A) \[\sqrt{{{A}_{1}}{{A}_{2}}}\]
B) \[\frac{{{A}_{1}}+{{A}_{2}}}{2}\]
C) \[{{\left( \frac{\sqrt{{{A}_{1}}}+\sqrt{{{A}_{2}}}}{2} \right)}^{2}}\]
D) None of these
Correct Answer: A
Solution :
\[{{m}_{{{A}_{1}}}}=\frac{{{A}_{1}}}{{{O}_{1}}}\,\,\,\,\,\,\,\,{{m}_{{{A}_{2}}}}\,\,=\,\,\frac{{{A}_{2}}}{{{O}_{1}}}\,\,\,\,\,\,\,\,\,\,({{O}_{1}}={{O}_{2}})\,\,\,\,\,\] \[{{m}_{{{A}_{1}}}}\times \,\,{{m}_{{{A}_{2}}}}\,=\,\,\left( \frac{{{A}_{1}}}{{{O}_{1}}} \right)\,\,\,\,\left( \frac{{{A}_{2}}}{{{O}_{1}}} \right)\,\,\,\,\,={{m}_{1}}^{2}{{m}_{2}}^{2}\,\,but\,\,{{m}_{1}}{{m}_{2}}\,=\,1\,\,\]\[so\,\,\,\frac{{{A}_{1}}}{{{O}_{1}}}\,\times \,\frac{{{A}_{2}}}{{{O}_{1}}}\,\,=\,\,1\,\,\Rightarrow \,\,{{O}_{1}}^{2}\,\,=\,\,{{A}_{1}}{{A}_{2}}\] Hence \[{{O}_{1}}={{O}_{2}}\,=\,\sqrt{{{A}_{1}}{{A}_{2}}}\]
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