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 = \[2\times \frac{10}{3}\] \[\Rightarrow \] \[AB=\]You need to login to perform this action.
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