The foot of the perpendicular from $D$ to $\overline{AC}$ is the midpoint of $\overline{AC},$ which is $6$ units from $B$

With $AD = a$ and $BD = b,$ the two right triangles give $a^2 - b^2 = 15^2 - 6^2 = 189;$ factor it

Let $AD = CD = a$ and $BD = b,$ and let $E$ be the foot of the perpendicular from $D$ to $\overline{AC}.$ Since $AD = CD,$ point $E$ is the midpoint of $\overline{AC},$ so $AE = 15$ and $BE = 15 - 9 = 6.$ The right triangles $DEA$ and $DEB$ share leg $DE,$ so $$a^2 - 15^2 = DE^2 = b^2 - 6^2,$$ that is $$(a+b)(a-b) = 189.$$

The factorizations $189 = 189 \cdot 1$ $= 63 \cdot 3$ $= 27 \cdot 7 = 21 \cdot 9$ give $(a, b) = (95, 94),$ $(33, 30),$ $(17, 10),$ and $(15, 6).$ The last is rejected: $b = 6$ would put $D$ on $\overline{AC}.$ Each valid pair gives a triangle with perimeter $2a + 30.$

Therefore $s = (190 + 30)$ $+ (66 + 30)$ $+ (34 + 30)$ $= 220 + 96 + 64$ $= 380.$