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
- 7.8
- 8.5
- 9.1
- 9.3
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 $\Delta AQB$
$\begin{align} & B{{Q}^{2}}=A{{B}^{2}}-A{{Q}^{2}} \\ & \Rightarrow B{{Q}^{2}}={{10}^{2}}-{{4}^{2}}=84 \\ & \Rightarrow BQ=\sqrt{84}\approx 9.1 \\ \end{align}$CAT Online Course
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