A) 12, 13
B) 14, 11
C) 15, 10
D) 16, 9
Correct Answer: D
Solution :
Draw a perpendicular OD from O to AB. \[\therefore \] \[AD=\frac{25}{2}=12.5\] In right angled \[\Delta \,\,ODA,\] \[O{{A}^{2}}=O{{D}^{2}}+A{{D}^{2}}\] \[{{(13)}^{2}}=O{{D}^{2}}+{{(12.5)}^{2}}\] or \[O{{D}^{2}}=169-156.25\] \[=12.75\] Again in right angled \[\Delta \,ODM,\] \[O{{M}^{2}}=O{{D}^{2}}+D{{M}^{2}}\] \[{{5}^{2}}=12.75+D{{M}^{2}}\] \[\therefore \] \[D{{M}^{2}}=25-12.75=12.55\] \[DM=\sqrt{12.25}=3.5\,cm\] \[\therefore \] \[AM=12.5+3.5=16\,cm\] and \[MB=9\,cm\]You need to login to perform this action.
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