Look at the picture.
[tex]ZY = 5+3=8[/tex]
For XY and ZX use the Pythagorean theorem.
[tex]XY^2=3^2+4^2\\\\XY^2=9+16\\\\XY^2=25\to XY=\sqrt{25}\\\\XY=5[/tex]
[tex]ZX^2=5^2+4^2\\\\ZX^2=25+16\\\\ZX^2=41\to ZX=\sqrt{41}[/tex]
The perimetero fo XYZ:
[tex]P_{\triangle XYZ}=8+5+\sqrt{41}=\boxed{13+\sqrt{41}}[/tex]
