From this Triangle
Taking Triangle XZH, wanting to find HZ
[tex]\begin{gathered} We\text{ have to make use of pythgoras Theorem} \\ (XZ)^2=(XH)^2+(HZ)^2 \\ 12^2=8^2\text{ + }(HZ)^2 \\ 144\text{ = 64 + }(HZ)^2 \\ 144-64\text{ = }(HZ)^2 \\ (HZ)^2\text{ = 80} \\ (HZ)\text{ = }\sqrt[]{80\text{ }}\text{ =8.94} \\ \end{gathered}[/tex]
From the Second triangle, ZHY
[tex]\begin{gathered} ZY^2=HZ^2+HY^2 \\ (x+3)^2\text{ =}8.94^2+x^2 \\ x^2+6x^{}\text{ + 9 = }80+x^2 \\ x^2-x^2\text{ + 6x = 80 -9} \\ 6x\text{ = 71} \\ x\text{ =}\frac{71}{6}\text{ =11.83} \\ \end{gathered}[/tex]
XY = x + 8
= 11.83 + 8
=19.83
