Answer :

The figure is attached above

A is the centre of the circle.

AB is the radius = 15 cm

CD is the chord = 18 cm.

We need to find the distance of the chord from the centre i.e. AE

In this circle, we draw the perpendicular

We know that perpendicular drawn from the centre to the chord, will bisect the chord, such that CE = ED = = 9 cm

Now,

In ΔAEC,

Applying Pythagoras theorem,

AC^{2} = AE^{2} + EC^{2}

⇒AE^{2} = AC^{2} – EC^{2}

⇒ AE^{2} = (15cm)^{2}–(9cm)^{2}

⇒ AE^{2} = 225 – 81 = 144

⇒ AE = √144

⇒ AE = 12 cm

∴The chord is 12cm away from the center of the circle.

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