question_answer

ABC is an equilateral triangle. P and Q are two points on AB and AC respectively such that \[PQ||BC.\]If \[PQ=5\,\,cm,\]then area of \[\Delta APQ\]is

A) \[\frac{25}{4}\,\,cm\]                

B) \[\frac{25}{\sqrt{3}}\,\,cm\]

C) \[\frac{25\sqrt{3}}{4}\,\,cm\]     

D) \[25\sqrt{3}\,cm\]          

Correct Answer: C

Solution :

Since, \[PQ||BC\]

\[\therefore \]\[\angle APQ=\angle AQP=60{}^\circ \]

\[\therefore \]\[\Delta APQ\]is an equilateral triangle.

So, \[\Delta APQ\,\sim \,\Delta ABC\]

\[\therefore \,Area\,of\,(\Delta APQ)\]

\[=\frac{\sqrt{3}}{4}{{(PQ)}^{2}}=\frac{\sqrt{3}}{4}\times 25=\frac{25\sqrt{3}}{4}c{{m}^{2}}\]