Let the two parallel chords be AB = 46 cm and CD = 40 cm. Radius of the circle, r = 25 cm. The centre O lies between the two chords. Draw perpendiculars from O to the chords (these perpendiculars bisect the chords). - Half of AB = 23 cm - Half of CD = 20 cm Distance from centre to each chord: For chord AB (46 cm): d\(_1 = \sqrt{r^2 - 23^2} = \sqrt{625 - 529} = \sqrt{96} = 4\sqrt{6} \approx 9.80 \text{ cm}\) For chord CD (40 cm): d\(_2 = \sqrt{r^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15 \text{ cm}\) Since the centre lies between the two chords, the distance between the chords is: d\(_1 + d_2 = 4\sqrt{6} + 15 \text{ cm}\) Approximate value: ≈ 24.80 cm