Answer :

Let, length of AB = 6cm and length of CD = 8cm

Radius of circle = 5cm

OB = OC = 5cm

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

AM = MB = 3cm

Also, CN = ND = 4cm

In ΔOMB,

⇒ OB^{2} = OM^{2} + MB^{2}

⇒ 5^{2} = OM^{2} + 3^{2}

⇒ OM^{2} = 25-9

⇒ OM^{2} = 16

⇒ OM = 4cm

In ΔONC,

⇒ OC^{2} = ON^{2} + CN^{2}

⇒ 5^{2} = ON^{2} + 4^{2}

⇒ ON^{2} = 25-16

⇒ ON^{2} = 9

⇒ ON = 3cm

Distance between AB and CD = OM + ON = 4 + 3 = 7cm

Hence the correct option is D.

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