Correct Answer: Either 8 cm or 22 cm

Solution : Given: Radius = 25 cm
Chords = 40 cm and 48 cm
Case 1: When chords lie on both sides of the centre.

$AB = 40$ cm
$CD = 48$ cm
$CE = DE = 24$ cm
$AF = BF = 20$ cm
$OA = OC = 25$ cm
In $\triangle AOF$,
⇒ $OF = \sqrt{OA^2 - AF^2}$
⇒ $OF = \sqrt{25^2 - 20^2}$
⇒ $OF = \sqrt{225}$
$\therefore OF = 15$ cm
In $\triangle$ COE,
$OE = \sqrt{OC^2 - CE^2}$
⇒ $OE = \sqrt{25^2 - 24^2}$
⇒ $OE = \sqrt{49}$
$\therefore OE= 7$ cm
$\therefore$ Required distance $= EF = OE + OF = (7 + 15) = 22$ cm

Case 2: When the chords lie on the same side of the centre.

$AF = 20$ cm
$CE = 24$ cm
$OC = OA = 25$ cm
In $\triangle OAF$,
$OF = \sqrt{OA^2 - AF^2}$
⇒ $OF = \sqrt{25^2 - 20^2}$
⇒ $OF = \sqrt{225} = 15$ cm
In $\triangle OCE$,
$OE = \sqrt{OC^2 - CE^2} = \sqrt{25^2 - 24^2}$
⇒ $OE = \sqrt{(25 + 24)(25 - 24)}$
⇒ $OE = \sqrt{49} = 7$ cm
$\therefore$ Required distance $= EF = OF - OE = 15 - 7 = 8$ cm
Hence, the answer is either 8 cm or 22 cm.