question_answer

Let ABC be an equilateral triangle and suppose KLMN be a rectangle with K, L on BC, M on AC and AN N on AB. If \[\frac{AN}{NB}=2\] and area of triangle BKN is 6, then area of triangle ABC is equal to

A) 48                                          

B) 54

C) 96                                          

D) 108

Correct Answer: D

Solution :

Here, \[\frac{AN}{2}=\frac{NB}{1}=\frac{AB}{3}\] \[\therefore \frac{NB}{AB}=\frac{1}{3}\]                                              ?(1) As, \[\Delta BKN-\Delta BDA\]   \[(A-A)\] So,\[\frac{ar.(\Delta BKN)}{ar.(\Delta BDA)}=\frac{{{(BN)}^{2}}}{{{(AB)}^{2}}}=\frac{1}{9}\] \[\Rightarrow ar.(\Delta BDA)=54\] \[\therefore ar(\Delta ABC)=108\]