question_answer

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

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