1. The cross-section of a canal is in the shape of an isosceles trapezium which is 3 m wide at the bottom and 5 m wide at the top. If the depth of the canal is 2 m and it is 110 m long, what is the maximum capacity of this canal? (Take π = 22/7)
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By: anil on 05 May 2019 02.16 am
The canal is 3 m wide at the bottom and 5 m at the top, thus the canal is in the shape of a trapezium whose volume will be equal to the product of area of trapezium and the width. Length of canal = 110 m Volume of canal = $$(frac{1}{2} imes $$(sum of parallel sides) $$ imes$$ height $$) imes$$ length => Volume = $$(frac{1}{2} imes (5+3) imes 2) imes 110$$ = $$8 imes 110 = 880$$ $$m^3$$ => Ans - (D)
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