Answer : (d) 4 or 28
Let AB and CD be chords of lengths 32 cm and 24 cm in a circle with centre O, on the same side of the centre.
Then OA = OD = 20 cm
Let the perpendicular from the centre intersect the chords AB and CD at E and F respectively. Then, E and F are the midpoints of AB and CD respectively.
(Perpendicular from the centre of the circle to a chord bisects the chord.)
Now in rt. ∆ OFD, OF = \(\sqrt{OD^2 -FD^2}\) = \(\sqrt{{20}^2 -{12}^2}\)
⇒ OF = \(\sqrt{400-144}\) = \(\sqrt{256
}\) = 16 cm.
In rt. ∆ OEB, OE = \(\sqrt{OB^2 -EB^2}\) = \(\sqrt{{20}^2 -{16}^2}\)
= \(\sqrt{400-256}\) = \(\sqrt{144
}\) = 12 cm.
∴ Required distance = EF = OF – OE = (16 – 12) cm = 4 cm.
So the option containing value 4 is correct. The other required distance is 28 cm when the chords lie on the opposite side of centre O.