Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Suppose C₁ and C₂ are two circles with the same center O.

And  AC is a chord touching C₁ at the point D

let’s join OD.

So, $OD\bot AC$

$AD=DC=4cm$  [perpendicular line OD bisects the chord]
Thus, in right angled ∆AOD,

$O{{A}^{2}}=A{{D}^{2}}+D{{O}^{2}}$  [Derived from Pythagoras theorem]
$D{{O}^{2}}={{5}^{2}}-{{4}^{2}}$

$\Rightarrow 25-16=9$

$DO=3cm$

So, radius of the inner circle OD is equals to $3cm$.