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Last updated at May 29, 2018 by Teachoo

Transcript

Example 5 Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see figure ). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m3 , and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the shed? Take π = 22/7) Total volume = Volume of cuboid + 1/2× Volume of cylinder Volume of cuboid Here, Length = 15 m Breadth = 7m Height = 8 m Volume of cuboid = Length × Bredth × Height = lbh = 15 × 7 × 8 = 840 m2 Volume of cylinder Diameter of cylinder = breadth of cuboid = 7 m ∴ radius = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 7/2 m Height of cylinder = Length of cuboid = 15 cm Volume of cylinder = 𝜋𝑟2ℎ = 22/7×(7/2)^2 × 15 = 22/7×7/2×7/2×15 = (11 × 7 × 15)/2 = 1155/2 = 577.5 m2 So, Total volume = Volume of cuboid + 1/2× Volume of cylinder = 840 + 1/2× 577.5 = 840 + 288.75 = 1128.75 m2 Hence , total volume of air that the shed will hold = 1128.75 m2 Further, suppose the machinery in the shed occupies a total space of 300 m3 , and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the shed Total space occupied by the machinery = 300 m3 Total workers = 20 Space occupied by 1 worker = 0.08 m3 Space occupied by 20 workers = 0.08 ×20 = 1.6 m3 Actual volume of air in the shed = Total volume of shed – Space occupied by machinery – Space occupied by 20 workers = 1128.75 – 300 – 1.60 = 1128.75 – (300 + 1.60) = 1128.75 – 301.60 = 827 .15 m3 Hence, Volume of air when there are machinery and workers = 827.15 m3

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Example 3 Important

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Example 5 You are here

Example 6 Important

Example 7 Important

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Example 9 Important

Example 10

Example 11 Important

Example 12 Important Deleted for CBSE Board 2022 Exams

Example 13 Deleted for CBSE Board 2022 Exams

Example 14 Deleted for CBSE Board 2022 Exams

Surface Area and Volume Formulas Important

Chapter 13 Class 10 Surface Areas and Volumes (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.