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Ex 11.2 , 7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. Let diameter of earth = x So, diameter of moon = ๐‘ฅ/4 Hence, Radius of earth = ๐‘ฅ/2 & Radius of moon = 1/2ร—๐‘ฅ/4 = ๐‘ฅ/8 Ratio of their surface area = (๐‘†๐‘ข๐‘Ÿ๐‘“๐‘Ž๐‘๐‘’ ๐‘Ž๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘š๐‘œ๐‘œ๐‘›)/(๐‘†๐‘ข๐‘Ÿ๐‘“๐‘Ž๐‘๐‘’ ๐‘Ž๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž) = (4๐œ‹(๐‘…๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘š๐‘œ๐‘œ๐‘›)^2)/(4๐œ‹(๐‘…๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž)^2 ) = (๐‘…๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘š๐‘œ๐‘œ๐‘›)^2/(๐‘…๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘“ ๐‘’๐‘Ž๐‘Ÿ๐‘กโ„Ž)^2 = (๐‘ฅ/8)^2/(๐‘ฅ/2)^2 = (๐‘ฅ^2/8^2 )/(๐‘ฅ^2/2^2 ) = ๐‘ฅ^2/8^2 ร— 2^2/๐‘ฅ^2 = ๐‘ฅ^2/64 ร— 4/๐‘ฅ^2 = 1/16

  1. Chapter 11 Class 9 Surface Areas and Volumes
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo