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