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

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

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.

References:

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