Open In App

Class 8 RD Sharma Solutions – Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) – Exercise 22.1 | Set 1

Last Updated : 19 Sep, 2024
Summarize
Comments
Improve
Suggest changes
Like Article
Like
Save
Share
Report
News Follow

Chapter 22 of RD Sharma’s Class 8 Mathematics textbook, titled “Mensuration III,” focuses on the surface area and volume of right circular cylinders. Exercise 22.1 | Set 1 specifically deals with calculating the curved surface area, total surface area, and volume of cylinders.

This exercise introduces students to the fundamental formulas and concepts needed to solve problems related to cylindrical objects. Students learn to apply these formulas in various real-world scenarios, enhancing their spatial reasoning and problem-solving skills.

Question 1: Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.

Solution: 

The details given about cylinder are:

Diameter of base of a cylinder = 7 cm

So, radius = (7/2)

Height of cylinder = 60 cm 

Curved surface area of a cylinder = 2 * (22/7) * r * h

                                                      = 2 * (22/7) * (7/2) * 60

                                                      = 1320 cm2  

Total surface area of cylinder = 2 * (22/7) * r * (h + r)

                                               = 2 * (22/7) * (7/2) * (60 + (7/2))

                                               = 22 * (60 * 2 + 7)/2

                                               = 22 * (127/2)

                                              = 1397 cm2

Question 2: The curved surface area of a cylindrical rod is 132 cm2. Find the length if the radius is 0.35 cm.

Solution: 

The details given about cylindrical rod are –

Curved surface area of a cylindrical rod = 132 cm2

Length of radius = 0.35 cm 

Let length of rod = h

Curved surface area of a cylinder = 2 * (22/7) * r * h

                                               132 = 2 * (22/7) * 0.35 * h

                (132 * 7)/(2 * 0.35 * 22) = h

                                                   h = 60 cm

Question 3: The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of cylinder.

Solution: 

The details given about right circular cylinder are –

Area of the base of a right circular cylinder = 616 cm2

Height of right circular cylinder = 2.5 cm  

Let radius of right circular cylinder = r

Area of base of a right circular cylinder = (22/7) * r2

                                                        616 = (22/7) * r2 

                                                       (616 * 7)/22 = r2

                                                                           196 = r2

                                                                            r = 14 cm

Curved surface area of cylinder = 2 * (22/7) * r * h

                                                  = 2 * (22/7) * 14 * 2.5

                                                  = 220 cm2

Question 4: The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved area and total surface area.

Solution: 

The details given about cylinder are –

Circumference of the base of a cylinder = 88 cm

Height of a cylinder = 15 cm

Let radius of cylinder = r

Circumference of the base of a cylinder = 2 * (22/7) * r

                                                           88 = 2 * (22/7) * r 

                                                          (88 * 7) = (2 * 22) * r

                                                                    r = 14 cm

Curved surface area of cylinder = 2 * (22/7) * r * h

                                                   = 2 * (22/7) * 14 * 15

                                                   = 1320 cm2

Total surface area of cylinder = 2 * (22/7) * r * (h + r)

                                               = 2 * (22/7) * 14 * (15 + 14)

                                               = 2 * (22/7) * 14 * 19

                                               = 2552 cm2

Question 5: A rectangular strip 25 cm * 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated.

Solution: 

The details given about rectangular strip are –

Dimension of rectangular strip = 25 cm * 7 cm

When the strip is rotated about its longer side,

Height of the cylinder  becomes = 25 cm

Radius of cylinder = 7 cm

Total surface area of cylinder = 2 * (22/7) * r * (h + r) 

                                               = 2 * (22/7) * 7 * (25 + 7)

                                               = 2 * (22/7) * 7 * 32

                                               = 1408 cm2

Question 6: A rectangular sheet of paper, 44 cm * 20 cm, is rolled along its length to form a cylinder. Find the total surface area of the cylinder thus generated.

Solution: 

The details given about rectangular sheet of paper are –

Dimensions of rectangular sheet = 44 cm * 20 cm

When the sheet of the paper is rolled along its length,

Height of the cylinder = 20 cm

Circumference of base becomes = 44 cm

Let radius of base = r

Circumference of base = 2 * (22/7) * r

                                44 = 2 * (22/7) * r

           (44 * 7)/(2 * 22) = r

                                  r = 7 cm

Total surface area of cylinder = 2 * (22/7) * r * (h + r)

                                               = 2 * (22/7) * 7 * (20 + 7)

                                               = 2 * (22/7) * 7 * 27

                                               = 1188 cm2

Question7: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of curved surface areas.

Solution: 

The details given cylinders are –

r1 / r2 = 2 : 3

h1 / h2 = 5 : 3

Curved surface area of cylinder 1/Curved surface area of cylinder 2 = (2 * (22/7) * r1 * h1 )/(2 * (22/7) * r2 * h2)

                                                                                                            = (2 * (22/7) * 2 * 5)/(2 * (22/7) * 3 * 3) 

                                                                                                            = 10/9

Question 8: The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Prove that its height and radius are equal.

Solution: 

Curved surface area of cylinder/Total surface area of a cylinder = 1/2

(2 * (22/7) * r * h)/(2 * (22/7) * r * (h + r) = 1/2

h/(h + r) = 1/2

2 * h = h + r

2h – h = r

h = r

Height = Radius

Hence, Proved.

Question 9: The curved surface area of cylinder is 1320 cm2 and its base has diameter 21 cm. Find the height of cylinder.

Solution: 

The details given about cylinder are –

Curved surface area of cylinder = 1320 cm2

Diameter of base = 21 cm

So, radius = 21/2

Let height of cylinder = h

Curved surface area of cylinder = 2 * (22/7) * r * h

                                          1320 = 2 * (22/7) * (21/2) * h

        (1320 * 7 * 2)/(2 * 22 * 21) = h 

                                               h = 20 cm

Question 10: The height of a right circular cylinder is 10.5 cm. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.

Solution: 

The details given about cylinder are –

Height of circular cylinder = 10.5 cm

Let radius of cylinder = r

Area of two bases of cylinder = 2 * (22/7) * r2

Area of curved surface of cylinder =  2 * (22/7) * r * h

3 * (2 * (22/7) * r2) = 2 * (2 * (22/7) * r * h))

6 * r = 4 * h

6 * r = 4 * 10.5

r = (4 * 10.5)/6

r = 7 cm 

Summary

Exercise 22.1 | Set 1 of Chapter 22 in RD Sharma’s Class 8 Mathematics textbook provides a comprehensive exploration of the surface area and volume of right circular cylinders. Through a variety of problems, students learn to apply the formulas for curved surface area (2Ï€rh), total surface area (2Ï€rh + 2Ï€r²), and volume (Ï€r²h) of cylinders. The exercise emphasizes the importance of unit conversion and precise calculations in solving real-world problems involving cylindrical objects. Students are challenged to work with different given parameters, such as radius, diameter, height, surface area, or volume, to find the missing measurements. This approach helps develop critical thinking and problem-solving skills, as students learn to analyze given information and determine the appropriate steps to solve each problem. By mastering these concepts, students gain a solid foundation in three-dimensional geometry, preparing them for more advanced topics in mathematics and practical applications in various fields of science and engineering.

FAQs on Surface Area And Volume Of Right Circular Cylinder

What is the difference between curved surface area and total surface area of a cylinder?

The curved surface area includes only the lateral surface of the cylinder (2πrh), while the total surface area includes both the lateral surface and the two circular bases (2πrh + 2πr²).

How do you find the height of a cylinder if you know its volume and radius?

Rearrange the volume formula V = πr²h to solve for h: h = V / (πr²).

Why is π (pi) used in cylinder formulas?

Ï€ is used because it represents the ratio of a circle’s circumference to its diameter, which is crucial in calculating the curved surface and circular bases of a cylinder.

How do you convert cubic centimeters to liters?

1 liter = 1000 cubic centimeters. To convert from cm³ to liters, divide by 1000.

Can these formulas be applied to any cylinder?

These formulas apply specifically to right circular cylinders, where the bases are perpendicular to the height. For oblique cylinders, different formulas are needed.



Previous Article
Next Article

Similar Reads

Class 8 RD Sharma Solutions- Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) - Exercise 22.2 | Set 1
Question 1. Find the volume of a cylinder whose (i) r = 3.5 cm, h = 40 cm (ii) r = 2.8 m, h = 15 m Solution: (i) Given that, r = 3.5 cm, h = 40 cm As we know that Volume of a cylinder = πr2h = 22/7 × 3.5 × 3.5 × 40 = 1540 cm3 (ii) Given that, r = 2.8 m, h =15 m As we know that Volume of a cylinder = πr2h = 22/7 × 2.8 × 2.8 × 15 = 369.6 m3 Question
11 min read
Class 8 RD Sharma Solutions- Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) - Exercise 22.2 | Set 2
Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) - Exercise 22.2 | Set 1Question 21. A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad. What is the height of the platform so formed? Solution: Given that, Depth of well = 20m
15+ min read
Class 8 RD Sharma Solutions - Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) - Exercise 22.1 | Set 2
Chapter 22 of RD Sharma's Class 8 Mathematics textbook delves into the fascinating world of mensuration, specifically focusing on the surface area and volume of right circular cylinders. This chapter builds upon students' previous knowledge of geometry and introduces them to more complex three-dimensional shapes. The chapter covers various aspects
9 min read
Class 9 RD Sharma Solutions - Chapter 19 Surface Area And Volume of a Right Circular Cylinder - Exercise 19.2 | Set 2
Question 17. 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: Given that Height of the cylinder = 10.5 m According to the question 3(2Ï€r2) = 2(2Ï€rh) 3r = 2h r = 2/3 x h r = 2/3 x 10.5 = 7cm So, the radius
14 min read
Class 9 RD Sharma Solutions - Chapter 19 Surface Area And Volume of a Right Circular Cylinder - Exercise 19.2 | Set 1
This geometric figure is known as Right Circular Cylinder, which has ample usages in day to day real life situation arises in Engineering, Architecture and Manufacture. It is well used in areas of estimations concerning materials as well as design works concerning the physical structure of an object. Area of regression: right circular cylinderThis
14 min read
Class 9 RD Sharma Solutions - Chapter 19 Surface Area And Volume of a Right Circular Cylinder - Exercise 19.1
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. Solution: Given, radius (r) = 0.7 m and Curved Surface Area(C.S.A) = 4.4 m2 By formula, C.S.A = 2πrh where, h = height of the cylinder Putting values in the formula, 4.4 m2 = 2 * (22/7) * 0.7 * h (using π = 22
4 min read
Class 9 RD Sharma Solutions- Chapter 20 Surface Area And Volume of A Right Circular Cone - Exercise 20.1 | Set 1
Question 1. Find the curved surface area of a cone, if its slant height is 6 cm and the radius of its base is 21 cm.2 Solution: According to the question, Slant height of cone, l = 60 cm and radius of the base of cone, r = 21 cm Since, curved surface area of cone = πrl = 22/7 x 21 x 60 =3960 cm2 Question 2. The radius of a cone is 5 cm and vertical
4 min read
Class 9 RD Sharma Solutions- Chapter 20 Surface Area And Volume of A Right Circular Cone - Exercise 20.1 | Set 2
Question 11. A joker's cap is in the form of a right circular cone of base radius 7cm and height 24 cm. Find the area of the sheet required to make 10 such caps.Solution: As per question, r = 7 cm and h = 24 cm. Therefore, slant height of cone, l = √r2 + h2 = √(72+242) = 25 cm Now, CSA of one cone = πrl = (22 x 7 x 25)/7 = 550 cm2 Therefore, area o
8 min read
Class 9 RD Sharma Solutions- Chapter 20 Surface Area And Volume of A Right Circular Cone - Exercise 20.2
In geometry, the right circular cone is a three-dimensional shape with a circular base and a pointed apex. Understanding the surface area and volume of the right circular cone is crucial in solving the problems related to the various real-life applications such as designing containers and calculating materials needed for conical structures. This ch
11 min read
Surface Area of Cylinder | Curved and Total Surface Area of Cylinder
Surface Area of a Cylinder is the amount of space covered by the flat surface of the cylinder's bases and the curved surface of the cylinder. The total surface area of the cylinder includes the area of the cylinder's two circular bases as well as the area of the curving surface. The volume of a cylinder is calculated using the formula V = πr2h and
10 min read
Class 8 RD Sharma Solutions - Chapter 21 Mensuration II (Volume and Surface Areas of a Cuboid and a Cube) - Exercise 21.1 | Set 2
Chapter 21 Mensuration II (Volume and Surface Areas of a Cuboid and a Cube) - Exercise 21.1 | Set 1Question 12: A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it beads of volume 1.5 cm3 each are to be made. Find the number of beads that can be made from the block. Solution: The details given about cuboidal block of s
7 min read
Surface Area and Volume of a Right Circular Cylinder Problem
Understanding the surface area and volume of a right circular cylinder is essential for solving various geometric problems. In this article, we learn about an important topic of class 9 Surface Area and Volume of a Right Circular Cylinder. This article will provide a clear understanding of the right circular Cylinder and surface area and volume of
7 min read
Class 9 RD Sharma Solutions - Chapter 18 Surface Area and Volume of a Cuboid and Cube - Exercise 18.2 | Set 2
Question 14. The dimensions of a cinema hall are 100 m, 50 m, 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air? Solution: Length of hall = 100 m Breadth of hall = 50 m Height of hall = 18 m So, the Volume of the hall = l × b × h = 100 × 50 × 18 = 90000 m3 Also, the volume of air required per person = 150 m3 So we ca
13 min read
Class 9 RD Sharma Solutions- Chapter 18 Surface Area and Volume of a Cuboid and Cube - Exercise 18.2 | Set 1
Question 1. A cuboidal water tank is 6 m long, 5 m wide, and 4.5 m deep. How many liters of water can it hold? Solution: Given, the length of water tank = 6 m The breadth of water tank = 5 m The height of water tank = 4.5 m The quantity of water that tank can hold = Volume of Cuboid = length × breadth × height = 6 × 5 × 4.5 = 135 m3 As we know 1 m3
8 min read
Class 9 RD Sharma Solutions - Chapter 21 Surface Area and Volume of a Sphere - Exercise 21.2 | Set 2
Question 17. A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimeters? Solution: Given that, Radius of the cylinder jar = r1 = 6 cm, Level to be raised = 2 cm, Radius of each iron sphere = r2 = 1.5 cm, Number of sphere =
9 min read
Class 9 RD Sharma Solutions- Chapter 21 Surface Area and Volume of a Sphere - Exercise 21.2 | Set 1
Chapter 21 of RD Sharma's Class 9 mathematics textbook explores the surface area and volume of a sphere, a fundamental concept in geometry. Exercise 21.2 Set 1 focuses on applying formulas and problem-solving techniques to calculate these measurements for spheres in various contexts, helping students develop a strong understanding of three-dimensio
11 min read
Class 8 RD Sharma Solutions - Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) - Exercise 21.1 | Set 1
Question 1: Find the volume of cuboid whose: i) length = 12 cm, breadth = 8 cm and height = 6 cm ii) length = 1.2 m, breadth = 30 cm and height = 15 cm iii) length = 1.5 dm, breadth = 2.5 dm and height = 8 cm Solution: i) The details given about cuboid are - Length of cuboid = 12 cm Breadth of cuboid = 8 cm Height of cuboid = 6 cm Volume of cuboid
8 min read
Class 8 RD Sharma Solutions - Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) - Exercise- 21.2 | Set 2
Question 11. How many bricks each of size 25 cm x 10 cm x 8 cm will be required to build a wall 5 m long, 3 m high, and 16 cm thick assuming that the volume of sand and cement used in the construction is negligible? Solution: Given, length of brick = 25 cm The breadth of brick = 10 cm The height of brick = 8 cm So, the volume of 1 brick = l × b × h
7 min read
Class 8 RD Sharma Solutions- Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.4 | Set 2
Question 11. A field is 150m long and 100m wide. A plot (outside the field) 50m long and 30m wide is dug to a depth of 8m and the earth taken out from the plot is spread evenly in the field. By how much is the level of field raised? Solution: Given, the length of field = 150 m The width of the field = 100 m So, the area of field = Length × breadth
8 min read
Class 8 RD Sharma Solutions- Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.2 | Set 1
Question 1. Find the volume in cubic meters (cu.m) of each of the cuboids whose dimensions are :(i) length = 12 m, breadth = 10 m, height = 4.5 m(ii) length = 4 m, breadth = 2.5 m, height = 50 cm(iii) length = 10 m, breadth = 25 dm, height = 25 cmSolution: i) length = 12 m, breadth = 10 m, height = 4.5 m Since, Length (l) = 12 m Breadth (b) = 10 m
6 min read
Class 8 RD Sharma Solutions - Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.4 | Set 1
This article provides step-by-step solutions to all the questions in the Exercise 21.4 | Set 1 of Class 8 RD Sharma Book of Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube). This exercise deals with the volumes and surface areas of cuboids and cubes, which are fundamental concepts in geometry and have practical applicati
8 min read
Class 9 RD Sharma Solutions - Chapter 18 Surface Area and Volume of a Cuboid and Cube - Exercise 18.1
Question 1: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm, and height 20 cm. Solution: Given, dimensions of cuboid are: Length(l) = 80 cm Breadth(b) = 40 cm Height(h) = 20 cm Formula for total surface area of cuboid: TSA(Cuboid) = 2[lb + bh + hl] Now, substituting the given values of in the formula,
7 min read
Class 9 RD Sharma Solutions- Chapter 21 Surface Area and Volume of a Sphere - Exercise 21.1
In this article, we will seeing how applications of finding the surface areas of various 3-D sphere, and it's variation Hemisphere. We will see various examples regarding how to find the surface area of the various 3-D solid figures. FINDING SURFACE AREA OF SOLID 3-D FIGURESQuestion 1: Find the surface area of a sphere of radius:(i) 10.5 cm (ii) 5.
8 min read
Class 8 RD Sharma - Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) - Exercise 21.3 | Set 1
Question 1. Find the surface area of a cuboid whose (i) length = 10 cm, breadth = 12 cm, height = 14 cm(ii) length = 6 dm, breadth = 8 dm, height = 10 dm(iii) length = 2 m, breadth = 4 m, height = 5 m(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm. Solution : i) length = 10 cm, breadth = 12 cm, height = 14 cm Given, the Length of cuboid = 10
8 min read
Class 8 RD Sharma - Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube) - Exercise 21.3 | Set 2
Question 10. A classroom is 11 m long, 8 m wide, and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows, etc.)? Solution: Given, the Length of classroom = 11 m The Breadth of classroom = 8 m The Height of the classroom = 5 m Area of floor = length × breadth = 11 × 8 = 88 m2 And, the Area of four walls (inc
7 min read
Mensuration - Volume of Cube, Cuboid, and Cylinder | Class 8 Maths
Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas, and volume. It uses geometric calculations and algebraic equations to compute the measurement of various aspects of objects like Surface Area, Volume, etc.  Volume is the measurement of the amount of space inside a 3D object that can be
4 min read
Volume and Surface Area of a Cylinder
The volume of a cylinder can be calculated using the formula V = πr2h. While the total surface area of a cylinder can be calculated using the formula: A=2πr(h+r). In geometry, a cylinder is a three-dimensional solid figure which contains two parallel circular bases joined by a curved surface, situated at a particular distance from the center of the
5 min read
Surface Area and Volume of a Right Circular Cone : Practice Problems
Understanding the surface area and volume of a right circular cone is essential for solving various geometric problems. In this article, we learn about an important topic of class 9 Surface Area and Volume of a Right Circular Cone. This article will provide a clear understanding of the right circular cone and surface area and volume of a right circ
7 min read
Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.1 | Set 1
Chapter 16 of RD Sharma's Class 10 textbook focuses on the Surface Areas and Volumes which are crucial concepts in geometry. This chapter deals with the measurement of the surface areas and volumes of various solid figures such as cubes, cuboids, cylinders, cones, and spheres. Understanding these concepts is essential for solving real-life problems
12 min read
Class 10 RD Sharma Solutions - Chapter 16 Surface Areas and Volumes - Exercise 16.2 | Set 2
Chapter 16 of RD Sharma's Class 10 textbook "Surface Areas and Volumes" explores the mathematical principles used to calculate the surface areas and volumes of the various geometric shapes. This chapter is crucial for understanding how to measure and interpret the size of three-dimensional objects which has practical applications in fields such as
13 min read
three90RightbarBannerImg