question_answer

The in circle of a \[\Delta ABC\] touches the sides \[AB,\,\,BC\] and \[AC\] at the points \[P,\,\,Q,\,\,R\] respectively then which of the following statements is/are correct:

I. \[AP+BQ+CR=PB+QC+RA\]

II.\[AP+BQ+CR=\frac{1}{2}\](perimeter of \[\Delta ABC\])

III. \[AP+BQ+CR=3(AB+BC+CA)\]

A)  I, II and III                   

B)  only I

C)  II and III only   

D)         I and II  

Correct Answer: D

Solution :

As the tangents drawn from an external point to a circle are equal. \[\therefore \]      \[AP=AR,\,\,BQ=BP\]and\[CR=QC\] \[\therefore \]\[AP+BQ+CR=BP+QC+RA\] Here I is correct. Perimeter of             \[\Delta ABC=AB+BC+AC\]             \[=AP+PB+BQ+QC+RC+RA\]             \[=2(AP+BQ+CR)\] \[\therefore \]      \[AP+BQ=CR=\frac{1}{2}\]                                        (Perimeter of \[\Delta ABC\]) II is correct.