question_answer

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

A) 19 cm                         

B)        20 cm             

C)        16 cm                         

D)        \[\sqrt{150}\,cm\]

Correct Answer: A

Solution :

Produce BD to meet the bigger circles at E. Join AE. Then  \[\angle AEB={{90}^{o}}\]  [Angle in a semicircle] \[OD\bot BE\]              [\[\because \]  BE is tangent to the smaller circle at D and OD is its radius] and, BD=DE [\[\because \] BE is a chord of the circle and \[OD\bot BE\]] \[\therefore \]    \[OD||AE\]  \[[\because \,\,\,\angle AEB=\angle ODB={{90}^{o}}]\] In \[\Delta AEB,\text{ O}\]and D are mid- points of AB and BE. Therefore, by mid-point theorem, we have\[OD=\frac{1}{2}AE\] \[\Rightarrow \]            \[AE=2\times 8=16\,cm\] In \[\Delta ODB,\]we have       \[O{{B}^{2}}=O{{D}^{2}}+B{{D}^{2}}\] [By Pythagoras Theorem]          \[\Rightarrow \]     \[132={{8}^{2}}+B{{D}^{2}}\] \[\Rightarrow \] \[B{{D}^{2}}=169-64=105\Rightarrow BD=\sqrt{105}\,cm\] \[\Rightarrow \]  \[DE=\sqrt{105}\,cm\]          \[[\because \,\,BD=DE]\] In  \[\Delta AED,\] we have      \[A{{D}^{2}}=A{{E}^{2}}+E{{D}^{2}}\] [By Pythagoras Theorem]            \[\Rightarrow \] \[A{{D}^{2}}={{16}^{2}}+{{(\sqrt{105})}^{2}}=256+105=361\] \[\Rightarrow \]            \[AD=19\,cm\]