question_answer

A circular ring with centre O is kept in the vertical position by two weightless thin strings TP and TQ attached to the ring at P and Q. The line OT meets the ring at E whereas a tangential string at E meets TP and TQ at A and B respectively. If the radius of the ring is 5 cm and OT =13 cm, then what is the length of AB?

A) \[\frac{10}{3}cm\]                    

B) \[\frac{20}{3}cm\]

C) 10 cm              

D) \[\frac{40}{3}cm\]

Correct Answer: B

Solution :

In \[\Delta \]OTQ, \[\begin{align} & O{{T}^{2}}=O{{Q}^{2}}+T{{Q}^{2}} \\ & \Rightarrow {{(13)}^{2}}={{(5)}^{2}}+{{(TQ)}^{2}} \\ \end{align}\] \[\Rightarrow \]   TQ2 = 169 ? 25 = 144 \[\Rightarrow \]   TQ = 12 cm Then, in \[\Delta \]TEB, \[{{\operatorname{TB}}^{2}}= E{{B}^{2}}+ T{{E}^{2}}\] \[\therefore \] (EB = BQ) (Common Tangent) \[{{\left( 12 -x \right)}^{2}}= B{{Q}^{2}}+T{{E}^{2}}\] \[144 +{{x}^{2}} 24x={{x}^{2}}+ {{\left( 8 \right)}^{2}}\] \[144 +{{x}^{2}}- 24x={{x}^{2}}+ 64\] \[\Rightarrow 24x=80\Rightarrow x=\frac{20}{6}=\frac{10}{3}m\] \[\therefore \] AB = 2EB \[\Rightarrow \] 2x = \[2\times \frac{10}{3}\] \[\Rightarrow \]   \[AB=\]