1. ABCD is a trapezium in which AD||BC and AB = DC = 10 m. then the distance of AD from BC is
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By: anil on 05 May 2019 01.49 am
Given : DC = 10 m and $$angle$$ DCE = $$45^circ$$ DE is the distance between AD and BC To find : DE = ? Solution : In $$ riangle$$ DEC => $$sin(45^circ)=frac{DE}{DC}$$ => $$frac{1}{sqrt{2}}=frac{DE}{10}$$ => $$DE=frac{10}{sqrt{2}}$$ => $$DE=5sqrt{2}$$ m => Ans - (C)
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