Chord AB of circle of radius 10 cm is at a distance 8 cm from the centre O. If tangents drawn at A and B intersect at P, then the length of the tangent AP (in cm) is:

7.5 

In the given figure,

\(\Delta \)DAO is similar to \(\Delta \)APO  (As \(\angle\)OAP = \(\angle\)ODA = \({90}^\circ\) and \(\angle\)AOD is common in two triangles)

So,

\(\frac{AD}{AP}\) = \(\frac{DO}{AO}\)

= \(\frac{6}{AP}\) = \(\frac{8}{10}\)   [6, 8, 10 pythagoras triplets]

= AP = \(\frac{60}{8}\)

= AP = 7.5 cm

Therefore, AP is 7.5 cm.