Last updated at March 3, 2017 by Teachoo

Transcript

Ex 13.5, 8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. Here cube of side 12 cm is divided into 8 cubes of side a cm. Given that Their volumes are equal Volume of big cube of side 12 cm = Volume of 8 cubes of side a cm (Side of big cube)3 = 8 × (Side of small cube) 3 (12)3 = 8 × a 3 (12 × 12 × 12) = 8 × a 3 1/8×12×12×12 = a 3 a 3 = 1/8×12×12×12 a3 = 1/(2 × 2 × 2)×12×12×12 a3 = 6 ×6×6 cm3 a3 = 63 cm3 a = 6 cm ∴ Side of small cube = 6 cm Ratio of their surface areas = (𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑢𝑏𝑒 𝑜𝑓 𝑠𝑖𝑑𝑒 12 𝑐𝑚)/(𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑢𝑏𝑒 𝑜𝑓 𝑠𝑖𝑑𝑒 6 𝑐𝑚) = 6(𝑠𝑖𝑑𝑒 𝑜𝑓 𝑏𝑖𝑔 𝑐𝑢𝑏𝑒)2/6(𝑠𝑖𝑑𝑒 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑐𝑢𝑏𝑒)2 = (6 × 12 × 12)/(6 × 6 × 6) = 4/1 So the ratio of surface area is 4 : 1.

Example 2
Important

Ex 13.1, 6 Important

Ex 13.1, 7 Important

Ex 13.2, 3 Important

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 You are here

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.