Question

$ABCD$ is a triangle in which $AB=4cm,BC=5cm$ and $AC=6cm$. A circle is drawn to touch side $BC$ at $P$, side $AB$ extended at $Q$ and side $AC$ extended at $R$. Then, $AQ$ equals

• A. $7.0cm$
• B. $7.5cm$
• C. $6.5cm$
• D. $15.0cm$

Answer 1

The correct option is A $7.0cm$

$\begin{array}{c}\text{Given that,}\\ \begin{array}{cc}AB& =4cm\\ BC& =5cm\\ CQ& =6cm\end{array}\end{array}$

$\begin{array}{c}\text{We know that tangent segments to a circle}\\ \text{from the same external point are congurent}\\ \therefore \text{We have}\\ AR=AQ,BQ=BP,CR=CP\\ \text{NOW,}AQ+BQ=4cm⟶-\left(1\right)\\ BP+CP=5cm⟶-\left(2\right)\\ AR+CR=6cm⟶\text{(3)}\end{array}$

$\begin{array}{c}\text{also,}BQ+CR=4cm-\text{(5)}\\ AR+CP=5cm\\ \text{Adding all these we get}\\ AQ+BQ+AQ+CR+BQ+CR=15\\ 2(AQ+BQ+CR)=15-\text{(4)}\\ AQ+BQ+CR=7.5\end{array}$

$\begin{array}{c}\text{Solving}\left(1\right)\text{and}4\\ \Rightarrow CR=3.5cm\\ \text{Solving}\left(5\right)\text{and}\left(4\right)\\ \Rightarrow AQ=3.5cm\end{array}$