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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. $9.3$
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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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