Perimeter of a trapezoid

Consider the diagram above.

Apply pythagorean theorem on $\triangle ABE$ ,

$h^2+x^2=64$ $\color{#D61F06}(1)$

Apply pythagorean theorem on $\triangle EBO$ ,

$h^2+(16-x)^2=256$ $\implies$ $h^2+256 -32x + x^2 =256$ $\implies$ $h^2+x^2-32x=0$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ ,

$64-32x=0$ $\implies$ $32x=64$ $\implies$ $x=2$

Therefore, $BC=32-2x=32-2(2)=28$

Finally, the perimeter is

$P=8+8+28+32=$ $\color{plum}\boxed{76}$