question_answer

A circular ring with centre O is kept in the vertical position by two weightless, this 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) 10/3 cm            

B) 20/3 cm

C) 10 cm              

D) 40/3 cm

Correct Answer: B

Solution :

In \[\Delta OTQ,\]\[O{{T}^{2}}=O{{Q}^{2}}+T{{Q}^{2}}\]

\[\Rightarrow \]   \[{{(13)}^{2}}={{(5)}^{2}}+{{(TQ)}^{2}}\]

\[\Rightarrow \]   \[T{{Q}^{2}}=169-25=144\]

\[\Rightarrow \]   \[TQ=12\,\,cm\]

Then, in \[\Delta TEB,\]

\[T{{B}^{2}}=E{{B}^{2}}+T{{E}^{2}}\]

\[\Rightarrow \]   \[{{(120-x)}^{2}}=B{{Q}^{2}}+T{{E}^{2}}\]

\[[\because EB=BQ]\]

\[\Rightarrow \]   \[144+{{x}^{2}}-24x={{x}^{2}}+{{(8)}^{2}}\]

\[\Rightarrow \]   \[144+{{x}^{2}}-24x={{x}^{2}}+64\]

\[\Rightarrow \]   \[24x=80\]\[\Rightarrow \]\[x=\frac{20}{6}=\frac{10}{3}\,\,cm\]

\[\therefore \]                  \[AB=2EB=2x=2\times \frac{10}{3}\]

\[\Rightarrow \]   \[AB=\frac{20}{3}\,\,cm\]