A) \[\sqrt{7}\,cm\]
B) \[3\sqrt{7}\,cm\]
C) \[4\sqrt{7}\,cm\]
D) \[(8+2\sqrt{7})\,cm\]
Correct Answer: D
Solution :
M is the mid-point of AB, \[\therefore \] \[AM=6\,cm\] \[AO'({{r}_{1}})=10cm,\] \[AO\,({{r}_{2}})=8\,cm\] AB is perpendicular to OO', then In \[\Delta AOM,\,\,100=36+O{{M}^{2}}\] [using pythagoras theorem] \[\Rightarrow \]\[OM=8\,cm;\] In \[\Delta AMO',\,64=36+M{{O}^{2}}\] \[\Rightarrow \] \[\sqrt{28}=MO'\,\,\,\Rightarrow \,\,\,2\sqrt{7}=MO'\] \[\therefore \] \[OO'=(2\sqrt{7}+8)\,\,cm\]You need to login to perform this action.
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