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

Correct Answer: 9.1 (Option 3)

AB is the diameter of a circle of radius 5 cm, so the length of AB is $10\text{cm}$.
P and Q are points on the circle. Since AB is the diameter, the angle subtended by AB at any point on the circle is a right angle.
Therefore, $\angle \mathrm{APB}=90°$ and $\angle \mathrm{AQB}=90°$.

In the right-angled triangle APB, using Pythagoras' theorem:

${\mathrm{AP}}^2+{\mathrm{PB}}^2={\mathrm{AB}}^2$

${\mathrm{AP}}^2+6^2={10}^2$

${\mathrm{AP}}^2+36=100$

${\mathrm{AP}}^2=64\Rightarrow \mathrm{AP}=8\text{cm}$

We are given that the length of AP is twice that of AQ:

$\mathrm{AP}=2\mathrm{AQ}\Rightarrow 8=2\mathrm{AQ}\Rightarrow \mathrm{AQ}=4\text{cm}$

Now, in the right-angled triangle AQB, using Pythagoras' theorem:

${\mathrm{AQ}}^2+{\mathrm{QB}}^2={\mathrm{AB}}^2$

$4^2+{\mathrm{QB}}^2={10}^2$

$16+{\mathrm{QB}}^2=100$

${\mathrm{QB}}^2=84$

$\mathrm{QB}=\sqrt{84}\approx 9.16\text{cm}$

The value nearest to 9.16 in the options is 9.1. Therefore, option 3 is correct.