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Ex 13.8, 4 - Chapter 13 Class 9 Surface Areas and Volumes (Term 2)

Last updated at March 6, 2017 by

Ex 13.8, 4 - The diameter of the moon is approximately - Ex 13.8

Ex 13.8, 4 - Chapter 13 Class 9 Surface Areas and Volumes - Part 2
Ex 13.8, 4 - Chapter 13 Class 9 Surface Areas and Volumes - Part 3

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Ex 13.8 , 4 The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? Let diameter of earth = x So, diameter of moon = π‘₯/4 Hence, Radius of earth = π‘₯/2 & Radius of moon = 1/2Γ—π‘₯/4 = π‘₯/8 Now, (π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘šπ‘œπ‘œπ‘›)/(π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘’π‘Žπ‘Ÿπ‘‘β„Ž) = (4/3 πœ‹(π‘…π‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘šπ‘œπ‘œπ‘›)^3)/(4/3 πœ‹(π‘…π‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘’π‘Žπ‘Ÿπ‘‘β„Ž)^3 ) = (π‘…π‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘šπ‘œπ‘œπ‘›)^3/(π‘…π‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘’π‘Žπ‘Ÿπ‘‘β„Ž)^3 = (π‘₯/8)^3/(π‘₯/2)^3 = (π‘₯^3/8^3 )/(π‘₯^3/2^3 ) = π‘₯^3/8^3 Γ— 2^3/π‘₯^3 = 2^3/8^3 = (2 Γ— 2 Γ— 2)/(8 Γ— 8 Γ— 8) = 1/(4 Γ— 4 Γ— 4) = 1/64 ∴ (π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘šπ‘œπ‘œπ‘›)/(π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘’π‘Žπ‘Ÿπ‘‘β„Ž) = 1/64 Volume of moon = 1/64 (Volume of earth) Hence, the volume of the earth is 1/64 of the volume of the earth.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.