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 \[\Delta TEB,\] |
\[T{{B}^{2}}=E{{B}^{2}}+T{{E}^{2}}\] |
\[\Rightarrow \] \[{{(120-x)}^{2}}=B{{Q}^{2}}+T{{E}^{2}}\] |
\[[\because EB=BQ]\] |
\[\Rightarrow \] \[144+{{x}^{2}}-24x={{x}^{2}}+{{(8)}^{2}}\] |
\[\Rightarrow \] \[144+{{x}^{2}}-24x={{x}^{2}}+64\] |
\[\Rightarrow \] \[24x=80\]\[\Rightarrow \]\[x=\frac{20}{6}=\frac{10}{3}\,\,cm\] |
\[\therefore \] \[AB=2EB=2x=2\times \frac{10}{3}\] |
\[\Rightarrow \] \[AB=\frac{20}{3}\,\,cm\] |
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