Discussion

Explanation:

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In right ∆AOC,
AC = $\sqrt{O{A}^2-O{C}^2}$ = $\sqrt{{26}^2-{16}^2}$ = $\sqrt{420}$ = 2$\sqrt{105}$.

⇒ AB = 4√105

Now, since AD is the diameter, hence ∠ABD = 90° [Angle subtended by diameter on circle is always right angle.]

∴ In right ∆ABD
AD² = AB² + BD²
⇒ 52² = (4√105)² + BD²
⇒ BD² = 2704 – 1680 = 1024
⇒ BD = 32.

Now in ∆CBD
⇒ CD² = CB² + BD²
⇒ CD² = 420 + 1024 = 1444
⇒ CD = 38

Hence, option (d).

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