Question : In the given figure, $\mathrm{CD}$ and $\mathrm{AB}$ are the diameters of the circle and $\mathrm{AB}$ and $\mathrm{CD}$ are perpendicular to each other. $\mathrm{LQ}$ and $\mathrm{SR}$ are perpendiculars to $\mathrm{AB}$ and $\mathrm{CD}$, respectively. The radius of the circle is $5\;\mathrm{cm}$, $\mathrm{PB:PA = 2:3}$ and $\mathrm{CN:ND = 2:3}$. What is the length (in $\mathrm{cm}$) of $\mathrm{SM}$?
Option 1: $\left [ \left ( 5\sqrt{3} \right )-3 \right ]$
Option 2: $\left [ \left ( 4\sqrt{3} \right )-2 \right ]$
Option 3: $\left [ \left ( 2\sqrt{6} \right )-1 \right ]$
Option 4: $\left [ \left ( 5\sqrt{6} \right )-3 \right ]$
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Correct Answer: $\left [ \left ( 2\sqrt{6} \right )-1 \right ]$
Solution :
Given that, $\mathrm{PB:PA = 2:3}$ and $\mathrm{AB}$ (is a diameter) $= 10 \;\mathrm{cm}$
Construction: Draw a line from $\mathrm{O}$ to $\mathrm{S}$.
$\mathrm{PA} = \frac{3}{5}\times10 = 6 \;\mathrm{cm}$
$\mathrm{PB} = 10-6 = 4 \;\mathrm{cm}$
$\mathrm{OP = NM}$
$\mathrm{OP }= 6-5 = 1 \;\mathrm{cm}$
$\mathrm{ND} = 6 \;\mathrm{cm}$ and $\mathrm{NC} = 4 \;\mathrm{cm}$
$\mathrm{NO} = 6-5 = 1 \;\mathrm{cm}$
Applying Pythagoras' theorem in $\triangle \mathrm{OSN}$
$\mathrm{OS^2=NS^2+NO^2}$
$\mathrm{25 = NS^2+1}$
$\mathrm{NS = 2\sqrt{6}}$
$\mathrm{SM = NS-NM }= [2\sqrt{6}-1]\;\mathrm{cm}$
Hence, the correct answer is $[(2\sqrt{6})-1]$.
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