Correct Answer: $\frac{\sqrt{265}}{4}$

Solution :
Given, AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle.
Let OM be perpendicular to AB and ON be perpendicular to CD.
Perpendicular from the centre to a chord bisects the chord.
So, AM = BM = 4 cm and CN = DN = 3 cm
Let the radius of the circle be r.
ON – OM = MN = 2 cm----(1)
In $\triangle$OAM,
OA ² = OM ² + AM ²
⇒ r ² = OM ² + 4 ² ----(2)
In $\triangle$OCN,
OC ² = ON ² + CN ²
⇒ r ² = ON ² + 3 ² ----(3)
From equation 2 and 3, we get,
OM ² + 4 ² = ON ² + 3 ²
⇒ ON ² – OM ² = 16 – 9
⇒ (ON – OM)(ON + OM) = 7
⇒ (ON + OM) = $\frac{7}{2}$----(4)
Adding equation (1) and (4), we get:
⇒ ON = $\frac{11}{4}$
Substituting this value in equation (3), we get,
⇒ r ² = $(\frac{11}{4})^2$ + 3 ²
⇒ r ² = $\frac{265}{16}$
$\therefore$ r = $\frac{\sqrt{265}}{4}$ cm
Hence, the correct answer is $\frac{\sqrt{265}}4$.