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Ex 13.4, 7 - The diameter of the moon is one fourth - Area Of Sphere

  1. Chapter 13 Class 9 Surface Areas and Volumes
  2. Serial order wise
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Ex 13.4 , 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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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