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

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 surfaceareas.

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 itsheight 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 facesis 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 (2πrh), total surface area (2πrh + 2πr²), and volume (πr²h) 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.