Question : In the given figure, $\mathrm{MNOP}$ is a square of side $6\;\mathrm{cm}$. What is the value (in $\mathrm{cm}$) of the radius of a circle?
Option 1: $4.25$
Option 2: $3.75$
Option 3: $3.5$
Option 4: $4.55$
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Correct Answer: $3.75$
Solution :
Given that $\mathrm{MNOP}$ is a square of side $6\;\mathrm{cm}$.
Construction: Join $\mathrm{QC}$ and $\mathrm{PC}$.
$\mathrm{OT = \frac{ON}{2} = 3\;\mathrm{cm}}$
$\mathrm{OP = 6\;\mathrm{cm}}$
$\mathrm{OT^2 = OR \times OP}$
$\mathrm{3^2 = OR \times 6}$
⇒ $\mathrm{OR = \frac{9}{6} = \frac{3}{2}\;\mathrm{cm}}$
$\mathrm{PR = OP - OR}$
$ \mathrm{PR = 6 - \frac{3}{2} = \frac{9}{2}\;\mathrm{cm}}$
$\mathrm{PQ = \frac{1}{2}× \frac{9}{2} = \frac{9}{4}\;\mathrm{cm}}$
$\mathrm{QC = \frac{1}{2} × 6 = 3\;\mathrm{cm}}$
Applying the Pythagorean theorem in $\triangle \mathrm{PQC}$,
$\mathrm{PC^2 = PQ^2 + QC^2}$
⇒ $\mathrm{ PC^2 = \left(\frac{9}{4}\right)^2 + 3^2 = \frac{225}{16}⇒ \mathrm{PC}= \frac{15}{4} = 3.75}\;\mathrm{cm}$
Hence, the correct answer is $3.75$.
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