Question: 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?
Options:
A) 10/3 cm
B) 20/3 cm
C) 10 cm
D) 40/3 cm
Show Answer
Answer:
Correct Answer: B
Solution:
- In $ \Delta OTQ, $ $ OT^{2}=OQ^{2}+TQ^{2} $
$ \Rightarrow $ $ {{(13)}^{2}}={{(5)}^{2}}+{{(TQ)}^{2}} $
$ \Rightarrow $ $ TQ^{2}=169-25=144 $
$ \Rightarrow $ $ TQ=12cm $
Then, in $ \Delta TEB, $
$ TB^{2}=EB^{2}+TE^{2} $
$ \Rightarrow $ $ {{(120-x)}^{2}}=BQ^{2}+TE^{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 $
BETA
