Using the Pythagorean theorem, we have three triangles for which we can use the equation
[tex]\begin{gathered} (4+9)^2=x^2+z^2\text{ \lparen equation 1: triangle with sides }z,x,\text{ and }(4+9)\text{\rparen} \\ z^2=y^2+9^2\text{ \lparen equation 2: triangle with sides }z,y\text{ and }9\text{\rparen} \\ x^2=y^2+4^2\text{ \lparen equation 3: triangle with sides }x,y\text{ and }4\text{\rparen} \end{gathered}[/tex]
Use equation 2 and equation 3, and solve in terms of y²
[tex]\begin{gathered} \text{Equation 1} \\ z^2=y^2+9^2 \\ z^2=y^2+81 \\ z^2-81=y^2 \\ y^2=z^2-81 \\ \\ \text{Equation 2} \\ x^2=y^2+4^2 \\ x^2=y^2+16 \\ x^2-16=y^2 \\ y^2=x^2-16 \end{gathered}[/tex]
Equate both y², and solve in terms of x²
[tex]\begin{gathered} y^2=y^2 \\ x^2-16=z^2-81 \\ x^2=z^2-81+16 \\ x^2=z^2-65 \end{gathered}[/tex]
Substitute x² into equation 1
[tex]\begin{gathered} \begin{equation*} (4+9)^2=x^2+z^2 \end{equation*} \\ 13^2=(z^2-65)+z^2 \\ 169=2z^2-65 \\ 169+65=2z^2 \\ 234=2z^2 \\ \frac{234}{2}=\frac{2z^2}{2} \\ 117=z^2 \\ z^2=117 \\ \sqrt{z^2}=\sqrt{117} \\ z=3\sqrt{13} \end{gathered}[/tex]
Therefore, the length of z is
[tex]z=3\sqrt{13}[/tex]
