Question

In the figure shown here, a circle touches the side $BC$ of a triangle $ABC$ at $P$ and $AB$ and $AC$ produced at $Q$ and $R$ respectively. What is $AQ$ equal to$?$

• A. One-third of the perimeter of $△ABC.$
• B. Half of the perimeter of $△ABC.$
• C. Two-third of the perimeter of $△ABC.$
• D. Three-fourth of the perimeter of $△ABC.$

Answer 1

The correct option is B Half of the perimeter of $△ABC.$

Here, $BQ=BP$ ....(i)
$PC=CR$ ....(ii)
$AQ=AR$ ....(iii)
(Lengths of the two tangents drawn from an external point to a circle are equal)
From (iii) we have
$AB+BQ=AC+CR$
$\Rightarrow$ $AB+BP=AC+CP$ (using (i) and (ii)) ....(iv)
Now perimeter of $△ABC=AB+BC+AC$
$=AB+(BP+PC)+AC$

$=(AB+BP)+(PC+AC)$
Perimeter of $△ABC=2(AB+BP)$ using (iv)
Perimeter of $△ABC=2(AB+BQ)$
$=2AQ$
Therefore, $AQ=\frac{1}{2}$(Perimeter of $△ABC$)