The largest area of a trapezium inscribed in a semi-circle of radius $R$ , if the lower base is on the diameter, is

$AD=AB\cos \theta =2R\cos \theta ,AE=AD\cos \theta =2R{\cos }^2\theta$  or 

$EF=AB-2AE=2R-4R{\cos }^2\theta$ 

$DE=AD\sin \theta =2R\sin \theta \cos \theta$

Thus, area of trapezium  $S=\frac{1}{2}(AB+CD)\times DE$

$=\frac{1}{2}(2R+2R-4R{\cos }^2\theta )\times 2R\sin \theta \cos \theta$

$=4R^2{\sin }^2\theta \cos \theta$

For maximum area,  $\frac{dS}{d\theta }=0 \text{or }{\tan }^2\theta =3$    or

$\tan \theta =\sqrt{3}(\theta \text{is acute}) \text{or }{S}_{max}=\frac{3\sqrt{3}}{4}{R}^2$