Question

Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

Answer 1

seg OP ⊥ chord CD            …[Given]

∴ l(PD) = `1/2 l(CD)`           …[Perpendicular drawn from the centre of a circle to its chord bisects the chord]

∴ l(PD) = `1/2 xx 48`           …[∵ l(CD) = 48 cm]

∴ l(PD) = 24 cm

In ΔOPD, 

∠ OPD = 90°

∴ [l(OD)]² = [l(OP)]² + [l(PD)]²             … [Pythagoras theorem]

∴ (25)² = [l(OP)]² + (24)²

∴ (25)² − (24)² = [l(OP)]²

∴ 625 − 576 = [l(OP)]²

∴ 49 = [l(OP)]²

∴ l(OP) = `sqrt(49)`           …[Taking square root of both sides]

∴ l(OP) = 7 cm

∴ The distance of the chord from the centre of the circle is 7 cm.