A chord 21 cm long is drawn in a circle of diameter 25 cm. The perpendicular distance of the chord from the centre is:

$\sqrt{46}$

Let AB be the chord of length 21 cm.

= OD be the perpendicular distance

= AO be the radius of the circle

\( {OA }^{2 } \) = \( {OD }^{2 } \) + \( {AD }^{2 } \)

= \( {25/2 }^{2 } \) = \( {OD }^{2 } \) + \( {21/2 }^{2 } \)

= \(\frac{625}{4}\) = \( {OD }^{2 } \) + \(\frac{441}{4}\)

= \( {OD }^{2 } \) = \(\frac{625}{4}\) - \(\frac{441}{4}\) = \(\frac{184}{4}\) = 46

Therefore, OD is \(\sqrt {46 }\).