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 \[{{S}_{\max }}=\frac{3\sqrt{3}}{4}{{R}^{2}}\]