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# CAT 2019 Quant Question with Solution 20

Question:
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

1. 7.8
2. 8.5
3. 9.1
4. 9.3

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}

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