We can draw, the below diagram:
Let $`x\text{’}$ be the length of $ \text{AQ},$ and $`y\text{’}$ be the length of $\text{QB}.$
Given that,
Radius of a circle $ = 5 \; \text{cm} $
And, $ \text{AB}$ is a diameter of a circle $ = 2 \times \text{radius of a circle} = 10 \; \text{cm} $
Let $ \text{P}$ and $ \text{Q}$ be two points on the circle.
So, the length of $ \text{PB} = 6 \; \text{cm} $
The length of $\text{AP}$ is twice that of length $\text{AQ}$.
So, the length of $ \text{AP} = 2x $
We can take out the $\triangle \text{APB},$ from the above diagram:
Apply the Pythagoras’ theorem $ : \boxed{\text{Hypotenuse}^{2} = \text{Perpendicular}^{2} + \text{Base}^{2}} $
$ (10)^{2} = (2x)^{2} + (6)^{2} $
$ \Rightarrow 100 = 4x^{2} + 36 $
$ \Rightarrow 100 – 36 = 4x^{2} $
$ \Rightarrow 64 = 4x^{2} $
$ \Rightarrow 16 = x^{2} $
$ \Rightarrow \sqrt{16} = x $
$ \Rightarrow \boxed{x = 4}$
Now, We can take out the $\triangle \text{AQB},$ from the first diagram:
Again apply the Pythagoras’ theorem :
$ (10)^{2} = (4)^{2} + (y)^{2} $
$ \Rightarrow 100 = 16 + y^{2} $
$ \Rightarrow 84 = y^{2} $
$ \Rightarrow \sqrt{84} = y $
$ \Rightarrow \boxed{y = 9.16} $
Correct Answer $: \text{C} $
$\textbf{PS:}$ Thales's theorem: if $AC$ is a diameter and $B$ is a point on the diameter's circle, the angle $ABC$ is a right angle.
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