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Answer :
Since tangent to a circle is perpendicular to the radius through the point of contact,
\(\therefore \) \( ∠ \ OTA \ = \ 90° \)
In \( ∆ \ OTA \), we have
\(\Rightarrow OA^2 \ = \ OT^2 \ + \ AT^2 \ \)
\( \Rightarrow \ 5^2 \ = \ OT^2 \ + \ 4^2 \)
\( \Rightarrow \ OT^2 \ = \ 5^2 \ - \ 4^2 \ = \ 25 \ - \ 16 \ = \ 9 \)
\( OT \ = \ \sqrt{9} \ = \ 3 \)
\(\therefore \) Radius of the circle is 3 cm.