A) \[{{r}^{2}}\]
B) \[\frac{3\sqrt{3}}{4}{{r}^{2}}\]
C) \[\frac{3\sqrt{3}}{8}{{r}^{2}}\]
D) \[\sqrt{3{{r}^{2}}}\]
Correct Answer: C
Solution :
See that figure since,\[OP\bot PY\] |
\[\therefore \angle APY=90{}^\circ -\theta \] |
Where \[OPA=\theta \] |
\[\therefore \angle PAY=\theta \] |
Now in \[\Delta OPA,\] |
\[A{{P}^{2}}={{r}^{2}}+{{r}^{2}}-2r.r\cos (\pi -2\theta )=4{{r}^{2}}{{\cos }^{2}}\theta \] |
\[\therefore AP=2r\cos \theta \] |
\[\Rightarrow PY=AP\sin \theta =r\sin 2\theta \] |
\[and\,AY=AP\cos \theta =2r\,{{\cos }^{2}}\theta \] |
\[\therefore \]Area of \[\Delta APY,\] |
\[\Delta =\frac{1}{2}PY.AY={{r}^{2}}\sin 2\theta {{\cos }^{2}}\theta \] |
\[\frac{d\Delta }{d\theta }={{r}^{2}}\left[ 2\cos 2\theta {{\cos }^{2}}\theta -{{\sin }^{2}}2\theta \right]=0\] |
\[\Rightarrow \theta =\frac{\pi }{2},\frac{\pi }{6}\] |
\[\because 0\ne \frac{\pi }{2}.\] |
Also \[\Delta \] is maximum at \[\theta =\frac{\pi }{6}\](check) |
\[\therefore {{\Delta }_{\max }}={{r}^{2}}.\frac{\sqrt{3}}{2}{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{3\sqrt{3{{r}^{2}}}}{8}\] |
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