question_answer

A circle is touching the side BC of \[\Delta ABC\] at P and is also touching AB and AC produced at Q and R, respectively. If \[AQ=6\,\,cm,\] then perimeter of \[\Delta ABC\] is

A) \[6\,\,cm\]                     

B) \[10\,\,cm\]

C) \[12\,\,cm\]

D) \[18\,\,cm\]

Correct Answer: C

Solution :

We know that, the length of tangents drawn to a circle from any external point are equal.

\[\therefore \]      \[AQ=AR\]                                (i)

\[BP=BQ\]                                (ii)

and       \[CP=CR\]                                 ... (iii)

Now. perimeter of \[\Delta ABC=AB+BC+AC\]

\[=AB+BP+CP+AC\]

Here,     \[BQ=BP\] and \[CP=CR\]

\[=AB+BQ+CR+AC\]

[using Eqs. (ii) and (iii)]

\[AB+BQ=AQ\]                         ... (iv)

\[AC+CR=AR\]                          ... (v)

Now, adding Eqs. (iv) and (v)

\[AB+BQ+AC+CR=AQ+AR\]

[Now, AQ = AR]

\[\Rightarrow \]   \[AB+BQ+AC+CR=2AQ\]

\[=2\times 6=12\]

Now, perimeter of \[\Delta ABC=2\times 6=12\]