A right angled triangle ABC of maximum area is inscribed in a circle of radius R, then the false statement of the following is

Right angled triangle  of maximum area inscribed in a
circle of radius .B is isosceles triangle  in a semicircle.
 

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}AB=AC=R\sqrt{2}\text{ and }BC=2R\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \therefore \text{ Area of }\mathrm{\Delta }=2R\cdot \frac{R}{2}=R^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[\right(\mathrm{a}\left)\text{ is correct }\right)]\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2s=2R\sqrt{2}+2R\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\therefore (\mathrm{d}\left)\text{ is correct }\right)]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}s=(\sqrt{2}+1)R\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\because (\mathrm{b}\left)\text{ is correct }\right)]\\ r=\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{\Delta }}{s}=\frac{R^2}{(\sqrt{2}+1)R}=R(\sqrt{2}-1)\end{array}$
(c) is not correct.