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] \[\Rightarrow \] \[A{{D}^{2}}={{16}^{2}}+{{(\sqrt{105})}^{2}}=256+105=361\] \[\Rightarrow \] \[AD=19\,cm\]You need to login to perform this action.
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