The correct option is
A 11 cm
We know that tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴OP⊥BC and OQ⊥AB⇒∠P=90∘ and ∠Q=90∘ Also, OP=OQ=r Now, BP=BQ [Tangents drawn from an external point to a circle are equal in lengths]
Given: ∠B=90∘. By angle sum property of a quadrilateral,∠OQB+∠QBP+∠BPO+∠POQ=360∘ 90∘+90∘+90∘+∠POQ=360∘∠POQ=360∘−270∘=90∘ All the angles measure 90∘.∴OPBQ is a square Hence, BP=BQ=r (Since, all sides of a square are equal in length) DR=DS=5 cm [Tangents drawn from an external point to a circle are equal in lengths]
Now, AR=AD−DR=23−5AR=18 cmAQ=AR=18 cm [Tangents drawn from an external point to a circle are equal in lengths]
BQ=AB−AQ=29−18=11 cm∴BQ=r=11 cm