Question 3 - Conversion of one shape to another - Chapter 12 Class 10 Surface Areas and Volumes

Last updated at April 16, 2024 by

Ex 13.3, 3 - A 20 m deep well with diameter 7 m is dug - Conversion of one shape to another

Ex 13.3, 3 - Chapter 13 Class 10 Surface Areas and Volumes - Part 2
Ex 13.3, 3 - Chapter 13 Class 10 Surface Areas and Volumes - Part 3

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Question 3 A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. Here, shape of well is in form of cylinder & shape of platform will be in shape of cuboid. Hence, Volume of well = Volume of platform So , Volume of cylinder = Volume of cuboid Volume of cylinder Diameter = 7 m radius = r = /2 = 7/2 m & Height = h = 20 m Volume of cylinder = 2 = 22/7 (7/2)^2 20 = 22/7 7/2 7/2 20 = 770 m3 Volume of cuboid Length = l = 22 m Breadth = b = 14 m Let Height = h m Volume of cuboid = lbh = 22 14 h Now, Volume of cylinder = Volume of cuboid 770 = 22 14 h 770/(22 14) = h h = 770/(22 14) h = 770/(22 14) h = 2.5 m Hence, height of platform = 2.5 m

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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