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

Ex 13.4, 7 - Chapter 13 Class 9 Surface Areas and Volumes - Part 2

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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.