Answer: A
Given that, AB=CD=16 cm and OB=OD=10 cm
From the right-angled triangle ∆POB,
OB\(^2\)=OP\(^2\)+PB\(^2\)
or, 10\(^2\)=OP\(^2\)+\((\frac{16}{2})^2\)
or, OP\(^2\)=100-64
or, OP=\(\sqrt{36}\)=6
∴PQ=6×2=12 cm
Given that, AB=CD=16 cm and OB=OD=10 cm
From the right-angled triangle ∆POB,
OB\(^2\)=OP\(^2\)+PB\(^2\)
or, 10\(^2\)=OP\(^2\)+\((\frac{16}{2})^2\)
or, OP\(^2\)=100-64
or, OP=\(\sqrt{36}\)=6
∴PQ=6×2=12 cm