical tank = 4.2 m

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

Question 1.
A soft drink is available in two packs – (i) a tin can with a rectangular base of a length of 5 cm and width of 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with a circular base of a diameter of 7 cm and height of 10 cm. Which container has greater capacity and by how much? [NCERT]
Solution:
In the first case, in a rectangular container of soft drink, the length of the base = 5 cm
and Width = 4 cm
Height = 15 cm
∴ Volume of soft drink = lbh = 5 x 4 x 15 = 300 cm³
and in the second case, in a cylindrical container, the diameter of the base = 7 cm

∴ The soft drink in the 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 of 10 m. How much concrete mixture would be required to build 14 such pillars? [NCERT]
Solution:
Radius of each pillar (r) = 20 cm

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:
The inner diameter of a cylindrical wooden pipe = is 24 cm

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:
The lateral surface area of a cylinder = 94.2 cm2

Question 5.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it? [NCERT]
Solution:
The capacity of a closed cylindrical vessel = is 15.4 l

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 have to prepare daily to serve 250 patients? [NCERT]
Solution:
The diameter of the cylindrical bowl = 7 cm
∴ Radius (r) = 72cm
Level of soup in it = 4 cm
∴ The volume of soup in one bowl for one patient

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 a hollow cylinder (A) = 63 cm
Girth = 440 cm

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

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:
The ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3
Let the radius of the first cylinder (r₁) = 2x
and radius of the second cylinder (r₂) = 3x
and height of first cylinders (h₁) = 5y
and height of the second cylinder (h₂) = 3y
(i) Now volume of the first cylinder = πr²h = π(2x)² x 5y = 20πx²2y
and volume of the 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
∴ The 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:
The ratio in curved surface area and total surface area of a cylinder =1:2
Total surface area = 616 cm²

Question 11.
The curved surface area of a cylinder is 1320 cm² and its base has a diameter of 21 cm. Find the height and the volume of the cylinder. [Use π = 22/7]
Solution:
The curved surface area of a cylinder = 1320 cm²
The diameter of the base = 21 cm

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:
The ratio between the radius and height of a cylinder = 2:3
Volume =1617 cm³
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr²h

Question 13.
A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.
Solution:
Length of sheet = 44 cm
Breadth = 20 cm
By rolling along the length, the height of a cylinder (h) = 20cm
and circumference of the base = 44cm

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:
The curved surface area of a pillar = 264 m²
and volume = 924 m³
Let r be the radius and It is the height, then 2πrh = 264

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

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 the curved surface
Let r be that radius,

Question 17.
How many cubic meters of the earth must be dug out to sink a well 21 m deep and 6 m in diameter? Find the cost of plastering the inner surface of the well at ₹9.50 per m².
Solution:
The diameter of a well = 6 m
∴ Radius (r) = 62 = 3 m
Depth (h) = 21 m
∴ The volume of earth dugout = πr²h

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

Question 19.
A cylindrical container with a 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:
The diameter of the cylindrical container = is 56 cm
∴ Radius (r) = 562 = 28 cm
Dimensions of a rectangular solid are 32 cm x 22 cm x 14 cm
∴ The volume of solid = lbs
= 32 x 22 x 14 = 9856 cm³
∴ The volume of water in the container = 9856 cm³
Let h be the level of water, then
πr2h = 9856

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) = 10.42 = 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

Question 21.
From a tap of an inner radius of 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:
The inner radius of a tap = 0.75 cm
Speed of flow of water in it = 7 m/s
Time = 1 hour
∴ Length of the flow of water (h)
= 7 x 60 x 60 m = 25200 m
∴ Volume of water = πr²h

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

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 the mouth of pipe = 5 cm²
∴ Volume = 1800 x 5 = 9000 cm³
The volume of water in litres = 9000 ml

Question 24.
Find the cost of sinking a tubewell 280 m deep, having a diameter of 3 m at the rate of ₹3.60 per cubic meter. Find also the cost of cementing its inner curved surface at ₹2.50 per square meter.
Solution:
Depth of well (h) = 280 m
Diameter = 3 m

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

Question 26.
A solid cylinder has a total surface area of 231 cm². Its curved surface area is 23 of the total surface area. Find the volume of the cylinder.
Solution:
The surface area of the solid cylinder = 231 cm²
and curved surface area = 23 of 231 cm²

Question 27.
A well with a 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:
The diameter of a well = 14 m
∴ Radius (r) = y = 7 m
Depth (h) = 8 m
∴ The volume of the earth dugout = πr²h

Question 28.
The difference between the 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 between 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

Question 29.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into a cylindrical tank. The radius of the base is 60 cm. Find the rise in the level of water in 30 minutes.
Solution:
The internal diameter of the pipe = 2 cm
∴ Radius (r) = 22 = 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

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 meters per second. In how much time the tank will be filled?
Solution:
The diameter of the cylindrical tank = 1.4 m
∴ Radius (r) = 1.42 = 0.7 m
and height (h) = 2.1 m
∴ The volume of water in the tank = πr²h

Question 31.
The sum of the radius of the base and the 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:
The sum of the radius and height of a cylinder = 37 m
Let r be the radius and h be the height, then r + h = 37m …(i)
The total surface area of a solid cylinder = 1628m³

Question 32.
A well with a 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) = 102 = 5 m
Depth (h) = 8.4 m
∴ The volume of earth dugout = πr²h

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

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

Question 2.
Write the ratio of the 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 the height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.
Solution:
The ratio in radius and height of the cylinder = 2 : 3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr²h

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:
The ratio of radii of two cylinders = 2:3
Let the radius of the first cylinder (r₁) = 2x
and second cylinder (r₂) = 3x
and ratio in their heights = 5:3
Let the height of the first cylinder (h₁) = 5y
and height of second (h₂) = 3y
∴ The 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 them,
= 20πx²y: 21πx²y
= 20: 27

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

Mark the correct alternative in each of the following:
Question 1.
In a cylinder, if the radius is doubled and the height is halved, the curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
Let the radius of the first cylinder (r₁) = r
and height (h₁) = h
Surface area = 2πrh
If the radius is doubled and the height is halved

∴ Their surface area remains the same (c)

Question 2.
Two cylindrical jars have their diameters in the ratio of 3:1, but their height is 1:3. Then the ratio of their volumes is
(a) 1: 4
(b) 1: 3
(c) 3: 1
(d) 2: 5
Solution:
Sol. The ratio in the diameters of two cylinders = 3:1
and ratio in their heights = 1:3
Let the radius of the first cylinder (r₁) = 3x
and radius of a second (r₂) = x
and height of the first (h₁) = y
and height of the second (h₂) = 3y
Now the 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 the 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.
A 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 the height of the cylinder the volume will be

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

Question 9.
In a cylinder, if the radius is halved and the height is doubled, the volume will be
(a) same
(b) doubled
(c) halved
(d) four times
Solution:
Let r be the radius and h be the height, then Volume = πr²h
If the radius is halved and the height is doubled

Question 10.
If the diameter of the base of a closed right circular cylinder is equal to its height h, then its whole surface area is

Solution:
Let the diameter of the base of a cylinder (r) = h
Then its height (h) = h

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:
The diameter of a cylindrical tunnel = 2 m
∴ Radius (r) = 22 = 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

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,

Question 13.
The radius of a wire is decreased to one-third. If the 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 the second case, the radius is decreased to one-third

∴ In the 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?

Solution:
Let r be the radius and h be the height then volume = πr²h
If the height is doubled and volume is the same and let x be the radius then πr²h = π(x)² x 2h

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?

Solution:
Let r be the radius and h be the height, then volume = πr²h
This volume is 14 of the volume of a rectangular box
∴ Volume of box = 4πr²h
Let the side of the base of the box = x and height h,
then volume = x²h
∴ 4πr²h = x²h

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?

Solution:
In a cylinder,
h = circumference of the cylinder

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 it is the height of a closed cylinder
The total surface area = 2πr(r + h ) (a)

Question 18.
The height of sand in a cylindrical shape can drop 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?

Solution:
Let h be the height and d be the diameter of a cylinder, then

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 the other having volume v₂ and height a₁. Then,

Solution:
Length of each sheet = a₁
and breadth = a₂
Volume of cylinder = πr²h
In the first case,
v₁ is volume and a₂ is the height

Question 20.
The altitude of a circular 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

Solution:
In the first case,
Let r be the radius and h be the height of the cylinder. Then,
∴ Lateral surface area = 2πrh
In the second case,

 

Surface Area and Volume of A Right Circular Cylinder Ex 19.1

Detailed Exercise-wise Explanation with Listing of Important Topics

RD Sharma class 9 chapter 19 exercise 19a:

This exercise includes topics related to the surface area of a cylinder. Solving problems on surface areas of a circular cylinder with the help of RD Sharma solutions assists the students to have a better understanding of each concept.

These solutions are framed by subject experts to assist the students in increasing their scoring potential in the final exam. The students can develop problem-solving skills to answer any type of question with ease.

RD Sharma class 9 chapter 19 exercise 19b

RD Sharma’s solutions for class 9 Maths are considered the perfect tools to prepare for the exam. Each question is solved in such a way that assists the students in understanding the topics easily.

Every student must consider these while practicing to prepare for the exam more effectively. Each student must practice these exercises regularly as this assists them to excel in the exams.

Important Topics from Class 9 Maths Chapter 19

RD Sharma Solutions Class 9 Maths Chapter 19 covers some important concepts that are listed below:

• Introduction of Right Circular Cylinder 
• Important terms definition of Base, Radius, Axis, Height, and Lateral Surface.
• Surface Area of a Cylinder
• The volume of a Cylinder

The students must practice RD Sharma Solutions CBSE Class 9 Maths Chapter 19 regularly to assist them in enhancing their confidence level to secure excellent ranks in the final exams. To know more about the Class 9 Maths exam, ask in the comments.

FAQs on RD Sharma Solutions Class 9 Maths Chapter 19