Por semelhança de triângulos:
[tex3]L-r\sqrt3+x+x=L+\frac{r}{\sqrt3} \implies
2x=r\sqrt3+\frac{r}{\sqrt3}\\
\text{porém tomando a base do triangulo AC=L}\\
L=\frac{r}{\sqrt3}+x+x+r\sqrt3=\frac{r}{\sqrt3}+r\sqrt3+2x=L=2r\left(\frac{1}{\sqrt3}+\sqrt3\right)\\
a\ altura\ será:\\H=\frac{\sqrt3}{2}.L=\frac{\sqrt3}{2}.2r\left(\frac{1}{\sqrt3}+\sqrt3\right) \implies
H=r\left(1+3\right)\\\therefore \boxed{\color{red}H=4r}
[/tex3]
(Solução: Jedi -
viewtopic.php?t=58342)