The least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is

$6\sqrt{3}r$

2s = AB + BC + CA = 2AB + 2BD

$\Rightarrow s =A B+B D=A F+2 B D=r \cot \alpha+2 A D \tan \alpha$

$=r \cot \alpha+2(r+r ~cosec \alpha) \tan \alpha$

$=r(\cot \alpha+2 \tan \alpha+2 \sec \alpha)$

Find  $\frac{d s}{d \alpha}$  and  $\frac{d s}{d \alpha}=0 \Rightarrow \alpha=\frac{\pi}{6}$

∴  $2 s=6 r \sqrt{3}$.