question_answer

Let D, E be the points on sides AB and AC respectively of a triangle ABC such that DE is parallel to BC. Let \[AD=2\text{ }cm,\] \[DB=1\text{ }cm,\] \[AE=3\text{ }cm\] and area of triangle\[ADE=3\text{ }c{{m}^{2}}\]. What is EC equal to?

A)  \[1.5\,cm\]                   

B)  \[1.6\,cm\]

C)  \[1.8\,cm\]                   

D)  \[2.1\,cm\]

Correct Answer: A

Solution :

 \[DE\text{  }\!\!|\!\!\text{  }\!\!|\!\!\text{ }BC\] so, \[\Delta \,ADE\text{ }\tilde{\ }\text{ }\Delta ABC\] (According to Thale's theorem) \[\therefore \,\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \frac{2}{1}=\frac{3}{EC}\Rightarrow EC=\mathbf{1}\mathbf{.5}\,\mathbf{cm}\]