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Answer :
Since QT is a tangent to the circle at T and OT is radius,
\(\therefore \ OT ⊥ QT \)
It is given that OQ = 25 cm and QT = 24 cm,
\(\therefore \) By Pythagoras theorem, we have
\( \Rightarrow OQ^2 \ = \ QT^2 \ + \ OT^2 \)
\(\Rightarrow OT^2 \ = \ OQ^2 \ - \ QT^2 \)
\( \Rightarrow \ OT^2 \ = \ 25^2 \ - \ 24^2 \)
\( \Rightarrow OT^2 \ = \ 49 \)
\( \Rightarrow \ OT \ = \ \sqrt{49} \ = \ 7 \)
\(\therefore \) radius of the circle is 7 cm
\(\therefore \) Option (A) is correct.