The triangle of maximum area the can be inscribed in a given circle of radius ‘r’ is :

$h=r\sin \theta +r$ 
 base  $=BC=2r\cos \theta$
 $\theta \in [0,\frac{\pi }{2})$
Area of  $\Delta ABC=\frac{1}{2}\left(BC\right).h$
$\Delta =\frac{1}{2}\left(2r\cos \theta \right).(r\sin \theta +r)$ 
 $=r^2\left(\cos \theta \right).(1+\sin \theta )$
 $\frac{d\Delta }{d\theta }=r^2[{\cos }^2\theta -\sin \theta -{\sin }^2\theta ]$
 $=r^2[1-\sin \theta -2{\sin }^2\theta ]$
 $=\underset{positive}{\underbrace{r^2[1+\sin \theta ]}}[1-2\sin \theta ]=0$
 $\Rightarrow \theta =\frac{\pi }{6}$               
$\Rightarrow \Delta$  is maximum where $\theta =\frac{\pi }{6}$ 
 ${\Delta }_{\max .}=\frac{3\sqrt{3}}{4}{r}^2=areaof\Delta with$
$BC=\sqrt{3}r.$