[tex3]\mathsf{△ATB (retângulo):∠ATB=90^∘.\\
Trace ~PE \perp BQ (E \in BQ) \implies AB=PD=2\sqrt{r.R} = 2\sqrt2(^*propriedade)\\
△QBH∼△QPD⟹QH=\frac{2}{3},BH=\frac{4\sqrt2}{3}\\
TH=TQ−HQ=\frac{4}{3}\\
BT=TH^2+BH^2=\frac{4}{\sqrt3}\\
inraio : \frac{a+b−c}{2} \\
\therefore Inraio △BTH=\frac{(2\sqrt2+1−\sqrt3)}{3}\\
△BTH∼△BTA, inraio △BTA \implies
r=\frac{\sqrt3}{2}.\frac{2(\sqrt2 +1 -\sqrt3)}{3}\therefore \boxed{\color{red}r \approx 0,55}}[/tex3]...