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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.

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According to question

Given : AC = 8 CM
OC = 5 CM
AC = AB+BC
8 = BC+BC    (\because AB =BC)
BC = 4CM
In \bigtriangleup OBC using Pythagoras theorem
H^{2}+B^{2}+P^{2}
\left ( OC \right )^{2}= \left ( BC \right )^{2}+\left ( OB \right )^{2}
\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( OB \right )^{2}
\left ( OB \right )^{2}= 25-16
OB= \sqrt{9}= 3cm
Hence radius of the inner circle is 3 cm.

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