A) Only (i)
B) Only (ii)
C) Only (iii)
D) None of these
Correct Answer: A
Solution :
\[{{A}_{1}}=\pi {{R}^{2}},\,\,{{A}_{2}}=\frac{\pi {{R}^{2}}}{4}\] \[\therefore \] \[{{A}_{3}}=\pi {{R}^{2}}-\frac{\pi {{R}^{2}}}{4}=\frac{3\pi {{R}^{2}}}{4}\] \[\therefore \] \[{{A}_{1}}{{A}_{3}}=\pi {{R}^{2}}\times \frac{3}{4}\,\pi {{R}^{2}}\] \[=\frac{3}{4}\,{{\pi }^{2}}{{R}^{4}}=3\pi {{R}^{2}}{{A}^{2}}\] \[=3\times 4{{A}_{2}}\times {{A}_{2}}=12A_{2}^{2}\] \[\Rightarrow \,\,{{A}_{1}}{{A}_{3}}<16\,A_{2}^{2}\]You need to login to perform this action.
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