Last updated at May 29, 2018 by Teachoo

Transcript

Ex 13.2, 3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Inner curved surface area, Inner radius of cylindrical pipe = r1 = ( )/2 = (4/2) cm = 2 cm Height (h) of cylindrical pipe = 77 cm Curved Surface Area of inner surface of pipe = 2 r1h = (2 22/7 2 77) cm2 = 968 cm2 Ex 13.2, 3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. (ii) Outer curved surface area, Outer radius of cylindrical pipe =r2 = ( )/2 = (4.4/2) cm = 2.2 cm Height of cylinder = h = 77 cm Curved Surface Area of outer surface of pipe = 2 r2h = (2 22/7 2.2 77) cm2 = (2 22 2.2 11) cm2 = 1064.8 cm2 Ex 13.2, 3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. (iii) Total surface area, r1 = 2 , r2 = 2.2 cm, h = 77 cm Total surface area = Curved Surface Area of inner cylinder + Curved Surface Area of outer cylinder + 2 Area of base Area of base = r22 r12 = 22/7 ((2.2)2 (2)2 ) = 22/7 (4.84 4) = 22/7 (0.84) = 2.74 cm2 Total surface area = Curved Surface Area of inner cylinder + Curved Surface Area of outer cylinder + 2 Area of base = 968 + 1064.8 + 2 2.74 = 2032.8 + 5.76 = 2038.08 cm2 Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2

Chapter 13 Class 9 Surface Areas and Volumes

Example 2
Important

Ex 13.1, 6 Important

Ex 13.1, 7 Important

Ex 13.2, 3 Important You are here

Ex 13.2, 4 Important

Ex 13.2, 9 Important

Example 6 Important

Ex 13.3, 3 Important

Ex 13.3, 5 Important

Ex 13.4, 3 Important

Ex 13.4, 8 Important

Ex 13.4, 9 Important

Ex 13.5, 6 Important

Ex 13.5, 8 Important

Ex 13.5, 9 Important

Example 14 Important

Ex 13.6, 3 Important

Ex 13.6, 7 Important

Ex 13.7, 5 Important

Ex 13.7, 6 Important

Ex 13.7, 7 Important

Ex 13.7, 8 Important

Example 18 Important

Ex 13.8, 4 Important

Surface Area and Volume Formulas Important

Class 9

Important Questions for Exam - Class 9

- Chapter 1 Class 9 Number Systems
- Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.