question_answer

The vertices of triangle are \[A(1,\,2),\,B(3,\,2)\,C\,(2,\,\sqrt{3}+2)\] a point P moves with a triangle such that it satisfies the relation \[(d(P,\,AB)\le d(P,\,AC)\] and \[(d(P,\,AB)\le d(P,\,BC)),\] then the area of traced by point P, where d(P, AB) denotes the distance of point P from line AB is

A)  \[\frac{2}{\sqrt{3}}\]                        

B)  \[\sqrt{3}\]

C)  \[\frac{1}{\sqrt{3}}\]                        

D)  \[\frac{\sqrt{3}}{2}\]

Correct Answer: C

Solution :

 \[\Delta ABC\] is an equilateral triangle, \[AB=BC=CA=2\] units. \[AD,\,BE\] are internal angle biscetors meet at a point I. then, Point P will inside the triangle \[IAB\] Area of \[\Delta IAB=\frac{1}{3}\Delta ABC\] \[=\frac{1}{3}\,\times 4\times \frac{\sqrt{3}}{4}=\frac{1}{\sqrt{3}}\] units