A) \[\sqrt{29}\,cm\]
B) \[2\sqrt{21}\,cm\]
C) \[\sqrt{21}\,cm\]
D) \[2\sqrt{29}\,cm\]
Correct Answer: A
Solution :
[a] According to the question, OB = OD = radius of sphere In \[\Delta OMB,\] \[{{(OB)}^{2}}={{(OM)}^{2}}+{{(MB)}^{2}}\] \[\Rightarrow \] \[{{r}^{2}}={{x}^{2}}+{{5}^{2}}\] (Let OM = x) \[\therefore \] \[{{r}^{2}}={{x}^{2}}+25\] (i) In \[\Delta {\mathrm O}\Nu D,\] \[{{(OD)}^{2}}={{(ON)}^{2}}+{{(ND)}^{2}}\] \[\Rightarrow \] \[{{r}^{2}}={{(x+3)}^{2}}+{{2}^{2}}\] \[\Rightarrow \] \[{{r}^{2}}={{x}^{2}}+9+6x+4\] \[\Rightarrow \] \[{{x}^{2}}+25={{x}^{2}}+6x+13\] [From Eq.(i)] \[\Rightarrow \] \[6x=25-13=12\] \[\therefore \] \[x=2\,cm\] Then, \[r=\sqrt{4+25}\]\[\Rightarrow \]\[r=\sqrt{29}\] \[\therefore \] \[2r=2\sqrt{29}\] |
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