CAT 2019 Quant Question with Solution 20

Correct Answer: Option: 3

Refer to the figure below:

$\angle APB=\angle AQB={{90}^{0}}$ {angle in a semicircle is a right angle}

Also, let AQ=x, so AP=2x

In Right $\Delta APB$

$A{{P}^{2}}=A{{B}^{2}}-B{{P}^{2}}$

$A{{P}^{2}}=A{{B}^{2}}-B{{P}^{2}}$ $\begin{align} & \Rightarrow A{{P}^{2}}={{10}^{2}}-{{6}^{2}}={{8}^{2}} \\ & \Rightarrow AP=8\Rightarrow 2x=8 \\ & \Rightarrow x=4 \\ \end{align}$

Similarly, in Right $\Delta AQB$

$\begin{align} & B{{Q}^{2}}=A{{B}^{2}}-A{{Q}^{2}} \\ & \Rightarrow B{{Q}^{2}}={{10}^{2}}-{{4}^{2}}=84 \\ & \Rightarrow BQ=\sqrt{84}\approx 9.1 \\ \end{align}$