Since ABC is a 90-60-30 triangle, it is half of an equilateral triangle. This means that BC is half AC, i.e. 17.
So, we have
[tex]AB = \sqrt{AC^2-BC^2}=\sqrt{AC^2-\left(\dfrac{AC}{2}\right)^2}=\dfrac{\sqrt{3}}{2}AC = \sqrt{3}\cdot BC =\sqrt{3}\cdot 17\approx 29.4[/tex]
So, the perimeter is
[tex]AD+DC+BC+AB\approx34+34+17+29.4\approx 114.4[/tex]
