question_answer

Given \[\Delta ABC\] with sides \[AB=6,\] \[AC=5,\] and \[CA=4.\]I is the incentre of \[\Delta ABC,\] and the internal angle bisectors of A B and C intersect the opposite sides.at D, E and F, respectively. The ratio of area of \[\Delta AIB\] to area of \[\Delta CID\] is                               

A) 2                    

B) 3          

C) 4                      

D) 5                                   

Correct Answer: B

Solution :

[b] Let r be the inradius. Area of \[\Delta AIB,\] \[{{\Delta }_{1}}=\frac{1}{2}\times r\times AB=\frac{1}{2}rc\] Area of \[\Delta \,CID\], \[{{\Delta }_{2}}=\frac{1}{2}\times r\times CD=\frac{1}{2}r\left[ \frac{ba}{b+c} \right]\] \[\therefore \,\,\,\,\frac{{{\Delta }_{1}}}{{{\Delta }_{2}}}=\frac{(b+c)c}{ab}=\frac{(4+6)6}{4\times 5}=3\]