Because $∆ABC$ is right at $B$
So $BC^2+BA^2=AC^2$ (Pythagorean theorem)
Or $BC^2+8^2=10^2$
Thus $BC=6(cm)$
So the perimeter of the base is: $AB+BC+AC=8+6+10=24(cm)$
B is the right angle so B' is the right angle too and AB=A'B'=8cm so ∆A'B'B is a right triangle
So $A'B'^2+B'B^2=A'B^2$
$⇔8^2+BB'^2=17^2$
$⇔BB'^2=225$
$⇔BB'=15cm$
Or the height of the prism is 15cm
Thus the lateral area of the prism is:
Height. Perimeter of the base/2=BB'.$24/2$=$15.24/2$=$180$cm^2