Question 18

ABCD is a trapezium in which AD||BC and AB = DC = 10 m. then the distance of AD from BC is

Solution

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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