A) \[\frac{3\sqrt{3}}{4}{{R}^{2}}\]
B) \[\frac{\sqrt{3}}{2}{{R}^{2}}\]
C) \[\frac{3\sqrt{3}}{8}{{R}^{2}}\]
D) \[{{R}^{2}}\]
Correct Answer: A
Solution :
[a] |
\[AD=AB\cos \theta =2R\cos \theta ,AE=AD\cos \theta \] |
\[=2R{{\cos }^{2}}\theta \] |
or \[EF=AB-2AE=2R-4R{{\cos }^{2}}\theta \] |
\[DE=AD\sin \theta =2R\sin \theta \cos \theta \] |
Thus, area of trapezium, \[S=\frac{1}{2}(AB+CD)\times DE\] |
\[=\frac{1}{2}(2R+2R-4R{{\cos }^{2}}\theta )\times 2R\sin \theta \cos \theta \] |
\[=4{{R}^{2}}{{\sin }^{3}}\theta \cos \theta \] |
\[\frac{dS}{d\theta }=12{{R}^{2}}{{\sin }^{2}}\theta {{\cos }^{2}}\theta -4{{R}^{2}}{{\sin }^{4}}\theta \] |
\[=4{{R}^{2}}{{\sin }^{2}}\theta (3\,\,co{{s}^{2}}\theta -si{{n}^{2}}\theta )\] |
For maximum area, \[\frac{dS}{d\theta }=0\] or \[{{\tan }^{2}}\theta =3\] or \[\tan \theta =\sqrt{3}\] |
(\[\theta \] is acute) or \[{{S}_{\max }}=\frac{3\sqrt{3}}{4}{{R}^{2}}\] |
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