jmc

Since $XY=YZ,$ then $△XYZ$ is isosceles. Draw altitude $YW$ from $Y$ to $W$ on $XZ.$ Altitude $YW$ bisects the base $XZ$ so that $$XW=WZ=\frac{30}{2}=15,$$as shown.



Since $\angle YWX=90^{\circ },$ $△YWX$ is right angled. By the Pythagorean Theorem, $17^2=Y{W}^2+15^2$ or $Y{W}^2=17^2-15^2$ or $Y{W}^2=289-225=64,$ and so $YW=\sqrt{64}=8,$ since $YW>0.$

We rotate $△XWY$ clockwise $90^{\circ }$ about $W$ and similarly rotate $△ZWY$ counter-clockwise $90^{\circ }$ about $W$ to obtain a new isosceles triangle with the same area. The new triangle formed has two equal sides of length $17$ (since $XY$ and $ZY$ form these sides) and a third side having length twice that of $YW$ or $2\times 8=16$ (since the new base consists of two copies of $YW$).

Therefore, the desired perimeter is $17+17+16=\boxed{50}.$